QUESTION:
A spaceship is 240,000 miles from the earth. At t = 0, the ship has an
initial velocity of 0 and starts to accelerate toward the earth due to
earths gravity. The ship is in free fall with respect to the earth.
How long will it take for the ship to reach earth?
What is the velocity of the ship as a function of time as the ship
approaches the earth?
What is velocity of the ship as a function of distance from the earth?
The problem I am having is that the gravitational acceleration is not 9.8
meters per second squared. The grav_acc is constantly changing as the ship
approaches the earth. All the equations I have found assume that the
grav_acc is constant.
My interest in this problem stems from the apollo command module returning
to the earth in free fall.

ANSWER#1: This is a pretty
mathematical question, so I will answer it in
pieces as I get the time to work it out. To find
the time it takes to fall, you start with
Newton's second law, ma=m [d^{2} r /dt ^{2} ]=-GMm /r ^{2} =dv /dt
and integrate it once to get v =dr /dt =√[2GM (R-r )/(Rr )]
where, for your question, R is the
distance from the center of the earth to the
point from which
it was dropped, r is where it is at the
time t (r ), G is the
universal gravitation constant, and M
is the mass of the earth. (The mass m
of the ship does not matter.) (You can more
easily get v (r ) from energy
conservation.) Next integrate v (r )=dr /dt to get
t (r )=-2.69x10^{5} {0.5tan^{-1} ((2x -1)/(2√[x (1-x )]-√[x (1-x )]-π /4}
where x=r /R ; this is not a simple
integral to perform, but you can find good
integrators on the web, for example
Wolfram
Alpha . Once you have gotten the indefinite
integral you are not finished, you need to find
the integration constant which will put the ship
at rest at x=r /R =1; that is
where the π /4 comes from. Putting in your
numbers, I find t =4.22x10^{5} s=117 hr=4.89 days. The
figure shows the graph of t (x ).

QUESTION:
My friends and I were messing around one night and posed the question "could a falcon break a human neck" and we tried to solve it but couldn't get anywhere with it (we are all theater majors not physicists lol) Could you solve that for me?

ANSWER:
A falcon weighs more than 1.5 lb and can fly at speeds
exceeding 200 mph. I think no fancy physics is required —surely
a 1.5 lb projectile hitting your head at 200 mph could
easily break your neck. (Wouldn't be too pretty for the
falcon either!) Just to make it a little more rigorous, I
calculate the average force exerted on the falcon as he (and
the head) move a distance d before stopping would
be about 25,000/d lb if d is measured in
inches; Newton's third law says that the head would
experience an equal and opposite force. So the average force
for the head moving 3 inches would be about 8,000 lb.
Information I could find indicated that 1000 lb should
be adequate to break a neck but that the actual force would
depend on circumstances —the individual, where
the force was applied and its direction, and whether the
neck was initially bent or not. I think that 8,000 would do
the trick for almost all situations.

QUESTION:
Do objects falling from the sky reorient themselves to reduce friction with air (even if just a tiny bit)?

ANSWER:
It depends on the object. For example, a parachute would not
work if it reoriented itself to minimize drag. Consider a
simple parachute having a rigid disc and the load hanging
from the center. Orienting such that the disc were in a
vertical plane would reduce the air drag but it does not
happen. Some things, though, will reorient to minimize drag.
For example, a cone will orient pointing down because if it
is falling sideways there is a torque trying to orient it
pointing down because the drag on the pointy end is smaller
than on the blunt end.

QUESTION:
Glad I found your site, my name is Ben, and this question is something I'm working on professionally, not homework. I use Autodesk Inventor's Dynamic Simulation to model collisions and some of the results for a specific simulation don't seem consistent. So, a colleague suggested we simplify the problem and work the numbers out by hand, but we can't figure it out. Here is a diagram:

A rod (blue) of length 2m and mass of 1kg can freely pivot about it's center (orange dot) which is connected to a frictionless track (purple) running in the Y direction. The rod is positioned at 45 degrees to the track, and a ball of mass 1kg travelling at 15 m/sec strikes the rod at 0.5m from the rod's center. Collision is perfectly elastic, friction is zero. How can I determine the final velocity of the rod along the track, and what is it's rotational speed? And more importantly, I need to know what effect the rod angle has on these two answers. Then I can solve and plot a chart for all angles from 0 to 90 and compare with my Inventor results. I'm really having trouble with the moment of inertia part, being that it's not struck in the center, or the end.

ANSWER:
There are important errors with the problem as you state it. The first
obvious error in this problem is that the direction and magnitude of the
final velocity of the ball is impossible. If the ball carries off all the
energy it came in with, the rod must end up with no energy. And,
conservation of linear momentum in the y direction (not conserved in the
x
direction) demands that the speed of the center of mass (COM) of the rod
along the rail would be 15 m/s meaning it would not be rotating. And, the
angular momentum relative to the COM of the incoming ball is equal and
opposite that of the outgoing ball, so the rod would have to be rotating to
conserve angular momentum. So I thought to simply redraw the picture but
with the final ball velocity having unknown components (2 unknowns). Two
other unknowns are the final speed of the COM and the angular velocity of
the rod about the COM. However, I only can see three equations: conservation
of energy, linear momentum (only y ), and angular momentum. I am still
pondering the question, but there is too little information or I am missing
something.

CONTINUED ANSWER:
Since the questioner is ultimately interested in a more
general solution than the special case, I will set it up
generally from the start and apply that solution to his
initial question; I will, however, set m _{1} =m _{2} =m
as soon as I have written the most general solutions to
simplify the algebra after setting up the problem. As noted
above, the problem begins with three conservation equations:

conservation of
linear momentum in the y -direction:
m _{1} v _{1} =m _{1} v _{2y} +m _{2} u _{2}

conservation of
energy: ½m _{1} v _{1} ^{2} =½m _{1} v _{2} ^{2} +½m _{2} u _{2} ^{2} +½Iω ^{2}

conservation of
angular momentum: m _{1} v _{1} S sinθ =Iω+m _{1} v _{2y} S sinθ-m _{1} v _{2x} S cosθ

The moment of inertia
of a thin rod about its COM of mass m and length
L is I=mL ^{2} /12. Therefore, if
m _{1} =m _{2} =m ,
the three conservation equations are:

v _{1} =v _{2y} +u _{2
}

v _{1} ^{2} =v _{2} ^{2} +u _{2} ^{2} +L ^{2} ω ^{2} /12

v _{1} S sinθ =L ^{2} ω /12+v _{2y} S sinθ-v _{2x} S cosθ

There
are now three equations and four unknowns, v _{2x} ,
v _{2y} , u _{2} , and ω.
To generate a fourth equation, consider the impulse
J delivered by the rod to the ball:
J = Δp =m (v _{2} -v _{1} )or
J _{x} =mv _{2x} and J _{y} =mv _{2y} -mv _{1} .
But, the impulse must be normal to the surface of the rod,
so tanθ=|J _{y} /J _{x} |=(v _{1} -v _{2y} )/v _{2x} .
The fourth equation is

v _{2x} =(v _{1} -v _{2y} )cotθ .

I first solved these
four equations for θ= 45^{0} , L =2
m, S =0.5 m, and v _{1} =15 m/s, the
conditions specified by the questioner; note that, because
it is a quadratic equation we are solving, we get two
solutions:

v _{2y} =(15,
8.33) m/s

v _{2x} =u _{2} =(0,
6.67) m/s

ω =(0,
14.15) s^{-1} =(0, 2.25) rev/s.

The meaning of the
first (trivial) solution is that there was no interaction
between the rod and the ball (you missed!). Another
situation for which we can check the solution intuitively is θ= 90^{0} ,
for which v _{2x} must be zero:

v _{2y} =(15,
4.1) m/s

v _{2x} =(0,
0) m/s

u _{2} =(0,
10.9) m/s

ω =(0,
16.35) s^{-1} =(0, 2.6) rev/s.

Finally, the general
solutions are:

v _{2y} =v _{1} (A -1)/(A +1)
where A =cot^{2} θ +1+[12S ^{2} /(L ^{2} sin^{2} θ )]

v _{2x} =(v _{1} -v _{2y} )cotθ= 2v _{1} cotθ /(A +1)

u _{2} =(v _{1} -v _{2y} )=2v _{1} /(A +1)

ω =[12(v _{1} -v _{2y} )S /(L ^{2} sinθ )]=[24v _{1} S /((L ^{2} sinθ )(A +1))].

The plots requested by
the writer are shown to the right.

ADDED NOTE:
An alternative way to specify the vector v _{2}
is to specify its magnitude, v _{2} , and the
angle φ it makes with the +x
axis:

ACKNOWLEDGMENT:
Thanks to "haruspex" at
Physics Forums for helping me realize how to get the
desperately sought fourth equation!

QUESTION:
This question concerns angular momentum as related to motorcycles. If a motorcycle rests atop a trailer, but sits on a roller system, the bike will stand upright without any supports or tethers if the wheels are rolling (throttle is locked in the
"on" position and wheels are spinning). If the trailer itself travels forward in a straight line, the bike should remain upright. But if the trailer takes a sharp turn, what happens? Does the bike fall, or does it instead do what a rider does to achieve a turn-countersteer and remain upright?

ANSWER:
I have waited a long time to answer this question because I
am bothered by the way the problem is stated. First of all,
if the throttle is locked on, only the rear wheel will be
spinning, so we can discuss the problem by looking only at
the wheel. It is certainly correct that if the truck goes
straight the wheel will continue running upright (assuming
that the center of gravity of the bike is in the vertical
plane passing through the center of gravity of the wheel).
Imagine the bike and rollers to be mounted on a big "lazy
Susan" the base of which is bolted to the truck bed. So, if
a north-bound truck turns to the west, the angular momentum,
experiencing no torque, will remain constant and continue
pointing in the same direction (originally either east or
west). Viewed from inside the truck it will appear that the
whole bike rotated through 90^{0} relative to the
truck.

Now,
if the rollers are attached to the truck bed, when the truck
turns the rollers turn and the wheel, trying to not turn,
will come off the rollers at some point. We first need to
understand the physics relationship between the torque and
the angular momentum of the wheel. The rotational form of
Newton's second law is τ =ΔL /Δt ,
torque equals the time rate of change of the angular
momentum. The first figure shows the wheel, as seen from
above, turning through some small angle which results in a
change of angular momentum from L _{1}
to L _{2} and ΔL =L _{2} -L _{1} .
So the torque which you must apply to make it turn this
way is in the direction of ΔL .

If
you think you can just steer it as if it were not rotating,
you would fail. The second figure shows what would happen if
you try to steer it like your intuition would have you do it
by exerting a force like F in the
figure. The torque points up and so the wheel would not turn
but lean in the opposite direction from the way you would
lean on the bike if you were riding it and making a turn.
You can find some videos showing this by googling
gyroscope in a suitcase video .

If you want the wheel
to turn with the truck, you need to have a torque which
causes that. One way I thought of to achieve this was to
have strings attached from the axles on each side and the
truck bed below. These need to have no tension on them when
the truck is going straight. When the truck turns as
indicated, the tension in the string on the right-side
string will be bigger than the other side and there will
therefore be a net force N on the wheel. This results in a
torque in the horizontal plane which will cause a change in
angular momentum in the direction consistent with the wheel
turning with the truck. Two strings are needed because the
truck might turn either left or right.

QUESTION:
Is there a formula for calculating the side-ways deflection wind has on a lawn bowl(over and above the bias deflection ) running at 12
s, the time a bowl takes from delivery to stop over a 26 m distance over
bowling green grass?

ANSWER:
Once again, doing Ask the Physicist has led me to learn
something new. I never really knew anything about lawn bowls
other than it is done on grass and rolling balls are
involved. For the benefit of others who are ignorant of the
game, let me summarize by describing the ball. (A good
article on the physics of lawn bowls balls can be found
here .) The ball is not a sphere but rather an oblate
spheroid which makes it sort of like a door knob but not so
extremely flattened; but it is slightly more flattened on
one side of the ball than on the other which results in a
center of gravity being displaced to one side of the
equatorial plane as shown in figure (a). This results in a
tendency for the ball to curve left if it is rolling the
angular velocity shown in the figures; this motion is the
"bias" referred to by the questioner which I am to ignore.
When rolling in the x direction (figure (b)), there
is a frictional drag force called, rolling friction
D , which opposes the motion (v )
and eventually brings the rolling to a halt. If there is a
wind, there is a force W due to
the wind which tries to make the ball roll to the right
(figure (a)) but if it does roll, there will also be rolling
friction trying to keep it from rolling. In order for the
wind to have any effect at all, it is clear that we must
have W>D ; if this is not the case, there will only
be static friction in the y direction which will be
equal and opposite to W . A lawn
bowls ball has a mass of about m =1.5 kg and a
radius of about R =6 cm=0.06 m.

To get the equations of
motion for the x and y motions, we first
need expressions for D and W . The rolling
friction may be expressed as D=-μmg where μ
is the coefficient of rolling friction and mg is
the weight of the ball. The force due to the wind may be
approximated as W ≈¼AV ^{2}
where A=πR ^{2} is the cross sectional
area of the ball and V is the speed of the wind;
this approximation is only correct if SI units are used. The
equations of motion in the x -direction are

Here t is the
time and v _{0 } is the speed of the ball at
t =0. If the ball is rolling in the y-direction
because of the wind, the equations of motion are:

It should be noted that
if (¼AV^{2} /m )<μg , these
equations imply that the ball will accelerate opposite the
direction of the wind, obviously not correct; hence the wind
will have no effect on the ball if V <√(4μmg /A ).
In that case, a_{y} =v_{y} =y =0.

So, having found the
general solutions, let us now apply the solutions to the
specific case from the questioner. We are told that when
t =12 s, v_{x} =0 and x =26 m.
With that information you can solve the x -equations
to get v _{0} =4.32 m/s and μ= 0.037,
reasonable values compared to numbers in the
article I read. The area is 3.14x0.06^{2} =0.0113
m^{2} . The first question we should ask is what is
the minimum speed of the wind to have any effect at all:
V _{min} =√(4x0.037x1.5x9.8/0.0113)=13.9
m/s=31 mph=50 km/hr; this is a pretty stiff wind, so the
wind probably has no effect on bowling under normal
conditions. So, just to complete the problem, consider V =15
m/s=34 mph=54 km/hr.

The^{ }
trajectory during the 12 seconds is shown in the graph
below; after 12 seconds the ball will continue accelerating
in the y direction.

So the bottom line is that unless you are playing in a gale-force wind, the wind has no effect on the ball if the wind has no component along the original direction of the ball (which I have called the
x -axis). You can tell if wind makes a difference by simply setting the ball on the ground—unless the wind blows the ball away, you need not worry about its effect. If the wind is blowing in the +x or -x direction, that is a whole different thing, but
the questioner asked for the sideways deflection.

ADDED THOUGHTS: This question continues
to intrigue me and I have carried my investigation further.
The question originally stipulated "over and above the bias
deflection" so my whole discussion totally ignored the fact
that the ball, owing to its off-center center of mass, will
curve. At the very end of my answer I noted that if the wind
is not perpendicular to the path of the ball, it would
be a different story; indeed for a spherically symmetric ball I showed that, except for very strong winds,
a wind perpendicular to the path has no effect at all.
However, for an actual lawn bowls ball, the path curves to
where a wind in the y -direction might have a
significant component along the path. I have calculated
(graphed below) the
x and y positions of a realistic path with no
wind using equations (10) and (11) of the
article referred to above. To do these I used all the
numbers used above (R , m , A ,
v _{0} ,
μ ); I used the moment of inertia
for a solid sphere ( I _{0} =I _{cm} +mR ^{2} =(7/5mR ^{2} ))
and chose the COM off-center distance to be d =1 mm. As you can see, the curving is
substantial, carrying the ball about 4 m from its original
direction. You can see that now a wind of any magnitude can
have an effect on the trajectory. The angle φ
which the tangent to the trajectory is given in the article
as φ =(2/p )ln(v _{0} /(v _{0} -μgt ))
where p is a constant also given in the article. As
can be seen, once the trajectory leaves the x-axis the wind
contributes with the component of its force along the
trajectory; this has the effect of reducing the effect of
the frictional force causing the ball to slow down less
rapidly. However, this is now like having a time dependent
force of friction which, I believe, will lead to equations
of motion which will not have an analytical solution but
would have to be solved numerically.

QUESTION:
How can rockets move in space. Because they have nothing to push against won't trust be useless. Like for example the reason the rocket gets out of the atmosphere is because of the thrust pushing against the earth then on the air in the atmosphere and if space has no air how can rocket move in space.

ANSWER:
The reason is that your statement that "the thrust pushing
against the earth then on the air" is the reason the rocket
accelerates is totally wrong. The reason a rocket
accelerates is that hot gas is expelled out the back at a
high velocity. The reason this propels the rocket can be
illustrated by the following example: an astronaut is
floating in empty space and happens to be holding a bowling
ball; she throws the ball and the result is the balls moves
away and she recoils so she is moving also. This is called
conservation of linear momentum —if the ball has
1/20^{th} as much mass as she does, she recoils with
1/20^{th} the speed the ball has. The rocket is not
"pushing against" anything.

QUESTION:
This is something I remember discovering when I was younger. If i would take a small cylindrical object (like a AA battery) , set it on a flat hard surface, then using my fingers I would apply downward pressure on the edge of the battery. This would create backspin on the battery but also shoot the battery across the floor. When the battery's finally stopped moving forward it would spin rapidly on its axis until stopping completely. Could you explain the physics behind movement of the battery?

ANSWER:
Problems like this are standard in intermediate-level
classical mechanics. Round objects can move translationally
on a horizontal surface by rolling without slipping or by
sliding. The two classic extremes of the slipping scenario
are a skid where the object is initially not spinning at all
(e.g . a bowling ball begins by approximately not
rotating but may end up rolling without slipping), or is at
rest but spinning (like "peeling out"); in both cases, the
problem is usually to find the time elapsed or distance
traveled before rolling sets in. In your case, the object is
initially both translating horizontally and rotating with a
backspin. What will happen depends on the initial conditions —the
initial speed and the initial angular velocity; also the
properties of the object will also matter—its mass
m , radius R and shape, and the coefficient of
kinetic friction μ . Three possibilities are that

it will reverse directions and come
back still spinning and slipping, eventually stop
slipping and roll (what I think you are remembering),

it will stop slipping and continue rolling in the
initial direction, or

it will stop dead.

In the figure there are three forces on the object: its
weight mg , the normal force
N up from the floor, and the
frictional force f which the floor
exerts on the sliding object; the object begins with
velocity v _{0} and angular
velocity ω _{0} . The
mass of the object is m and its radius is R .
The friction slows down both the velocity and the angular
velocity. Sliding ceases when v=-Rω (see
footnote*). If v _{0} is very small and ω _{0}
is very large, possibility #1 will happen; if v _{0}
is very large and ω _{0} is very
small, possibility #2 will happen; if v _{0}
and ω _{0} are just right, possibility
#3 will happen. I think that this gives you a good
qualitative overview of the physics of the problem.

For those interested in
the quantitative solution, I will give it here for a uniform
solid cylinder. For the translational motion the two
equations (taking +x as to the right, +y
up) are -f=ma and N-mg =0; since f=μN=μmg
we can write a=-μg . For rotational motion about
the center of mass (choosing the direction it is initially
spinning as positive), only the frictional force exerts a
torque, so -fR =-μmgR=Iα where
I is the moment of inertia about the center of mass
and α is the angular acceleration. For a
cylinder I =½mR ^{2} , so α=- 2μg /R.
So, we can write the equations for the velocity and
angular velocity as functions of time t : v=v _{0} +at =v _{0} -μgt
and ω=ω _{0} +αt=ω_{0} -2μgt /R.
Now, the time when slipping ceases will be when
v=-Rω ; solving, t =(v _{0} +Rω _{0} )/(3μg ).
Finally, put this into the equation for v to get
v (t )=(v _{0} -Rω _{0} )/3.
So if Rω _{0} >v _{0} ,
possibility #1 will happen; if Rω _{0} <v _{0} ,
possibility #2 will happen; and if Rω _{0} =v _{0} ,
possibility #3 will happen.

*The condition v=-Rω
is a little tricky. The negative sign is because of my
choice of the positive direction for ω which
is opposite that which would be the case for no slipping.

QUESTION:

Imagine a cart moving on a frictionless surface with a cannon aimed opposite the direction of travel. At 40 km/hr the cannon fires a projectile accelerating the cart-cannon to 50 km/hr. A second projectile (identical to the first) is then fired accelerating the cart to 60 km/hr. The impulse (and thus recoil) against the cart should be the same no matter what the moving velocity of the cart was. After all, the cart is only moving within my frame of reference. If I selected a frame of reference in which the cart was stationary, an impulse acceleration of 10 km/hr would be the same irregardless of whether in another frame of reference the cart was initially moving at either 40 or 50 km/hr. It's true that the second firing of the cannon would be acting on slightly less mass, after the first cannon ball was no longer part of the system—but that would seem a trivial difference. I also understand that the momentum of the total system is the same before and after the cannon is fired.

Does it take more energy to accelerate an object from 40 km/hr to 50 km/hr than it takes the same object to accelerate from 50 km/hr to 60 km/hr ? I know that the Kinetic Energy at 60 km/hr is 60/50 squared or 1.44 times greater than at 50 km/hr so the answer is undoubtedly, yes. What am I missing about Kinetic Energy - does it depend on your frame of reference?

ANSWER:

I have rearranged
your question so we can talk about one question at a
time. Your first question, not really a question,
basically says that, although the momentum of any object
is dependent on your frame of reference, that the change
of momentum if you do something to it (impulse) is not
frame dependent. Why is that? The reason is that
Newton's second law may be written as F =dp /dt
and the rate of change of momentum must be frame
independent because we know that the force is frame
independent.

Now, when you change the momentum you also change the
kinetic energy because either the speed or the mass is
changing. But does everyone see the same change in
kinetic energy? The answer is no. And your example shows
very clearly that the change in kinetic energy does
depend on the frame of reference. A more general
example, calculating the work done by a constant force
as seen in different frames, can be found in an
earlier answer .

QUESTION:
For a snow plow that is very heavy, is there an advantage to having the
connection point of the winch line up high so that there is less weight to
pull or is there no difference?
I can send a picture for clarification.

ANSWER:
Yes send me a picture. You are talking about a winch which is used for what?
Pulling a stuck vehicle? Lifting and lowering the plow? What?

FOLLOWUP QUESTION:
Yes, lowering and lifting a plow. The problem is that the plow is out very far out and that my winch is mounted pretty low on my machine. So instead of the winch simply lifting the plow up, right now it is mostly pulling backwards and then that is making the plow come up.
I attached a picture of what the manufacturer recommends but I haven't had too good of luck with them in the past. The last two pictures are of my machine. You can see how far out the plow is and how low my winch is. Would that pulley and cable system helped at all?

ANSWER:
You may not want to get the full physics explanation here,
so I will first give you a qualitative explanation. The
tension T in the strap is what is
lifting plow and any part of it which is horizontal (T_{H} )
is wasted. Your gut feeling is right, "…it is mostly
pulling backwards…" Anything which you can do to
increase the vertical part (T_{V} ) will
make the lift easier, and moving the winch up is a good way
to do this but it might be easier to have a pulley
higher up which then brings the strap back down to the
winch.

The figure shows all
the forces on the plow assembly: the weight W
which acts at the center of gravity (yellow X), the tension
T in the strap (where I have shown
vertical (T_{V} ) and horizontal pieces (T_{H} )), and the force the
truck exerts on the support which I have represented as its
vertical and horizontal parts, V
and H respectively. (Note that the
various forces are not drawn to scale since T has
to be much larger than W to lift the plow.) Suppose T
is just right that the plow is just about to lift. Then the
sum of all the forces must add to zero, or V+T_{V} -W =0
and H-T_{H} =0. The sum or torques must also
add to zero; summing torques about the point of attachment
to the truck (light blue X), WD+T_{H} s-T_{V} d =0.
Note that the T_{V} is trying to lift the
plow but T_{H} is trying to push it down.
Now, in order to get a final answer for the unknowns (which
are T , V , and H ) we note that
T_{V} =T sinθ and T_{H} =T cosθ
where θ is the angle which the strap to the
winch makes with the horizontal. The final answers I get
are:

T=DW /(d sinθ-s cosθ )

V =W-T sinθ

H=T cosθ

I put in some
reasonable numbers just to get an idea of the answers, W =500
lb, θ =20^{0} , D =2m, d =1.8
m, and s =0.1 m. Then T =1920 lb, H =1800
lb, and V =-157 lb. The negative value for V
means that V is down, not up. In
this scenario, the pulling force has to be nearly four times
greater than than the weight being lifted.

Now
we need to look at whether the manufacturer's suggestion
will be better than a straight shot to the winch. Now there
are two forces pulling up, the tension T
from the pull point to the winch and the tension
P from the pull point to some anchor
higher up. Of course the magnitudes of these two tensions
are the same, P=T . The picture shows only the
pulling forces, the rest are the same as in the picture
above. There are still three unknowns, T , V ,
and H . I will call the angle that P
makes with the horizontal φ . I will not show
the details, just give the final results:

As a numerical example,
I will use the same numbers as above and add φ= 40^{0} .
Then T =624 lb, H =1070 lb, and V =-115
lb. It is definitely advantageous to use the manufacturer's
suggestion here which, in my numerical example, reduced the
force the winch needed to exert by a factor of about 3.

QUESTION:
We are a seventh grade science class, and we are discussing friction. We understand that life as we know it would end if there were no friction, but we were wondering what, exactly, would happen to your body without friction.

ANSWER:
It
would have been impossible for your body to form in the
first place. When two surfaces are in contact with each
other they exert two forces on each other. As an example,
think about a block which is sitting at rest on a slight
incline. The part of the force which the incline exerts on
the block which keeps the block from breaking through the
incline and that is called the normal force. The part of the
force which the incline exerts on the block which keeps the
block from sliding down the incline and that is called the
frictional force. With no frictional force, the block would
slide down the incline. Now imagine a single bone cell on
the surface of your leg bone. It has a weight which makes it
want to slide down your leg to the ground; friction between
it and its neighbors with which it is in contact keeps that
from happening. With no friction, all the cells in you body
would slide down to the floor and spread across the floor.

QUESTION:
What is the ratio between the height H of a mountain and depth h of a
mine,if a pendulum swings with the same period at the top of the mountain
and at the bottom of the mine?

ANSWER:
This must be a homework question where you assume that the density of the
earth is uniform.

QUESTION:
No it's not. I am a student who is trying to crack NEET. While I was solving questions, I saw this one but I couldn't get the explained answer.
The answer I got was 1,but the answer given here was 1/2. I just wanted to know was my answer correct.

ANSWER:
There is no way to solve this problem without an assumption
regarding the density of the earth. The standard assumption
in introductory physics is to assume that it is uniform
(which is
not a good approximation); I will do that. Also, I
will assume that it is a simple pendulum with a small angle
amplitude so that the period is T ≈2 π √(L /g )
where L is the length and g=MG /R ^{2} is the
acceleration due to gravity. At a distance H above
the earth's surface g _{H} =MG /(R+H )^{2} ;
at a distance h below the surface of a
uniform-density earth, g _{h} =MG (R-h )/R ^{3}
(see footnote*). Now, for the periods to be equal it is
necessary that g _{h} =g _{H} ;
a little algebra shows that equivalently (1-(h /R ))=1/(1+(H /R ))^{2} =g _{H(h)} /g .

It is instructive to
look at h /R as a function of H /R :
h /R =1-1/(1+(H /R ))^{2
} shown in the first graph above. When h=H =0
you are at the surface of the earth; when h /R =1,
H /R =∞ and g _{h} =g _{H} =0.

Next, look at the
plots of g _{H(h)} /g as a function of H /R
(black) and of h /R (Red). There
are only two locations where g _{h} =g _{H} ,
at the surface where h /R =H /R =0
corresponding to g _{h} =g _{H} =g
and near h /R =H /R =0.6.
The third plot shows a closer look around 0.6 showing h /R =H /R =0.618
corresponding to g _{h} =g _{H} =0.382g .

This was a pretty
long-winded answer, but the upshot is that you were right
and the answer key was wrong: h /H =1.

*M is the mass
of the earth, R is the radius of the earth, and
G is the universal gravitational constant.

ADDED NOTE: Actually, I did not need to
solve this problem graphically, I could have solved it
analytically. If you assume (guess) that h /R=H /R≡x ,
then the equation to determine x is 1-x =1/(1+x )^{2}
or (1-x )(1+x )^{2} -1=0=(1-x ^{2} )(1+x )-1=
x ^{3} +x ^{2} -x=x (x ^{2} +x -1)=0.
One solution is x =0 (the surface) and the positive
solution to the quadratic equation is x = ½( √(5)-1)=0.618.
I guess I shied away from this because I know I am not very
good at solving cubic equations, but I can handle this one!

QUESTION:
We have been talking in science class about air pressure and vacuums,and how a straw even in perfect vacuum will only suck water up to ~10m, my question is why does the diameter of the straw not effect the height to which the water is pulled as a greater diameter would mean more mass.

ANSWER:
The mass m of the water in the tube is the density
of water (ρ =10^{3}
kg/m^{3} ) times the volume of water (V=Ah );
therefore the weight of the water is mg =ρgAh ≈10^{4} Ah
(taking g ≈10 m/s^{2} ). The pressure at
the bottom of the tube is atmospheric pressure, P _{bottom} ≈10^{5}
N/m^{2} and at the top the pressure is zero;
therefore there is a force F up on the tube which
is F =(P _{bottom} -P _{top} )A ≈10^{5} A
N. But F is holding up the weight, so F=mg or 10^{5} A= 10^{4} Ah
or h ≈10 m. A cancels out!

QUESTION:
hi, so, there is a fairly recent
video going around the internet of nasa ISS astronaut Randy Bresnik spinning a fidget spinner, in space, in a quite low gravity environment and then apparently grabbing hold of the spinner then video cuts and we, presumably some short time later, see the entire body of the astronaut holding the spinner spinning fairly rapidly in space.
Perhaps it is a quite elementary question, but, I wonder if that video might have been faked, wondering if the what must be a relatively small amount of energy in the spinner could cause the entire body of mass of astronaut plus the spinner to spin in the fashion observed in the video. That is to say: wouldn't the energy from the spinner, say, be absorbed by the astronauts body or the muscles in his arm, or some such, upon grabbing the spinner; or, if there was movement, wouldn't it be far far less than what is demonstrated in the video?

ANSWER:
You are right, I think the astronauts are messing with you
here! You have looked at it from the perspective of energy
conservation; however, energy would not be conserved here
and energy would actually be lost, not gained as it appears.
So, your reasoning was sound —where did the
energy come from? What is conserved is the angular momentum.
If the moments of inertia are I _{toy} and
I _{man} and the original angular velocity
of the toy is ω _{1} , then the
initial angular momentum is L _{1} =I _{toy} ω _{1}
and the final angular momentum is L _{2} =(I _{toy} +I_{man} ) ω _{2}
where ω _{2} is the angular velocity
of the man+toy. Conserving angular momentum and solving for
ω _{2} , ω _{2} =[I _{toy} /(I _{toy} +I_{man} )]ω _{1} <<ω _{1} .
Now, I actually found the moment of inertia of a fidget
spinner,
I _{toy} ≈7x10^{-5} kg·m^{2}
and I estimate
I _{man} ≈70 kg·m^{2} .
This means that ω _{2} =ω _{1} /1,000,000!
The astronauts have angular velocity of about 1 revolution
per second and you know perfectly well that the fidget
spinner did not have a speed of a million revolutions per
second.

Finally, if you are interested, you can show that energy is
not conserved. The expression for kinetic energy is E =½Iω ^{2
} so E _{1} =½I _{toy} ω _{1} ^{2
} and E _{2} =½[I _{toy} /(I _{toy} +I_{man} )]ω _{1} ^{2} ≠E _{1} .

QUESTION:
How would i explain bus or a truck(high CG)flip over, when driving in a curve with high speed, from a inertial reference frame? If centrifugal force is fictitious and bound to non-inertial reference frame, what torque causes truck to flip over observed from an inertial ref. frame?

ANSWER:
I have attached figure from the
earlier answer
where I did essentially this problem in the noninertial
frame; this saves me having to draw the whole picture again.
For your problem, the vector labelled C
is zero. Part of what makes your problem harder
than in the accelerating frame is that you must sum torques
about the center of mass (green x). For equilibrium, the sum
of torques must be zero; summing torques gives 0=H (f _{1} +f _{2} )+LN _{1} -LN _{2} .
(Translational equations are f _{1} +f _{2} =mv ^{2} /R
and N _{1} +N _{1} -W=0.)
Now, when the truck is about to tip over the left wheel is
about to leave the ground so N _{1} =f _{1} =0,
f _{2} =mv ^{2} /R ,
and N _{1} =W ; therefore Hmv ^{2} /R-LW= 0.
So, if v > √[LWR /(Hm )]
the truck will flip over.

QUESTION:
A Newton's Cradle is some times used to demonstrate the Law of Conservation of Momentum but, I noticed anomaly. Specifically, whenever the steel balls collide, I hear a clicking sound and that requires energy. This tells me some of the kinetic energy of a moving ball transforms into thermal and accoustic. How do physicists explain that momentum is conserved when a ball leaving a collision is moving slightly slower? If there is less kinetic energy but no change in the total energy, it seems logical to think a certain quantity of motion has vanished after any collision. Both momentum and kinetic energy use the same variables. I also understand that when I pull one ball back, after a collision one ball emerges. If two balls are pulled back and released, two balls pop out.

VIDEO

ANSWER:
You are right, energy is not conserved in the collisions.
And, you can tell that it is not conserved because the
machine eventually stops. But momentum is always conserved
in an isolated system (no external forces on the colliding
bodies). I commend you for thinking about this which seems
paradoxical. To illustrate, let's choose the simplest
situation, the collision of two balls of equal mass m ,
one at rest the other with speed v , in one
dimension. As you know, if the collision is perfectly
elastic the first ball ends up at rest, the second moving
with speed v , so the momentum is conserved because it is
mv before and after the collision. Next you say that if
the second ball emerges with speed u=v-δ that
momentum has been lost. But you have assumed that the first
ball is at rest, but it cannot be if momentum is conserved:
mv=mu'+m (v-δ ) and so the first ball
has a speed u'= δ after the collision.
In the
video I have inserted above, look closely at the first
demo, with one ball striking four and one popping out; you
will clearly see that the next to last ball is not at rest
after the collision(s).

QUESTION:
If I threw a ball straight up into the air and it stayed there for 6hrs would it drop back to excactly the same spot?

ANSWER:
Any object thrown vertically will not fall back exactly to
where it started. To understand why, see a
recent answer . Your question
has a slightly different twist: when this type of problem is
calculated it is usually assumed that g , the
acceleration due to gravity, is constant; if your ball stays
in the air for six hours, that will not be the case.
Nevertheless, the ball would always have a Coriolis force on
it so it would be deflected westward by some amount. If you
had said something like the ball goes up 1000 m, you could
have done a pretty good calculation since 1000 m is very
small compared to the radius of the earth and it would be a
reasonable approximation to say that g =9.8 m/s^{2}
the whole time (which would be about 28 s). It can be shown
that, for a latitude of λ , on the way up the
Coriolis deflection is [(4ω /3) √(8h ^{3} /g )]cosλ
west and on the way down it is
[(ω /3) √(8h ^{3} /g )]cosλ
east, so the net deflection is
x=ω √(8h ^{3} /g )cosλ
west; the angular velocity of the earth is
ω= 7.27x10^{-5} s^{-1} , so
x =2.1 m west at the equator (λ= 0).

Finally we should note that at the poles (cosλ =0)
there is no deflection due to the earth's rotation, but
since the earth revolves around the sun you are still in a
frame where there would be a Coriolis force but it would be
much smaller; it would be smaller yet because the angular
velocity of the motion around the sun is almost parallel
(about 23^{0 } away) to the vertical at the poles.
The deflection for the example above (h =1000
m)would be about 2 mm compared to 2 m; the direction would
be opposite the direction the earth is moving in its orbit.

All of the above
ignores air drag.

QUESTION:
Is the graviitational force on Earth the same everywhere or is the force different near the south or North Pole?

ANSWER:
If the earth were a perfect sphere with a spherically
symmetric mass distribution, the gravitational force would
be everywhere the same. Since it is not, there are small
variations as you move around the surface. Even if it were,
it would appear to vary as you moved from the poles
to the equator because of the earth's rotation. A scale
would read your weight (gravitational force on you)
correctly at the poles but would, because of the centrifugal
force (a "fictitious" force), read slightly less at the
equator.

QUESTION:
Why does a satellite shot straight up from the equator deflect to the west?

ANSWER:
There are two ways to look at this.

If you view it from outside the
earth, the satellite will continue moving directly away
from the center of the earth.
But, viewing the earth
from above the north pole, the earth rotates
counterclockwise, west to east.
So the earth is actually rotating under it which gives
it the appearance of deflecting to the west.

The second is a little fancier and harder to understand.
If you view Newton's laws from a rotating coordinate
system (the rotating earth with you on it) you find that
they are wrong; these "laws" of physics only work in
inertial frames of reference. However you can force
Newton's laws to be correct if you invent just the right
fictitious forces. I will not go into all the complexity
here, but one of the fictitious forces is called the
Coriolis force and can be written as F =-2m ω xv
. Since the vector ω
points along the earth's axis and out
through the north pole and the velocity vector
v points radially outward, the
negative of their cross product points west. Hence, it
will appear that the satellite is pushed west when
viewed from the ground.

QUESTION:
I am a safety coordinator in a warehouse operation. I am attempting to impress the need for safety with my co-workers. To that end, I am trying to answer the following:
A 10,636 lb. forklift (4825 kg) traveling at a speed of 8 mph (13 kph) strikes a fixed object (metal pole). How much force is transferred during this collision?
.
Secondly, same forklift, same traveling speed strikes a 150 lbs. (68 kgs) person. How much force is transferred to the person in this collision?

ANSWER:
There is no accurate way to calculate the forces. They depend on the details of the collisions.
In particular, how long do the actual collisions last? Also, how do the forklift and the person move after the collision has occurred (assuming the metal pole stays stationary)?
I could make some rough estimates to get an idea:

For the first problem, let's say that the forklift stops
dead and that it plus the pole deform by 1 cm so that
the distance traveled during the collision time is d ≈1
cm. Since these are very rough calculations, I will let
the mass of the forklift be m _{f} ≈5000
kg and the initial speed be v =8 mph≈3.6
m/s. If the forklift stops uniformly, it will stop in a
time t =2d /v =0.0056 s. The
force F experienced (by both the pole and the
forklift) will be F=m _{f} v /t ≈3.2x10^{6}
N=720,000 lb.

For the man-forklift collision, assume the forklift
keeps right on moving with the same speed but with the
man stuck to the front. Suppose that the man's body
compresses 3 cm during the collision, so the time of
collision will be t=0.06/3.6=0.017 s. The force now is
F=m _{m} v /t =68x3.6/0.017=1.44x10^{4}
N=3200 lb.

QUESTION:
what is the terminal velocity of a soccer ball sized piece of hail?

ANSWER:
I don't think there is hail that large! But, the calculation
is straightforward. I usually estimate the air drag of
something of cross section A having a speed v
to be ¼Av ^{2} =0.038v ^{2} .
When this is equal to the weight of the object, mg =(4πR ^{3/} 3) ρg =402
N, the object will have the terminal velocity; so v =√(402/0.038)=103
m/s=230 mph. I have used R =0.22 m, ρ= 920
kg/m^{3} , and g =9.8 m/s^{2} .

QUESTION:
A swing is suspended by nylon ropes and connected to a limb that is 30 feet in the air. The problem is that the kids cannot build any momentum in order to swing. They pull and kick their legs but the swing will not swing. Is there a way to fix this swing?

ANSWER:
30 ft is a very long pendulum. The period T of a
pendulum (the time it takes for one swing back to where it
started) is approximately T ≈2π √(L /g )
where L is the length and g =32 ft/s^{2}
is the acceleration due to gravity. In your case, L =30
ft, T ≈6 s. Now, if you watch a kid swing a much
shorter swing, you will see that they "pump" once per cycle;
this is called a driven oscillator and, driving with the
same frequency as the natural frequency of the swing is
called resonance and each pump will increase the amplitude.
No doubt the kids, being used to a much shorter swing, are
pumping with a period much shorter than 6 s, far off
resonance, therefore not having the effect of increasing the
amplitude. I suggest that you start them with a good push
and tell them to pump each time they reach the bottom of the
arc and moving forward.

QUESTION:
If a bullet of mass m moving at v strikes a stationary rod of mass
M at its center and passes through exiting with u , there is simple conservation of momentum if the rod is free to move/slide:
m (v-u )=MU . What happens if the bullet strikes and passes through the rod at its end (same
v and u ), causing the rod to move forward while at the same time start to spin?

ANSWER:
Your question is essentially the same as an
earlier question except that in
that question the collision was stipulated to be elastic. In
your case, there is no conservation of energy equation and
so there are only two equations —conservation of
linear momentum and angular momentum. But, there are only
two unknowns, U and ω , because you
stipulate u to be known. (Also, d=L /2 for
your question.)

ADDED THOUGHTS:
I thought it might be interesting to look at the energy
change for these collisions. It is straightforward algebra
to show that U=m (v-u )/M and ω= 6U /L .
The energy before the collision is E _{1} =½mv ^{2} .
Referring to the earlier answer, the energy after the
collision is ½mu ^{2} +½MU ^{2} +½I ω ^{2} .
For the simple case of hitting the center (ω= 0)
I find E _{2} =½m (u ^{2} +(m /M )(v-u )^{2} ),
and for the case of hitting the end I find E _{2} ' =½m (u ^{2} +4(m /M )(v-u )^{2} ) .
Since E _{2} ' >E _{2} ,
less energy will be lost in the case where the collision
happens off center (assuming u and v are
the same in each case). For example, if m=M /4 and
u =½v , I find ΔE =-(11/ 32)mv ^{2}
and ΔE' =-¼mv ^{2} . In
both cases, energy is lost in the collision.

QUESTION:
I have four axle stands, well made but not certified to any particular
weight.
If a vehicle is supported on four stands will each stand need to support
only 25% of the total weight?

ANSWER:
The answer is no, probably not. How much weight each stand
will carry depends on where the center of gravity of the car
is located. The singlest biggest weight of the car is the
engine so, in a front-engine car, the center of gravity is
going likely to be closer to the front axle than the rear;
so the front axle stands will bear a larger fraction of the
total than the rear. Usually the center of gravity is
equidistant from the left and right sides of the car, so the
left and right stands of both front or rear axles will
bear the same weight. To make this more quantitative,
suppose each stand in the front (rear) exerts an upward
force of F (R )
on the car as shown in my figure; the weight W
acts at the center of gravity. Now, if the center of gravity
is a distance d _{F} from the front axel and
a distance d _{R} from the rear axel, it may
be shown that F= ½Wd _{R} /(d _{R} +d _{F} ).
So, if d _{R} =d _{F} ,
F=W /4 or if d _{R} = 2d _{F} ,
F=W /3.

By the way,
F and R are
forces the stands exert on the car. Newton's third law
tells you that the car exerts equal and opposite forces on
the stands (action/reaction).

QUESTION:
I am constructing a floating dock out of pressure-treated lumber and 55-gallon empty plastic barrels and want to ensure that I provide sufficient buoyancy. Although the dock comprises a ramp and a platform, they will be only loosely connected (with slack rope and eye bolts), and the buoyancy calculations for the platform are easy; it's the ramp that is giving me headaches. Specifically, one end of the ramp will be sitting on the shore and the other end will be supported by the plastic barrels. The ramp is T-shaped, with the lakeside end wider to accommodate the barrels. Please give me some guidance as to how I might analyze a T-shaped structure with the narrow end on shore and the wide end kept afloat by barrels, so that a weight of w placed at any point on the ramp will not submerge the barrels more than 45% (because, although I think I did a good job sealing off the barrel caps, I'd rather not have to find out). Although I've meant for this question to be somewhat general to allow for design modifications, I will mention that I've built part of the ramp already, with the walkway being 3' wide and 8' long, and the cross of the T being 5' wide and 3' long and covering two empty barrels; I'm willing to add another section with more barrels if needed. Any help in analyzing buoyancy of a T-shaped ramp with the narrow end resting on land would be greatly appreciated!

ANSWER:
I will not be able to do any quantitative calculations without knowing
the weights of the ramp and dock. Also, will the level of the water
remain fairly constant?

FOLLOWUP :

I must mention that I modified the design to make the walkway eight feet longer, so that it is now 16 feet long. Also, I had another empty 55-gallon barrel on hand, and placed it lengthwise under the walkway at the end where the walkway joins the cross of the tee. The salient data for the components are as follows:

T-shaped ramp (total weight 648 lb)

Walkway (463 lb) is 3 feet wide and 16 feet long; a 55-gallon drum is placed under the walkway on the lake end.

Cross of tee (185 lb): 5 feet wide and 3 feet long, covering two 55-gallon drums.

The dock platform weighs 515 lb and is 8'x8'. Four 55-gallon drums support it.

My dock will be located in a tributary of the Potomac River. As such, the water is tidal, varying in depth between 2 and 5 feet. (This is the reason I did not join the ramp and the platform rigidly, as I anticipate that the angle of surface of the ramp relative to that of the platform will fluctuate, given that one end of the ramp rests on the
shore.)

In order to help visualize the situation, I am attaching three JPG images which I created in Sketchup. Please note that these images do not include the latest modification, whereby I lengthened the walkway and placed a barrel under its far end. Also, I recognize that you will probably make certain simplifications in order to expedite the analysis.
That is fine with me; I only wish to obtain a rough idea of the limits of motion of the ramp and dock as I walk upon them.

At high tide the walkway and platform are approximately
horizontal. At low tide the water level is about 2-5 feet
down.

ANSWER:

Some preliminaries:

The volume of each
barrel is 55 gallons=7.35 ft^{3} ;

the density of
water is 64.2 lb/ft^{3} ;

each barrel is 45%
submerged, so each barrel provides a buoyant force of
7.35x64.2x0.45=212 lb;

each barrel has a
weight of 21.5 lb;

I will first do the
platform which is easiest if you assume that the load is at
the center.

The weight of the
platform is 515 lb;

the weight of the
four barrels is 21.5x4=86 lb;

the weight of the
load will be denoted as W ;

the buoyant force
of the four barrels will be 4x212=848 lb.

Therefore,
W = 848-515-86=247
lb.

If
the load is not in the center, the total buoyant force will
still have to be 848 lb but the platform will tilt so that
two of the barrels will be submerged more than 45% and the
other two less than 45%. I made an estimate of how much the
platform would tilt if the load were moved over to 1 ft
from one side edge. Without going into details, the heavy side
would go down by about 2.8 inches and the other side would
go up by the same distance; the corresponding tilt would be
about θ =4.50. This would make two
of the barrels 65% submerged, beyond your desired limit.

Next I will look at the
walkway.

the weight of the
tee and two barrels under it as well as the net buoyant
force of those barrels act at 17.5 ft from the shore ;

the weight and
buoyant force of the third barrel act at 14.5 ft from
the shore

the maximum weight
W acts at a distance x from the shore;

there is a force
F exerted up by the shore which we will not
need to know;

I have assumed that
the ramp has no interaction with the platform, since
reference is made to "slack ropes".

All this is shown in my
diagram. If one now sums the torques about the point of
shore contact and sets that sum equal to zero, the product
Wx can be solved for.

0=212x14.5+424X17.5-228x17.5-21.5x14.5-Wx -463x8=2488-Wx .

Wx =2488 ft⋅lb.
So, for x =19 ft, the end of the ramp, W =131
lb. I am guessing that this result does not make you happy!

I am not sure how
rigidly coupled the platform and walkway are ("slack ropes"), but suppose
that they are coupled as if, when horizontal, they were
rigidly attached. So now the summed torque equation is
0=212x14.5+424X17.5-228x17.5-21.5x14.5-Wx -463x8+848x23-601x23=8169-Wx .
So W =8169/x . So, for x =19, W =430
lb and for x =27, W =303 lb. It would seem
that it is important that the coupling be designed such that
the platform can help hold up the walkway. I figure that the
walkway will only go down a maximum of about 15^{0}
at low tide; this should
not significantly alter the estimates I made for the horizontal
situation. So you
should allow a fairly rigid coupling like some kind of
hinge.

QUESTION:
If you have two metal spheres of the exact same volume, however with differing masses, say one sphere at 1kg and one at 10kg, attached each to an identical parachute, will they fall at different speeds? We know that two spheres of the same volume with differing mass will fall at a nearly identical speed, as the drag is identical. I have had it put forward to me that somehow adding a parachute into the system dramatically affects the outcome.

ANSWER:
"We know that two spheres of the same volume with
differing mass will fall at a nearly identical speed …"
Sorry, that statement is wrong. It is true, as you state,
that, since they have the same size and shape, the drag
forces will be the same on both. But, that force will have a
much bigger effect on the less massive sphere because it has
much smaller inertia. Adding identical parachutes will still
result in the drag forces being identical, but the more
massive ball+parachute will still fall faster. (You can
understand this intuitively. Imagine a bowling ball and a
balloon the same size. Drop them from a height of 2 m and
surely the bowling ball will hit the floor first.)

To be a bit more
quantitative, the drag force is of the form f _{D} =Cv ^{2}
where C is a constant determined by geometry and
v is the speed the object if falling. As the object
falls from rest, v gets bigger and bigger.
Eventually f _{D} will equal the weight
W and thereafter it will fall at constant speed because
the two forces add to zero: W =Cv ^{2} ,
so v _{t} =√(C /W )
where v _{t} is called the terminal
velocity.

QUESTION:
I have a 40' ladder that weighs 300 pounds.
Standing it up with it's foot
against a wall. And walking it up. It gets heaverier and heavier; at the
half way point it feels the heaviest. How much does it weigh half way stood
up?

ANSWER:
The weight of the ladder does not change if you go up it. It is always 300
lb. Why do you think it gets heavier?

FOLLOWUP QUESTION:
With the feet of the ladder against the wall..with ladder not extended...each section is like..21and a half feet long... Starting at the inter end of the ladder..standing it up you have to go ring by ring...it just feels like it weighs a ton before it is stood all the way up..feels
heaviest at a 35-40 degree angle . it takes all you can do to stand it up

ANSWER:
There are four forces acting on the ladder: the weight
W of the ladder, the force
F which you apply, the force N
which the floor exerts up, and the force f
which the wall exerts to hold the ladder from sliding. I
have assumed that F is applied
perpendicular to the ladder. At any angle θ
the force necessary to hold it can be determined by summing
torques about the corner of the wall and floor: Στ =0=FD- ½LW cosθ.
So, the force you must exert to hold up the ladder is
F =½W [(L /D )cosθ ].
Now, this is not so simple because as the ladder gets
higher, cosθ gets smaller but L /D
gets bigger. So, we have to make a further assumption about
how the ladder is lifted. Having done this, my recollection
is that I start off lifting the end straight up over my head
and then raise it by walking toward the wall lifting as I
go. So F is always applied at a
constant height h above the ground; then sinθ =h /D
and
cosθ= √[1-(h /D )^{2} ].
So, F =½W (L /D )√[1-(h /D )^{2} ].
So we should put in some numbers, W =300 lb, L =21',
h =7': F =(3150/D )√[1-(7/D )^{2} ].
The graph illustrates how F varies: starting at
D =21' the force you must apply increases from a little
less than half the total weight until you get about halfway
(10', just as you perceived qualitatively) where it is about
225 lb. For the rest of the way F decreases rapidly
to zero when the ladder is upright. The angle for the
maximum F is about 35^{0} , also about what
you observed.

ADDED COMMENT:
To determine in general the location of the maximum value of
F , set its derivative equal to zero and solve for
D . dF /dD =LW (2h ^{2} -D ^{2} )/[2D ^{4} √(1-(h /D )^{2} )]=0.
The root of interest to us is D=h √2, so for
h =7', D =9.90'. (The other real roots are
at D=-h √2 and D =± ∞.)

QUESTION:
If a wheel is rolling without slipping on an inclined plane, friction force is the one that provides the torque, why does the torque provided by friction increase as the angle of the inclined plane is increased yet the friction force decreases as the angle of inclination is increased?

ANSWER:
You are wrong about how f changes with θ .
You are probably thinking that f= μN=μmg cosθ ,
but this is static friction and the correct expression is
that f ≤μN , so in this problem
f is whatever it has to be in order to satisfy Newton's
laws, as long as it does not exceed μN , in which
case the wheel will not roll without slipping. This problem
has three unknowns, so you must generate three equations:

-f+mg sinθ=ma

N-mg cosθ= 0

fR=Iα=Ia /R=mRa

To simplify the algebra I have assumed that the wheel is a
hoop of radius R , so the moment of inertia is
I=mR ^{2} . So, from equations 1 and 3,
f=ma=m (g sinθ-ma )
or a = ½g sinθ
and f =½mg sinθ. You
can see that f increases with θ. The
third unknown is easily found, N=mg cosθ.

You can also calculate
the largest angle for which the hoop can roll without
slipping since the maximum static friction you can get is
f _{max} =μN=μmgcos θ _{max} =½mg sinθ _{max} .
Solving, θ _{max} =tan^{-1} (2μ ).
For example, if μ= 0.5, θ _{max} =45^{0} .

QUESTION:
Glad I found your site, my name is Ben, and this question is something I'm working on professionally, not homework. I use Autodesk Inventor's Dynamic Simulation to model collisions and some of the results for a specific simulation don't seem consistent. So, a colleague suggested we simplify the problem and work the numbers out by hand, but we can't figure it out. Here is a diagram:

A rod (blue) of length 2m and mass of 1kg can freely pivot about it's center (orange dot) which is connected to a frictionless track (purple) running in the Y direction. The rod is positioned at 45 degrees to the track, and a ball of mass 1kg travelling at 15 m/sec strikes the rod at 0.5m from the rod's center. Collision is perfectly elastic, friction is zero. How can I determine the final velocity of the rod along the track, and what is it's rotational speed? And more importantly, I need to know what effect the rod angle has on these two answers. Then I can solve and plot a chart for all angles from 0 to 90 and compare with my Inventor results. I'm really having trouble with the moment of inertia part, being that it's not struck in the center, or the end.

ANSWER:
There are important errors with the problem as you state it. The first
obvious error in this problem is that the direction and magnitude of the
final velocity of the ball is impossible. If the ball carries off all the
energy it came in with, the rod must end up with no energy. And,
conservation of linear momentum in the y direction (not conserved in the
x
direction) demands that the speed of the center of mass (COM) of the rod
along the rail would be 15 m/s meaning it would not be rotating. And, the
angular momentum relative to the COM of the incoming ball is equal and
opposite that of the outgoing ball, so the rod would have to be rotating to
conserve angular momentum. So I thought to simply redraw the picture but
with the final ball velocity having unknown components (2 unknowns). Two
other unknowns are the final speed of the COM and the angular velocity of
the rod about the COM. However, I only can see three equations: conservation
of energy, linear momentum (only y ), and angular momentum. I am still
pondering the question, but there is too little information or I am missing
something.

CONTINUED ANSWER:
Since the questioner is ultimately interested in a more
general solution than the special case, I will set it up
generally from the start and apply that solution to his
initial question; I will, however, set m _{1} =m _{2} =m
as soon as I have written the most general solutions to
simplify the algebra after setting up the problem. As noted
above, the problem begins with three conservation equations:

conservation of
linear momentum in the y -direction:
m _{1} v _{1} =m _{1} v _{2y} +m _{2} u _{2}

conservation of
energy: ½m _{1} v _{1} ^{2} =½m _{1} v _{2} ^{2} +½m _{2} u _{2} ^{2} +½Iω ^{2}

conservation of
angular momentum: m _{1} v _{1} S sinθ =Iω+m _{1} v _{2y} S sinθ-m _{1} v _{2x} S cosθ

The moment of inertia
of a thin rod about its COM of mass m and length
L is I=mL ^{2} /12. Therefore, if
m _{1} =m _{2} =m ,
the three conservation equations are:

v _{1} =v _{2y} +u _{2
}

v _{1} ^{2} =v _{2} ^{2} +u _{2} ^{2} +L ^{2} ω ^{2} /12

v _{1} S sinθ =L ^{2} ω /12+v _{2y} S sinθ-v _{2x} S cosθ

There
are now three equations and four unknowns, v _{2x} ,
v _{2y} , u _{2} , and ω.
To generate a fourth equation, consider the impulse
J delivered by the rod to the ball:
J = Δp =m (v _{2} -v _{1} )or
J _{x} =mv _{2x} and J _{y} =mv _{2y} -mv _{1} .
But, the impulse must be normal to the surface of the rod,
so tanθ=|J _{y} /J _{x} |=(v _{1} -v _{2y} )/v _{2x} .
The fourth equation is

v _{2x} =(v _{1} -v _{2y} )cotθ .

I first solved these
four equations for θ= 45^{0} , L =2
m, S =0.5 m, and v _{1} =15 m/s, the
conditions specified by the questioner; note that, because
it is a quadratic equation we are solving, we get two
solutions:

v _{2y} =(15,
8.33) m/s

v _{2x} =u _{2} =(0,
6.67) m/s

ω =(0,
14.15) s^{-1} =(0, 2.25) rev/s.

The meaning of the
first (trivial) solution is that there was no interaction
between the rod and the ball (you missed!). Another
situation for which we can check the solution intuitively is θ= 90^{0} ,
for which v _{2x} must be zero:

v _{2y} =(15,
4.1) m/s

v _{2x} =(0,
0) m/s

u _{2} =(0,
10.9) m/s

ω =(0,
16.35) s^{-1} =(0, 2.6) rev/s.

Finally, the general
solutions are:

v _{2y} =v _{1} (A -1)/(A +1)
where A =cot^{2} θ +1+[12S ^{2} /(L ^{2} sin^{2} θ )]

v _{2x} =(v _{1} -v _{2y} )cotθ= 2v _{1} cotθ /(A +1)

u _{2} =(v _{1} -v _{2y} )=2v _{1} /(A +1)

ω =[12(v _{1} -v _{2y} )S /(L ^{2} sinθ )]=[24v _{1} S /((L ^{2} sinθ )(A +1))].

The plots requested by
the writer are shown to the right.

ADDED NOTE:
An alternative way to specify the vector v _{2}
is to specify its magnitude, v _{2} , and the
angle φ it makes with the +x
axis:

ACKNOWLEDGMENT:
Thanks to "haruspex" at
Physics Forums for helping me realize how to get the
desperately sought fourth equation!

QUESTION:
Because the earth is rotating, is there a centrifugal force that is acting against gravity. If the earth stopped rotating, would a mass on the earth's surface weigh more?

ANSWER:
Yes, there is a centrifugal force (except at the poles). A
scale you are standing on would indeed read more if the
earth stopped rotating, but the increase would be too small
to notice. (It is customary to call "weight" the force which
the earth's gravity exerts on something, so in that context
you would not weigh more even though the scale would read
more.)

QUESTION:
I understand the practical aspect of buoyancy of a helium balloon and the Archimedes principle. But I don't understand why the balloon goes up. I understand that if it were in water, the the force on the bottom of of a 12 inch balloon would be approx 0.5 psi more than the top of the balloon and there would be a force upward. But in the atmosphere, the pressure differential would be very small. So why does it go up?

ANSWER:
It is exactly the same as in water, but the change in
pressure from the bottom to the top of the balloon is much
less. A balloon filled with air will have a buoyant force
less than the weight of the baloon so it will fall rather
than rise. Filled with hydrogen or helium, though, the
balloon will rise if the weight of the balloon plus contents
is smaller than the weight of the air it displaces; of
course a lead balloon will not rise even if filled with
helium. A hot-air balloon rises because if you heat air it
expands and becomes less dense.

QUESTION:
This question concerns angular momentum as related to motorcycles. If a motorcycle rests atop a trailer, but sits on a roller system, the bike will stand upright without any supports or tethers if the wheels are rolling (throttle is locked in the
"on" position and wheels are spinning). If the trailer itself travels forward in a straight line, the bike should remain upright. But if the trailer takes a sharp turn, what happens? Does the bike fall, or does it instead do what a rider does to achieve a turn-countersteer and remain upright?

ANSWER:
I have waited a long time to answer this question because I
am bothered by the way the problem is stated. First of all,
if the throttle is locked on, only the rear wheel will be
spinning, so we can discuss the problem by looking only at
the wheel. It is certainly correct that if the truck goes
straight the wheel will continue running upright (assuming
that the center of gravity of the bike is in the vertical
plane passing through the center of gravity of the wheel).
Imagine the bike and rollers to be mounted on a big "lazy
Susan" the base of which is bolted to the truck bed. So, if
a north-bound truck turns to the west, the angular momentum,
experiencing no torque, will remain constant and continue
pointing in the same direction (originally either east or
west). Viewed from inside the truck it will appear that the
whole bike rotated through 90^{0} relative to the
truck.

Now,
if the rollers are attached to the truck bed, when the truck
turns the rollers turn and the wheel, trying to not turn,
will come off the rollers at some point. We first need to
understand the physics relationship between the torque and
the angular momentum of the wheel. The rotational form of
Newton's second law is τ =ΔL /Δt ,
torque equals the time rate of change of the angular
momentum. The first figure shows the wheel, as seen from
above, turning through some small angle which results in a
change of angular momentum from L _{1}
to L _{2} and ΔL =L _{2} -L _{1} .
So the torque which you must apply to make it turn this
way is in the direction of ΔL .

If
you think you can just steer it as if it were not rotating,
you would fail. The second figure shows what would happen if
you try to steer it like your intuition would have you do it
by exerting a force like F in the
figure. The torque points up and so the wheel would not turn
but lean in the opposite direction from the way you would
lean on the bike if you were riding it and making a turn.
You can find some videos showing this by googling
gyroscope in a suitcase video .

If you want the wheel
to turn with the truck, you need to have a torque which
causes that. One way I thought of to achieve this was to
have strings attached from the axles on each side and the
truck bed below. These need to have no tension on them when
the truck is going straight. When the truck turns as
indicated, the tension in the string on the right-side
string will be bigger than the other side and there will
therefore be a net force N on the wheel. This results in a
torque in the horizontal plane which will cause a change in
angular momentum in the direction consistent with the wheel
turning with the truck. Two strings are needed because the
truck might turn either left or right.

QUESTION:
Is there a formula for calculating the side-ways deflection wind has on a lawn bowl(over and above the bias deflection ) running at 12
s, the time a bowl takes from delivery to stop over a 26 m distance over
bowling green grass?

ANSWER:
Once again, doing Ask the Physicist has led me to learn
something new. I never really knew anything about lawn bowls
other than it is done on grass and rolling balls are
involved. For the benefit of others who are ignorant of the
game, let me summarize by describing the ball. (A good
article on the physics of lawn bowls balls can be found
here .) The ball is not a sphere but rather an oblate
spheroid which makes it sort of like a door knob but not so
extremely flattened; but it is slightly more flattened on
one side of the ball than on the other which results in a
center of gravity being displaced to one side of the
equatorial plane as shown in figure (a). This results in a
tendency for the ball to curve left if it is rolling the
angular velocity shown in the figures; this motion is the
"bias" referred to by the questioner which I am to ignore.
When rolling in the x direction (figure (b)), there
is a frictional drag force called, rolling friction
D , which opposes the motion (v )
and eventually brings the rolling to a halt. If there is a
wind, there is a force W due to
the wind which tries to make the ball roll to the right
(figure (a)) but if it does roll, there will also be rolling
friction trying to keep it from rolling. In order for the
wind to have any effect at all, it is clear that we must
have W>D ; if this is not the case, there will only
be static friction in the y direction which will be
equal and opposite to W . A lawn
bowls ball has a mass of about m =1.5 kg and a
radius of about R =6 cm=0.06 m.

To get the equations of
motion for the x and y motions, we first
need expressions for D and W . The rolling
friction may be expressed as D=-μmg where μ
is the coefficient of rolling friction and mg is
the weight of the ball. The force due to the wind may be
approximated as W ≈¼AV ^{2}
where A=πR ^{2} is the cross sectional
area of the ball and V is the speed of the wind;
this approximation is only correct if SI units are used. The
equations of motion in the x -direction are

Here t is the
time and v _{0 } is the speed of the ball at
t =0. If the ball is rolling in the y-direction
because of the wind, the equations of motion are:

It should be noted that
if (¼AV^{2} /m )<μg , these
equations imply that the ball will accelerate opposite the
direction of the wind, obviously not correct; hence the wind
will have no effect on the ball if V <√(4μmg /A ).
In that case, a_{y} =v_{y} =y =0.

So, having found the
general solutions, let us now apply the solutions to the
specific case from the questioner. We are told that when
t =12 s, v_{x} =0 and x =26 m.
With that information you can solve the x -equations
to get v _{0} =4.32 m/s and μ= 0.037,
reasonable values compared to numbers in the
article I read. The area is 3.14x0.06^{2} =0.0113
m^{2} . The first question we should ask is what is
the minimum speed of the wind to have any effect at all:
V _{min} =√(4x0.037x1.5x9.8/0.0113)=13.9
m/s=31 mph=50 km/hr; this is a pretty stiff wind, so the
wind probably has no effect on bowling under normal
conditions. So, just to complete the problem, consider V =15
m/s=34 mph=54 km/hr.

The^{ }
trajectory during the 12 seconds is shown in the graph
below; after 12 seconds the ball will continue accelerating
in the y direction.

So the bottom line is that unless you are playing in a gale-force wind, the wind has no effect on the ball if the wind has no component along the original direction of the ball (which I have called the
x -axis). You can tell if wind makes a difference by simply setting the ball on the ground—unless the wind blows the ball away, you need not worry about its effect. If the wind is blowing in the +x or -x direction, that is a whole different thing, but
the questioner asked for the sideways deflection.

ADDED THOUGHTS: This question continues
to intrigue me and I have carried my investigation further.
The question originally stipulated "over and above the bias
deflection" so my whole discussion totally ignored the fact
that the ball, owing to its off-center center of mass, will
curve. At the very end of my answer I noted that if the wind
is not perpendicular to the path of the ball, it would
be a different story; indeed for a spherically symmetric ball I showed that, except for very strong winds,
a wind perpendicular to the path has no effect at all.
However, for an actual lawn bowls ball, the path curves to
where a wind in the y -direction might have a
significant component along the path. I have calculated
(graphed below) the
x and y positions of a realistic path with no
wind using equations (10) and (11) of the
article referred to above. To do these I used all the
numbers used above (R , m , A ,
v _{0} ,
μ ); I used the moment of inertia
for a solid sphere ( I _{0} =I _{cm} +mR ^{2} =(7/5mR ^{2} ))
and chose the COM off-center distance to be d =1 mm. As you can see, the curving is
substantial, carrying the ball about 4 m from its original
direction. You can see that now a wind of any magnitude can
have an effect on the trajectory. The angle φ
which the tangent to the trajectory is given in the article
as φ =(2/p )ln(v _{0} /(v _{0} -μgt ))
where p is a constant also given in the article. As
can be seen, once the trajectory leaves the x-axis the wind
contributes with the component of its force along the
trajectory; this has the effect of reducing the effect of
the frictional force causing the ball to slow down less
rapidly. However, this is now like having a time dependent
force of friction which, I believe, will lead to equations
of motion which will not have an analytical solution but
would have to be solved numerically.

QUESTION:
This is something I remember discovering when I was younger. If i would take a small cylindrical object (like a AA battery) , set it on a flat hard surface, then using my fingers I would apply downward pressure on the edge of the battery. This would create backspin on the battery but also shoot the battery across the floor. When the battery's finally stopped moving forward it would spin rapidly on its axis until stopping completely. Could you explain the physics behind movement of the battery?

ANSWER:
Problems like this are standard in intermediate-level
classical mechanics. Round objects can move translationally
on a horizontal surface by rolling without slipping or by
sliding. The two classic extremes of the slipping scenario
are a skid where the object is initially not spinning at all
(e.g . a bowling ball begins by approximately not
rotating but may end up rolling without slipping), or is at
rest but spinning (like "peeling out"); in both cases, the
problem is usually to find the time elapsed or distance
traveled before rolling sets in. In your case, the object is
initially both translating horizontally and rotating with a
backspin. What will happen depends on the initial conditions —the
initial speed and the initial angular velocity; also the
properties of the object will also matter—its mass
m , radius R and shape, and the coefficient of
kinetic friction μ . Three possibilities are that

it will reverse directions and come
back still spinning and slipping, eventually stop
slipping and roll (what I think you are remembering),

it will stop slipping and continue rolling in the
initial direction, or

it will stop dead.

In the figure there are three forces on the object: its
weight mg , the normal force
N up from the floor, and the
frictional force f which the floor
exerts on the sliding object; the object begins with
velocity v _{0} and angular
velocity ω _{0} . The
mass of the object is m and its radius is R .
The friction slows down both the velocity and the angular
velocity. Sliding ceases when v=-Rω (see
footnote*). If v _{0} is very small and ω _{0}
is very large, possibility #1 will happen; if v _{0}
is very large and ω _{0} is very
small, possibility #2 will happen; if v _{0}
and ω _{0} are just right, possibility
#3 will happen. I think that this gives you a good
qualitative overview of the physics of the problem.

For those interested in
the quantitative solution, I will give it here for a uniform
solid cylinder. For the translational motion the two
equations (taking +x as to the right, +y
up) are -f=ma and N-mg =0; since f=μN=μmg
we can write a=-μg . For rotational motion about
the center of mass (choosing the direction it is initially
spinning as positive), only the frictional force exerts a
torque, so -fR =-μmgR=Iα where
I is the moment of inertia about the center of mass
and α is the angular acceleration. For a
cylinder I =½mR ^{2} , so α=- 2μg /R.
So, we can write the equations for the velocity and
angular velocity as functions of time t : v=v _{0} +at =v _{0} -μgt
and ω=ω _{0} +αt=ω_{0} -2μgt /R.
Now, the time when slipping ceases will be when
v=-Rω ; solving, t =(v _{0} +Rω _{0} )/(3μg ).
Finally, put this into the equation for v to get
v (t )=(v _{0} -Rω _{0} )/3.
So if Rω _{0} >v _{0} ,
possibility #1 will happen; if Rω _{0} <v _{0} ,
possibility #2 will happen; and if Rω _{0} =v _{0} ,
possibility #3 will happen.

*The condition v=-Rω
is a little tricky. The negative sign is because of my
choice of the positive direction for ω which
is opposite that which would be the case for no slipping.

QUESTION:
For a snow plow that is very heavy, is there an advantage to having the
connection point of the winch line up high so that there is less weight to
pull or is there no difference?
I can send a picture for clarification.

ANSWER:
Yes send me a picture. You are talking about a winch which is used for what?
Pulling a stuck vehicle? Lifting and lowering the plow? What?

FOLLOWUP QUESTION:
Yes, lowering and lifting a plow. The problem is that the plow is out very far out and that my winch is mounted pretty low on my machine. So instead of the winch simply lifting the plow up, right now it is mostly pulling backwards and then that is making the plow come up.
I attached a picture of what the manufacturer recommends but I haven't had too good of luck with them in the past. The last two pictures are of my machine. You can see how far out the plow is and how low my winch is. Would that pulley and cable system helped at all?

ANSWER:
You may not want to get the full physics explanation here,
so I will first give you a qualitative explanation. The
tension T in the strap is what is
lifting plow and any part of it which is horizontal (T_{H} )
is wasted. Your gut feeling is right, "…it is mostly
pulling backwards…" Anything which you can do to
increase the vertical part (T_{V} ) will
make the lift easier, and moving the winch up is a good way
to do this but it might be easier to have a pulley
higher up which then brings the strap back down to the
winch.

The figure shows all
the forces on the plow assembly: the weight W
which acts at the center of gravity (yellow X), the tension
T in the strap (where I have shown
vertical (T_{V} ) and horizontal pieces (T_{H} )), and the force the
truck exerts on the support which I have represented as its
vertical and horizontal parts, V
and H respectively. (Note that the
various forces are not drawn to scale since T has
to be much larger than W to lift the plow.) Suppose T
is just right that the plow is just about to lift. Then the
sum of all the forces must add to zero, or V+T_{V} -W =0
and H-T_{H} =0. The sum or torques must also
add to zero; summing torques about the point of attachment
to the truck (light blue X), WD+T_{H} s-T_{V} d =0.
Note that the T_{V} is trying to lift the
plow but T_{H} is trying to push it down.
Now, in order to get a final answer for the unknowns (which
are T , V , and H ) we note that
T_{V} =T sinθ and T_{H} =T cosθ
where θ is the angle which the strap to the
winch makes with the horizontal. The final answers I get
are:

T=DW /(d sinθ-s cosθ )

V =W-T sinθ

H=T cosθ

I put in some
reasonable numbers just to get an idea of the answers, W =500
lb, θ =20^{0} , D =2m, d =1.8
m, and s =0.1 m. Then T =1920 lb, H =1800
lb, and V =-157 lb. The negative value for V
means that V is down, not up. In
this scenario, the pulling force has to be nearly four times
greater than than the weight being lifted.

Now
we need to look at whether the manufacturer's suggestion
will be better than a straight shot to the winch. Now there
are two forces pulling up, the tension T
from the pull point to the winch and the tension
P from the pull point to some anchor
higher up. Of course the magnitudes of these two tensions
are the same, P=T . The picture shows only the
pulling forces, the rest are the same as in the picture
above. There are still three unknowns, T , V ,
and H . I will call the angle that P
makes with the horizontal φ . I will not show
the details, just give the final results:

As a numerical example,
I will use the same numbers as above and add φ= 40^{0} .
Then T =624 lb, H =1070 lb, and V =-115
lb. It is definitely advantageous to use the manufacturer's
suggestion here which, in my numerical example, reduced the
force the winch needed to exert by a factor of about 3.

QUESTION:
What is the ratio between the height H of a mountain and depth h of a
mine,if a pendulum swings with the same period at the top of the mountain
and at the bottom of the mine?

ANSWER:
This must be a homework question where you assume that the density of the
earth is uniform.

QUESTION:
No it's not. I am a student who is trying to crack NEET. While I was solving questions, I saw this one but I couldn't get the explained answer.
The answer I got was 1,but the answer given here was 1/2. I just wanted to know was my answer correct.

ANSWER:
There is no way to solve this problem without an assumption
regarding the density of the earth. The standard assumption
in introductory physics is to assume that it is uniform
(which is
not a good approximation); I will do that. Also, I
will assume that it is a simple pendulum with a small angle
amplitude so that the period is T ≈2 π √(L /g )
where L is the length and g=MG /R ^{2} is the
acceleration due to gravity. At a distance H above
the earth's surface g _{H} =MG /(R+H )^{2} ;
at a distance h below the surface of a
uniform-density earth, g _{h} =MG (R-h )/R ^{3}
(see footnote*). Now, for the periods to be equal it is
necessary that g _{h} =g _{H} ;
a little algebra shows that equivalently (1-(h /R ))=1/(1+(H /R ))^{2} =g _{H(h)} /g .

It is instructive to
look at h /R as a function of H /R :
h /R =1-1/(1+(H /R ))^{2
} shown in the first graph above. When h=H =0
you are at the surface of the earth; when h /R =1,
H /R =∞ and g _{h} =g _{H} =0.

Next, look at the
plots of g _{H(h)} /g as a function of H /R
(black) and of h /R (Red). There
are only two locations where g _{h} =g _{H} ,
at the surface where h /R =H /R =0
corresponding to g _{h} =g _{H} =g
and near h /R =H /R =0.6.
The third plot shows a closer look around 0.6 showing h /R =H /R =0.618
corresponding to g _{h} =g _{H} =0.382g .

This was a pretty
long-winded answer, but the upshot is that you were right
and the answer key was wrong: h /H =1.

*M is the mass
of the earth, R is the radius of the earth, and
G is the universal gravitational constant.

ADDED NOTE: Actually, I did not need to
solve this problem graphically, I could have solved it
analytically. If you assume (guess) that h /R=H /R≡x ,
then the equation to determine x is 1-x =1/(1+x )^{2}
or (1-x )(1+x )^{2} -1=0=(1-x ^{2} )(1+x )-1=
x ^{3} +x ^{2} -x=x (x ^{2} +x -1)=0.
One solution is x =0 (the surface) and the positive
solution to the quadratic equation is x = ½( √(5)-1)=0.618.
I guess I shied away from this because I know I am not very
good at solving cubic equations, but I can handle this one!

QUESTION:
For a book pushed horizontally against a vertical wall, we know that the friction force is equal to the weight of the book. We also know that the friction force is equal to the normal force multiplied by the coefficient of friction. So, technically speaking, the harder you push against the book, the greater the normal force. Therefore, the friction should get bigger the harder the you push, but we know that isn't true since friction is equal to the weight of the book, which doesn't change when you press the book harder to the wall. Thus the only way to appease the equation: (mu)(normal force) = (weight of book), with an increase in normal force (caused by an increase in applied force to the book), that would mean the (mu) or coefficient of friction has to change. So my question is, why does the coefficient of friction change when we apply more force to the book on the wall? Its still the same two surfaces!

ANSWER:
You should have stopped after the first sentence which is
correct! You misunderstand static friction. Although it is
true that f= μ _{k} N
for kinetic friction, for static friction the equation
f= μ _{s} N
is not true. The correct formula is an inequality rather
than an equality:
f ≤ μ _{s} N.
For one single force you may write an equation, f _{max} =μ _{s} N
which is the largest frictional force you can
get for a given N . If you do not push your book
with a force equal to or larger than N=mg /μ _{s}
it will fall to the floor. (The Q&A right
below yours is closely related to your question.)

QUESTION:
hi, so, there is a fairly recent
video going around the internet of nasa ISS astronaut Randy Bresnik spinning a fidget spinner, in space, in a quite low gravity environment and then apparently grabbing hold of the spinner then video cuts and we, presumably some short time later, see the entire body of the astronaut holding the spinner spinning fairly rapidly in space.
Perhaps it is a quite elementary question, but, I wonder if that video might have been faked, wondering if the what must be a relatively small amount of energy in the spinner could cause the entire body of mass of astronaut plus the spinner to spin in the fashion observed in the video. That is to say: wouldn't the energy from the spinner, say, be absorbed by the astronauts body or the muscles in his arm, or some such, upon grabbing the spinner; or, if there was movement, wouldn't it be far far less than what is demonstrated in the video?

ANSWER:
You are right, I think the astronauts are messing with you
here! You have looked at it from the perspective of energy
conservation; however, energy would not be conserved here
and energy would actually be lost, not gained as it appears.
So, your reasoning was sound —where did the
energy come from? What is conserved is the angular momentum.
If the moments of inertia are I _{toy} and
I _{man} and the original angular velocity
of the toy is ω _{1} , then the
initial angular momentum is L _{1} =I _{toy} ω _{1}
and the final angular momentum is L _{2} =(I _{toy} +I_{man} ) ω _{2}
where ω _{2} is the angular velocity
of the man+toy. Conserving angular momentum and solving for
ω _{2} , ω _{2} =[I _{toy} /(I _{toy} +I_{man} )]ω _{1} <<ω _{1} .
Now, I actually found the moment of inertia of a fidget
spinner,
I _{toy} ≈7x10^{-5} kg·m^{2}
and I estimate
I _{man} ≈70 kg·m^{2} .
This means that ω _{2} =ω _{1} /1,000,000!
The astronauts have angular velocity of about 1 revolution
per second and you know perfectly well that the fidget
spinner did not have a speed of a million revolutions per
second.

Finally, if you are interested, you can show that energy is
not conserved. The expression for kinetic energy is E =½Iω ^{2
} so E _{1} =½I _{toy} ω _{1} ^{2
} and E _{2} =½[I _{toy} /(I _{toy} +I_{man} )]ω _{1} ^{2} ≠E _{1} .

QUESTION:
How would i explain bus or a truck(high CG)flip over, when driving in a curve with high speed, from a inertial reference frame? If centrifugal force is fictitious and bound to non-inertial reference frame, what torque causes truck to flip over observed from an inertial ref. frame?

ANSWER:
I have attached figure from the
earlier answer
where I did essentially this problem in the noninertial
frame; this saves me having to draw the whole picture again.
For your problem, the vector labelled C
is zero. Part of what makes your problem harder
than in the accelerating frame is that you must sum torques
about the center of mass (green x). For equilibrium, the sum
of torques must be zero; summing torques gives 0=H (f _{1} +f _{2} )+LN _{1} -LN _{2} .
(Translational equations are f _{1} +f _{2} =mv ^{2} /R
and N _{1} +N _{1} -W=0.)
Now, when the truck is about to tip over the left wheel is
about to leave the ground so N _{1} =f _{1} =0,
f _{2} =mv ^{2} /R ,
and N _{1} =W ; therefore Hmv ^{2} /R-LW= 0.
So, if v > √[LWR /(Hm )]
the truck will flip over.

QUESTION:
Suppose I am in space, in a 100 meter barrel strapped to one end. The barrel and I have mass 100 kg (I'm pretty light). I throw a 100 kg ball down the center of the barrel towards the other end, accelerating it to 1m/s velocity. Presumably, it takes the ball 100 seconds to reach the other end and it sticks because it is covered in velcro. I presume the barrel started in motion in space when I threw the ball from the reaction from the force I applied to the ball and it stopped when the ball hit the other wall. Could I keep moving the barrel by throwing more balls? How is this possible without violating Newtons laws? Surely this is not a case of propellantless thrust?

ANSWER:
These 100 kg balls just materialize from nothing, do they?
So you must have a supply of them which means that the total
mass of the entire barrel and its contents is much bigger
than 100 kg which means you go much less far with each
throw. Eventually you will run out of balls, but you will
have gone some distance. What propelled the barrel? You did.

FOLLOWUP QUESTION:
Thank you so much for you previous answer. My question about throwing balls at the back of a barrel in space was inspired by what I had read about some experts criticizing the "EM Drive" for violating Newton's Laws. Basically the writers quote experts as saying that you can't have forward thrust without throwing mass out the back " no such thing as propellantless thrust". I suppose these experts are simplifying their response for the lay press because qualifying their answer would take up to much column space. In retrospect, I should have known that some net movement was possible as I believe NASA uses large gyroscopes to adjust the pointing of some space telescopes.
This did get me to thinking though. Suppose I attach a laser to one end of the barrel and fire a single photon towards the other end. Presumably, momentum is imparted to the wall attached to the laser and the barrel moves through space as the photon moves to the opposite wall where it is absorbed. Now I wonder why I can't keep firing photons and moving the barrel. Is the hitch in my "propellantless thrust" scheme come from the absorption process? Is the heat generated through absorption dissipated out of the barrel preferentially so as to counteract my momentum drive?

ANSWER:
The
situation is similar to your first question since, as you
apparently know, photons have energy and momentum. But, to
operate your laser you have to supply energy somehow.
Suppose that you use a fusion reactor to generate that
energy. Then as you shoot more and more photons, the whole
ship gets less and less massive. But, energy has to be
conserved (mass is a form of energy, mc ^{2} )
and when the photons get absorbed, the heat generated will
"retrieve" the lost mass. Eventually you will run out of
fuel for your reactor and you will be right back to being at
rest. (I am assuming that the rules of the game forbid
letting the heat at the back radiate into space.)

QUESTION:
Why does a satellite shot straight up from the equator deflect to the west?

ANSWER:
There are two ways to look at this.

If you view it from outside the
earth, the satellite will continue moving directly away
from the center of the earth.
But, viewing the earth
from above the north pole, the earth rotates
counterclockwise, west to east.
So the earth is actually rotating under it which gives
it the appearance of deflecting to the west.

The second is a little fancier and harder to understand.
If you view Newton's laws from a rotating coordinate
system (the rotating earth with you on it) you find that
they are wrong; these "laws" of physics only work in
inertial frames of reference. However you can force
Newton's laws to be correct if you invent just the right
fictitious forces. I will not go into all the complexity
here, but one of the fictitious forces is called the
Coriolis force and can be written as F =-2m ω xv
. Since the vector ω
points along the earth's axis and out
through the north pole and the velocity vector
v points radially outward, the
negative of their cross product points west. Hence, it
will appear that the satellite is pushed west when
viewed from the ground.

QUESTION:
I am a safety coordinator in a warehouse operation. I am attempting to impress the need for safety with my co-workers. To that end, I am trying to answer the following:
A 10,636 lb. forklift (4825 kg) traveling at a speed of 8 mph (13 kph) strikes a fixed object (metal pole). How much force is transferred during this collision?
.
Secondly, same forklift, same traveling speed strikes a 150 lbs. (68 kgs) person. How much force is transferred to the person in this collision?

ANSWER:
There is no accurate way to calculate the forces. They depend on the details of the collisions.
In particular, how long do the actual collisions last? Also, how do the forklift and the person move after the collision has occurred (assuming the metal pole stays stationary)?
I could make some rough estimates to get an idea:

For the first problem, let's say that the forklift stops
dead and that it plus the pole deform by 1 cm so that
the distance traveled during the collision time is d ≈1
cm. Since these are very rough calculations, I will let
the mass of the forklift be m _{f} ≈5000
kg and the initial speed be v =8 mph≈3.6
m/s. If the forklift stops uniformly, it will stop in a
time t =2d /v =0.0056 s. The
force F experienced (by both the pole and the
forklift) will be F=m _{f} v /t ≈3.2x10^{6}
N=720,000 lb.

For the man-forklift collision, assume the forklift
keeps right on moving with the same speed but with the
man stuck to the front. Suppose that the man's body
compresses 3 cm during the collision, so the time of
collision will be t=0.06/3.6=0.017 s. The force now is
F=m _{m} v /t =68x3.6/0.017=1.44x10^{4}
N=3200 lb.

QUESTION:
We know that if a body rotates about an axis that is not the axis of symmetry then the angular momentum about a point on the axis does not point along the axis. If we rotate such body about such axis at a constant angular speed then a contradiction arises: since angular acceleration is zero, the torque about that axis has to be zero but since the angular mometum vector is changing (since the axis is not the axis of symmetry) a net torque is required. Where am I getting wrong? Please help.

ANSWER:
Your question is a little ambiguous since you do not clearly
specify which axis you are talking about after the first
sentence of your question; if you just say axis, I do not
know if you mean the symmetry axis or the rotation axis. In
the laboratory frame, though, the angular momentum points
along the rotation axis and does not change. As you state,
there are no torques on the system about the rotation axis
so the angular momentum is not expected to change. But, you
are probably looking at the problem from the body frame and
see that the angular momentum is changing in that frame; but
the body frame is not an inertial frame and therefore
Newton's laws are not applicable (in particular, torque is
not necessarily equal to the rate of change of angular
momentum). That is probably what you are getting wrong.

QUESTION:
I've been thinking about this one, and cannot find a good way to reason it out. if you are in an elevator that is going down and you jump in there while its going down would your head hit the elevator's ceiling??

ANSWER:
If the elevator is moving with constant speed, it is exactly
as if it were standing still —if you give
yourself an adequately large velocity you will hit the
ceiling. If d is the distance from your head to the
ceiling at the instant that your feet leave the floor, then
you will hit the ceiling if the speed v you launch
with is greater than √(2gd ) where g
is the acceleration due to gravity. If the elevator has an
acceleration a , the critical speed will be √(2(g-a )d ).
For example, if a=g you are in free fall and even
the tiniest velocity will cause you to hit the ceiling
(assuming that the elevator does not get to the ground
before you get to the ceiling).

QUESTION:
A swing is suspended by nylon ropes and connected to a limb that is 30 feet in the air. The problem is that the kids cannot build any momentum in order to swing. They pull and kick their legs but the swing will not swing. Is there a way to fix this swing?

ANSWER:
30 ft is a very long pendulum. The period T of a
pendulum (the time it takes for one swing back to where it
started) is approximately T ≈2π √(L /g )
where L is the length and g =32 ft/s^{2}
is the acceleration due to gravity. In your case, L =30
ft, T ≈6 s. Now, if you watch a kid swing a much
shorter swing, you will see that they "pump" once per cycle;
this is called a driven oscillator and, driving with the
same frequency as the natural frequency of the swing is
called resonance and each pump will increase the amplitude.
No doubt the kids, being used to a much shorter swing, are
pumping with a period much shorter than 6 s, far off
resonance, therefore not having the effect of increasing the
amplitude. I suggest that you start them with a good push
and tell them to pump each time they reach the bottom of the
arc and moving forward.

QUESTION:
i'm traveling at the speed of sound and a gunshot is fired at the exact moment I am passing the gun, what is the resulting sound that I hear and for how long?

ANSWER:
The figure shows the plane at three times:

just after the gun has been shot;

at the time when the sound reaches
where the plane had been at time 1;

at a time twice as long as 2.

Also shown are the corresponding spherical wave fronts of
the sound from the shot. As you can see, the wave fronts
never catch up with the plane. You will never hear the gun.

QUESTION:
I am a nurse at a long term care facility. My back hurts every time I get finished pushing a medication cart for the 8 hour shift. My question is...If the medication cart weighs 220 pounds when assembled, how much weight am I pushing given the fact it is on wheels?

ANSWER:
You may assume that the wheels almost remove the frictional force of moving forward; in other words, it takes very little force to keep it moving once it is up to speed. The times you need to exert a significant force on the cart would be when it speeds up or slows down. The more quickly you bring it up to speed or bring it to rest, the larger force you need to exert. So plan ahead and speed it up or slow it down gradually. Here is an example: if you sped the cart up to a speed of about 6 ft/s in 1 s, you would need to exert a force of about 40 lb, whereas if you took 2 s, the needed force would be about 20 lb.
Also, when you turn a corner, go slowly since it takes a
force to turn the cart also and the faster you take the
corner, the greater the required force.

QUESTION:
If a bullet of mass m moving at v strikes a stationary rod of mass
M at its center and passes through exiting with u , there is simple conservation of momentum if the rod is free to move/slide:
m (v-u )=MU . What happens if the bullet strikes and passes through the rod at its end (same
v and u ), causing the rod to move forward while at the same time start to spin?

ANSWER:
Your question is essentially the same as an
earlier question except that in
that question the collision was stipulated to be elastic. In
your case, there is no conservation of energy equation and
so there are only two equations —conservation of
linear momentum and angular momentum. But, there are only
two unknowns, U and ω , because you
stipulate u to be known. (Also, d=L /2 for
your question.)

ADDED THOUGHTS:
I thought it might be interesting to look at the energy
change for these collisions. It is straightforward algebra
to show that U=m (v-u )/M and ω= 6U /L .
The energy before the collision is E _{1} =½mv ^{2} .
Referring to the earlier answer, the energy after the
collision is ½mu ^{2} +½MU ^{2} +½I ω ^{2} .
For the simple case of hitting the center (ω= 0)
I find E _{2} =½m (u ^{2} +(m /M )(v-u )^{2} ),
and for the case of hitting the end I find E _{2} ' =½m (u ^{2} +4(m /M )(v-u )^{2} ) .
Since E _{2} ' >E _{2} ,
less energy will be lost in the case where the collision
happens off center (assuming u and v are
the same in each case). For example, if m=M /4 and
u =½v , I find ΔE =-(11/ 32)mv ^{2}
and ΔE' =-¼mv ^{2} . In
both cases, energy is lost in the collision.

QUESTION:
Lately I've been dealing with the Work, Force and Energy thing. I can't deny the fact that understanding some of the equations included in most of the textbooks has been one hell of a challenge to me. I'm currently trying the understand something here: we know that the work done (W) is equal to the change of kinetic energy. Is it also the same when it comes to potential energy? Does it amount to the done work as well? I never see the work of the gravity force included in the equations. I think I know the answer but I just want to make sure. There's something more bugging though--when we write the conservation of energy equations for a certain setting, do we include the work there or is it already included? For example, if there's a body on the top of an inclined plane and we take friction into account, should the equation be "potential energy at the top + kinetic energy at the top = potential energy at the bottom (==0) + kinetic energy at the bottom + the work done by the friction"? Getting an answer would mean a lot to me as an engineer-to-be. :)

ANSWER:
This question verges on a violation of a "concise,
well-focused" question as required by site groundrules. I
will do a couple of examples to try to clarify work-energy
for you. The "guiding principle" which is always true is
that the the total work W _{total} done by
all forces on an object of mass m is equal to
the change in kinetic energy, ΔK =½mv _{2} ^{2} -½mv _{1} ^{2} .
The picture is a mass m on an incline θ
being pushed by a constant force F up the incline
and moving a distance s ; for simplicity, I choose
m
to be at rest when you start pushing. Other forces on m
are its own weight W=mg , vertically down, and a
possible frictional force f pointing down the
plane. For the time being I will assume f =0. Only
the component of W along the plane W_{s} =-mg sinθ
will do work. The net work w (sorry if it is confusing to
have big W and little w ) is w=Fs-W_{s} s=Fs-mgs sinθ= ΔK= ½mv^{2}
and so v =√[2s (F -mg sinθ )/m ].

Rather than derive the
gravitational potential energy, U=mgy , I will
assume that you already know that. I will now do the problem
over but using potential energy. w=Fs=E _{2} -E _{1} =(K _{2} -K _{1} )+(U _{2} -U _{1} )=½mv^{2} +mgh= ½mv^{2} +mgs sinθ=Fs.
Solving this, you find exactly the same answer: v =√[2s (F -mg sinθ )/m ].
So here is how I look at it: potential energy is a very
clever bookkeeping device to keep track of the work done by
a force which is always present like gravity. It just makes
life a lot easier to not always have to calculate the work
done by some force that you know is always there. So, here
is the so-called work-energy theorem: the change in total
energy, kinetic energy plus any potential energy you have
included, is equal to the work done by all external
forces, E _{2} -E _{1} =W _{ext} ;
here an external force is any force for which you have not
introduced a potential energy function. I like to rearrange
the work-energy theorem as E _{2} =E _{1} +W _{ext} —what
you end up is what you started with plus what you added (or
subtracted if W _{ext} <0).

Finally, let's include the frictional force, f=μW_{N} =μmg cosθ.
(μ
is the coefficient of kinetic friction.) The work which the friction does is negative because it
is opposite the direction of s , w _{f} =-μmgs cosθ.
Therefore, W _{ext} =Fs -μmgs cosθ=½mv^{2} +mgs sinθ ;
solving, v =√[2s (F -mg (sinθ+μ cosθ ))/m ].
You might wonder why we did not introduce a potential energy
function for f . The reason is that there are two kinds of
forces in nature, conservative forces and nonconservative
forces and the latter kind cannot have a meaningful
potential energy function; friction is a nonconservative
force. But this answer has rambled on long enough and that
is a topic for another day!

ADDED NOTE:
I see that I did not answer your question ("…if
there's a body on the top of an inclined plane and we take
friction into account, should the equation be "potential
energy at the top + kinetic energy at the top = potential
energy at the bottom (==0) + kinetic energy at the bottom +
the work done by the friction"?" ) explicitly.
You have it wrong. It should be that the total energy at the
bottom equals the total energy at the top plus the work done
by friction; don't forget that, as in my example going up
the plane, the work done by the friction is negative. Some
books like to write it your way replacing the friction part
by the work done against
the friction ; I actually do not like that one bit.

QUESTION:
Work done by ship's engine = KE of ship + Work done to overcome frictional forces. Work done by ship's engine - work done to overcome frictional forces = KE of ship Eventually when the ship travels at constant speed, and all work done by engine is used to overcome frictional forces, then mathematically,
Work done by engine - work done to overcome frictional forces = 0
But KE of the ship is not zero.
So where is the source of the ship's KE coming from?

ANSWER:
While the ship is accelerating, the engine is both overcoming drag and accelerating the ship. The drag is dependent on the speed, gets larger as the speed gets larger. Eventually, the drag will become equal the force the engine applies so the net force will be zero so the ship will move at constant speed. But I do not understand your question since the work done by engine was clearly adding kinetic energy to the ship during the acceleration time. Your error was to assume that the work to overcome friction was always the same which is clearly not the case.

QUESTION:
I would like to perform a calculation of a man descending a tower using cords and subjected to the action of the winds. I have been trying to find some equations but it is somewhat difficult due to the drag force. My main objective is to calculate the maximum horizontal distance
x that the man could reach due to the wind action against the technician descending the tower using cords.
Tower height: 78 m (please consider up tower as a zero reference).
wind speed: 20 m/s
Mass of man: 70 kg
Man descending with constant speed and slowly.
So, please what is the maximum distance when the man is at 66 m from the top of the tower ?

ANSWER:
As shown in the figure, there are three forces on the man, his weight
mg , the tension in the cords T ,
and the force of the wind F . The equations of
equilibrium are F-T sin θ =0 and mg-T cosθ =0.
Solving, tanθ =x /y =F /(mg ),
so x=y tanθ =Fy /(mg ). Now, how
can we get F ? There is a very good approximation to the drag
force by air at sea level moving with speed v : F ≈¼Av ^{2}
where A is the area of the object presents to the onrushing air
(and which works only for SI units). Finally, x =Av^{2} y /(4mg ).
If I approximate g ≈10 m/s^{2} and A ≈1
m^{2} and use your numbers, x ≈9 m.

ADDED
COMMENTS:
I should have emphasized that air drag calculations are only rough
calculations, probably accurate to maybe ±20%. Also, for
your situation, the general approximation for x as a function
of y is x≈y /7. Also, if the horizontal displacement seems
too large, keep in mind that 20 m/s is a very strong wind, about 45 mph
which is gale force.

QUESTION:
Hi. I'm creating an aluminum can crusher and I need to know the weight required in order to crush the can from a set height knowing the amount of Pa required to crush the can.
I'm crushing a regular sized coke (375ml) can and need to get it to about 4mm long once crushed. On a pneumatic crusher I only needed 600000 Pa or 6 Bar to achieve this result. However I am trying to create another method of crushing the can using a weight (unknown) that drops 0.5m on top of the can, theoretically crushing it to the required length.
My question is: what weight in Kg is required to crush a can when dropped 0.5m to reach a crushed can length of 4mm, or a conversion of 6 bar? And maybe the equation used to work it out?

ANSWER:
I looked up the geometry of the can: the height is about 128 mm and the
radius is about 30 mm, so the crush distance is s =128-4=124
mm=0.124 m and the area of the top is about π (0.03)^{2} =2.8x10^{-3}
m^{2} . The force which the press exerted on the can was
F=PA =6x10^{5} x2.8x10^{-3} =1680 N. This means if
you put a mass 1680/9.8=170 kg it will crush the can. Your idea, I
presume, is to drop a smaller mass from some height (0.5 m) above can.
The way I will attack this problem is to first calculate the work done
by the hydraulic press W _{1} and equate it to the work
done by the dropped mass W _{2} ; I can then solve for
the mass for any height. I will do it in general so you can estimate the
mass from any height. W _{1} =Fs =1680x0.124=208
J; v =√(2gh ) where g =9.8 m/s^{2}
and h is the height (0.5 m in your case); W _{2} =½mv ^{2} -mgs =mgh-mgs =208
J. Solving, m =208/[g (h-s )]. In your case,
h =0.5 m and s =0.124 m, m =56 kg.
Keep in mind that this is just an
estimate. I would be curious to know if it was close. There is an
earlier answer which might
be of interest to you; it might give you an idea of how you might
improve your crusher.

QUESTION:
If a ball impacts a rod off center from the rods center of mass how does
the rotational momentum and energy gained by the rod affect the amount of
gained translational momentum and energy? For example, if the ball impacted
at the rods center of mass the equations for conservation of momentum and
kinetic energy could be used to determine the velocities of the objects
after the collision, but with energy and momentum being transferred to the
rotation of the rod would the ball loss more or less velocity or would the
energy transferred remain the same? Would the rotational energy and momentum
for the objects after the collision be solved for separately from the
translational transfer?

ANSWER:
No homework.

REPLY:
Thank you for the response. I'm unsure if this could be a homework question for someone else, but I've posted in forums with no response or had the post deleted. This is an attempt to gain an understanding of basic physics concepts and I was given this site as a possibility to find answers from a local university. Would it be possible for me
to ask the question again in a couple of months to certify that this is not a homework question or does "no homework" amount to no homework "type" questions that could require tutoring or classes?

ANSWER:
Well, you seem sincere about wanting to learn the physics, so I am happy
to answer your question. The algebra to get the final solution is very
tedious, but it seems you are most interested in getting the physics
right, so I will set the problem up and if you really need all the final
answers for a particular situation, I will leave that to you. Since you
refer to conservation of energy, I assume the problem of interest is an
elastic collision. The system is shown before and after the collision in
the figure. A point mass m with velocity v
approaches a uniform thin rod of mass M and length L ;
v is normal to the rod and the collision
occurs at a point a distance d from the center of mass of the
rod. After the collision, the center of mass of the rod has a velocity
U , the point mass has a velocity u ,
and the rod has an angular velocity ω
about the center mass. Because the rod is uniform, its moment of
inertia about the center of mass is I=ML ^{2} /12; the
angular momentum and kinetic energy of a rigid body are L=I ω
and K =½I ω ^{2}
respectively. There are no external forces or torques on this system, so
angular momentum and linear momentum are both conserved. We now have
three equations:

energy
conservation: ½mv ^{2} =½mu ^{2} +½MU ^{2} +ML ^{2} ω ^{2} /24

linear
momentum conservation: mv=mu+MU

angular
momentum conservation: mvd=mud +ML ^{2} ω /12

Take stock of
what we have: three equations and three unknowns—u , U ,
and ω. The physics is done, only algebra remains.

QUESTION:
I have attached some images to show the question. In one, a heavy
package is affixed to the drone firmly.
In the other it hangs by a rope that is affixed on the same level as the engines.
The weight of the drone is 5 lb and the weight of the load is 7 lb. Where has the center of gravity moved?

ANSWER:
In the diagram the yellow X s are the centers of gravity
(COG) for the unladen drone and the load; the load COG is a distance
L below the drone COG. The red X is the COG for
the combination. Suppose the weight of the drone is W _{drone}
and the weight of the load is W _{load} . Then the
distance d by which the new COG is below the COG of the drone
is d=LW _{load} /(W _{drone} +W _{load} ).
In your example, d =(7/12)L . Clearly, when L
is increased d is increased. And, contrary to your expectation,
the COG goes farther down when you hang the load from a rope increasing
L . What might be a problem for stability, though, is that if
the drone is not horizontal the rope will still be vertical and so the
COG will no longer be on the center line of the drone.

QUESTION:
I take my chocolate Lab to the river to play. The ball-flinger (Chuck-it) imparts backspin to the ball, and water flied off. F=MA - the ball is losing mass; is it accelerating?

ANSWER:
The mass is not disappearing, so the total mass of the system is not
changing. You should actually think of Newton's second law as
F =dp /dt where
p =mv is the linear
momentum, so if there is no net force the momentum does not change.
Then, if you watch the water coming off one drop at a time you can
calculate the new speed of the ball due to the expulsion of the drop;
therefore the ball does accelerate in that sense but not because the
mass is changing but that the momentum stays the same. As a simple
example, let the mass of the ball plus one drop be M and the
mass of the drop be m; the velocity u of the
drop is in the same direction as the original velocity V
of M and the velocity of the ball after the drop has left (now
of mass M-m ) is U . The momentum
before the drop leaves is p _{1} =MV and after
the drop leaves is p _{2} =mu +(M-m )U.
Since p _{1} =p _{2} , the new speed
of the ball is U =(MV-mu )/(M-m ). Notice that
if m is very small compared to M , U ≈V .

QUESTION:
If a person were in a closed container filled with water and the container was accelerated at high speeds, would the person in the container feel the g-forces the same as if they were not in the container?

ANSWER:
You would move backward until you hit the back wall and then the back
wall would exert a force forward on you. If there were no water, the
back wall would exert a force F _{1} =ma _{
} on you where m is your mass. If there were water, the
water would exert a force F _{2} toward the back on you
and the wall would exert a force F _{3} forward on you.
So now, F _{3} -F _{2} =ma so the
force on you from the wall would be bigger (F _{2} +ma )
than if there were no water. Plus the water would be trying to
crush you.

QUESTION
ABOUT THIS ANSWER:
You recently answered a question regarding the effect of acceleration on a person in a closed container of water. You suggested the result would be movement to the back of the container and an increase in the force experienced by the person.
Buoyancy is dependent on relative densities, so a person will float with the same percentage immersed regardless of the local gravity/acceleration. This implies that the victim would, at least at moderate levels of acceleration, be forced to the front of the container.
Initially I thought that if the container was not full, it would be quite a comfortable experience since the accelerating force would be evenly distributed. However, humans are not of uniform density, so the persons chest with its air-filled lungs would be forced to the front and his bony, less buoyant extremities would be dragged the the back. Unfortunately, the questioner filled the container completely so the person would be pressed uncomfortably against the front wall.
At higher acceleration the body would be compressed sufficiently that he would become denser than the water, only then would he move to the back of the container to be further crushed by the water column.

ANSWER:
You are right, the motion depends on the density of the astronaut
relative to the density of the water. If the ship is in empty space (no
gravity) and not accelerating, there would be no buoyant force in any
direction and the astronaut would float either at rest or at constant
velocity until he hit something. The equivalence principle says that
there is no experiment you can do to distinguish between being in a
uniform gravitational field with associated acceleration g and
having an acceleration g in zero gravitational field. If the
ship had an acceleration a forward it would be the same as being in a
gravitational field pointing backward in the ship. Therefore, there
would be a buoyant force which would cause objects with smaller density
than water to move forward ("float") and objects with larger density
than water to move backward ("sink"); you correctly point that out and
in my original answer I was assuming an astronaut whose overall density
is larger (quite possible if he were wearing a heavy space suit, for
example). In the back wall case, my original answer was correct. In the
front wall case, the water would be pushing you forward and the wall
backward, so F _{2} -F _{3} =ma
or F _{2} =F _{3} +ma ; now the
water pushes on you with a force greater than the wall pushes back.

QUESTION:
Having a bit of a debate about whether this tennis ball would've landed in with a tennis player and we have a $100 bet on it. The ball machine fed the ball from the other side of the court at the baseline the player that hit the ball is a top ranked junior player...
the ball hit the ball machine edge 5 inches off the ground at the top of the wheel base and the player claims that it would've landed on the line if it had not hit the ball machine of which the picture demonstrates the point of impact is 5 inches above ground at the back edge of the line and there are no external elements such as wind as we are playing indoor.
the ball was traveling at approximately 30 mph at time of impact. We have attached an image for reference. We appreciate any clarity you could provide :-)
[Note that the angle relative to the horizontal is specified to be 60^{0} -70^{0}
in the photograph attached by the questioner.]

ANSWER:
My first reaction was to say that, of course, it would not hit the line.
That was because, as physicists often do, I was thinking of the ball as
a point and ignoring its size. You can see from the figure that there
could easily be a combination of h and v _{0}
where the ball would strike the line had the obstruction not been there.
As best as I could tell, some part of the ball must touch the line so if
we calculate where the bottom-most point of the ball strikes the ground
(y =0) the ball will be in if x >0 in my coordinate
system. My guess is that if we simply assumed that the bottom point went
in a straight line in the direction of its initial velocity we would get
the right answer; but the ball is actually a projectile and moves in a
parabolic path so, since this is a $100 bet, I better do it right! The
equations of motion are

x (t )=x _{0} +v _{0x} t

y (t )=y _{0} +v _{0y} t- ½gt ^{2}

where t is the time it takes to hit the ground and g =9.8
m/s^{2} is the acceleration due to gravity. Being a scientist, I
prefer to work in SI units, so I will do that. OK, let's summarize
everything we know. I will assume θ =70^{0} since
that gives the best chance of hitting.

x _{0} =R =2.7"=0.0686 m

y _{0} =h-R =5"-2.7"=2.3"=0.0584 m

v _{0x} =-v _{0} cosθ=- 30cos70^{0}
mph=-4.583 m/s

v _{0y} =-v _{0} sinθ=- 12.59
m/s

y (t )=0.

So now the task is to put these into the equations above, solve the
y equation for t and put that value of t into the
x equation to find x . I find that t =0.00463 s
and x =0.047 m=1.9". Because x is positive, the ball
will hit the line. Going through the same procedure for θ =60^{0} ,
I find t =0.00502 s and x =1.4", again hitting the line.
When the bet is settled, don't forget to
reward The
Physicist !

ADDED
THOUGHT:
The curvature of the parabolic path is, as I had speculated, miniscule.
For θ =70^{0} above I found x =0.0474
m and if you just assume the ball went in a straight line to the ground
you find that x =0.0473 m.

FOLLOWUP QUESTION:
For clarity sake I am sending you two more pictures from the exact set up of which we have not moved.
Previously you only were sent the Sideview so I have included below sent a top view showing that the ball machine is actually out and the court view from the approximate angle that the ball was struck showing the approximate ballpark at the ball was hit from approximately contact was made 3 feet off of the ground.
I still can't wrap my head around if an apparatus is actually out and a ball strikes it from 5 inches above the court going in the opposite direction that it can still be in unless air resistance is involved of which there are no air elements. Just want to make sure that you still think the ball Woodland in from more data provided before I pay $100?

ANSWER:
The top view is helpful in verifying what I had assumed in the original
answer—the surface which is hit is aligned with the outer
edge of the line. The best way that I can convince you that it is
possible that my first answer is correct is to show two figures, one
showing a trajectory of the ball which lands it in bounds and one out:

Each figure shows the ball at the instant of impact with the machine and
the instant of impact with the floor had the machine not been there. The
dashed red line shows the trajectory of the center of the ball and the
solid red line shows the trajectory of the bottom of the ball. The
diameter of a tennis ball is 5.4". The ball with the steeper trajectory
(which would include your 60^{0} -70^{0} trajectory) is
clearly in bounds by standard tennis rules.

FOLLOWUP QUESTION:
We did not know how to define the arc so we just said 60 70 as an ignorant guesstimate but we are ok with the actual real representation of the arc that we sent you in the last image...we just want to understand based on the ACTUAL real flight path arc that the ball really took when struck and its possibility of landing on the line of which the image with the actual art provided is a more accurate visual representation... and we are not sure of how to define the angle of approach based on that visual.
Our original explanation of the ball angle degree of approach to the baseline could have been lost in translation somewhat we are not certain... that's why we contacted you and that's why I sent a VISUAL representation because we really aren't sure of how to define that...we reckon the more information we provide you the more accurately u can clarify our understanding.
In saying that.... what approximate degree is the arc approaching the baseline from the image we provided this morning... assuming the ball was struck at about 3 feet off the ground traveling approximately 3 4 feet over the net? The distance from the net to the baseline is 39 feet and the player was approx 3 feet behind the baseline at time of contact so that would be approx 42 feet from the net at the time the contact was made. Does this change whether or not the ball lands on the line or is it irrelevant?

ANSWER:
This Q&A is turning into quite a tome! With the information you sent, I
can calculate a pretty good estimate of the trajectory of the ball
ignoring air drag. It turns out that your guess of 60^{0} -70^{0}
for the trajectory angle was way off. The algebra is tedious, so I will
just give the final results. To check that I made no algebra errors, I
plotted the trajectory, shown in the figure. I have worked in meters.
The ball is hit at x =0 from a height of 1 m, passes 1 m above
the 1 m high net at x =13 m, and then hits the machine at x =25
m just above the ground (y =0 m). You can see that my result
describes the path you specified quite well. Do not be deceived by the
picture, though, because the scales of the two axes are very different,
only 2.5 m for vertical motion compared to horizontal motion of 25 m, a
factor of 10. The inset shows the trajectory as it would look to the
eye. As you can see, the angle is much smaller than you estimated. The
analytical solution is that θ =15.6^{0 } and t =1.1 s; your estimate of the initial speed was pretty
good, 23.5 m/s=52.6 mph but the final speed (although it is not
important) is 23.9 m/s=53.5 mph not 30 mph.

Now I am compelled to recalculate with the best possible numbers whether
the ball hits in bounds or not. However, given all that I learned above,
I can make it brief: there will be a critical angle
θ _{c} =tan^{-1} (y _{0} /x _{0} )=tan^{-1} (2.3/2.7)=40.4^{0} ;
any angle smaller than θ _{c } will be out of
bounds. For 15.6^{0} , x =2.7-(2.3/tan15.6)=-5.5", 5.5
inches out of bounds.

Finally, just for completeness, I would like to estimate the effect of
air drag. Using the approximation in an
earlier answer I find that t =1.13
s (0.03 s longer) and v =22.1 m/s (1.8 m/s slower). This would
correspond to the angle being a bit bigger, about 17^{0} , but not nearly
enough to cause the ball to drop in bounds.

QUESTION:
How much lateral force is needed to damage bearings in the hub
motors of a skateboard when carving? So...if you roll in a straight
line, on a skateboard, you exert a radial load on the eight bearings
contained in the four wheels. But skating is more fun when you ride in
wavy lines - carving! So the bearings start getting a lateral or axial
load. How "hard" would the carve have to be (let's assume a rider of 100kg) to
break the weakest of the bearings? The
rotor is supported by two bearings, one that has an radial maximum force of 3.5
kN and the other of 7 kN. So the maximum axial forces would half those.

(The questioner and I had several
exchanges. The bearings are cylindrical and he was only guessing that
the axial force was half the radial force based on data for similar
bearings. I have edited his several emails to get the gist of things in
the question above.)

ANSWER:
The way I see the problem is shown to the right. The forces on the
skater plus skateboard are his weight mg vertically down, the
normal forces on the inner (N _{2} ) and outer (N _{1} )
wheels, and the corresponding frictional forces f _{1}
and f _{2} . Whatever the rated axial (along the wheel
axis) maximum force is, that is what we want to use as the to find the
limiting conditions for the "carve"; the determining factors will be the
mass m of the skater plus board, the speed v he is
going, and the radius R of the path. For the lean, d
and h are the distances, respectively, of the center of mass
horizontally and vertically from the front wheels; note tanθ =d /h .
The wheel base is s . The normal forces and frictional forces
shown each represent the forces on two wheels. Note that the inner
wheels carry the most force.
The easiest way to do this problem, since it is an accelerating system,
is to introduce a fititious centrifugal force C =mv ^{2} /R
pointing outward. Newton's equations are N _{1} +N _{2} -mg =0
(vertical forces), f _{1} +f _{2} -C =0
(horizontal forces), and N _{1} s+mgd-Ch =0
(torque about inner wheels). Now, from the torque equation, there will
be a maximum speed you can go for a given m and R
before the outer wheels leave the ground. At this time N _{1} =0=(Ch-mgd )/s
or v ^{2} /(gR )=d /h =tanθ.
Also, f _{1} =0, so f _{2} =C=mv ^{2} /R .
Now, the inner two wheels each are experiencing the axial force of F =½mv ^{2} /R .
Just to do an example, let v =10 mph=4.47 m/s, R =5 m,
and m =100 kg; then F =0.2 kN. If your guess that F _{max} ≈1.75
kN is correct, you should be ok. For this example, θ =tan^{-1} [v ^{2} /(gR )]=22^{0} . If you execute the turn with
a smaller lean angle, all four wheels will share part of the load. To do
the general solution where both wheels have N ≠0 would
not be two difficult but would be quite a bit messier algebraically and
probably not all that useful.

FOLLOWUP QUESTION:
Thanks for the explanation on your website. It wasn't really what I was looking for because it doesn't really give the answer something layman can understand.
Also, you compare your estimated lateral force of 1.96 kN
(0.2 kN, see below) to 1.75 kN, although they would be the breaking limit for only one bearing.
You assumed that just the 2 inner wheels receive the load, so isn't that a total of 4 bearings... shouldn't the breaking limit be 4*1.75?

ANSWER:
I am glad you asked this question because it got me looking more
carefully at what I had done late last night and I found an extraneous
factor of g had crept into the final stage of my calculation, ("…so f _{2} =C=mg v ^{2} /R …")
so my final answer was too big by a factor of 9.8 meaning that the
lateral force per wheel is F= 0.2 kN; this has been corrected in the
original answer. So you should have no problem after all. I was in the
process of writing an email trying to "verbalize" my answer somewhat to
make it more accessible when I discovered this. Here is what I wrote:

You asked me How hard would the carve have to be (let's assume a rider of 100kg) to break the weakest of the bearings? That is what I gave you. I assumed that the weaker of two bearings in a wheel can break even if the stronger does not.
(I assume they are coaxial, one inside the other.) Sorry if the explanation was too technical, but that is as basic as I can get to convey the details. A few comments to try to clarify:

What I call
C , the centrifugal force, determines how "hard" the carve is.
C=mv ^{2} /R . For the example I did, C =100x4.47^{2} /5=400 N.

When carving, the harder you carve the greater the load will be carried by the inner wheels on each axel. Eventually, the outer wheels will just lift off the ground which necessarily results in the inner wheels taking all the load, both lateral and radial.
Since you asked me for how to calculate the maximum lateral force
any wheel would experience for a given m , v , and
R , that is what I gave you.

I do not
understand your last question but I do know that if two wheels
experience a force of 0.4 N then each experiences a force of 0.2 N
(assuming they equally share the load).

ADDED
THOUGHT: I guess I have not really fully answered your question
"How 'hard' would the carve have to
be…to break the weakest of the bearings?" I suggest that the
appropriate equation would be 1.75x10^{3} =½mv ^{2} /R.
Here are a couple of examples:

What is
the fastest a 100 kg rider could go in a curve of radius 10 m? v =√(2x10x1.75x10^{3} /100)=18.7
m/s=67 km/hr=42 mph.

What is
the tightest curve a 100 kg rider could turn at 30 mph=48 km/hr=13.4
m/s? R =½x13.4^{2} x100/1.75x10^{3} =5.1
m.

These, of
course, assume 1.75 kN is the strength of the weakest bearing.

QUESTION:
If I am moving 55 MPH East (or West) at the equator how much weight would I gain (or lose) due to the Eötvös Effect. Thank You in advance. I am 73 years old and too dumb to figure this out myself.

ANSWER:
First of all, weight is the force the earth exerts on you so you
never gain or lose weight when you are moving; you might want to say
"apparent weight" which is the force which would be measured by a scale
you were standing on. You experience two real forces, your weight
W down
and the normal force N (a scale, for example) up. One way to solve this
problem is to note that an object with mass m with speed v
moving in a circle of radius R has an acceleration a=v ^{2} /R
which points toward the center of the circle; then apply Newton's second
law, F=ma =mv ^{2} /R=W-N and solve for
N to get your apparent weight of W -mv ^{2} /R ,
smaller than your actual weight. This is the
"Eötvös effect".
But there is another way to approach the problem. Rather
than solving the problem from the outside the earth, we might want to
solve it here on the earth. But Newton's laws are not valid in an
accelerating reference frame (accelerating because it is rotating). You
can force Newton's second law to work, though, by inserting a fictitious
force which I will call E for
Eötvös but it is more
commonly known as the centrifugal force; E =mv ^{2} /R
pointing radially out. Newton's first law now applies, N+E-W =0,
so, again, N=W -mv ^{2} /R=W [1-v^{2} / (gR )]
where g =32 ft/s^{2} . In the figure above you have a velocity
v=v _{Earth} +v _{man} . If you are at
rest, v=v _{Earth} =1040 mph=1525 ft/s and R =3959
mi=2.09x10^{7} ft; so N=W (1-0.00348) and a scale will
read 0.348% smaller than your actual weight. If you move with a speed of
55 mph=81 ft/s in an east direction, v =1525+81=1606 ft/s and
N=W (1-0.00386), 0.386% smaller than your actual weight. If you
move with a speed of 55 mph=81 ft/s in a west direction, v =1525-81=1444
ft/s and N=W (1-0.00312), 0.312% smaller than your actual
weight.

FOLLOWUP QUESTION:
I made a donation, but the way I read your answer you have both directions being SMALLER, if I read it right. I know that West is an increase and East is a decrease, just thought you would like to know.

ANSWER:
Thanks for your support! Very generous, particularly because you
say that I am wrong! I must stand by my calculations, though. It is
certainly not unheard-of that I make an error, but this answer is right.
Since I define weight to be what a scale would read if the earth were
not rotating (or at north or south poles), you will see that the
apparent weight (what the scale reads) increases if you go west and
decreases if you go east, just the same as what you "know"! All apparent
weights are smaller than the actual weight; the only exception is if you
go west with speed of v _{Earth} in which case the
actual and apparent weight will be the same. If you want to compare to
your apparent weight at rest, it is 0.386-0.348=0.038% lighter going
east, 0.348-0.312=0.036% heavier going west.

QUESTION:
If I'm driving and hit the gas and turn left would the angle
between the velocity vector and acceleration vector be less than, greater
than, or equal to 90 degrees. I would think it's greater.

ANSWER:
You would think wrong! The velocity vector
v points
straight ahead. The tangential acceleration
a _{t} points
parallel to the velocity vector because you are speeding up. The
centripetal acceleration vector a _{c}
points toward the center of the circle you are turning. As you can see
in the figure, the total acceleration vector
a makes an acute (less than 90^{0} )
angle with v .

QUESTION:
I have an old fashion balance scale, center fulcrum and two dishes on either side of equal weight. If I place two weights on either side and the weights are nearly the same, the heavier side dips slightly, if the difference between the weights are large, the heavier side dips much more. I do not understand why this is so. Logic says that if there is any difference at all, the heavier side should continue to drop until it reaches a barrier to the fall no matter what the difference in the weight.
I asked a physics instructor and he did not have the answer either.

ANSWER:
I have received this question before and, alas, I have not found
the answer. I have asked readers to suggest answers but nobody ever has.
See an earlier answer .

ADDED
ANSWER:
I have found the answer. The center of mass
⊗ of the scale
itself must be below the fulcrum (suspension point). Then, as shown
above, if the beam is off horizontal for the empty scale or equal
weights in each pan, there will be a restoring torque to force a balance
only for the horizontal beam.

QUESTION:
I made a statement to somebody that a plane hitting a building was the same as if the building hit the plane at exactly the same speed,the plane now stationary. The results would be the same. In other words, if a man with large hands slapped my hand at 50 mph, it would be same as me slapping his hand at 50 mph.....its interchangable.....the other person said, no, the mass of the building and hand would have different results...

ANSWER:
Either you or your friend could be right depending on what you
mean by "different results". Let me try to set up a simple example to
demonstrate why.

Imagine we have a 2 lb ball of putty moving
with a speed of 5 mph striking and sticking to a 18 lb bowling ball
at rest; the time it takes to collide is 0.1 s. After the collision,
the two move together with a speed of v _{1} . To
find v _{1} , use momentum conservation: 2x5=(18+2)v _{1} ,
v _{1} =0.5 mph.

Next, imagine we have a 18 lb bowling ball
moving with a speed of 5 mph striking and sticking to a 2 lb ball of
putty at rest; the time it takes to collide is 0.1 s. After the
collision, the two move together with a speed of v _{2} .
To find v _{2} , use momentum conservation:
18x5=(18+2)v _{2} , v _{2} =4.5 mph.

So, you see
that the two scenarios have different speeds after the colliision. But,
suppose that you were the putty ball. During the collision you feel a
force and the force is what is going to hurt you. Do you get hurt as
badly, not as badly, or equally as badly during the collision? What
determines the force you feel is the acceleration you experience during
the collision, how quickly your velocity changes, which is your final
velocity minus your initial velocity divided by the time of the
collision.

For the
putty ball moving initially, (v _{final} -v _{initial} )/t =(0.5-5)/0.1=-45
mph/s.

For the
bowling ball moving initially, (v _{final} -v _{initial} )/t =(0-4.5)/0.1=-45
mph/s.

You could go
through through the exact same process to find that the bowling ball
experienced exactly the same force regardless of who moved initially. A
physicist would say that you were right, but the ambiguity of your
statement means that the other guy could split hairs. As far as physics
is concerned, the only thing which matters is the relative
velocities of the two before the collision. If the putty ball were
moving 105 mph and the bowling ball were moving 100 mph in the same
direction, the result of the collision which matters (the force) would
be the same.

QUESTION:
My question is about the maximum tension experienced by a bow string. I'm
specifically concerned with a traditional or recurve bow NOT a compound bow
with pullies. I want to know the max tension compared to the draw weight so
I have an idea how strong to make my strings. So here's the scenario, what's
the maximum tension in the string for a recurve with a 70 lbs draw weight
and a physical weight of 25 oz? I'm assuming the maximum tension is when
it's at brace (not full draw), right after the arrow leaves. I think this
because not only does the string have to oppose the restoring force of the
bow limbs but it also has to stop the momentum of the limbs that isn't
transferred to the arrow.

ANSWER:
To compute the tension I would need to know the geometry of the bow. I can
tell you that the tension will be at the maximum draw for a simple bow, not
where the arrow leaves the string.

REPLY:
The bow is 64 inches long and the string is about 4-5 inches shorter than the bow. It Is braced at 6 inches and has tips that are 3 inches recurved behind the handle. Its draw length is 28 inches and has an elipitcal/circular tiller shape at full draw.

ANSWER:
(An incorrect answer was posted earlier. This is a reposting, correct
now, I hope!)
When researching the physics of archery I discovered that this can be a
very complicated problem requiring very sophisticated numerical
calculations on computers if you want precision descriptions of all the
details. You, however, require only a rough calculation for estimating
the strength of the string. I can do that and it is more appropriate for
the spirit of this site —to solve problems with simple physics
concepts. This problem requires facility with trigonometry,
understanding of Hooke's law, and application of Newton's first law. The
simple model I will use was one used before the advent of computers; the
bow is modeled as two straight rods (purple) the ends of which move on a
circle as the string (red) is drawn. With this simple model, most of the
details of your bow are not necessary. When the string is braced
(undrawn) there is a certain tension in the string and this tension will
increase as the bow is drawn. So, the maximum will be at the maximum
draw. The figure shows, roughly to scale, the situation. Using simple
trigonometry (law of cosines), I find β =59.6^{0} .
The point where the draw weight W is being applied must be in
equilibrium, W -2T cosβ =0; solving, T =69.2
lb. I think that your concern about the string having to "stop the
momentum of the limbs" is misplaced because bows tend to be quite
elastic so that nearly all the energy imparted to the bow by drawing it
is imparted to the arrow and the limbs will end nearly at rest. Not so
if you draw and release without an arrow, though; what I have read is
that in that situation you are more likely to break your bow than the
string. I am working on a general solution which I will later add here
but thought I would post the part of the solution which answers your
question regarding the tension in the string.

GENERALIZED
SOLUTION:
To get a better understanding of this problem it is worthwhile to
find an analytical solution for the tension as a function of draw
distance. My research showed me that a traditional or recurve bow
behaves, to an excellent approximation, like a simple spring
(Hooke's law), the draw weight being proportional to the draw distance,
i.e . W≈kx where x is the distance the
string is drawn and k is the spring constant. In this case, since W =70 lb when x =28
in, k =2.5 lb/in. Using the law of cosines, cosβ =(L ^{2} +(x+d )^{2} -R ^{2} )/(2L (x+d )).
Again, the point where the force W is applied
is in equilibrium so W -2T cosβ =0 or T (x )=kx /(2cosβ ).
Now, note that in the limit where x → 0, β → 90^{0}
and cosβ → x /L . Therefore T (0)=kL /2.
Using your numbers, T (0)=37.5 lb and the angle and tension for
all points are plotted below. At full draw, β= 56^{0}
and T =64 lb. It was interesting to me that in order for the
calculated values to be correct at zero draw, very precise relative values of
R and L
had to be used because otherwise the expression for cosβ would not be 0 exactly when x =0. The value was
R =30.59411708 inches for L =30.

QUESTION:
Suppose a block is moving with constant velocity towards right on a frictionless surface and during its motion another block of slightly smaller mass lands on top of it from a negligible height.
I argue that the lower block will eventually start moving to the left and upper block will end up moving towards right provided that there is friction between the blocks but not between lower block and ground . My friends can't accept my reasoning. Am I wrong? Please help!

ANSWER:
I hate to tell you, but you are wrong. This is actually a simple
momentum conservation problem. Call the masses of the upper and lower
blocks m and M , respectively. Before they come
together the momentum is Mv where v is the incoming
speed of M . When the masses come in contact they will slide on
each other but, because there is friction, they will eventually stop
sliding and both will move with a velocity u ; the linear momentum will
now be (M+m )u . Conserving momentum, u =[M /(M+m )]v .
They both end up going with speed u and move to the right.

You can also determine the time it takes for the sliding to stop. m
will feel a frictional force to the right of magnitude f= μmg
and M
will feel a frictional force to the left of magnitude f= μmg
(Newton's third law). So, choosing +x to the
right, the acceleration of m is a =μg and
the acceleration of M is A =-μg (m /M ).
The velocities as a function of t are v _{m} =μgt
and v _{M} =v -μgt (m /M );
we are interested in the time when v _{m} =v _{M} ,
so solving for t , t=Mv /[μg (M+m )].
If you substitute this back into v _{m} or v _{M} ,
you will find the same value we found for u above: v _{m} =v _{M} =u =[M /(M+m )]v .

FOLLOWUP QUESTION:
If the block of mass M is not too long, i.e., the total distance that the upper block can slide is less that the distance it could move in your calculated time " t" , wouldn't the two blocks get separated?

ANSWER:
Well, of course the block has to be big enough, otherwise m
will drop down on the frictionless surface. It would be a good exercize
for a student to calculate how far the block would slide for some
μ . And then, if this is greater than the size of the bottom
block, how fast will each be moving after separating.

QUESTION:
I do not want a theoretical answer, but has any experimentalist ever put a very sensitive weight balance below a vacuum chamber before and after vacuating it? Does it get lighter or... heavier? I do not have a sensitive balance nor a vacuum chamber.
The reason I ask is that it would say something about the density, or absence of density, of the vacuum.
If I understand pressure correctly, the scale would read a smaller weight value, due to less gas being in the column of air directly above it, but there might also be new physics there, if it is not the case. I simply do not know.

ANSWER:
You do not want a "theoretical answer" but you clearly do not understand
the physics so I am obliged to give you one anyway. Let us assume the
simplest possible "weight balance" so that I do not have to worry that
it might operate differently in a vacuum. Envision just a simple string
with a tiny butcher's scale which will measure the tension in the string
and then hang an unknown weight of mass M and volume V
from the string. Besides the string, there are two forces on the object
being weighed, its weight Mg and the buoyant force B= ρVg
where ρ is the density of air (about 1 kg/m^{3} at
atmospheric pressure) and g =9.8 m/s^{2} is the
acceleration due to gravity. The scale will read W=Mg-B , an
incorrect measure of the weight. Putting the whole device in a vacuum
will change B to zero because the air is gone, so W=Mg ,
the correct weight. To get an idea of how important this is, consider
weighing a solid block of iron whose mass is 1 kg. The density of iron
is ρ _{iron} =7870 kg=M /V , so the
volume of the block is V =1/7870=1.27x10^{-4} m^{3} .
So, the true weight is W =9.8 N and the measured weight in air
is (9.8-1.27x10^{-4} ) N=9.7999 N, an error of 0.0013%. However,
there are certainly examples where the effect of buoyancy would be very
important. For example, consider an air-filled balloon. I did a rough
calculation and estimated that the volume of an inflated balloon is
about 5x10^{-3} m^{3} so it contains about 5x10^{-3
} kg of air; the mass of an uninflated balloon is about 5 gm=5x10^{-3}
kg, so the total weight of the inflated balloon is about 9.8x10^{-2}
N. But if you weighed it in air you would only measure half that amount.
Your question, has anybody ever actually observed this, is a no brainer:
since the existence of a buoyant force has been known and understood for
well over 2000 years (Archimedes' principle), anyone wanting to make an
extremely accurate measurement of a mass would either correct for it or
eliminate it.

QUESTION:
Several years ago, I was caught in a massive windstorm in a skyscraper. I was on the 54th floor (approx. 756 feet from street level, full building height is 909 feet) , pulling cable, and I stopped for a break. I left a cable pulling string hanging from the ceiling (48 inches free hanging length) in the office, with a 1/4 lb weight attached, and when the storm hit, the weight began swinging like a pendulum. The arc was 16 inches (eyeballing it), and traversing the length of the arc took about 1 second. How can I calculate how far (full arc) the skyscraper was moving by observing what the pendulum in the building was doing?

ANSWER:
A 48" pendulum has a period of about 2.2 s, the time to swing over the
arc and back. Since you were estimating, the pendulum was swinging with
about the period it would if the building were not moving at all. I
would conclude that either the pendulum got swinging somehow and the
building was not perceptibly moving or that the period of the building's
motion was about the same. If the building was swinging with a period
significantly different from 2 s, the pendulum would be swinging with
that same period; that is called a driven oscillator.

QUESTION:
Let's say you have three 20-foot putts. Same stimp meter (pace on the green). Let's say in all cases, the green is flat after the hole. I.e. don't worry about things like the ball running away or anything.

Putt 1 travels flat and then shortly before the hole, it rises 6 inches.

Putt 2 has an rise of 6 inches right after you hit it, and then travels flat to the hole the rest of the way.

Putt 3 rises 6 inches gently from the ball to the hole at a constant angle.

Do you hit all three putts the same initial speed?
Follow up question : What if you have the same 20 foot putt but halfway to the hole the ground rises 8 inches, and then drops 2 inches and then flattens out the rest of the way to the hole. . .same speed?

ANSWER:
Refer to the figure.
If there were no friction, it would make no difference how the rise
occurred, only the amount of rise. In that case v _{hole} = √(v _{putt} ^{2} -2gh ).
However, there is friction which will slow the ball further and the
frictional force on a slope (f=μmg cosθ ) will
be different than on level ground (f=μmg ); here μ
is the coefficient of friction (essentially, I believe, what is
measured by the "stimp meter"), m is the mass of the ball, and
g is the acceleration due to gravity. Since the frictional
force is smaller on the slope, you might think that less energy is lost
to friction on the slope. But, what matters is not the force but its
product with the distance s over which it acts, specifically
the work done by friction is W _{f} =-fs where
the negative sign indicates that the friction takes energy away from the
ball. When the ball is moving forward along along the segment L _{2} ,
the distance it travels is s=L _{2} /cosθ
and so the work done is W _{f} =fs=(μmg cosθ )(L _{2} /cosθ )=μmgL _{2} ,
exactly the same as if the slope was not there. The total work done by
friction, regardless of the path, is W _{f} =μmgL.
The final velocity is then
v _{hole} = √[v _{putt} ^{2} -2g (h+μL )].

QUESTION:
I'm coordinating a piece of action for a tvc (tv ad), which requires me to build a rig to raise an actor 3 m off the ground, as if he is being taken away by aliens! It's been a long time since school for me, however I feel that there would be a formula that would assist me in calculating how much counterweight is required on a pulley system with 2:1 MA to raise an 80kg actor 3m in 1.5 seconds?
Are you able to clarify the principals to be used in calculating this problem?
This is definitely not homework!

ANSWER:
The actor will experience an accelleration a such that he moves a
distance y in a time t : y= ½at^{2}
, so for this case a =2y /t ^{2} =2x3/(1.5)=2.7
m/s^{2} . Choosing the man plus pulley above him as the body
(assuming the mass of the pulley is negligible), Newton's second law is
2T-mg =ma or T =½m (g+a )=½x80x(9.8+2.7)=500
N where g =9.8 m/s^{2} is the acceleration due to
gravity. Finally sum forces for the weight Mg which has the
same magnitude of acceleration as the actor: Mg-T=Ma , so
M=T /(g-a )=500/(9.8-2.7)=70.4 kg. You may want to keep in
mind that the speed of the actor at the end of the 1.5 s is v=at =2.7x1.5=4.05
m/s; if the weight is stopped after the 1.5 s, he will continue moving
upwards until he has risen another h =½v ^{2} /g =0.84
m and then fall back.

QUESTION:
I've been trying to figure out how to weigh my boat on its trailer, and it seemed to me that putting a scale under each of the two wheels and the tongue and adding the three weights together might not yield a correct result, but your
table weight answer , makes me think maybe it could be.

ANSWER:
Assuming that you do not have three scales which could be all engaged at
once, lifting one wheel at a time will introduce errors. Refer to the figure above
(noting that I have made the boat on the trailer invisible!): I have
notated 2S as the distance between the wheels, L the
distance from the axle to the tongue support, q the distance
from the ground to the center of mass (COM
© ) of the boat plus trailer, d
the distance of COM forward of the axle, and W the weight of
the boat plus trailer. The normal forces N _{i}
are the forces up on each of the three points of contact; if an ideal
scale of zero thickness were placed under a wheel, the magnitude of the
appropriate normal force is what it would read and N _{1} +N _{2} +N _{3} =W. I have also assumed that the
COM is centered laterally. The geometry and algebra are a bit tedious,
so I give only final answers. In each case I imagine lifting one point
of contact by the angle θ and then measuring the normal
force which I will call N' _{i} . N' _{3}
is the easiest since it does not depend on θ , N' _{3} =Wd /L .
The other two are, by symmetry, the same: N' _{1} =N' _{2} = ½W [1-(d /L )-(½q /S )tanθ ].
If you now compute the sum, N' _{1} +N' _{2} +N' _{3} ≡W'=W [1-(q /S )tanθ ].
So your measurement of the weight will be wrong. As an example, suppose
that S =3 ft and q =5 ft and you lift a wheel h =4
in=1/3 ft off the ground; then tanθ =(1/3)/(2x3)=1/18 and
the measured weight will be W (1-(5/3)/18)=0.91W ,
nearly a 10% error. The culprit here is that the center of gravity is so
far off the ground. To get a more accurate measurement, block the
opposite wheel so that it is also h above the ground (the
tongue support does not need to be blocked); you only have to do this
once since N _{1} =N _{2 } and N _{3}
does not depend on its height.

FOLLOWUP QUESTION:
Also, in your
table weight example , wouldn't elevating one end of the table with a 2x4 and putting a scale beneath it shift (rotate) the table's weight (CoG) somewhat toward its other end and lessen the measured weight of the weighed end, and error be particularly noticeable if the span of the table is short and the scale is several inches high? If I could remember my trig I could pose this question more intelligently, but the extreme, if hypothetical, case would be if the table end were elevated all the way to vertical, where it would weigh nothing.

ANSWER:
Congratulations, you have caught one of The Physicist's rare
errors! I must admit that I was thinking small angle here and that is
the only case where it is a good approximation to say that elevating one
end of the table does not change the normal force. And, again, I did not
really appreciate the importance of the fact that the COM is not
colinear with the other two forces until I started puzzling over your
boat trailer question. You can see an example of a case where this is
not an issue in an earlier question about a
boat dock . The correct
answer to the
table weight
example would be W' =W [1-(q /S )tanθ ].
(I have changed the notation and assumptions from original problem just
to make the solution more concise. In particular, I have assumed that
the COM is a distance q above the floor and equidistant from
the ends and sides; I have neglected the weight w of the 2x4,
and ½W=W _{1} =W _{2} ; S
is the length of the table.) So, if q =3 ft, S =10 ft,
and h =1/3 ft (COM above floor), W'=W [1-(3/10)(1/3)/10]=0.91W .

QUESTION:
I would like to know the formula to calculate the final speed of a
mass sliding down a steep slope and hitting a less steep slope to the bottom.
Obviously I cannot calculate the speed on each slope separately and
add them. Here are the numbers.

First slope : 1.721 Meter high, 35 degrees, friction coefficient
0.14.

Second slope : 2.536 Meter high, 25 degrees, friction
coefficient
0.14.

Obviously the final speed will not be the total of both speeds, so how
do I add the speed of the first slope to the second slope to obtain
final velocity?

ANSWER:
Well, I started to work this out and found that the length of the two
runs are exactly 3 m and 6 m. I therefore surmise that this is
homework and this site is not for homework help.

FOLLOWUP QUESTION:
Hahaha, thank you for answering. At 51, I am surely not going back to school :) In fact the lenght I gave you are not precisely the one I will use for the slides. I want to understand the way to make the right calculations because the first incline has to be steeper to reduce the overall length of the slide. I am a conceptor mostly in transportation but this project is for a waterpark.

ANSWER:
It is easiest to do this using energy concepts. Total energy of a mass
m moving with speed v and at a height h above
some arbitrarily chosen h =0 level is E = ½mv ^{2} +mgh ;
g =9.8 m/s^{2} is the acceleration due to gravity. For a
given system the energy is conserved (the same everywhere) as long as
there is no agent adding or subtracting energy from the system. If there
were no friction in your slides, energy would be conserved. so, if you
chose h =0 at the bottom and started from the top with v =0,
the initial energy would be E _{1} =mgh and the
final energy would be E _{2} =½mv ^{2} ;
setting them equal and solving for v , v =√(2gh );
note that the mass does not matter. In your case, v =√(2x9.8x(1.721+2.536))=9.13
m/s.

Alas, there is friction and the
frictional force f for a mass m sliding on an incline
θ is f=μmg cosθ. The amount
which a force changes the energy if it acts over a distance s
is called the work done and the work done by friction is W _{f} =-μmgs ·cosθ ;
the minus sign indicates that friction takes energy away from the
system. Now, instead of writing E _{2} =E _{1} ,
we write E _{2} =E _{1} +W —the
energy you end up with equals the energy you started with plus what you
added (negative for the case of friction).

Now we can address your specific
case. ½m v^{2} =m gh-μm gs_{1} cosθ _{1} -μm gs _{2} cosθ _{2} =9.8x[(1.721+2.536)-0.14( 3cos35^{0} +6cos25^{0} )]=31.6
m^{2} /s^{2} . Solving for the velocity, v =7.95
m/s.

The formula you seek if you are
doing a slide with two different slopes is v =√{2g [h-μ (s _{1} cosθ _{1} +s _{2} cosθ _{2} )]}.

QUESTION:
I was asked by my son if you get more speed on a flat (fixed slope) skate ramp or a curved ramp (1/4 pipe) of equal height (12ft). I think the flat ramp would result in greater speed as there is no change in energy force and the fixed straight ramp is constant, the 1/4 pipe is vertical and needs to change direction to horizontal resulting in directional change and loss of energy

ANSWER:
Your reference to "change in energy
force" has no meaning in physics. If friction is neglected, the only
thing which matters is the vertical distance fallen and so the speed at
the bottom would be identically the same regardless of the shape of the
ramp. Refer to the figures on the left. Without friction, the speed at
the bottom is v = √(2gh ) where h is the
height. Including friction complicates things.

The ramp is easiest to do: the three forces on
the skater are his weight mg , the normal
force N , and the frictional force
f . The magnitude of the frictional force
is f= μN where μ is the
coefficient of friction. Newton's equations are ΣF _{y} =N-mg cosφ =0
and ΣF _{x} =mg sinφ -μN =ma ;
the solutions are N=mg cosφ and a=g (sinφ -μ cosφ ). Now,
notice that the acceleration depends on the angle of the incline
which means that the speed at the bottom will also depend on the
angle, the steeper the faster. Therefore it is not really fair to
compare the flat ramp with the quarter pipe because the quarter pipe
will always have its horizontal distance equal to its vertical fall.
So the only fair comparison is for the flat ramp to have φ= 45^{0} .
In that case, a=g (1-μ )/√2. I calculate that
the speed at the bottom is v _{flat_ramp} =√(2gh (1-μ )).

For the quarter pipe, the situation is much harder to calculate
because the normal force gets larger as the skater falls thereby
increasing the friction. Newton's equations are ΣF _{x} =mg cosθ -μN =ma
and ΣF _{y} =N-mg sinθ =mv ^{2} /R
where a is the tangential acceleration and R
is the radius of the pipe. I have no idea how to solve these
because they are coupled, second-order, nonlinear differential
equations. I am guessing that there is not a closed-form
analytical solution and the equations would have to be solved
numerically. For this case, though, we are interested in cases
where the friction is low, μ<< 1. We can calculate the
velocity v as a function of θ for the
no-friction case, μ =0 and then use that v to
approximate the normal force as a function of theta and use that
to find the energy lost to friction: v (θ )≈√(2gR sinθ ),
N ≈3mg sinθ , f ≈3μmg sinθ.
Using these, I find v _{¼pipe} ≈√(2gh (1-3μ)).

So, to do a
numerical example, R=h =12 ft, μ= 0.1, g =32
ft/s^{2} , v _{flat_ramp} =26.3 ft/s, v _{¼pipe} ≈23.2
ft/s. (v =27.7 ft/s for zero friction.)

ADDED
COMMENT:
The person who suggested approximating the velocity as the zero-friction
velocity for the ¼ pipe performed numerical solutions for the
differential equations.

His post : "Out of curiosity, I simulated the setup. With μ =0.001 and
m=R=g =1 I got E =0.99701, in agreement with your result.
μ =0.01 leads to
E =0.9704. Even with μ =0.1, the approximation is not bad:
E =0.732. The mass stops at the center for μ >0.60. Tested with 100, 200 and 500 steps: The critical value is somewhere between 0.603 and 0.605. There is no obvious mathematical constant in that range."

Interpreting, his E is
the calculated value of the kinetic energy divided by mgh. For
this quantity, my approximation was [½mv ^{2} /(mgh )]=(1-3μ ).
Comparing the computed values with the approximated values: 0.99701≈0.997
for μ= 0.001, 0.9704≈0.97 for μ= 0.01, and
0.732≈0.7 for μ= 0.1. Thanks to
mfb for
his interest and calculations!

QUESTION:
I was told that when a car is moving on a circular banked road. If the car reaches a very high velocity then it may move up the slope of the banked road! Is this possible and if so how?

ANSWER:
The
figure shows the car on the track. Three forces on it are the weight
W , the normal force N from the road, and the
frictional force f
from the road. Imagine that it is at rest so that all the forces are in
equilibrium: ΣF _{y} =-W+N cosθ -f sinθ =0
and ΣF _{x} =N sinθ -f cosθ =0.
The solutions are f=W sinθ and N=W cosθ. Now, if the car has some speed v , the car is still in
equilibrium in the y direction, but now has a centripetal acceleration
in the +x -direction of a=v ^{2} /R where R is
the radius of the track. So now, ΣF _{y} =-W+N cosθ -f sinθ =0
and ΣF _{x} =N sinθ -f cosθ =mv ^{2} /R.
Solving these equations, f=m (g sinθ-v ^{2} cosθ /R ).
Now, notice that if v _{0} =√(gR tanθ ), f =0; at
this speed the car can go around the track even if it is perfectly
slippery. For speeds greater than v _{0} , f will be
negative which means it must point down the slope. As the speed
increases beyond v _{0} , the magnitude of the frictional
force will get bigger and bigger down the incline; but, there is a
maximum value which f can be, f _{max} =μN
where μ is the coefficient of static friction. The
corresponding velocity may be
shown to be v _{max} =√[gR (sinθ +μ cosθ )/(cosθ -μ sinθ )].

Not
every angle of incline or every μ will result in the car
slipping, however. Whenever (cosθ -μ sinθ )]=0,
v _{max} =∞ and the car will not slip no matter how
fast it goes. For any given μ there will be a minimum angle θ _{min} = tan^{-1} (1/μ )
beyond which slipping will never occur; for example, for μ= 0.5,
θ _{min} = tan^{-1} (2)=63.4^{0} .
The graph shows the behavior of v _{max} as a function
of angle for several values of μ.

If you want a way to understand
this qualitatively and you have worked with fictitious forces in
accelerated frames, you can look in the frame of the car and there will
be a centrifugal force pointing in the -x direction; this force
will have a magnitude of mv ^{2} /R and, if it is greater than N sinθ -f cosθ=N (sinθ -μ cosθ )
the car will move up the incline.

QUESTION:
There's an e-skate (electric skateboard) that I'm looking at that says it has 7 Nm of constant torque
and the radius of the wheel is 40 mm=0.04 m. How many seconds would it take to bring me (85 kg) to its top speed of 40 km/h? Their 4WD version says it has double torque, so 14 Nm. How long would that take to get to top speed? The e-skate is Mellow Drive. Other brands have lots of initial torque, but it drops as speed increases. Is this possible to calculate in the same way?

ANSWER:
The torque on the drive wheels will result in a frictional force between
the wheels and the road which drives the board forward. I am assuming
that that is the net torque to the two wheels, not the torque to each
wheel. So if the torque is 7
Nm, the force accelerating you is F =7/0.04=175 N. Ignoring all
friction, the
resulting acceleration would be a =175/85=2.1
m/s^{2} . If the final velocity is v =40 km/hr=11.1 m/s,
the time is t=v /a =5.4 s. Now, is friction really
negligible? First consider the frictional force f from the
wheels which could be approximated as f=μmg where μ
is a coefficeint kinetic friction and mg =85x9.8=833 N. A
reasonable approximation for μ would be μ ≈0.1;
with this μ the unpowered board would stop in about about 20
m if it had a speed of 2 m/s. The frictional force would then be about
83 N, not at all negligible. Redoing the calculations above for a net
force of 175-83=92 N, I find t =10.3 s. Finally we should talk
about air drag. The drag force D may be approximated (only in
SI units) as D≈ ¼Av ^{2} where
A is the cross-sectional area you present to the wind, let's say
about A ≈1 m^{2} ; so the biggest D gets
is about 30 N. Since most of the time spent is at lower speeds where
D is much smaller, I will not calculate the the time including
D (a much harder calculation because the force is not constant)
since all this is only a rough caclulation anyway; just think of 10 s as
a lower limit. You can, though, estimate the maximum speed if the torque
really did remain constant by finding the speed for which D =92
N: v _{max} ≈√(4x92/1)≈19 m/s; this is
called the terminal velocity. For
constant torque, doubling it would result in a net force (again
neglecting D but including f ) of 267 N and a time of
3.5 s. If you know how the
torque varies with speed, you could do a similar calculation but it
would be trickier because the force and therefore the acceleration would
change with time.

ADDED
COMMENT:
Just for fun I added the air drag D for the two-wheel drive
case. I found that 11.1 m/s is reached in 11.5 s, only a very small
increase as I anticipated. The graph of v vs. t is
shown; you can see that the curve is nearly linear up to 11.1 m/s and
approaches 19.2 m/s at large times. The solutions are t =17.5·tanh^{-1} (v /19.2)
and v =19.2·tanh(t /19.2). Don't take any of this
too seriously because of the numerous approximations I have made; I
would guess there would be about a 30-40% uncertainty —probably
5-15 s would give the experimental time. If you get one of these, let me
know how long it actually takes.

FOLLOWUP QUESTION:
Someone said the mechanical power needed to ride at 40 kmh was 1900 KW. Is that true?

ANSWER:
The power P delivered by a torque τ at
angular velocity ω is given by P=τω .
In your case τ =7 N·m and ω=v /r= 11.1/0.04=278
s^{-1} , so P =1943 W, not kW.

QUESTION:
Hello, I'm looking for a physics expert to help me with a problem. We were learning Simple Harmonic Motion in our class today and I was struck with curiosity when my professor showed us a ball rolling without slipping on a turntable. The motion that the ball took surprised me because I never imagined that it would move in that fashion. I then started to wonder if it was possible to describe the ball motion in terms of (x,y) for a function of time. I talked to my professor about this problem and it had him stumped. Can you please help me out with this problem and explain in detail how you got your solution.
Here’s the full question I’m proposing and all the conditions:
A spherical ball of mass “m”, moment of inertia “I” about any axis through its center, and radius “a”, rolls without slipping and without dissipation on a horizontal turntable (of radius “r”) describe the balls motion in terms of (x,y) for a function of time.

**The turntable is rotating about the vertical z-axis at a constant unspecified angular velocity.

**Radius and mass of the turntable and the ball are unspecified.

ANSWER:
I was unfamiliar with this problem, but it is interesting. Although the
path of the ball depends on the initial conditions, apparently if you
just set it on the turntable it will
move in a circle
which is at rest in the laboratory frame. I will tell you at the outset
that this is a quite advanced classical mechanics problem and is well
beyond the purposes of this site. However, it was interesting enough to
me that I did a little research to try to get a feeling for how it can
be approached. I will not provide details, but they can be found in a
1979 article in American Journal of
Physics . I do not know your level, but if you are a high school
student, the math is probably beyond your grasp, all the vector
calculus. As you can see from the article, trying to write the solution
in Cartesian coordinates, as you have specified, is not the way to go.
Using the vectors r and ω
as the author does implies using cylindrical polar coordinates (r ,
θ , z ).

QUESTION:
Ok so I am in AP Physics 1 and we are learning about Circular Motion. When
I was doing a problem I noticed a problem in which a car is set at a
constant speed and it asks what the free body diagram looks like at the top
of the hill. As I'm doing multiple questions I think about what the free body
diagram would look like just after passing the top of the hill so not at the
top but not fully down the hill. I tried to do the trick where you set it to
an x and y plane but there is no force inwards from the Normal Force and
then of course there's the Weight Force.... If a more skilled Physcisist could
help me it would help so much in furthering my understanding in physics.

COMMENT:
This student and I had a spirited debate about what constitutes homework
or tutoring help, forbidden on Ask the Physicist . His argument
was "…the
question I brought to your attention does not exsist…" because he
could not find it in a book somewhere and therefore not classifiable as
homework; that, of course, is nonsense because this is a very standard
question. Nevertheless, since he is so avid to learn, I am going to
answer it!

ANSWER:
The question is kind of ambiguous because it refers to
constant speed; but if the car is to maintain constant speed down the
hill, there will have to be some drag friction. If a constant speed
v is maintained, the free body diagram is shown on the left. I will
choose a coordinate system with +y pointing along the direction
opposite the normal force and +x tangent to the circle and
pointing downhill. The car is inequilibrium in the x direction and has
an acceleration of magnitude v ^{2} /R in the
positive y direction. Newton's laws are ΣF _{x} =0=mg sinθ-f
and ΣF_{y} =mv ^{2} /R=mg cosθ-N.
The solutions are simply obtained: f =mg sinθ
and N =m [g cosθ-v^{2} /R ].
An interesting result is that when θ =cos^{-1} (v^{2} /(gR )),
N =0 and the car will leave the road.

Another possible case is that there is no friction and the car has speed
v
at the top of the hill; the car will therefore speed up as it goes down
the hill. Using energy conservation it is easy to show that its speed
when at angle
θ is v' ^{2} =v ^{2} +2gR (1-cosθ )
and so the centripetal acceleration is a_{y} =v ^{2} /R +2g (1-cosθ ).
Notice that there will also be a tangential acceleration, a_{x} ,
an unknown at this stage. Now the equations to solve are ΣF _{x} =ma_{x} =mg sinθ
and ΣF_{y} =m (v^{2} /R+ 2g (1-cosθ ))=mg cosθ-N.
The solutions are a_{x} =g sinθ
and N =m (v^{2} /R+g (2- 3cosθ )).
Again, there will be an angle for which N =0 which is where the car would
leave the road, the determination of which I will leave to my AP
student.

QUESTION:
Say you have a conveyor belt with constant speed v. I put a bottle on the belt suddenly.
How do I know that it would not tip and whether there is a critical speed for the conveyor belt so tipping would never happen?

ANSWER:
I believe that it depends only on the geometry of the bottle and on the
coefficient of kinetic friction μ between the bottle
and the belt. In the figure, H is the position of the center of
mass ⊗ of the
bottle above the base and R is the radius of the base. When the
bottle touches the belt there will be a frictional force f
in the same direction as the velocity of the belt which will cause an
acceleration of the bottle to bring it up to the belt speed. But, as you
note, the bottle might tip over because of the torques it experiences
due to the forces on it. Because the bottle is trying to tip, the normal
force on the bottle should be expressed as two different forces
N _{1} and N _{2}
acting on the leading and trailing edges of the base. Since the bottle
is accelerating, it is imperative to calculate the torques about the
center of mass of the bottle. The bottle is in equilibrium in the
vertical direction so N _{1} +N _{2} =mg .
The frictional force is f =μ (N _{1} +N _{2} )=μmg.
The torques will sum to zero if the bottle is not tipping, 0=fH+RN _{1} -RN _{2} .
Now, if the bottle is just about to tip over, N _{1} =0
and so the torque equation becomes 0=fH -RN _{2} =μmgH-Rmg
or μ _{max} =R /H ; this is the
maximum that the coefficient of friction could without the bottle
tipping. The result does not depend on either m or v .
For example, if R =3 cm and H =6 cm, then μ ≤0.5.

ADDED
NOTE:
The normal force could have been handled differently by having a single
force N acting a distance d from the leading edge of the base of the
bottle. Then when d =2R the bottle would be about to
tip. You would get the same result as above.

QUESTION:
I am standing on a scale on carriage that is going down the slope without friction. The slope angle is
theta. What will the scale show in two cases:

The carriage is shaped in a way that the scale is horizontal with respect to ground (that is is a special edge shaped carriage)

The carriage is regular so the the scale is not paralleled to the ground.

I am trying to understand how a special wedge or insoles in ski show would effect weight felt by a person. In case 1 I think the answer should be mg(1-(sin(theta)^2)) but I am not sure.
In case 2. I have no idea. I think scales can misfunction if the slope is too strong.

ANSWER:
Before doing either case, imagine that we choose the man and the
carriage as the body to look at. It is the standard introductory physics
problem of something sliding down a frictionless incline. The principle
result is that the acceleration of the man and the carriage are the same
and have a magnitude a=g sin θ down the
slope.

The figure
on the left
shows the forces on the man standing on the horizontal scale. There
is his own weight, mg and the scale exerts
a normal force N vertically up; the scale
will read N . The contact between the man and the scale
cannot be frictionless or else the man would not accelerate along
with the carriage, so there is also a frictional force f .
Newton's equations for the man are ma=m g sinθ=-N sin θ+f cosθ+mg sin θ
and 0=N cosθ -f sinθ-mg sinθ .
Solving, I find N=mg cos^{2} θ and
f=mg sinθ cosθ .

The figure
on the right shows the man on the scale parallel to the incline. Now
no friction is needed and the scale, again, will read N but
N will be different. Newton's equations are 0=-mg cosθ+N
and ma=mg sinθ , just the same as the classic
object on the frictionless incline. So the scale reads N=mg cosθ.

The graph
compares the two cases.

QUESTION:
If two people are carrying a canoe parallel to the ground, how much weight are they each carrying? would it be half of the weight of the canoe, more than half, or less than half?

ANSWER:
It all depends on two things —where the center of gravity is
and where the two people exert their forces. In the figure the weight
W of the canoe acts at the center of gravity
and the two people exert forces F _{1}
and F _{2} at distances at distances
d _{1} and d _{2} from the center of
gravity. The two people must hold up all the weight between them, so
F _{1} +F _{2} =W . The canoe is not
rotating about its center of gravity so the net torque must be zero, so
F _{1} d _{1} =F _{2} d _{2} .
Solving, F _{1} =W [d _{2} /(d _{1} +d _{2} )]
and F _{2} =W [d _{1} /(d _{1} +d _{2} )].
If d _{1} =d _{2} , each holds up half the
weight.

QUESTION:
I'm shopping for a pole chainsaw and trying to figure how heavy a 7 lbs saw at the end of a 8' pole would "feel", at about a 45 degree angle.

ANSWER:
It depends on how you hold it and where the center of gravity is. Refer
to the figure. The weight W acts at the center
of gravity which I have chosen to be at a diatance c from your
right hand. The position of your left hand I have denoted as being a
distance d from your right hand. For simplicity I have assumed
that you will exert a force L with your left
hand perpendicular to the pole; your right hand exerts a force with
horizontal and vertical components H and
V respectively. You specify 45 °, so
the horizontal and vertical components of L
are equal and of magnitude L /√2. The two equations for
translational equilibrium are ΣF _{vertical} =0=-W +L /√2+V
and ΣF _{horizontal} =0=-L /√2+H .
These may be simplified to H =L /√2 and V=W-H .
There are three unknowns and only two equations, so we need a third
equation; the sum of torques must also be zero. Summing torques about
the right hand, Στ=dL-cW /√2=0 or L =cW /(d √2).
Therefore, H =cW /(2d ) and V=W [1-cW /(2d )].
For example, suppose W =7 lb, d =3 ft, c =4 ft:
I find L =6.6 lb, H =4.7 lb, V =2.3 lb. The
force R exerted by your right hand is R =√(V ^{2} +H ^{2} )=5.2
lb. The key to easy handling is to choose the saw with the smallest
c which will minimize the torque you must exert.

QUESTION:
How does a torque applied to a wheel cause translational motion? Please don't say the ground pushes on the wheel.
The wheel is stationary at the road surface. From my understanding friction creates a torque equal to the driving torque so where does the net accelerating force come from and where does it act? If you could explain with a free body diagram showing all the forces present that would greatly enhance my understanding.

ANSWER:
If you are asking someone to explain something you do not understand,
don't tell them what not to say! You have, in essence, asked me to not
answer your question because it is the friction which accelerates your
car forward. Let me see if I can explain it to you. Suppose you have a
car with its brakes on at rest on a road. You get behind the car and
push as hard as you can but, alas, it will not budge. What force is
preventing it from moving? The road exerts a static frictional force on
the car, even though it is "stationary"; then, clearly, you cannot say
that the road cannot push a surface in contact with it just because the
surfaces are not sliding on each other. Suppose that there were no
friction between the tire and the road and the car was at rest. No
matter how hard you push your engine which, through the transmission and
various linkages, exerts a torque on the wheels, you would go nowhere,
the torque just spinning the wheels. But, if there is friction, the
wheel, at the point of contact with the road, would exert a force on the
road (opposite the direction you want to go). But Newton's third law
tells you that if the wheel exerts a force on the road, the road exerts
an equal and opposite (forward) force on the wheel. That is the force
which accelerates that car forward.

QUESTION:
A steel pole weighing 1000kgs and is 9 metres long what would the weight of the pole be on the end if I tried to lift it from the end?

ANSWER:
First, let's get one thing clear—the weight of something is the
force which the earth exerts on it and never changes. Also, technically
a kilogram is a unit of mass but I will treat it as a weight since many
countries do. What you want, I believe, is the force you need to exert
to lift it.
It is not really clear what it means to lift "from the end". Two
scenarios suggest themselves to me:

Lift it
from one end while the other end rests on the ground

In order for the rod to be
just barely lifted off the floor at one end, the torque due to
F must be equal in magnitude to the torque due to W =1000,
9xF =4.5x1000=4500 or F =500 kg. The floor holds up the
other half of the weight.

Grab onto
one end and lift the whole rod keeping it horizontal.

It is
impossible to lift it horizontally using only a force at the end because
you could not exert enough torque to keep the rod from rotating due to
the torque caused by the weight. You might try to do it by using two
hands, one at the end pushing down and the other some distance d
from the end pushing up. In that case (see the figure above), summing
the torques about the end you find that F _{up} =4500/d ;
for example, choosing d =20 cm=0.2 m, F _{up} =22,500
kg. You can also find F _{down} by summing all forces to
zero, F _{up} -W-F _{down} =0=22,500-1000-F _{down}
or F _{down} =21,500 kg.

QUESTION:
According to third law of motion a bird sitting on a tree branch cannot fly due to reaction then how does it fly?

ANSWER:
Why would you think that? The so-called reaction force is exerted down
on the branch, not on the bird. Let's dissect the physics of this bird.
The eagle on the left is sitting on a branch. There are two forces on
him —his weight W and the force
the branch exerts up on him F . These forces
are equal and opposite because of Newton's first law. But, what
about the reaction partner of F , the equal and
opposite Newton's third law partner? That force is the force which the
eagle exerts down on the branch and is not a force on the eagle
(which is why I did not even draw it in the force diagram of the eagle).
The picture on the right shows the eagle just as he is lifitng off. Now
there is an additional force L due to the lift
generated by the flapping wings. Since L+F>W , the eagle is no
longer in equilibrium and will accelerate upward. As soon as the talons
leave the branch, F disappears.

QUESTION:
This past weekend I was in Vermont and was walking down a relatively steep gravelly road when I started gaining speed unintentionally and it seemed as thought my feet couldn't keep pace with the top half of my body. I kept accelerating for about 20-25 paces and saw a tree looming in front of me when I fell down flat.
Is there a name for what happened and is there
anyway I could have stopped it once it began? I ask because this happened to me once before in almost the same conditions years ago and, at the time, I ended up with 2 missing front teeth, a black eye, and bruises/cuts on both forearms.

ANSWER:
I have had this happen to me and know the helpless feeling of not being
able to stop running. The red arrow in the picture represents your
weight, acting at your center of gravity somewhere in your trunk. This
force exerts a torque about your leading foot which tends to rotate you
in a clockwise direction. Imagine that this runner were not running but
just standing still. The only way to keep from falling forward (rotating
about that leading food) would be to exert an opposite torque and this
could only be done by your foot/ankle; but the requisite muscles are not
strong enough to exert the necessary torque. Think about it —if
you were just standing on level ground, how far could you lean forward
without falling forward? Not very far! And it is even worse if you are
running because you have a forward momentum which also is tending to
topple you forward if you try to stop. All that I can think of that you
can do to avoid a fall is to try to slow enough that you can get
yourself more vertical so that your center of gravity is vertically
above your feet. If the hill is not too long, you can keep running until
you get to the bottom, but you will be speeding up the whole way.

QUESTION:
How many g forces will a driver experience accelerating from 0 to 200 mph
in 1/4 mile, straight line acceleration.

ANSWER:
No homework. But there are lots of advertisers on this page which will help
with homework.

FOLLOWUP:
This is not homework.
It happens to come from one of the
country's top 100 trial lawyers who would have gone to medical school
instead if he had the math ability to figure this out myself.
Of course, I have lost two trials in 28 years and my time goes for $650 an h)p8
I will just have an associate do it for me tomorrow--if it is still of any concern.
This is the first time that I have ever used one of these "Ask Jeeves" sites and will be the last.
And my new Vette has a g force display so if I really wanted to know I would have gotten in and fkoored it instead of relying on some bottom feeder like you.

ANSWER:
Wow! I am so excited to get a question from one of the country's top 100
trial lawyers. I am so honored, even though I am just a tiny "bottom
feeder" in the presence of such a paragon. Sorry for the sarcasm, dear
readers, but this is a good opportunity for me to emphasize that one of
the important things about Ask The Physicist is that it is not a
homework help or tutoring site. I feel very strongly about this because,
having been a teacher for 40 years, I feel strongly that students should
do their own work and the internet gives too many opportunities for
"cheating" in that regard. I reject what I judge to be homework
questions and, inevitably, I occasionally make a mistake. If you think
my ethical stand on this issue makes me the scum of the earth, so be it.
When I do make a mistake, I usually answer the question and I will do
that here. The equations of motion for uniform acceleration for the
object starting at rest are v=at and d = ½at ^{2}
where, in this case, v =200 mph=89.4 m/s and d =¼
mile=402 m. The time can be eliminated using the first equation, t =89.4/a
and substituting into the second equation, 402=½a (89.4/a )^{2} =3996/a
or a =9.94 m/s^{2} . The acceleration due to gravity
is 9.8 m/s^{2} and so this is just about 1 g of acceleration,
a =1.01g . I figure that anyone who charges $650/hour
for his services can afford to send a little of that my way!

QUESTION:
From the physics theories and calculation, how does a crane fall of a bridge when trying to pull out a bus from the water? (crane = 28 ton, bus = 14 ton)

ANSWER:
It's all about the torques. In the figure, if the crane tips over it
will rotate about the rear wheel. When this is just about to happen the
front wheel has zero normal force from the road on it. At this time, the
road exerts a 42 ton force up on the rear wheels (they are holding up
all the weight). The weights each act at the center of gravity of the
crane and the bus. If I sum torques about the rear wheel, they must
equal zero, that is 28D =14d or d /D =2.
So, if d is greater than 2D , the whole system will
rotate about the rear wheels and topple off the bridge.

QUESTION:
If you had two point masses m_{1} and m_{2} in space separated by a distance d, how long would it take for them to collide under gravity alone? I'd imagine it's a changing force/acceleration, so some calculus must be involved.

ANSWER:
I have solved this problem many times before but always for specific
masses and distances. To avoid the calculus you refer to, the trick is
to use Kepler's laws for planetary motion. You can find references on the
FAQ page. The one you
might find most useful is this
link . You will see that
the relevant time to collision is t=T /2=√(πμa ^{3} /K)
where μ =m _{1} m _{2} /(m _{1} +m _{2} )
is the reduced mass, a=d /2, and K=G m _{1} m _{2} .
Therefore, t = √[πd ^{3} /(8G (m _{1} +m _{2} )].

QUESTION:
A tire rolling on a level surface at a linear speed of 10 MPH rolls on to a conveyor belt which is also moving at 10 MPH in the same direction. How will the tire's speed change? Will it be 20 Mph? 10 MPH? Will it's rotation stop?
Reverse?

ANSWER:
This problem is a little trickier than I had anticipated. On the other
hand, the final answer is much simpler than I had anticipated. Shown in
the figure to the left is (top) the situation when the rolling tire
first touches the conveyer belt. It is rolling without sliding so the
point of contact with the floor is at rest, the center is moving forward
with a speed v (your 10 mph), and the top is moving forward
with speed 2v ; the belt is moving forward also with speed v .
I find this problem much easier to do if I transform into a coordinate
system which is moving with speed v to the right; in that
coordinate system the belt (upper surface) and tire are both at rest and
the top and bottom edges of the tire have speeds v as shown.
Before getting into the hard part, there is a special case which we can
get out of the way first: if there is no friction the tire will continue
on its merry way unchanged, both its speed and its angular velocity
unchanged because there are no net forces or torques on it.

Now,
as soon as it gets on the belt there will be a frictional force
f trying to accelerate it to the right so it will
start sliding along the belt (at rest in the frame we are using). The
frictional force will be f=μmg where μ is the
coefficient of kinetic friction, m is the mass of the tire, and
g is the acceleration due to gravity. Call v' the
speed which the center acquires in some time t . Then Newton's
second law for translational motion is m Δv=μmgt=mv' ,
so v'=μgt.

There
will also be a torque τ=fr=μmgr which acts opposite the
direction the tire is rotating. There will come a time when the bottom
edge of the tire will be at rest relative to the belt because of this
torque. At that instant, we can transform back into the original
coordinate system and the tire will be moving forward with speed
v+v' . Newton's second law for rotational motion is ΔL =τt=- (μmgr )(v' /μg )=-mv'r .
where L is the angular momentum of the system. The angular
momentum is the angular momentum about the center of mass plus the
angular momentum of the center of mass. L _{1} =Iω _{1} +0=Iv /r
where I is the moment of inertia about the center of mass;
L _{2} =Iω _{2} +mrv' =(v' /r )(I+mr ^{2} ).
Therefore, (v' /r )(I+mr ^{2} )-Iv /r=-mv'r
or, v'=Iv /(I+ 2mr ^{2} ). This is
surprisingly simple, particularly surprising that it does not depend on
μ. However, the time to stop sliding does depend on μ ,
t=v' /μg ; the slipperier the surface, the longer it
takes to stop slipping, as expected.

Finally
we need to transform back into the original coordinate system by simply
adding v as shown in the final figure. As an example, suppose
we model the tire as a uniform cylinder of mass m and moment of
inertial I =½mr ^{2} ; then v'=v /5,
or in your case, 2 mph, so the tire is rolling without slipping with a
speed of 12 mph.

ADDED
NOTE:
Note that the new angular velocity of the tire is ω'=v' /r
only a small fraction of the original value of ω=v /r.
This drop in angular velocity can be clearly seen in a
Mythbusters
episode showing a car driving onto a ramp from a moving truck, a
very similar situation to the conveyor belt problem in this question.

QUESTION:
I was wondering if, when an object enters earth's atmosphere from space, does it experience a lateral force as it enters the atmosphere? In other words, does it suddenly get "pushed" sideways due the earth's rotation?

ANSWER:
This had never occurred to me, but certainly the meteorite will be
deflected because of the "wind" of the rotating atmosphere. But, how big
an effect is this? The average speed of a meteorite is about 17 km/s;
and I calculate the speed of the "wind" to be about 0.47 km/s at the
equator. So, even if the meteorite acquired all of the speed of the
atmosphere, it will still be trivially small compared to the vertical
speed. Another way to look at it is to look at the drag forces the
meteorite experiences when it encounters the atmosphere; these are proportional
to the squares of the speeds, so the ratio of the horizontal force to the
vertical force will be about 0.47^{2} /17^{2} =0.0008. Compared to the force slowing down the vertical motion, the force
speeding up the horizontal motion is tiny.

QUESTION:
I have a question related to determining the force at impact.
Here's the question in two parts:

If a firefighter adds 45 pounds of gear to his overall weight, does it
increase the impact force if he has no choice but to jump out a window and
if so to what degree?

What is the effect on impact force of jumping out a second floor window
vs. 3rd floor and above?

This actually isn't homework. I'm 54 years old and advising a charitable organization that provides safety systems to firefighters. One of the challenges they face is fire departments who don't have high rise buildings and feel they don't need the bailout systems. So I'm trying to figure out what the impact is for a firefighter forced to jump out a second story or third story window. This will help inform how they talk to prospective donors.
If you could help me with that (understanding/identifying the increase in impact force from 1st to 2nd to 3rd floors and then up) it would be greatly appreciated. I've done a lot of googling around calculating impact and g forces, but its not making sense to me (I'm not being lazy, just not understanding).

ANSWER:
The "force at impact" depends, essentially, on the time to stop. If she
has a speed v , a mass m , and stops in a time t ,
the average force F during the stopping time is given by
F=m (g+v /t ) where g= 32 ft/s^{2} is the acceleration due to
gravity. So, yes, if you increase the mass you increase the force
proportionally; I guess I would toss that extra 45 lb over before I
jumped! Regarding your second question, call the height of one floor
h . Jumping from the second floor would result in a speed of v _{2} = √(2gh );
jumping from the third floor would result in a
speed of v _{3} =2 √(gh ) which is
about 1.4 times larger than v _{2} . In general, jumping
from the n ^{th} floor would result in a speed v _{n} =√[2(n- 1)gh ].

Maybe some numerical examples would be useful to you. You prefer
Imperial units, which makes things a little complicated. The quantity
mg is the weight. I will take mg =160 lb and h =12
ft for my numerical calculations, so when I need the mass I will use 160
lb/32 ft/s^{2} =5 lb·s^{2} /ft. (The unit of mass
in Imperial units is called the slug, 1 slug=1 lb·s^{2} /ft.) The speeds from
second and third floors will then be v _{2} =27.7
ft/s=18.9 mph and v _{3} =39.2 ft/s=26.7 mph; these
speeds are independent of the mass. Finally we must approximate the
times to stop. If she lands feet first, she could extend, as
parachutists do, her stopping time by bending her knees into a squat; I
will estimate the distance s she will travel while stopping as
s =3 ft. The time may be shown to be t =2s /v
so t _{2} =2x3/27.7=0.22 s and t _{3} =2x3/39.2=0.15
s. Finally, the average forces during impact are F _{2} =160+5x27.7/0.22=790
lb and F _{3} =160+5x39.2/0.15=1467 lb. These are
approximately 5 and 9 times her weight (g-forces of about 5g
and 9g ).

QUESTION:
Well
I hope this doesn't seem off the wall but I want to know how much force
it would take to tip a 7.62 m tall,6.40 m wide and 22 ton Transformer over? It is for a story I'm doing about a comic book character.

ANSWER:
It depends on where and how the force F is
applied; if applied as shown in the diagram, it will be the smallest
possible.
I will assume that the center of gravity of the transformer is in the
geometrical center, so the weight W acts
there. When the force is just about to tip it over, all the force from
the floor, the normal force N and the
frictional force f , act on the edge opposite
where the force is applied. It will be in equilibrium, so you can easily
see that F=f and N=W . But, what you need is F
and to get it you need to sum the torques around the edge where N
and f act and sum to zero: Fh-Ww /2=0 or F=Ww /(2h ).
Using your numbers, F =22x6.40/(2x7.62)=9.24 tons. If you want
the force in Newtons, assuming that the 22 ton mass is metric tons
(22,000 kg), F =9.8x9240=90,500 N.

QUESTION:
I f you tow a boat from a tow path which is the best point to tie the rope onto the boat?

VIDEO

ANSWER:
The rope will exert a force on the boat, obviously. This force will tend
to do three things: exert a torque which will tend to rotate the boat
about a vertical axis through the center of gravity (you don't want
this), have a component along the bank which will pull it along the
canal (that is what you want), and have a component which will pull the
boat toward the shore (you don't want this). To minimize the tendency of
the boat to turn, attach the rope close to the center of gravity of the
boat. To minimize the tendency to drift (not turn) to the bank, make the
rope as long as possible so that most of its force will be exerted along
the bank. Some tiller will be needed to make small corrections, but
those should be minimal.QUESTION:
I am an avid sports fan and I have often wondered if the experts may be wrong about the myth of the rising fastball in the game of baseball. I played baseball for over 20 years and I can tell you that the ball does appear to rise when certain pitchers throw hard put a heavy backspin on the ball. I have been told that experts say it is nothing more than a visual trick your eyes play on you because a rising fastball is considered to be physically impossible. I can tell you first hand that a softball pitcher I know can throw a ball that rises after being thrown on a straight trajectory. I suspect the Magnus effect may have something to do with the anit-gravitational behavior of the ball. Do you think this could be what causes the ball to appear to rise as it travels, or is it just our perception?

ANSWER:
There is such a thing as a rising fastball, but it does not actually
rise; it simply falls more slowly than a nonspinning ball does. An
experienced hitter knows intuitively what a normal fastball does and
when presented with a rising fastball he will swear that it rose because
it actually fell less. Incidentally, by rise or fall, I am talking about
the direction of the acceleration. So a ball which is thrown at an angle
above the horizontal is obviously rising but it rises at a
decreasing
rate of rise until it reaches the peak of its trajectory and then begins
back down; the rising fastball will actually rise farther. Then why do
we say it is a myth? It is easiest to understand by looking at a ball
thrown purely horizontally. Can spin cause the ball to actually go
upwards? To answer this, you need to think about all the forces on a
pitched ball. There is the weight, F _{G} , which points
vertically down and causes the ball to accelerate downward (a
horizontally thrown 90 mph fastball falls about 4 feet on the way to the
plate); there is the drag F _{D } which points opposite
the direction of flight and tends to slow the ball down (a
90 mph fastball
loses about 10 mph on the way to the plate); and there is the magnus
force F _{M} which, for a ball with backspin ω
about a horizontal axis points perpendicular to the velocity and upward.
If the ball is moving horizontally the only way it could rise is for the
Magnus force to be larger than the weight. Measurements have been done
in wind tunnels and it is found that if the rotation is 1800 rpm, about
the most a pitcher could possibly put on it, the ball would have to be
going over 130 mph for the Magnus force to be equal to the weight. When
you say "thrown on a straight trajectory", you cannot mean it left his
hand horizontally because it would hit the ground before it got to the
plate; a fast pitch like that is impossible to accurately judge the
initial angle of the trajectory.

QUESTION:
In lifting an object to a higher level directly over its original location, the energy I expend increases potential energy. But, does some of the energy used in lifting it also go to accelerating it to higher rotational velocities as the circumference of its "orbit" increases as it is raised over its original position? Does this add kinetic energy and mass to the object whereas increasing potential energy does not?

ANSWER:
(Preface: all my calculations below assume that the height lifted is
much smaller than the radius of the earth. I also neglect the change in
the gravitational force over the distance the mass is lifted. Also, to
simplify things, all my calculations are at the equator.) Yes, work is
done to increase the kinetic energy. As viewed from an inertial frame,
watching the mass M get lifted to h , I estimate that
the kinetic energy changes by ΔK≈ 2hK _{initial} /R=MhRω ^{2}
where R =6.4x10^{6} m is the radius of the earth and ω =7.3x10^{-5}
s^{-1} is the angular velocity of the earth. For example,
lifting 1 kg a height of 1 m requires 0.03 J of work to increase the
kinetic energy. But wait a minute! Once we acknowledge that the earth is
rotating, we have to recognize that the mass, being in a circular orbit,
has a centripetal acceleration a _{c} =Rω ^{2}
and therefore the net force on M is Mg-MRω ^{2} .
Therefore, the net work done is W ≈(Mg-MRω ^{2} )h +MhRω ^{2} =Mgh .

QUESTION:
If I weigh 200lbs and am riding a kick scooter that weighs 14lbs, and I am riding at a speed of 10 mph and I jump the scooter off a curb, say 6 inches, what is the force in terms of pounds, that I am applying to the scooter as it lands?

ANSWER:
Usually it is not possible to answer this kind of question because what
is needed is to know how long the collision between you and the ground
lasted. In this case, though, we can estimate the time of this
collision. As you may know, paratroupers are trained to not land with
stiff legs, rather to bend at the knees during the landing; the purpose
is to prolong the landing time and this reduces the average force on the
legs during the landing. If a mass M hits the ground with some
speed V and stops in time t , the average force over
the time is F=W+MV /t where W is the weight,
200 lb; to convert the
weight to mass, divide by the acceleration due to gravity, g =32
ft/s^{2} : M=W /g =(200 lb)/(32 ft/s^{2} )=6.25
ft ·lb/s^{2} . The speed at impact can be determined
from V =√(2gh )=√[2x(32
ft/s^{2} ) x(½ ft)]=5.7 ft/s. If we approximate that
the distance S over which your legs bend on landing as S =1
ft, the time to stop is t= 2S /V=(2x1 ft)/(5.7
ft/s)=0.35 s. So, finally, F =200 lb+(6.25
ft ·lb/s^{2} )(5.7 ft/s)/(0.35 s)=302 lb. This is
the force the scooter exerts on you which, by Newton's third law, is
equal the the force you exert on the scooter (but in opposite
direction). If you stop in ½ ft, the force would be 404 lb. Keep
in mind that this is a very approximate estimate of the average force.

QUESTION:
I am having a debate with my brother about climbing on an incline.
I understand the basics of climbing a hill on a diagonal. If you climb diagonally, you can avoid taking larger vertical steps at the cost of more horizontal movement. This makes each step take less energy while increasing the overall work and time needed.
However, it seems as though this rule does not work for stairs. Stairs do not allow for shorter vertical steps (you either make 100% progress on a step or 0%). Do I have this correct? Am I missing something?

ANSWER:
Slaloming up the incline will increase time spent but not increase work
done. This assumes no frictional forces are important, the only work you
do is the work lifting you. Since work done does not change but elapsed
time does, the average power you are generating going straight up is
greater than zigzaging. Going up steps, though, if you go across a step
you do no work, the only work done is lifting you to the next step. The
only way to get the equivalent lowered average power output as you do by
slaloming up the slope is to rest between steps.

QUESTION:
How strong would a man have to be to push a 16,000 lb bus on a flat surface?

ANSWER:
That depends on how much friction there is. And not just the friction on
the bus, but more importantly, the friction between the man's feet and
the ground. Newton's third law says that the force the man exerts on the
bus is equal and opposite the force which the bus exerts on the man (B
in the picture). Other forces on the man are his weight (W ),
the friction the the road exerts on his feet (N ),
and the force that the road exerts up on him (N ).
If the bus is not moving, N=W and f=B , equilibrium.
The biggest that the frictional force can be without the man's feet slipping is
f= μN where μ is the coefficient of
static friction between shoe soles and road surface. A typical value of
μ for rubber on asphald, for example,
μ≈ 1, so the biggest f could be is
approximately his weight W ;
this means that the largest force he could exert on the bus without
slipping would be
about equal to his weight. Taking W ≈200 lb, if the
frictional force on the bus is taken to be zero, the bus would
accelerate forward with an acceleration of a=Bg /16000=200x32/16000=0.4
ft/s^{2} where g =32 ft/s^{2 } is the
acceleration due to gravity; this means that after 10 s the bus would be
moving forward with a speed 4 ft/s. If there were a 100 lb frictional
force acting on the bus, the acceleration would only be a=0.2 ft/s^{2} .
If there were a frictional force greater than 200 lb acting on the bus,
the man could not move it.

QUESTION:
I'm confused? One moment I'm reading about "inertial reference frames" and that "acceleration due to gravity" is unaffected by mass. This is followed up with examples such as the Bowling ball and feather. All good. Maths seems clear enough.
But then we start talking about a particular body/objects "acceleration due to gravity" at or near the surface of the Earth as being 9.8m/s^{2} which is calculated using the masses of earth and the "falling" body and Newton's Law. Similarly I read that if I go and stand on the moon the "acceleration due to gravity" will be different because the masses are different? Again seems to be clear.
What am I missing? How is it that the mass of two object does not affect the acceleration one moment but the accelerations of the moon and the earth on a body have differing values due in part (large part) to their mass?

ANSWER:
Look at the figure. Two masses, M and m , are separated
by a distance r . M exerts a force F _{mM}
on m and m exerts a force F _{Mm}
on M ; because of Newton's third law, these forces are equal and
opposite, F _{mM} =-F _{Mm} .
Because of Newton's law of universal gravitation the forces have
magnitude F _{Mn} =F _{mM} ≡F=mMG /r ^{2} .
Then, using Newton's second law, each mass will have an acceleration
independent of its own mass of a =[mMG /r ^{2} ]/m=MG/r^{2}
and A =[mMG /r ^{2} ]/M=mG/r^{2} .
Note that I have been viewing this from outside the system; this frame
of reference is is called an inertial frame of reference. It is
important that we view the system from an inertial frame, because
otherwise Newton's laws are not correct.

So, what you
have been taught is correct only if you view things from outside the
two-body system. But what you have also probably been taught is that you
measure this constant acceleration relative to the surface of the earth
and that is technically incorrect because the earth is accelerating up
to meet the falling mass and is therefore not an inertial frame.
However, the earth's acceleration is extremely tiny, too small to
measure. If M is much much larger than m , which is
certainly the case for the earth and the moon, what you have been taught
is, for all intents and purposes, correct.

QUESTION:
I am trying to explain to my brother why on a spinning wheel a point farther out is going faster than a point closer to axis, though the wheel is spinning at the same rpms. But he just cant figure it out. Could you give an explanation a 2 year old could understand?
PS My brother is 21.

ANSWER:
I could probably not convince a two-year old, but if your brother is
just a little smarter than one, I can probably convince him. The speed
of something is defined as the distance traveled divided by the time it takes to
travel that distance. For example, a car going around a circular race
track which has a total circumference of 2 miles takes 2 minutes to go
around once, its speed is (2 miles)/(2 minutes)=1 mile/minute=60 mph.
Suppose a wheel has an angular speed of 10 rpm and has a circumference
of 2 m. Then the distance a point on the rim will go in 1 minute is 20 m
because the wheel goes around 10 times; the speed of that point is
therefore 20 m/min. Now look at a point halfway from the axle to the
rim; it will move in a circle of circumference only 1 m so the distance
it travels in 1 min is only 10 m so its speed is therefore 10 m/min. In
a nutshell, a point near the center travels a shorter distance than a
point far from the center in the same time.

QUESTION:
Some car drivers reason that since tractor trailers have more tires, they
should be able to stop quicker. How much work does each car tire need to do
to stop it, compared with how much work each truck tire needs to do to stop
it? For ease of reference let's say a car is 4000 pounds and the truck is
80000 pounds while the speed involved is 65 miles per hour. Finally is the
work done proportional to the size difference of truck brakes versus car
brakes?

ANSWER:
I replied that this is not a homework
solving site.

FOLLOWUP
QUESTION :
I didn't realize I had posed this in the form of a homework-ish question. I'm a 47yo truck driver, tired of hearing all the lame-@$$ rhetoric of the motoring public. Long, quiet, empty night highway helped me come at this from a different angle. Without too much detail, one truck wheel wrangles slightly more than a single car's worth of mass; is why trucks do NOT stop better despite having more wheels.

ANSWER:
I have never heard this but it is pretty nonsensical. How quickly any
vehicle can stop is determined by the maximum force which can be applied
by braking; if the coefficient of static friction between the tires and
the road is μ , the maximum force on a level surface
is μW . where W is the weight of the vehicle. The
distance s traveled will be determined by the work done by this
force, μWs , which will equal the initial kinetic
energy, ½Mv ^{2} =½(W /g )v ^{2} ,
where M is the mass and g =32 ft/s^{2} . So,
s =(1/(2μg )v ^{2} . So here is the
interesting thing: the distance traveled does not depend on the weight
of the vehicle, only by the initial speed and the condition of the road.
The force applied does depend on the weight and the force per wheel
would be 0.7x4000/4=700 lb for the car and 0.7x80000/18=3111 lb; but the
larger force is simply because of the larger weight and the only thing
which is important is the distance to stop. Of course, this also depends
on whether you have antilock brakes which let you get the maximum amount
of friction when you are just on the verge of skidding; if you lock the
brakes and you skid, you go farther before stopping. For your example,
v =65 mph=95 ft/s and μ is approximately 0.7 for
tires on a dry road, so s ≈200 ft. I have neglected all
other forms of friction like rolling friction and air drag, but these
should be relatively unimportant. Bottom line—stopping distance
does not depend on number of tires nor on weight.

QUESTION:
I am a volunteer guide at South Foreland historic lighthouse in the UK. We have an optic weighing approximately 2 tons, floating in a close fitting trough containing only approx. 28 litres of mercury. What is the theory which enables this optic to float as it does not appear to fit within the basics of Archimedes principle.

ANSWER:
To float 2 metric tons (2000 kg) you must displace M =2000 kg of
mercury. The density of mercury is ρ =13,600 kg/m^{3} ,
so the volume you must displace is V=M /ρ =0.15 m^{3} =150
l; this, I presume, is what is bothering you since only 28 l are used.
Suppose that the reservoir for the mercury is a cylinder of radius 1 m
and depth d ; to contain 0.15 m^{3} , the depth of the
container would have to be d =0.05 m=5 cm (estimating
π ≈3). So, I will make a container 6 cm high for an
extra 20%, 180 l and I will buy that 180 l to fill it up. Now, let's
make the pedestal on which the lens sits be a solid cylinder of radius
99 cm so that when you put it into the reservoir there will be 1 cm gap
all around. So as you lower it into the reservoir, mercury will spill
out the top and you will be sure to capture it. When you have captured
150 l, the whole thing will be floating on the mercury. You return the
extra 150 l and have floated the lens with only 180-150=30 l. Of course,
in the real world you would only buy 30 l, put the pedestal into the
empty reservoir, and add mercury until it floats. (I realize that the
shape of the pedestal and reservoir are probably not full cylinders,
since you said "trough", but my simple example wouldn't be so simple
with more complicated volumes. The idea is the same, though.)

QUESTION:
Rod tied to a string tilts vertically but when it is rotated it becomes horizontal Why ???
Can't seem to find any answer

ANSWER:
The easiest way to see this is to introduce the (fictitious) centrifugal
forces shown in the figure above. As you can see, both exert a torque
about the suspension point which will tend to make the rod horizontal.
When the rod becomes horizontal, the forces are still there but no
longer exert torques.

QUESTION:
What is the effect of mass on torque? A wind turbine fan's blades are commonly very long to increase torque and to decrease speed.
How can I decrease speed using MASS? Or, can I increase torque, by increasing of mass, without increasing length of the blade? (without losing the energy.)
What is the formula applicable here?

ANSWER:
You are asking many questions here with no simple answers.

The simplest
place to start is your first question: does the mass of the rotor have an
effect on the torque on it? Typically, the turbine has three blades. I
will just analyze a single one and the same arguments could be made for
the other two. Call the length of the blade L and assume that
the force on it due to the wind is approximately uniform along the
length of the blade (the force on a tiny piece of the blade near the
center is the same as the force on an identical tiny piece near the
end). Then the total force F due to the wind will depend on the
length of the blade, but the force per unit length, Φ=F /L
will be more useful because it will depend only on how hard the wind is
blowing. It is now pretty easy to show that the torque due to the wind
is τ _{wind} =½ΦL ^{2} .
So, the answer to your question is no, mass does not affect the
torque; the torque depends only on how hard the wind is blowing and how
long the blade is.

Your second question is how can
you decrease speed by changing the mass M . If I model the blade
as a uniform thin stick of length L , its moment of inertia is
I=ML ^{2} /3. If it has an angular velocity ω _{1} ,
its angular momentum is L _{1} =Iω _{1} =Mω _{1} L ^{2} /3.
If you increase the mass to M+m , the moment of inertia will
increase to I' =(M+m )L ^{2} /3 and its
angular velocity will change to ω _{2} . But, the
angular momentum will not be changed, Iω _{1} =I'ω _{2} ;
you can then solve this for the new angular velocity, ω _{2} =(I /I' )ω _{1} =[M /(M+m )]ω _{1}
which is smaller. However, the rotational kinetic energy E of
the blade is now lower, E _{1} =½Iω _{1} ^{2}
and E _{2} =½I'ω _{2} ^{2} =½I {(M+m )/M}{[M /(M+m )]ω _{1} }^{2
} or E _{2} =[M /(M+m )]E _{1} .
On the other hand, if you wanted to add mass but keep the energy the
same, E _{2} /E _{1} =1=I'ω _{2} ^{2} /Iω _{1} ^{2}
or ω _{2} =ω _{1} √[M /(M+m )];
in this case, the angular momentum will have changed.

Your third question is moot since we have established that torque does
not depend on mass.

QUESTION:
If I stood beside a small operating hovercraft with a sail built into the front of it and blew air into the sail with a leaf blower I know that the craft would move forward. Now the question.If I sat down on the hovercraft with the leaf blower in hand and we became one with the hovercraft and I then blew air into the sail would we move forward or would action and reaction of the leaf blower neutralize the forward motion?

ANSWER:
It is easiest to understand if you think first of using a stick
instead of a leaf blower. Standing on the ground and pushing with the
stick on the sail, there is an unbalanced force acting on your
hovercraft (the stick). Now, if you stand on the hovercraft, the stick
exerts a backward force on you (part of the hovercraft, now) and the
stick exerts a forward force on the hovercraft and these cancel out. Or,
if you like, the only forces which have any effect on a system are
external forces and by becoming part of the system what you do is no
longer an external force. The leaf blower is a little trickier, but I
believe even worse! The leaf blower will exert a backward force on you
(like a little jet engine) and the stream of air will exert a forward
force on the sail; but some of the force from the stream of air will be
diminished by the air slowing down on its way to sail because of
interaction with the still air. So, the net effect would be for the
whole hovercraft to move backwards; probably not noticible because of
friction and the smallness of the loss of power due to the still air.

QUESTION:
If basketball (A) weighs 1 lb and is tossed upward to goal (A) at a height
of 8 ft (5 ft above my daughter's head) and basketball B is 6lbs, At what
height should goal (B) be to generate the same force to toss?

FOLLOWUP QUESTION :
Actually this is not homework, let me explain the situation. My daughter is 5 years old and playing Kindergarten basketball. Only one person on her team can toss the ball high enough to score, the ball weighs roughly 1 lb and the goal is roughly 8 ft high. I purchased a weight trainer ball that is exactly the same diameter as the regulation ball she uses but it's a 6 lb ball.
My theory is that I can build her a goal in the house that is not as tall but would require the same energy to make the basket. therefore making her stronger. And when it comes time to shoot the lighter ball in the taller goal, she shouldn't have any problems.

ANSWER:
For the 1 lb ball, the energy which must be supplied is 1x5=5 ft·lb,
assumning that she releases the ball at the level of the top of her
head. The 6 lb ball, if sent vertically with an energy input of 5 ft·lb,
will rise to a height of h =5/6=0.83 ft=10 inches above her head. All this
assumes that the ball is thrown straight up. Note that I have not really
answered your question because you asked for force and the energy input
depends both on force F and the distance s over which it is
applied. The energy input W could be written as W=Fs ,
So, if you assume that she throws it the same way and pushes as hard as
she can, the force need not be known.

QUESTION:
I am trying to figure this out for a dear Uncle on Vancouver Island as a challenge. I have suffered a concussion so trying is difficult. He used to live in S. Wales and dropped stones done old coal shaft. He was a teacher.
Wonder if you could help me please.
"A stationary rock is allowed to drop down an 800 foot shaft. Without compensating for air resistance, how far does it fall during
the sixth second of its descent? This is the formula. Assume gravity value to be 32 feet per second per second.
Please set out your answer clearly showing your thought process, line by line. Use words as well as numbers.
I'm afraid your answer so far is incorrect.
If needed, the formula we used was S = ut + half gt^{2} ."
Thank you very much. I am in Gr 5!! I want to by a physicist.

ANSWER:
I will assume that this is not a homework problem (forbidden on
this site!); at least if it is you went to a lot of trouble to disguise
it! Your equation is right except since we will start the clock (t =0)
when you let go of the stone, u =0 because u in your
equation is the speed at t =0. Also this equation assumes that
S =0 at t =0 and that S increases in a downward
direction. So, at the end of 5 seconds (the beginning of the 6^{th}
second) the position is S _{5} = ½x32x5^{2} =400
ft; at the end of 6 seconds the position is S _{6} = ½x32x6^{2} =576
ft. So the total distance traveled is 176 ft. I trust you will not
present this work to your uncle as your own.

QUESTION:
What would be the estimated terminal velocity be of a 4,300lb car falling from 30,000ft above sea level be?

ANSWER:
The terminal velocity, v _{t} , does not depend
on the altitude from which you drop your car. This can be a very tricky
problem because v _{t} does depend on the density of the
air which changes greatly from sea level to 30,000 ft. So to get a first
estimate, I will just assume sea level density everywhere. There is an
estimate for the drag force in sea level air which is good for a rough
estimate, F _{D} = ¼Av ^{2}
where A is the cross sectional area and v is the
speed. From this you can show that v _{t} =2 √(mg /A ).
In SI units, m =4300 lb=1950 kg, g =9.8 m/s^{2} ,
and A ≈2x4=8 m^{2} (estimating the car as 2 m wide
and 4 m long). Then v _{t} ≈100
m/s=224 mph.

I guess we should now ask
whether we expect it to reach terminal velocity before it hits the
ground. Actually, it will technically never really reach terminal
velocity, only approach it—see an
earlier answer . I will
calculate how far it falls before it reaches 99% of v _{t} .
In the earlier answer , I
show that the height from which you must drop it for it to reach
terminal velocity with no air drag is h ^{no drag} =v _{t} ^{2} /(2g ),
and the height from which you drop it for it to reach 99% of terminal
velocity with air drag is h ^{drag} =1.96v_{t} ^{2} /g
(derived from the expression v /v _{t} =0.99=√[1-exp(-2gh /v _{t} ^{2} )].
So, for your case, h ^{no drag} ≈510 m and h ^{drag} ≈2000
m. At 2000 m (around 6000 ft) the air is about 85% the density of
sea-level air, so I believe that my approximation assuming constant
density is pretty good and the car would probably reach 99% of the
terminal velocity by the time it hit the ground. To actually put in the
change in density with altitude would make this a much more difficult
problem.

QUESTION:
Does a round and square object, the same weight, fall the same?

ANSWER:
If air drag is negligible, like if you drop them from a few
feet, yes. If they fall fast enough for air drag to be important, they
will fall differently and, if they are about the same size, say a sphere
and a cube, the sphere will fall faster. That may be all you want, but I
will go on and explain in a bit more detail. The drag force F _{D}
on an object may be approximated for every day objects, masses, and
speeds as F _{D} = ½C _{D} ρv ^{2} A
where ρ is the density of the air, v is the speed,
A
is the cross-sectional area presented to the onrushing air, and C _{D}
is called the drag coefficient. C _{D } depends only on
the shape of the object. C _{D} ≈0.5 for a sphere
and C _{D} ≈0.8 for a cube (falling with one face
to the wind). So, if their areas are about the same, the drag on the
sphere will be smaller and it will go faster. Of course if the sphere
area were ten times bigger than the cube area, that would be more
important than the somewhat smaller drag coefficient and the cube would
win.

QUESTION:
What happens gravitationally when the center of mass can no longer be considered a point but is instead an area? Specifically, suppose the Sun was to "explode" or supernova; ignoring the obvious destruction of the solar system, what would happen to the planetary orbit of Earth? I presume it would be roughly akin to letting go of the string at the end of which I have a ball spinning around me.

ANSWER:
The center of mass is always a point. If the sun were to
"explode", the center of center of mass would continue to be at the
center of where it was before the explosion. A star explodes
approximately isotropically, that is, material goes out at the same rate
in all directions. So, until the material reached the earth's orbit, the
orbit would be unchanged. But, as material gets outside the earth's
orbit, only the material inside would contribute to the force felt by
the earth (this is Gauss's law). So the earth would behave as if there
were a star of constantly decreasing mass at the original center of
mass.

QUESTION:
I bought a 400lb gun cabinet and need to pull it on a 2 wheel hand cart up
a 12ft ramp at about 35degrees to the horizontal How much load does 1 or 2
people have to carry and how much is borne by the wheel. I am trying to make
sure we can be comfortably safe!

ANSWER:
I could make a rough estimate but would need to know the dimensions of the
cabinet, if the center of gravity is near the geometrical center. I would
assume that the cabinet was parallel to the ramp when being pulled.

FOLLOWUP QUESTION:
It is 20X29X55. It would not be parallel to the ramp but about 20 degrees from the ramp (which is about 35 degrees to the ground (thus avoiding 4 steps).

ANSWER:
Since only an approximation can be reasonably done here, I will
essentially model the case as a uniform thin stick of length L
with weight W , normal force N
of the incline on the wheel, and a force F
which you exert on the upper end. In the diagram above, I have resolved
F into its components parallel (x )
and perpendicular (y ) to the ramp. Next write the three
equations of equilibrium, x and y forces and the
torques; this will give you the force you need to apply
to move it up the ramp with constant speed.

ΣF _{x} =0=F _{x} -W sinθ ΣF _{y} =0=F _{y} +N-W cosθ
Στ= 0=½WL cos(θ+φ )-NL cos(φ ).

I summed torques about the end where you are
pulling. Putting in W =400 lb, θ= 35º, and
φ= 20º, I find F _{x} =229 lb, F _{y} =206
lb, and N =122 lb. Note that you do not need to know the length
L . The net force you have to exert is F =√[(F _{x} )^{2} +(F _{x} )^{2} ]=308
lb. If someone were at the wheel pushing up the ramp with a force
B , that would reduce both F _{x}
and F _{y} . This would change the equations to

ΣF _{x} =0=F _{x} -W sinθ +B ΣF _{y} =0=F _{y} +N-W cosθ
Στ= 0=½WL cos(θ+φ )-NL cos(φ )+BL sin(φ ).

For example, if B =100 lb, the solutions
would be F _{x} =129 lb, F _{y} =169 lb,
and N =159 lb; so your force would be F =213 lb.

QUESTION:
Would it be physically possible to create a parachute capable of delivering a main battle tank safely to a theatre of war?
Like how huge would it have to be?

ANSWER:
A reasonable estimate of the force F of air drag on an
object of mass m , speed v , and cross sectional area A is F = ¼Av ^{2} ;
this works only in SI units. The speed v when F=mg is called
the terminal velocity. I estimated a reasonable terminal velocity would
be the speed the tank would have if you dropped it from about 10 m,
v ≈14 m/s. The mass of the
MBT-70 (KPz-70)
is about 4.5x10^{4} kg. Putting it all together, I find that
A ≈10^{4} m^{2} , a square about 100 m on a side or a
circle of radius about 30 m. You could do a more accurate calculation
but this gives you a reasonable estimate.

QUESTION:
When solving questions involving two identical springs being stretched to points A and B to create a total length and the natural length of the springs are given. What is a consistent way to calculate the amplitude?
Is it half of the length of AB subtract the natural length of the spring?

ANSWER:
I assume that the springs are in series—one attached to
the end of the other. Suppose that you exert some force F such
that the springs are stretched by a distance s . Then each
spring will be stretched by ½s . If k is
the common spring constant, F =½ks . Therefore
the two together behave like a single spring with spring constant ½k.

QUESTION:
In my AP Physics class, we have had a class-wide debate over a physics
problem for the last few days. The problem asked how much work it would take
to move a satellite that was orbiting Earth at a certain height to a greater
height. Some of us say that the work equals the change in potential energy,
while others say that the work is the change in the total mechanical energy.
The total energy method gives an answer of exactly half of the amount the
potential energy method gives. Who is right?
Both orbits are circular.

ANSWER:
Since the speeds in the two orbits are different, the kinetic
energies will be different, so the correct way to do this is ½mv _{2} ^{2} -mMG /r _{2} ^{2} =W +½mv _{1} ^{2} -mMG /r _{1} ^{2} .
Of course you will have to figure out what the velocities are in terms
of the radii, but that should be a piece of cake for AP students!

QUESTION:
A car is travelling 45 mph, with the road being at a 5-8% incline. How
long would it take a car to slow down without using breaks? This is not homework. I am a claims examiner that is trying to get some information as to the rate of speed in which a car decelerates.

ANSWER:
You are a claims examiner and you spell brakes "breaks"?! I could do this if I assumed no friction whatever, but it would not be predictive of the real world because there is plenty of friction acting on a moving car even without brakes applied. I could make a better estimate if you could tell me how far this particular car, starting at 45 mph, traveled on level ground with brakes not applied. Also, is the car in gear? In neutral? Engine running?
If there were absolutely no friction, the car would keep going until it had gone
vertically up a distance of about 20 m ≈66 ft regardless of
the grade of the incline. For a 6% incline, the distance traveled by the
car would be about 1100 ft. In the real world it would be way less than
this.

QUESTION:
I knew that when a car which travels very fast and is brake very sudden, the
car will like "fly away". But is there any theory or principle or rule can
explain this?

ANSWER:
I do not know what "fly away" means.

FOLLOWUP QUESTION:
To explain, an illustration is made. Imagine a bicycle that travels
in very high velocity and it is braked suddenly and hardly, what will
happen. The bicycle will like "flying up" caused by the momentum. It is
also related to the principle of acceleration and deceleration.

ANSWER:
OK, I get it now. Refer to the figure above. The easiest
way to do this is to introduce a
fictitious force . If
the car is accelerating with acceleration a (which points
opposite the velocity v when braking), Newton's first law will
be valid in the car frame if a force F=-ma acting at the center
of mass (COM) is introduced. The "real" forces on the car are a
normal forces up on each wheel by the road, the frictional forces
backwards on each wheel by the road, and the weight mg which
acts at the COM. In the drawing I have only labeled the weight, the
normal force on the rear wheel, and the fictitious force because those
are all you need to answer your question. If you wish, you could find
both normal forces and the sum of the two frictional forces in terms of
a ; if all wheels were locked you could find the individual
frictional forces if you knew the coefficients of kinetic friction. Now,
I will sum torques τ about the point where the front wheel
touches the ground, Στ=mah+NL-mgs= 0 where h
is the height of the center of mass above the ground, L is the
horizontal distace from the front axle to the COM, and s is the
horizontal distance between the wheels. You can now solve for N ,
N =(mgs-mah )/L . Now think about N ; if
N <0, the road would have to pull down on the back wheels to
hold the car from rotating forward about the front wheels. N
will be zero when the car is just about to "fly up"; therefore, if a >g (s /h ),
the car will "fly up".

QUESTION:
Ships are often built on ways that slope down to a nearby body of water. often a ship is launched before most of its interior and superstructure have been installed and is completed when a float.
Is this done because the added weight would cause the ship to slide down the ways prematurely?

ANSWER:
Friction can be a tricky business, but the simplest behavior is
that the frictional force increases proportional to the weight. But the
force of the gravity trying to slide the weight down the slope is also
proportional to the weight. Therefore, doubling the weight of the ship
should not increase its tendency to slide down. Besides, if this were a
concern you could always temporarily block the path down the slope like
placing blocks in front of a vehicle on a slope to keep it from rolling.
I suspect the real reason is that the structure of the ramp is probably
not strong enough to support the full weight of the ship.

QUESTION:
I am doing an experiment on factors affecting the travel distance of a toy car from down a ramp and thought it would be a good idea to understand how a physicist thinks of things. My question is, how do you think weight of an object affects the distance it will travel after going down a ramp?

ANSWER:
The best discussion I have seen of the physics of pinewood derby
races is this
youtube video . You will see that what matters more than the added
weight is where you put it. (Thanks to my son Andy for pointing me to
this video; his son and my grandson Finn placed second in the Cub Scout
pinewood derby last year on the strength of the tips here!)

QUESTION:
What would happen if you threw a baseball at the speed of sound?

ANSWER:
At such a large speed, air drag has an enormous effect on ball.
To see the mathematical details see earlier answers for a
lacross ball
and a baseball . I
will make the same assumption that I did in those answers that the
amount by which the ball will fall will be very small compared to its
horizontal distance and the speed acquired in the vertical direction
will be very small compared to its horizontal speed. So I will ignore
the small effect which the vertical motion will have on the horizontal
motion. As discussed in the
earlier answer ,
the horizontal distance x and speed v are given by
v=v _{0} /(1+kt ) and x =(v _{0} /k )ln(1+kt )
where v _{0} is the initial velocity and k is a constant
determined by the mass, geometry, and initial velocity of the ball. For
v _{0} =340 m/s (speed of sound) and a baseball
(mass=0.15 kg, diameter=0.075 m) these become v =340/(1+2.8t )
and x =121ln(1+2.8t ) and are plotted below.

In one second the ball goes about 160 m and slows down to
less than 100 m/s. During the same time, the ball will fall
approximately 5 m and so, if launched horizontally from a height of 5 m
will hit the ground in one second as shown below. Be sure to note the
difference in horizontal and vertical scales; an insert shows the trajectory drawn to
scale. The small distance fallen is the justification for my
approximations above.

QUESTION:
Hi this may be a hard question but If I wanted to run 2200 gpm through a 2500 foot run with 50 feet of fall what size pvc pipe would this take?
This is all gravity.

ANSWER:
At first I just did a calculation with no corrections for
viscosity or drag. I found the velocity had to be about 17 m/s and the
diameter of a pvc pipe would have to be about 0.1 m ≈4 in.
But then I worried about the fact that a pipe that long is likely to
have significant drag over its length. It is a pretty complicated
engineering calculation and I was unfamiliar with many of the
parameters. But I did find a
web
site which seems to have made it easy for me by including a
calculator. Frankly, I have no idea what the roughness coefficient is,
but it suggests a value of 150 for plastic. The result is below. As you
can see, to get a flow rate of about 2200 gpm would require a pipe with
diameter of about 10 in.

QUESTION:
If I place a liquid filled container on a scale and suspend a mass with greater density than the liquid within the liquid and then release the mass, will the scale register the full weight of the mass while the mass is in motion (falling) as compared to when the mass has settled on the bottom?
Will the scale read the same while the mass is accelerating as when it has achieved terminal velocity?

ANSWER:
Your second question indicates that you understand the answer
will be different depending on whether the falling mass is accelerating
or not. The figure shows that the weights of the fluid and the container
will act down on the scale. Now look at the falling object. In addition
to its weight there are two upward forces, the buoyant force
B and the drag force D ;
these are both forces which the fluid exerts on the object. But Newton's
third law says if the fluid exerts a force on the object, the object
exerts an equal and opposite force on the fluid. Therefore the scale
will read W _{f} +W _{c} +B+D . If
the object is falling with constant speed, it is in equilibrium and so
B+D=W _{o} and the scale reads the total weight of
container, object, and fluid. If the object is accelerating down,
B+D<W _{o} and the scale reads less than the total weight of
container, object, and fluid. There is an
earlier question
similar to yours except the object is rising instead of
sinking.

QUESTION:
Have scientists done experiment on what is the value of gravity below the earth surface as depth increases? if done pl. provide chart g vs depth.

ANSWER:
The
deepest hole ever drilled is only about 12 km deep. I could
not find any reference to attempts to measure g at various depths
down
this hole. Since the radius is about 6.4x10^{3} km, you would
only expect about a 0.2% variation over that distance. There are models
of the density of the earth, though, which have been determined by
observing waves transmitted through the earth during earthquakes or
nuclear bomb tests; these are believed to be a pretty good
representation of the radial density and can be used to calculate g .
The two figures above show the deduced density distribution and the
calculated g .

Usually in introductory physics
classes we talk about the earth as having constant density, but as you
can see, that is far from true—the core is much more dense than
the mantles and crust. If it were true, g would decrease
linearly to zero inside the earth. Instead, it increases slightly first
to around 10 m/s^{2} and remains nearly constant until you are
at a depth of around 2000 km. There is little likelihood that g will
ever actually be measured deep inside the earth because the temperature
increases greatly as you go deeper, already to near 200ºC at 12 km.
However, if you have detailed information on density distribution, there
is really no need to measure g .

QUESTION:
As part of our business we bag wrap passengers bags / suitcases prior to flying at the major UK airports.
We use and have used for many years a power pre stretch cast film —17 micron nano with a 300% capability.
Recent feedback from Heathrow airport suggests some of the passengers bags are sticking to the conveyer belts and are being miss-directed. I am being asked for the
'coefficient of friction' for the film we are using.
I have advised our supplier of this, they have sent through the data spec sheet but there is no mention of COF, on speaking with them they have never had this question raised before. Personally, I do not think this Is an issue with our film but more where customers themselves are wrapping their own bags with home use film. However, I need to provide proof that the film we are using does not have any adhesive properties.
My question is—would the COF affect this and how do I get the actual information on the film?

ANSWER:
The force of friction f depends on only two things:
what the surfaces which are sliding on each other are (conveyer material
and your plastic film) and the force N which presses the two
surfaces together; normally, on a level surface, the force N is
simply the weight of the object (suitcase in your case). There are two
kinds of COFs, kinetic and static. The kinetic coefficient, μ _{k} ,
allows you to determine the frictional force on objects which are
sliding. In that case, f =μ _{k} N.
The static coefficient, μ _{s} , allows you to
determine how hard you have to push on the suitcase in order for it to
start sliding; in this case f _{max} =μ _{s} N
where f _{max } is the greatest frictional force you
can get. Since you are being asked to prove that it is not too "sticky",
it is the static, not the kinetic, coefficient which you need; measuring
μ _{s } is quite easy. The only problem is that μ _{s
} depends on the surfaces so you must have a piece of the material
from which the conveyer belts are made to make a measurement. Once you
have that, use it as an incline on which to place a wrapped suitcase.
Slowly increase the slope of the incline until the suitcase just begins
to slide. Your COF (μ _{s} ) is equal to the tangent
of the angle of the incline which (see diagram above) is simply μ_{s} =H /L .

QUESTION:
A 1500lb 8'x8' box that is 3'6" tall is lifted at one of the four side so that the opposite side acts as a pivot on the ground (like a strong man flipping a giant tire in a a world strongest man competition). How much actual weight is being lifted? assuming that the weight distribution of the box is perfectly even.

ANSWER:
Well, that depends on how you lift it. Let's assume that you
lift it so that you cause it to rotate with uniform speed. One way that
you could accomplish this is to push in a direction perpendicular a line
drawn from where you are pushing and the edge on the box remaining on
the floor. Referring to the figure above, the equilibrium conditions are
N +F cosθ -W=0, f-F sinθ =0,
and ½LW -LF cosθ =½W -F cosθ =0;
here F is the force you exert, W is the weight, N
is the force the floor exerts vertically, and f is the
frictional force exerted on the floor. Solving these I find that N =½W ,
F =½W /cosθ , and f =½W tanθ .
I suspect that the case you are interested in is when you first lift it
off the ground, θ= 0. F =750 lb, half the total
weight. Note that this analysis is valid as long as the floor is not too
smooth, that is the box does not start sliding at some angle; the angle
is less than θ= tan^{-1} (L /H )
because at that angle the center of gravity is directly over the pivot
side.

Of course there are lots of other ways you could
lift it which would be more efficient if your aim was to tip it over;
for example, you could start pushing horizontally once you got it off
the ground so that the floor would hold up all the weight rather than
half the weight. There is an
old answer very
similar to yours that you might be interested in.

QUESTION:
I am an engineer working on the reconstruction of a traffic accident where it is alleged that a car traveling over a railroad crossing became airborne at a speed lower than the posted speed of the road. The information that I have available includes the type and make of the car and the geometry of the road. Nothing in the engineering literature that I can find addresses this issue. Despite the five quarters of physics that I took long ago, I am having trouble finding the information that I need to model this.
Any suggestions?

ANSWER:
This is sort of a classic introductory physics problem. The idea
is: when does an object lose contact with the surface on which it is
moving, usually taken to be a segment of a circle. I will do an
approximation that the shape is a circle and that the car is a point
mass. You can then generalize to your case from there or else give me
more information. In the sketch above, the radius of the circle is R ,
the weight of the car is mg , the angle specifying the current
position of the car is θ , the force which is
causing the car to move with some constant speed v is F ,
and the normal force of the road on the car is N . Note that
although the car has a constant speed, it has a centripetal acceleration
a=v ^{2} /R toward the center of the circle.
Applying Newton's laws, F-mg cosθ =0
and mg sinθ -N =mv ^{2} /R.
The first equation tells you what force you need to keep it moving at a
constant speed, F=mg cosθ which is really not of
interest to you; note that the force is in the direction of v
on the way up and opposite on the way down since the cosine changes sign
at 90^{0} . The second equation tells you what N is for
any position of the car, N=mg sinθ-mv^{2} /R ;
note that if N is negative it points toward the center but the
road cannot pull down, only push up, so the car could not stay on the
road at that speed and angle. What is really of interest is under what
conditions would N =0=g sinθ-v^{2} /R
or v ^{2} =Rg sinθ ; this would
tell you the speed (angle) at which a car with a particular angle
(speed) would leave the road. Note that it is independent of the mass.
Notice also that v ^{2} /Rg< 1 because the sine
function cannot be larger than 1.0. For example, for what speed will the
car leave the road at 45^{0} if gR =300 (m/s)^{2}
(R ≈30 m)? v =√(300x0.707)=14.6
m/s=32.7 mph. Or, at what angle will a car with speed 35 mph=15.6 m/s
leave the road?
θ= arcsin(15.6^{2} /300)=54.2^{0} .

One thing which occurs to me though is, since you know the car and
railroad crossing, why don't you just do the experiment and drive it
over the crossing?

QUESTION:
Say there is a cylinder on a ramp and the friction force from the ramp cancels out the parallel component of gravity. Therefore, the cylinder should be in linear equilibrium. However, from the reference point of the center axis of the cylinder, there is a net torque exerted by the friction force. Additionally, there is also a net torque exerted by the gravitational force from the reference point of the point of contact of the cylinder and the ramp. Therefore, it is not in rotational equilibrium and should start to rotate, correct? How is this possible, because if the cylinder starts to roll how can it also be in linear equilibrium?

ANSWER:
There is a simple answer to your question: the frictional force
is not equal to the component of the weight along the incline. Rather,
-f+mg sinθ=ma where θ is
the angle of the incline and a is the acceleration of the center of mass
down the incline.

FOLLOWUP QUESTION:
Thank you very much for your response. However, I think you may have misunderstood my question. I was asking what would happen in a case where the frictional force is set to cancel out the parallel component of weight. It seems as if the center of mass cannot move, but the cylinder needs to rotate. Therefore, it would appear as if the only outcome of this situation would be a cylinder rotating in place on a ramp, which does not seem possible. I think that the cylinder would have to roll down the ramp, but I can't see how this would be consistent with linear equilibrium.

ANSWER:
I did not misunderstand your question. You cannot simply adjust
the friction to be what you want it to be. You can, however, simulate
what you want to happen by wrapping a string around the cylinder and
pulling up on the string with a force mg sin θ
where
θ is the angle of the incline and m is the mass
of the cylinder; imagine that the incline is smooth (frictionless). Now
there will be a net torque about the center of mass of mgR sinθ=Iα
where I is the moment of inertia and
α is the angular acceleration of the cylinder. The
cylinder will spin in place. Your hand will have an acceleration of
a =mgR ^{2} sinθ /I ; for a uniform
solid cylinder with I =½mR ^{2} ,
a =2g sinθ .

ADDED
THOUGHT:
If the coefficient of kinetic friction is exactly equal to
μ _{k} = tanθ ,
the cylinder will slide down the incline with constant speed because the
frictional force will be f=μ _{k} N =(tanθ )(mg cosθ )=mg sinθ.
So, if you start it sliding and not rolling, it will begin spinning
about its center of mass because of the torque due to the friction and
have an angular acceleration α=fR /I ; it will continue
sliding down the plane with constant speed, though.

FOLLOWUP QUESTION:
Thanks again but I am still a bit confused. It makes sense that the center of mass will move at a constant velocity while the cylinder is rolling, but how did it acquire that velocity in the first place if I start the cylinder at rest and not sliding as you wrote in the additional thought. In other words, what happens if the coefficient of static friction is equal to mgsinθ and the cylinder starts with no velocity of any kind?

ANSWER:
If you start the cylinder
at rest on the incline and
μ _{k} = tan θ
(coefficient of static friction will be larger),
the cylinder will roll without slipping. If you solve the dynamics for
the cylinder rolling without slipping you will find that (see the figure
above) f= (mg sinθ )/3 and a =(2g sinθ )/3
where f is the frictional force and a is the
acceleration of the center of mass. Since f <mg sinθ ,
if you simply let it go, it will roll without slipping. So, if you want
it to slide down with constant speed, you must give it a shove to start
it slipping.

QUESTION:
If I were to drop two round balls of different mass under water would they both fall to the bottom at the same velocity or would one reach the bottom first?

ANSWER:
The forces on a ball are its own weight mg down; the
buoyant force B up which would be equal in magnitude to the
weight of an equal volume of water; and the drag force f up
which would depend on the size the ball and its speed. The net force
F would be F=-mg+B+f ; as the ball went faster and faster,
f would get bigger and bigger until eventually F =0 and
the ball would go down with a constant speed called the terminal
velocity. The larger the mass, the larger the terminal velocity. Without
specifying the the sizes of the balls, your question cannot be answered.
If they had identical sizes, the heavier ball would reach the bottom
first because it would have a larger terminal velocity. Of course, if
B>mg the ball would float!

QUESTION:
I'm trying to find a fast/easy way to test whether a sealed, consistently-dimensioned rectangular box is sufficiently "stable" for transport on a (small) 2-wheeled bicycle trailer. The box is pretty tall. If it's over-weighted and top-heavy, it'll flip the trailer around turns (which are sufficiently tight/quick). I figure there might be a quick, static "tip test" with a combination pull gauge, inclinometer and scale, but my math skills are primitive. Is there a simple way to ascertain whether, for a given object, a target stability threshold is met?

ANSWER:
There is an
earlier
answer about a bicycle making a turn. It would be helpful for you to
read that first. The easiest way
to do this problem of your trailer turning a curve is to introduce a fictitious
centrifugal force which I will call C , pointed
away from the center of the circle; the magnitude of this force will be mv ^{2} /R where
m is the mass of the box plus trailer, v
is its speed, and R is the radius of the curve. The picture to the right
shows all the forces on the box plus trailer: W is the weight and the green
x is the
center of gravity (COG) of the box plus trailer; f _{1} and f _{2} are the
frictional forces exerted by the road on the inside and outside wheels
respectively; N _{1} and N _{2} are the normal
forces exerted by the road on the inside and outside wheels respectively;
the center of gravity of the box plus trailer is a distance H above the road and the wheel base is
2L (with the center of gravity halfway between the wheels). Newton's
equations yield:

f _{1} +f _{1} =C
for equilibrium of horizontal forces;

N _{1} +N _{2} =W
for equilibrium of vertical forces;

CH+L (N _{1} -N _{2} )=0
for equilibrium of torques about the red x .

If you work this out, you find the normal forces
which are indicative of the weight the wheels support: N _{1} =½(W-C (H /L ))
and N _{2} =½(W+C (H /L )). A few things to
note are:

the outer wheel supports more weight,

if C =0 (you are not turning), the inner
and outer wheels each support half the weight,

at a high enough speed C will become so large
that N _{1} =0 and if you go any faster you will tip over, and

if the road cannot provide enough friction you
will skid before you will tip over.

Now we come to your question. You first want the
maximum speed without tipping. Solving for v in the N _{1} =0
equation gives

v _{max} = √[RWL /(mH )]=√[RgL /H ]

where g =9.8 m/s^{2} =32 ft/s^{2}
is the acceleration due to gravity. For example, suppose that R =7
ft, L =17 in=1.42 ft, H =30 in=2.5 ft. Then
v _{max} = √[7x32x1.42/(2.5)]=11.3 ft/s=7.7
mph.

Be sure to note that the assumptions of a level
road (not banked) and wheels not slipping are used in my calculations.
Also be sure to note that W is the weight
of both box and trailer and 2L is the wheel base, not
the box width.

One more thing is that you might not know how to
find the COG of the trailer plus box. If the COG of the trailer is H _{trailer
} above the ground (probably close to the axle) and the COG of the
box is H _{box} above the ground , then H =(H _{box} xW _{box} +H _{trailer} xW _{trailer} )/W .

ADDED
THOUGHT:
When just about to tip, all the weight is on the outer wheel and so
N _{2} =W
and f _{2} =μN _{2} =μW ,
where
μ is the coefficient of static friction. If you work it out,
the minimum value the μ must have to keep the trailer from
skidding is μ _{min} =L /H . For the
example worked out above, μ _{min} = 0.57. For
comparison, μ for rubber on dry asphalt is about
0.9, so the trailer would not skid.

QUESTION:
Will a heavier ball roll down a small slope faster than a lighter ball?
Or will a lighter ball roll down a small slope faster than a heavier ball?

ANSWER:
If both are solid balls of the same radius but of different masses, they
will take equal times if air drag is neglected as might be appropriate for a
"small slope". If they get going fast enough that air drag becomes
important, the heavier ball will win the race.

QUESTION:
Gasoline contains 40 megajoules of energy per kilogram and gasoline trucks have around a hundred tons of it. So how does a gasoline truck exploding not produce an explosion similar to a small nuke?

ANSWER:
Since it is always a little ambigous what is meant by a ton, I did my own
calculation using the volume of a tanker truck of about 10,000 gallons and
the density of gasoline of about 2.7 kg/gallon. I got the total energy
content of about 10^{12 } J. The energy of the Nagasaki bomb, a small
bomb by today's standards, was about 10^{14} J, 100 times bigger.
Two things to consider are:

T he
bomb number represents total energy delivered whereas I would guess
that likely less than half the energy content of the gasoline would
actually be delivered.

The time over
which the energy is delivered is likely much longer for the gasoline
explosion than the nuclear explosion. As an example, just to
illustrate the importance, suppose the bomb exploded in 1 ms=10^{-3}
s and the tank in 1 s. Then the power delivered by each is 10^{8}
GW for the bomb and 10^{3} GW for the tank. As a result the
destructive power of the bomb would be much bigger.

QUESTION:

What exactly is momentum and how is it different from force?

I understand that p = mv and F = ma but they seem so similar in application that I haven't fully wrapped my head around it.

Additionally, what does it mean for something to have momentum in the first place, and why must it be conserved?

How does light have momentum if it, by nature, has 0 mass (I presume E =mc^2 comes into play somewhere here)?

Finally, what is the significance in the fact that light does have momentum (if it does)?

ANSWER:
You have lots of questions, really. I have rearranged your question to
delineate it into parts:

For your first question, just say that p=mv .
My answer to #2 should clarify how force and linear momentum are
related. Force and momentum have to be different because they are not
even measured in the same units—momentum is mass*length/time and
force is mass*length/time^{2} .

What is acceleration? It is rate of change of velocity,
a= (v _{2} -v _{1} )/t
where t is the time to change speed from v _{1}
to v _{2} . So one could write that F= (mv _{2} -mv _{1} )/t=(p _{2} -p _{1} )/t ;
so force may be thought of as the rate of change of momentum. Newton
actually stated his second law this way, not as F=ma . It is
the second law which is a fundamental law of physics, momentum is just
defined because of its simple and natural relationship to the second
law.

It doesn't "mean" anything for something to have
momentum, it is just a definition. However, consider the second law if
there is no force acting; then F =0=(p _{2} -p _{1} )/t.
In other words, p _{2} =p _{1 }
which simply means that momentum does not change (is conserved). The
condition for conservation of momentum of a system is that there be no
external forces on it. For example; suppose you look at an isolated
galaxy which has billions of stars in it all interacting with each
other and there are negligible forces from the outside; if you sum up
all the momenta of all the stars today and in 10 years from now, you
would get the same answer even though the shape and orientation of the
galaxy would change.

It turns out that if v is very large,
comparable to the speed of light c , Newtonian mechanics is
incorrect. (You could say that Newtonian mechanics is wrong but a
superb approximation for low speeds.) If you say that p=mv it
turns out that momentum is not conserved for an isolated system at
very large speeds. However, since momentum conservation is such a
powerful way to solve problems, we redefine momentum (in the theory of
special relativity), to be p=mv /√(1-(v ^{2} /c ^{2} )),
momentum is conserved again and we still have p ≈mv
for small v . It also turns out that, as a result of this new
definition of p , we can write that E =√(p ^{2} c ^{2} +m ^{2} c ^{4} )
where E is the energy of a particle of mass m . So if
the particle is at rest, p =0 and E=mc ^{2} ;
if the particle has no mass, p=E /c .

No more
significant than if a billiard ball has momentum—it just does.

I have deleted
your question about angular momentum—it is off topic.

QUESTION:
I work in a high school where this question was posed by one of the pupils in a class I support. The question is this....
If you could attach a rope to a rocket, which would also be attached to earth, and sent it into space (out of our atmosphere) until the rope went taught and then cut the string. Would it stay in space or would it fall back to earth?

ANSWER:
This is one of those problems which I had fun with and I hope it will not be
too exhaustive an answer.
I will make the following assumptions:

the rocket
always goes straight up;

the rocket
stops moving vertically when the rope is taught;

the rope
is cut the instant that the rocket stops;

fuel and
the weight of the rope are not issues; and

the launch
is from the equator. This makes things much simpler and I will briefly
talk about a similar launch from some latitude at the end.

My
view of this problem, therefore, is the same as if the rocket were on top
of a very long stick vertically straight up and the stick is suddenly
removed.

The
thing to appreciate is that even though the rocket goes straight up, it
will have the same angular velocity ω as the earth so its
speed will be ω (L+R ) where L is the
length of the rope and R is the radius of the earth. The angular
velocity is ω =[(2π radians)/(24 hours)]x[(1 hour)/(3600
seconds)]=7.27x10^{-5} s^{-1} . If L is just
right, the rocket will assume an orbit like the
geosynchronous communication satellites ; this turns out to be if L =5.6R .

So, if the rope happens to be 5.6 times larger than the radius of the
earth, the rocket will remain (apparently) stationary above its launch
point; it is actually going in a circular orbit with a period of 24 hours.
(The animation above illustrates this, although not to the correct scale.)
For any other L the orbit will be an elipse with the center of
the earth being at the focus.
Visualize .

If L >5.6R , the starting
point of the rocket will be the perigee (closest point to the earth) of
its orbit. Visualize .

If L <5.6R , the starting point of the rocket
will be the apogee (farthest point from the earth) of its orbit.
Visualize .

For L <5.6R , though, there will be some critical distance
L=L_{c} when the perigee of that orbit is
exactly equal to R; in that case the orbit will just skim the surface of
the earth. After some laborious algebra I found that L _{c} ≈3.7R ,
about 2.2 earth radii inside the geosynchronous orbit.
Visualize .

For L<L_{c} ,
the rocket will crash back into
the earth but not where it was launched from because it is a projectile
which has a horizontal speed greater than that of the earth's surface.
Visualize .

Finally,
if the horizontal speed of the rocket ω (L+R )
is greater than or equal the escape v _{e} =√[2MG /(R+L )],
where M is the mass of the earth, the rocket will escape the
earth and never come back. I calculated this to be when L =7.6R ,
two earth radii beyond the geosynchronous orbit.
Visualize .

For
any other latitude θ the speed of the satellite will be ω (L+R )sinθ.
The resulting orbits, while all elliptical, are much more difficult
to visualize and maybe we should save that for another day! However, the
simplest launch of all would be from the north or south pole (θ= 0^{0}
or 180^{0} ) because it acquires no horizontal velocity (ω (L+R )sinθ= 0)
where the rocket would fall straight back down regardless of how high
it went.

QUESTION:
Will two objects traveling in the same direction ever collide? Assume the objects are on earth, unmanned and their mass, volume, weight, density and speed are the same. All variables that can come in to play should be assumed that they are equal, for example no hills, curves bumps and no change in surfaces to have any effect on the coefficient of friction. Just a question my wife and I were wondering about.

ANSWER:
I'm
not really sure what you are getting at here. If you are just wondering
whether two parallel lines ever intersect, the answer is no. But you seem to
want to know about material objects originally moving on parallel lines. To
make the situation simpler, let's just have the two objects, originally with
parallel velocities, move in otherwise empty space. If the original
velocities are equal and they are traveling side by side, they will
eventually collide because there is a gravitational attraction between them
which will eventually bring them together. However, if they are originally
side by side and traveling with unequal speeds, they would not collide if
their relative velocities were greater than the escape velocity. The escape
velocity for equal masses m originally separated by a distance
r is v _{escape} =2 √(mG /r ) where G =6.67x10^{-11}
N·m^{2} /kg^{2} is the gravitational constant.

QUESTION:
A 400lb person jumps up 2-inches on earth. If same person jumps up on the
moon, how high would the jump be?

ANSWER:
I will assume that whatever the jumper does will add the same energy on both
earth and the moon. The gravitational potential energy U at the
highest point must be equal to that added by the jump, and U=mgh
where m is the mass (400 lb), h is the height (2 in on earth), and
g is the acceleration due to gravity (9.8 m/s^{2}
on earth, 1.6 m/s^{2} on the moon). The mass is the same both
places, so h _{earth} g _{earth} =h _{moon} g _{moon} . Solving,
h _{moon} =12.25 in.

QUESTION:
Hi, I was wondering what are the chances of survival from falling from the ninth floor of a building, going over the science of that how does surface affect the fall, body weight and trajectory. What is the difference from falling from a third story window as opposed to a higher up one?

ANSWER:
Someone else also asked this question; apparently it refers to a recent actual
incident of a student falling out a dorm room window about 85 ft ≈26
m high; the student survived without serious injury. The second person also
wanted to know if I could estimate the force experienced on impact. First I
will calculate the speed he would hit the ground if there were no air drag.
The appropriate equations of motion are y (t )=26-½gt ^{2} . and
v (t )=gt where y (t ) is the height above the ground
at time t , and g is
acceleration due to gravity which I will take to be g ≈10
m/s^{2} . The time when the ground (y =0) is reached is found
from the y equation, 0=26-5t ^{2} or t =√(26/5)=2.3
s. Therefore v =10x2.3=23 m/s (about 51 mph). The terminal velocity of a falling
human is approximately 55 m/s, more than double the speed here, so the
effects of air drag are small and can be neglected for our purposes of
estimating. (If there is air drag, terminal velocity is the speed
which will eventually be reached when the drag becomes equal to the weight.)

Estimating the force this guy experienced when he hit the ground is a bit
trickier, because what really matters is how quickly he stopped. Keep in
mind that this is only a rough estimate because I do not know the exact
nature of how the ground behaved when he hit it. The main principle is
Newton's second law which may be stated as F=m Δv /Δt
where m is the mass, Δt is the time to stop, Δv= 23
m/s is the change in speed over that time, and F is the average force
experienced over Δt. You can see that the shorter the time,
the greater the force; he will be hurt a lot more falling on concrete than on a
pile of
mattresses. I was told that his weight was 156 lb which is m =71 kg
and he fell onto about 2" of pine straw; that was
probably over relatively soft earth which would have compressed a couple of
more inches. So let's say he stopped over a distance of about 4"≈0.1 m. We
can estimate the stopping time from the stopping distance by assuming that
the decceleration is constant; without going into details, this results in
the approximate time Δt ≈0.01 s. Putting all that into
the equation above for F , F≈71x23/0.01=163,000 N≈37,000
lb. This is a very large force, but keep in mind that if he hits flat it
is spread out over his whole body, so we should really think about
pressure; estimating his total area to be about 2 m^{2} , I find
that this results in a pressure of about 82,000 N/m^{2} =12 lb/in^{2} .
That is still a pretty big force but you could certainly endure a force of
12 lb exerted over one square inch of your body pretty easily.

Another possibility is that the victim employed some variation of the
technique parachuters use when hitting the ground, going feet first and
using bending of the knees to lengthen the time of collision. Supposing
that he has about 0.8 m of leg and body bending to apply, his stopping
distance is about eight times as large which would result in in an eight
times smaller average force, about 5,000 lb.

Falling from a third story window (about 32 feet, say) would result in a
speed of about 14 m/s (31 mph) so the force would be reduced by a factor
of a little less than a half.

ADDED NOTE:
A rough estimate including air drag would have his speed at the ground be
about 21 m/s rather than 23 m/s as above. Given the rough estimates in all
these calculations, this 10% difference is indeed negligible.

QUESTION:
Gravity makes Earth orbit sun, Milky Way and Adromeda collide with each other,etc...
My question is, what can provide the necessary energy for a system like this for billions of years?

ANSWER:
The earth orbiting the sun has energy, but it takes zero energy to keep it
orbiting. The Milky Way and Andromeda galaxies have energy but no additional
energy is added as they move toward each other and eventually collide.

QUESTION:
It might look like a homework question, but it is not. Please help me. I have asked this question everywhere I could, but everybody seems to ignore it.
So, the problem is: Let us say we have two bodies A and B in contact with each other, with A lying at the back of B, and the system is on a friction-less horizontal surface. Let A have mass 5 kg and B 10 kg. Now let's say I apply a force of 45 N on A with my hand, then the system begins to accelerate at 3 m/s^2 and the net force on B by A is then 30 N, and B in reaction applies a net force of -30 N on A. Thus, the net force on A is 15 N. What I do not understand is why A is not applying a force of 45 N on B? If it is due to the reaction of B on A, how does A know in the first place that it is to exert a force of 30 N on B so that it receive a reaction of -30 N from B? Is not the reaction force of B on A some kind of a function of the action of A on B, and if it is, then how is the magnitude of the action of A on B is first determined? What is it that I do not understand about Newton's Third Law of Motion?

ANSWER:
OK, I will take your word for it that it is not homework. It is important to
be able to solve these kinds of problems. My method is to choose a body and
look only at that body. I choose first (as you did) to choose both masses as
the body, so M =15 kg and F =45 N and therefore a=F /M =3
m/s^{2} . Next I choose B as the body. The only force on it is the
force which A exerts on it, F _{BA} . Since we know that m _{B} =10
kg and a _{B} =3 m/s^{2} , F _{BA} = m _{B} a _{B} =30
N. Finally choose A as the body. Two forces act on A, F= 45 N and the
force which B exerts on it, F _{AB} =-F _{BA} =-30
N; its mass is m _{5} =5 kg and therefore a =(F+F _{AB} )/m _{5} =(45-30)/5=3
m/s^{2} . The reason that there is not a 45 N force on B is because
your finger is not touching B, only A is touching B. The reason A "knows" to
exert a force on A is that it has no choice since A's acceleration and mass
are already fixed. Once you know the "reaction" force you automatically know
the "action" force because of Newton's third law, F _{AB} =-F _{BA} =-30
N for this problem. You could also have chosen A as the body before you
chose B as the body. Two forces act on A, F= 45 N and the force which
B exerts on it, F _{AB} . Its mass is m _{5} =5 kg
and its acceleration is 3 m/s^{2} ; therefore (45+F _{AB} )= m _{5} a= 15
N or F _{AB} =15-45=-30 N.

QUESTION:
If I am running at average sprinting pace for an 17 year old male and
I jump off of a 80 meter drop, how far forward will I land from the jumping point? Assuming I am around 11 stone.

ANSWER:
80 m is pretty high, so you will be going very fast when you hit;
therefore, neglecting air drag might introduce significant error. But
air drag is pretty tricky to calculate and I will neglect it; the answer I
get will be somewhat bigger than what would really happen. I will take your
speed to correspond to running a 100 m dash in 15 s, about v ≈6.7
m/s. The equations of motion are x =6.7t and
y =-4.9t ^{2}
where x and y are the horizontal and ver tical positions
relative to the edge of the cliff and t is the time after jumping.
Solving the y equation for t when y =-80 m (the ground),
t =√(80/4.9)=4.04 s, so x =6.7x4.04=27.1 m. That is the answer
neglecting air drag. Note that it is independent of the mass m .

QUESTION:
What is the purpose of utilizing a percentage of body weight to determine how much weight to bench press/push/whatever? I know this seems like a fitness question and not a physics question, but what I am interested in is WHY weight would be used to determine how much one could (or should be able to) lift/push?
For example: A gym teacher wants to grade his students on their strength. He decides to use abililty to push a weighted sled across the floor as the measure. He wants to make the task equally difficult for every student in order to make the grading fair. So, he decides that each student will push 2x his/her body weight for 5 minutes and the grade will be based on how FAR the student is able to push.
So, Student A weighs 170 pounds and pushes 340 pounds (including the weight of the sled) for a total of 160 yards. Student B weighs 240 pounds and pushes 480 pounds (again, including the weight of the sled) for 80 yards. Student A pushed farther and gets a better grade, but Student B complains that he had to push much more weight so he should not get a worse grade.
Does Student B have a legitimate complaint or does his heavier weight contribute somehow to his ability to push that doesn't have anything to do with his strength? As in, does his weight help push the sled in some way?
Sorry, I don't know enough about physics to ask this question using proper physics terms like force, mass, etc. I hope you will still answer my question!

ANSWER:
I cannot comment on the rationale for correlating weight to strength. I can
certainly comment on the physics of your particular example of sled pushing.
I would first of all comment that this example is certainly not one solely
of strength because, since it is a timed activity, endurance as much as
strength is being tested; if one student, for example, were a heavy smoker,
he would likely become exhausted more easily. As a physicist, I would equate
"strength" with force. The specific example you give, though, seems to me to
be more related to energy (work done by the student) or power (rate of
energy delivered) than strength; purely in terms of strength, the heavier
student exerts more force. The force F which each student must exert
depends on the weight w he is pushing and the coefficient of friction
#956;
between the sled and the ground, F = μw .
The work W done in
pushing the sled a distance d is W =Fd = μwd .
The power generated if W
is delivered in a time t is P=W /t =μwd /t .
Both students have the same
μ
and
t , so W _{A} /W _{B} =P _{A} /P _{B} =d _{A} w _{A} /d _{B} w _{B} =(160x340)/(80x480)=1.42.
So student A did 42% more work, generated 42% more power, than student B.
From a physics point of view, B demonstrated more strength, A demonstrated
more power. I would judge that this is not a fair way to assign a grade. It
would be interesting to see if A (B) could move B's (A's) sled 80 (160)
yards.

QUESTION:
the Coriolis effect..
on a freely falling body... is to east...?
motion of earth is towards east right.? so we must feel that falling body deflects to the west isn't it..???

ANSWER:
I cannot give the full derivation of the motion of a particle in a rotating
coordinate system, it is much too involved. I can give you the results,
though. The coordinate system we will use is the coordinate system (x',y',z' )
shown to the right. The Coriolis force is given by 2mv' xω
where v' is
the velocity of m and
ω is the angular
velocity of the earth ( ≈ 7.3x10^{-5}
s^{-1} ). Now, for a body dropped from some height h , the
direction of the velocity is in the negative z' direction, so the
direction of
v' xω
is east. This is not what
you would intuitively suspect (as you note), but it is true. An expression
for how far eastward it would drift before hitting the ground is x' =[( ω· cosλ )/3]√(8h ^{3} /g )
where λ is the latitude. For example, at the equator where λ =90^{0} ,
and you drop it from 100 m, the deflection would be x' =2.2x10^{-2}
m=2.2 cm.

QUESTION:
Two bodies A and B are at rest and in contact with each other. Now if some arrangements made body A exerts a pressure of 10 NEWTON on body B then according to NEWTON'S 3rd law of motion body B will also exert an equal force of 10 NEWTON in opposite direction so the resultant force should be zero, while practically we see that there will be a resultant force of 10 NEWTON acting on body b and if mass and other conditions of body B are such that it moves by applying 10 NEWTON force on it. Then it will start moving. HOW?

ANSWER:
You have this all confused. But, you are not alone! When doing this kind of
problem, you must choose a body to focus on; only forces on that body
affect its motion. So, if you are interested in what body B will do, you
look only at that body. The force which B exerts on A is not a force on B.
So, if B has a mass of, say, 2 kg and there are no other forces on B, it
will have an acceleration of a _{B} =10/2=5 m/s^{2} .
Since the 10 N force continues, A and B remain in contact, so a _{A}
must also be 5 m/s^{2} . One force on A is the 10 N force from B
which is in the opposite direction as the acceleration. But the net force on
A must surely be in the direction of the acceleration, so there must be some
other force (probably due to you pushing on A) on A which is bigger than 10
N and points in the direction of the acceleration. For example, if the mass
of A is 4 kg, F -10=4x5=20 N, so F =30 N. Finally, you could
look at A and B together as the body. In that case the forces they exert on
each other do cancel, the acceleration is 5 m/s^{2} , and the total
mass is 6 N; therefore, there must be some external force (you again)
causing this 6 kg to have an acceleration of 5 m/s^{2} : F=ma =6x5=30
N. All three ways of looking at this problem are consistent with each other.

QUESTION:
i love physics and this question i asked to my teacher and principal but they couldn't answer it so my question is about third law of motion "every reaction has an equal and opposite reaction" so when a truck moving with constant speed hits a stationary car so according to newton's 3rd law of motion truck shoud be stopped after collision because car applies equal force on truck which it have during collision.

ANSWER:
Why should it be stopped? Certainly the truck experiences the force which
the car exerts on it, but every force does not have the effect of stopping
the object which experiences a force. If you are running toward a fly which
is hovering at rest in your path, you feel the force of the fly but it does not
stop you.

QUESTION:
Supposing a weightless container is filled with water. I am sure the pressure at the bottom of liquid, P1 = atmospheric pressure + height of liquid x density of liquid x g = Patm + hdg, where Patm is atmospheric pressure and d is density of liquid. We can calculated this pressure as if liquid in region A and Region C does not exist.
But how about the the pressure at the base of the container, that is P2. Is P2 same as P1? For P2, do we need to consider the whole weight of the liquid, that is inclusive of the weight of water in region A and C?

ANSWER:
It depends on what the force on the bottom is. If the container is in
equilibrium, imagine it sitting on a table. The table would exert an upward
force equal to the weight W of all the water, so P _{2}
would be W /A _{bottom} where A _{bottom}
is the area of the bottom of the container. This assumes that atmospheric
pressure is the same everywhere in the vicinity of the container; in other
words, I have ignored the buoyant force due to the air on the whole
container because it will surely be much smaller than W .

FOLLOWUP
QUESTION:
Indeed the container is resting on the table. For P1, I use h x d x g. But for P2 you use W / Abottom, So can I say P1 not equal to P2 ?

ANSWER:
Yes, but I have to admit that my answer was misleading in that I gave you
the gauge pressure, the pressure above atmospheric. So I should have said
that P _{2} =P _{atm} +W /A _{bottom} .
There is no problem that P _{2} ≠ P _{1}
because the force which the container exerts on the table is not P _{1} A _{bottom} .
Think about it—the
sides of your container exert a downward force on the bottom of the
container.

QUESTION:
Two of us disagree on part of a sol'n given by two people with Physics background, and I want to know if I am correct, or if I am missing something in the analysis of the problem...in case I have to explain it to a student.
Question concerning Forces/impulse..... 50kg person falling @15m/s is caught by superhero , and final velocity up is 10m/s.
Find change in velocity.
Find change in momentum .
It takes 0.1 sec to catch them.....ave Force is ?
answers are: vel = 25 m/s change in mom.. 1250 kg*m/s, and ave Force = 12,500 N.
Here's where we disagree: Person B says that 12,500 N is equiv. to 25 g ?????
They try to explain that 250 m/s^2 accel. corresponds to 25g.... I said it makes No sense at all, [ I know the accel. is 250, but that doesn't in any way imply a 25 g "equivalence" to me ]. They then went further to "prove" their point........Here is their argument...
500 N/g = 12500N / ( )g ..... I agree the ( ) = 25, but say there is No justification for the 500 N / g in the first place...... any ideas where it comes from , or how to justify that value ?
BTW I teach physics on and off at the HS level.... person B is an Engineer , I think

ANSWER:
Person B is wrong but has the right idea. (As you and your friend have
apparently done, I will approximate g ≈10
m/s^{2} .)
We can agree that the acceleration is a =250 m/s^{2} and that
is undoubtedly 25g . Now, we need to write Newton's second law for the
person, -mg+F=ma =-500+F =12,500, so F =13,000 N. This is
the average force by the superhero on the person as she is stopped, so the answer that the
average force is 12,500 N is wrong. When one expresses a force as "g s of
force", this is a comparison of the force F to the weight of the
object mg , F (in g s)=F (in N)/mg =13,000/500=26
g s; this simply means that the force on the object is 26 times the
object's weight. So neither of you is completely right, but if there is any
money riding on this, your friend should be the winner because the only
error he made was to forget about the contribution of the weight to the
calculation of the force. I am hoping that superman knows enough physics to
make the time be at least 0.3 s so that Lois does not get badly hurt!

QUESTION:
I have something I have been wondering about.maybe you can answer. I recently drank a glass bottle of rootbeer and then set the empty bottle on the hood of my car that was slanted slightly (1980's model). Within seconds the empty bottle started to vibrate slightly and then"walked" down the hood of my car. The engine was turned off. I was trying to tell my girlfriend that it was from heat convection, but she disagreed. How can this happen? I did it a second time and the empty bottle did it again. Not my imagination.

ANSWER:
Here is what I think: The bottle probably had condensation (water) which was
fairly cool and the hood of your car was warm or even hot. The viscosity of
water depends quite sensitively on its temperature. If the water is heated
from 25^{0} C to 50^{0} C, for example, the
viscosity
decreases from about 890 to 547
μPa·s which will result in a
much slipperier contact. (Think of a very heavy oil and a very light oil to
get an intuitive feeling for how viscosity would affect the frictional
force. Or, think of the old expression "…as slow as molasses in January…".)

QUESTION:
I don't know if you get this question alot, but a man standing on the ground is subjected to 2 forces. Gravity and normal reaction force. Do they form an action-reaction pair of forces?

ANSWER:
This is one of the most misunderstood aspects of Newton's laws. Let me state
Newton's third law for your man in a couple of ways:

The
earth exerts a downward force on the man (his weight) and
therefore the man exerts an upward force on the earth of the same
magnitude as his weight.

The
ground exerts an upward force on the man (the normal force) and
therefore the man exerts a downward force on the ground of the same
magnitude as his weight.

Both of these are correct
statements of Newton's third law. Notice that the pairs of forces
("action-reaction") are never on the same object. The two forces you give
are both on the same object (the man) and can therefore not be an
"action-reaction" pair. So why are they equal and opposite? Because Newton's
first law states that if the man is in equilibrium the sum of all forces on
him must be zero; therefore the ground must exert an upward force to balance
the weight. Your example has nothing to do with Newton's third law.

QUESTION:
Does a submarine have to work harder to travel at the same speed in deeper water?

ANSWER:
The drag
force in water is approximately proportional to the speed; it also
depends on the shape and on the viscosity of the water. As you go deeper
into the water, the pressure increases and the temperature decreases.
Viscosity only very weakly
depends on
pressure but increases significantly with temperature. After remaining
constant around 20^{0} C until a depth of about 200 m, temperature
rapidly decreases to about 4^{0} C at a depth of about 1000 m; the
viscosity, and therefore the drag,
nearly doubles at
that depth.

QUESTION:
if an object ( like a bullet ) is fired vertically at constant speed. Where does it land?

ANSWER:
This is a standard intermediate-level classical mechanics problem for
accelerated reference frames. All the mathematical background is much too
complicated to include here but the answer is that it lands a distance
d =4ωv _{0} ^{3} cosλ /(3g ^{2} )
west of where it was shot; here ω= 7.27x10^{-5} s^{-1}
is the angular velocity of the earth, v _{0} is the initial
velocity of the bullet, λ is the latitude where the gun is located,
and g =9.8 m/s^{2} is the acceleration due to gravity. This,
of course, neglects any air drag or wind corrections. For example, for λ= 45^{0}
and v _{0} =390 m/s (typical 9 mm muzzle speed), d≈60 m.
At the poles (λ= 90^{0} ), it goes up and comes back down
perfectly vertically.

QUESTION:
what is the cause of buoyancy.?
we faced many confusions about then we worked on it and did some experiments
and found an another reason which is satisfying all the conditions and
aspects...but we wanted to consult about it......

ANSWER:
The reason for buoyancy is simply that the force due to the fluid up on the
bottom of something is greater than the force of the fluid down on the top
because of the pressure difference.

QUESTION:
If I have two hoses a varying diameters such as a garden hose and a firehose that are vertical and approximately 20 feet long that are filled with the same amount of water and have the same size opening at the bottom, using just gravity, will water flow through each hose in the same amount of time?
This question came up during a recent visit with a customer. I sell feeding supplies for neonatal patients. The current product being used is a large bore tube compared to the smaller bore that my company sells. However the opening at the distal tip of the feeding tube is the same size for both. Again, using just gravity, shouldn't the speed at which formula flows through the distal end be the same since it bottlenecks there?

ANSWER:
Your first question ("… will
water flow through each hose in the same amount of time?") is ambiguous, so
let me answer the question by finding how the speed of the delivered formula
(labelled V _{bottom} in the figure) depends on the geometry.
The operative physics principal is Bernoulli's equation, P+ ЅρV ^{2} +ρgy =constant
where P is pressure, ρ is fluid density, y is the
height above some chosen reference level, and g =9.8 m/s^{2}
is the acceleration due to gravity. In your case, P is atmospheric
pressure both at the top and at the bottom, I will choose y =0 at the
bottom so y=h at the top. Therefore Bernoulli's equation becomes V _{top} ^{2} + 2gh =V _{bottom} ^{2}
or V _{bottom} =√(2gh +V _{top} ^{2} ).
There are two ways that V _{bottom}
will be independent of the geometry: (1) if h is held constant by
replenishing the formula at the top or (2) if the area of the bore is much
larger than the area of the distal tip A>>a . Both of these result in V _{top} ≈0 so V _{bottom} ≈√(2gh ). If
neither of these is true, it is a much more complicated problem.

QUESTION:
I sent you $20 before I even
asked the question. This should tell you that I am very interested in the
answer. I am way passed school age. This is just for me.

This is a velocity question: The projectile is 25 inches long, 1/2 inch in
diameter and weighs 1750 grains. Absolutely the most aerodynamic it can be.
It will have a guidance system that will include a very small fin in the
front and back. I know this is going to mess up the aerodynamics a bit but I
am looking for an

answer as close as you can approximate.

How high would this "Arrow" need to be dropped in order to reach terminal velocity and just how fast is terminal velocity? What if the projectile weighed 3500 grains, would this change the velocity or would that only change the kinetic energy on impact? I know barometric pressure, humidity, crosswinds, temperature and the like will affect the speed. Just use an average day...which there aren't any...but something like neutral.

As I read that over, I realized I am asking what the fastest terminal velocity could be of the most aerodynamic object ever in free fall.

ANSWER:
It is ill-advised to make a donation before receiving an answer because if a
question makes no sense, I cannot answer it and have no means of returning
your money. In your case, there are some misconceptions and you need to
understand that the key word in your question is "approximate". All air-drag
questions can only have approximate solutions. Your main misconception is
that there is no such thing as " the most aerodynamic object ever". A
second misconception is that you will ever reach the terminal
velocity; the analytical solution to the problem has the terminal velocity
reached only after infinite time and I will use 99% of the terminal velocity
as having reached full speed. Since I have no information at all about the
detailed shape of the object, I will simply use a generic form for the air
drag F which can be assumed to be well approximated as a quadratic
function of velocity for this case, F=-cv ^{2} where c
may be approximated as c≈ јA , A being the cross
sectional area presented to the wind (the negative sign indicating that F
is opposite v ). This approximate value of c is valid only if
SI units (meters, seconds, kilograms) are used.

Now, the solutions. I will not give full details since they are fairly
mathematical. Convert, first, to SI units: L =25"= 0.635 m, R =0.25"=0.00635
m, m _{1} =1750 grain=0.1134
kg, m _{2} =3500
grain=0.2268 kg, A=πR ^{2} =1.267x10^{-4} m^{2} ,
F =(3.167x10^{-5} N)v ^{2} . To calculate the
terminal velocity v _{t} , find when the drag force is equal to
the weight mg (g =9.8 m/s^{2} ): mg=cv _{t} ^{2} ,
v _{t} =√(mg /c )=√(4mg /A ); I find
v _{t1} ≈187 m/s and v _{t2} ≈265 m/s. The height
h from which you would have to drop these to achieve this speed with no
air drag is given by h=v _{t} ^{2} /(2g ), so h _{1} ^{no
drag} =1780 m and h _{2} ^{no drag} =3560 m. Now, if
air drag is included in the calculations, it can be shown (see any
intermediate-level classical mechanics text) that v /v _{t} =√[1-exp(-2gh /v _{t} ^{2} )].
Now, setting v /v _{t} =0.99 and solving for h , I
find h _{1} =6989 m and h _{2} =14,034 m. I have
shown a graph with the velocity relative to the terminal velocity v /v _{t}
as a function of the height h from which it was dropped. Keep in mind
that these calculations are approximations as explained above; I have tried
to give enough details so that you could vary the parameters if you like, in
particular the drag coefficient c and the ratio of v /v _{t} .

ADDED
COMMENTS:
I came across a more detailed discussion of the drag forces for cylinders if
you are interested. You will be referring to the picture to the left above
for the following discussion. First, I need to discuss the definition of
c above more carefully. In more detail, c =ЅρAC _{d}
where ρ is the density of air which is about 1.2 kg/m^{3} at
standard pressure and temperature and C _{d} is a number
(called the drag coefficient) which depends only on the geometry. Note that
for my calculations above the drag coefficient was C _{d} = јA /(Ѕx1.2xA )=0.6.
The graphs above show that the drag coefficient depends on the ratio of L /d=L /(2 R )
which has a value of 50 for your projectile. So, if your cylinder has a
blunt end,
C _{d} ≈0.81
(c ≈0.6xA x0.81=0.49·A ≈ Ѕ A )
and, for a rounded end,
C _{d} ≈0.4
(c ≈0.6xA x0.4=0.24·A ≈ јA ).
So, the calculation above is likely a good approximation to your
" … most
aerodynamic object…" My whole treatment of this problem, though, assumes
that c is a constant. However, as you can see above, c depends
on the air density and the air density depends on altitude. Surely at values
of h around 7 km or 14 km, the air density is much smaller than at
sea level and therefore c is much smaller and the terminal velocity
is much larger; this makes the problem considerably more difficult requiring
numerical solutions, beyond the scope of this web site.

QUESTION:
So if two cars were to hit each
other head on, what would be the best way to minimize danger? Would you slow
to a stop, go faster, or match the other cars speed? My sister says that you
would match the speed, because if you stopped you would absorb all of the
impact. My dad says you would stop because then the only force being put out
would be the other car.

ANSWER:
What hurts you is the force you feel. What determines the force you feel is
your acceleration, your rate of change of velocity during the collision. To
minimize acceleration, you should minimize the relative velocity between you
and the other car which means stopping would be best, but certainly any
braking before the collision will help. This assumes that the duration of
the actual collision does not depend on the relative speed which is probably
roughly true.

QUESTION:
I am interested in the effects of gravity on objects of the same mass in different configurations. Example:
Three long objects of the same mass, for the sake of example, three 6 foot 2x4 boards.
If I take two of the boards and lay them parallel, the gravitational attraction between them should be equal at all points.
However, if I arrange the three boards in a U shape, does the additional mass connecting the bottom of the two boards at the U cause the attraction to increase by only the amount of the additional mass (the third board), in other words, does the base of the U and the tips of the U retain the same gravitational attraction with regard to each other?
Or does the addition of the connecting mass at the base of the U change the attraction of the parallel boards and if so is it in decreasing effect along the parallel boards from the connected base to the tips of the U?

ANSWER:
First of all, it is not really clear what you mean when you say "… the
gravitational attraction between them should be equal at all points…"
The force felt by one little piece of mass on one board due to the other
board should be the same magnitude as the other board experiences; their
directions, however, will not be opposite each other except at the centers.
If we approximate your boards as thin sticks (pretty good since 2"<<72" and
4"<<72"), it is fairly easy to calculate the force experienced by a very
small piece of mass m in either board due to presence of the other
board for your parallel configuration; it is a little messy, so I will
not write it here. This force will have both an x -component and a
y -component. The graph to the left shows F_{x} and F_{y}
as well as the magnitude of the force, F =√(F_{x} ^{2} +F_{y} ^{2} )
for sticks of mass 1 kg, length 1 m, and separated by 1 m to illustrate; I
have chosen x =0 at the left end of the stick and x =1 at the
right end. I have normalized the forces to the gravitational constant G .
On the right you can see the forces on each stick at the ends and centers
(red shows the components). The addition of other masses, like your third
board, will not alter the forces between these two sticks but will certainly
alter the force felt by any mass on these sticks; I guess the answer to your
question is that the force which one board exerts on another is not changed
by adding more boards. Still, we are normally interested in the net
gravitational force on something, not just the force which one other object
of many exerts on it. (Incidentally, note that m will also experience
a force due to the rest of its own board.) You might be interested in an
earlier answer concerning gravitational
fields of cylinders

QUESTION:
Is tension a scalor or a vector
qunatity? While considering a rope why do we take into account its magnitude
and not direction? Plz clarify.

ANSWER:
Tension is the force which a rope or string exerts. Since it is a force, it
is a vector. Probably the reason you are confused is that, unlike most
unknown forces, you always know the direction of the tension force: it is
always in the same direction as the string and always pulls, never pushes.
So, when you solve a problem involving tension, the only unknown thing about
a tension force is its magnitude which is a scalar. So it might seem like
you are treating tension like a scalar, but you are not.

QUESTION:
This is a classic point of contention for pilots, and has recently come up again. I'd like to know your take on this. It seems more often than not, that the answer is the plane will take off. However, no one from that side of the argument addresses the force of the conveyer in the opposite direction canceling forward force from the engine thrust. I don't see how the aircraft could accelerate from a starting velocity of 0kts.
This question has grown men lobbing profanity-laced insults back and forth. The answers are generally not backed by aerodynamics or physics equations.
I am graduating from Embry-Riddle in August with a major in Aeronautics. I'm well past my aerodynamics class, so this is certainly not homework for me, but it is most interesting nonetheless. Without further delay, here is the question:
Imagine a 747 is sitting on a conveyor belt, as wide and long as a runway. The conveyor belt is designed to exactly match the speed of the wheels, moving in the opposite direction. Can the plane take off?

ANSWER:
My goodness, this is pretty basic physics and should not be a puzzle at a
school which has aviation as its main focus. In order for an airplane to
fly, lift must be generated. During takeoff, lift is almost entirely the
result of air passing over the wings and if the airplane is not moving
relative to the air, there is no lift. The engines, if you could point them
vertically, could provide lift and cause the airplane to "take off" if their
thrust were greater than the weight of the airplane. I looked up the maximum
thrust for a 747 and it is only a little greater than
ј the weight. This is no
different than an airplane revving up its engines with the brakes locked. It
will certainly not be able to take off. Vertical takeoff planes like the
Harrier (pictured above) can do this, not conventional ones. A final
thought: if you had your 747 on the conveyer belt pointed into a very high
wind, it could take off, but since the takeoff speed of a 747 is around 180
mph, there are no such winds even in a hurricane.

QUESTION:
If im holding a box connected to a non stretching rope to my hand and i start accelerating upwards ( gravity pulls down ) How would i describe a tension in a string if my acceleration is increasing with time ( defined by some function ). This is not homework i swear. I just cant seem to equate all this.

ANSWER:
If the mass of the box is M and its acceleration is a , then
T-Mg=Ma (Newton's second law) so T=M (g+a ).

QUESTION:
If a circular disk has one half of heavier metal and other one of heavier metal. Would the centre of gravity and centre of mass different? Gravity is uniform. And in uniform gravity , does same mass bodies have their c.g and c.o.m equal?

ANSWER:
The center of mass and center of gravity are coincident if the gravitational
field is uniform (the same everywhere across the object). The center of mass
of a uniform half disk of radius R is a distance d =4R /(3π )
from the center of the straight edge diameter. If the mass density is ρ
and the thickness of the disk is t , the mass is M =ρtπR ^{2} /2.
So you can treat the two halves as two point masses M _{1} and
M _{2 } separated by a distance 2d . So, the distance
from the center of the disk to the center of mass is (dM _{1} -dM _{2} )/(M _{1} +M _{2} )=[(d ρ _{1} tπR ^{2} /2)+ (dρ _{2} tπR ^{2} /2)]/[(ρ _{1} tπR ^{2} /2)+(ρ _{2} tπR ^{2} /2)]=d (ρ _{1} -ρ _{2} )/( ρ _{1} +ρ _{2} );
this will always be on the more massive side. For example, if
ρ _{1} =2ρ _{2} ,
the center of mass is a distance d /3= 4R /( 9 π )≈0.14R
from the center.

QUESTION:
I have seen in your site these
questions about guns portrayed in
science fiction and you have done calculations for the recoil of these
things. So I have a question on how do I calculate the Gs that the recoil
causes? Let's use an example. If I have a ship with a mass of three million
tons and it has a six hundred meter long mass accelerator gun that fires a
twenty kilogram projectile at 2000km/s, how much recoil do I get from that?

ANSWER:
The speed, 2x10^{6} m/s, is very small compared to c =3x10^{8}
m/s, so we can use classical mechanics and galilean kinematics. Assuming
that the projectile accelerates uniformly, 600=Ѕat ^{2}
and
2x10^{6} =at ; solving, a =3.3x10^{9} m/s^{2}
and t =6x10^{-4} s. The force on the projectile during the
6x10^{-4} seconds is F =20x3.3x10^{9} =6.6x10^{10}
N. Newton's third law says that the ship (mass 3x10^{6} metric
tons=3x10^{9} kg) will experience an equal and opposite force during
this time, so a _{ship} =6.6x10^{10} /3x10^{9} =22
m/s^{2} =2.2g . The speed acquired by the ship is (20/3x10^{9} )x(2x10^{6} )=1.3
cm/s.

QUESTION:
Is there an equation that tells the period of rotation of pendulum with two masses placed at different distances from point of pivit?

ANSWER:
You must mean period of oscillation, not of rotation. The period of any
pendulum can only be approximated for small angle oscillations. Your problem
would have M _{1} a distance L _{1} from the
pivot and M _{2} a distance L _{2} from the
pivot. The moment of inertia of the pendulum is I =M _{1} L _{1} ^{2} +M _{2} L _{2} ^{2} .
It is a standard introductory calculation to show that the period T
of a pendulum is T =2π √[I /(MgL )]
where M is the total mass and L is the distance from the pivot
of the center of mass. In your case, M= (M _{1} +M _{2} )
and L =(M _{1} L _{1} +M _{2} L _{2} )/(M _{1} +M _{2} ).
Putting it all together,
T =2 π √[(M _{1} L _{1} ^{2} +M _{2} L _{2} ^{2} ) /(g ( M _{1} L _{1} +M _{2} L _{2} ))].

QUESTION:
I have a question about the CoM on a rider and a bicycle.
The popular narrative or meme is that while riding a bike, with your hands on top of the bars, if you then move your hands to the 'drops' on road bike handlebars, that this lowers the CoM.
Personally, I do not believe it because we are talking about the riders shoulders and hands moving downward about 2-3".

ANSWER:

First of all, any drop at all of part of the mass of the object (even 2-3
inches) will result in a drop of the center of mass of the object.
However, usually the drop is substantially more than just the distance
from the top to the bottom of the handlebars because the cyclist also
bends his arms at the elbows as shown in the figures above. Of course
there is also the reduced air drag when in the lower position.

QUESTION:
I've seen the "battleship in a bathtub" physics illustration, in which we imagine a battleship in a bathtub just barely big enough to hold it (a one foot clearance on the sides and bottom). We are told that indeed the battleship will float with this seemingly minimum amount of water surrounding it. If this is true, how does that coincide with the fact that an object will float only if it displaces more than its own weight in water?

ANSWER:

Just because you cannot see the displaced water does not mean water has not
been displaced to make room for the battleship. If the water in the tub
were at the same level but the battleship missing, the volume occupied
by the battleship would now be filled with water. Now, put the
battleship in the tub and the displaced water will spill over the side
of the tub.

QUESTION:
I am a state trooper here in NC. I was struck by a commercial full sized bus on December 24 2014. This bus was the size of a (trailways bus) . They pulled the recorder from the bus and determined the speed to be 69.5 mph. After impact my vehicle which was stationary in the road (hit from behind) traveled 180 feet after impact. Could you tell me how many g's that I endured. I was knocked unconscious.

QUERY:
It is impossible to do more than a rough estimate with this information. I do need to know were you in gear or parking brake on so that you skidded rather than rolled?

RESPONSE:
Photo of my 2007 Chevy Tahoe which
would have had about an extra 700 pounds of equipment. This was a 2 wheel
drive vehicle. Parking brake was off I believe. In reference to my question
about how many g's did I experiance in the accident. Struck from behind by
commercial bus. My Tahoe traveled 170 feet after impact. Final rest was on
rt grassy shoulder of roadway. I was hospitalized for 6 days at Duke
Hospital.

ANSWER:

First, be clear that what I can do is an order-of-magnitude calculation,
making reasonable approximations. It should not be considered as an
accurate calculation, for example something which would be used in a
court of law. The bus would have a weight of about 28,000 lb≈12,700 kg
and your Tahoe would have a mass of about 6000 lb≈2700 kg. Your car slid
d =170 ft=52 m; the coefficient of sliding friction between rubber
and dry asphalt I will take to be about μ ≈0.65. First estimate
the speed v which your car acquired in the collision: change in
kinetic energy=work done by friction, Ѕmv ^{2} =μmgd ;
solving this I find v ≈26 m/s≈58 mph. Now, what is going to matter
regarding the acceleration of the car (and you) is the time t
which the collision lasted. Assuming that the acceleration was
approximately constant over this time, a≈v /t ≈26/t .
To approximate t , use kinematics of uniform acceleration, s =Ѕat ^{2} =Ѕx(26/t )t ^{2} =13t
where s is the distance the car moved during the collision
which I will estimate from your picture to be the amount by which the
car lost length during the collision, maybe about 10% of the approximate
length of a Tahoe of about 5 m or about 0.5 m. So, t ≈0.5/13≈0.038
s and so a ≈26/0.038=421 m/s^{2} ≈421/9.8≈43 g. Again, keep
in mind that there are lots of uncertainties, for example:

If the road were
wet or much of the sliding was done in the grass, μ would have
been smaller which would have resulted in the speed after the collision
being slower which would have resulted in a smaller acceleration. If
μ were only half as large, a ≈33 g.

If the distance
over the time of the collision were larger, the time would have been
larger and the acceleration smaller. If s ≈1 m, a ≈22 g.

If both 1 and 2
are applied, a ≈17 g.

In any case, you
suffered quite an acceleration! Note that the folks on the bus experienced
only about 2700/12,700=21% the acceleration you did.

QUESTION:
Is there a formula or equation something for vertical mass versus hortizontal mass. I work for a granite company, we have a forklift that's weight limit is 6,000 Lbs that is not a problem holding the slabs vertically however we have to place slabs horizontally on a flat table with suction cups. We bought a 7 foot boom to attach to the forklift. I just want an illustration of the inverse relationship of vertical and horizontal weight supported by physics.

ANSWER:

There is certainly not anything like horizontal and vertical mass. Mass is
mass.
What is going to matter to you for your problem is where the center of
gravity (COG) of your load is and that will determine what the torque which will
be exerted on the forklift; this torque will tell you whether you can
lift the load without tipping the car. What you want not to happen is
what is shown in the left-hand picture. I think that you can intuitively
tell that the farther out the load is located, the likelier the forklift
is to tip. The picture on the right shows how you need to think about
the problem to plan for the car not to tip over the front wheels. What
you need to know is where the COG of the car without
the load is and where the COG of the load is; in the picture, these
locations are where the little pinwheels are. The weight of the load is
W _{L} and its COG a distance D _{L} from
the front axle; the weight of the car is W _{C} and its
COG a distance D _{C} from the front axle. The torque due
to the load is T _{L} =W _{L} D _{L
} and the torque due to the car weight is T _{C} =W _{C} D _{C} .
As long as T _{C} is greater than T _{L} ,
the forklift will not tip over. Now you can see why things are different
for a long slab loaded vertically and horizontally—since the COG of a
uniform slab is at its geometrical center, D _{L} , and
therefore T _{L} , is much bigger for horizontal.
Many cranes and forklifts have a lot of weight added toward the back to
allow the load to have bigger torques; note in the figure that the COG
of the car is near the rear implying added weight back there.

QUESTION:
From what I understand, when an elastic spring is stretched, the energy done to stretch it is stored as elastic potential energy in the spring. Also, there is a limit of proportionality where the spring will no longer go back to its original shape. Therefore, when I stretch a spring and the energy is stored, when I let go, the energy stored is transferred to kinetic energy when it goes back to its original shape. My question is, what happens to the stored energy in a spring when the spring has reached its limit of proportionality? The energy stored isn't transferred to kinetic energy as it doesn't go back to its original shape so where does it go?

ANSWER:

The spring is comprised of atoms which are bound together by molecular
forces. You might think of each pair of atoms as being connected by tiny
springs. When you stretch the spring, you are pulling pairs of atoms
farther apart and each tiny stretch takes work and therefore stores
energy. When you release it, each pair of stretched atoms pulls back to
its original distance. But there are limits to how far you can stretch a
bonded pair of atoms before the bond will be broken. The "lost" energy
was used to break bonds between atoms. If you want to get more
microscopic, just focus your attention on one pair of iron atoms, with
total mass M . You pull them until they break apart; you have done
some amound of work, call it W . But, you started with and ended
with two atoms at rest, so what happened to the energy you put in? It
turns out that if you had a sensitive enough scale you would find that
the mass of the two separated atoms was no longer M but was
slightly bigger, M+m . And, invoking Einstein's most famous
equation, m=W /c ^{2} . No energy in a closed system
is ever lost!

QUESTION:
You may have answered this before, but I could not find it by searching the site. Assume there are two objects in space separated by some relatively large distance. Assume object 1 to have the mass of the sun and object two the mass of a typical 100 meter iron asteroid. Assume the separation distance to be approximately 100 million kilometers. There is currently a fictional force preventing object 2 from accelerating towards object 1 and the relative velocity between the two objects in any other direction is zero. Now the fictional force disappears and object 2(asteroid) starts accelerating towards object 1(sun). I am wondering what the formula would be to determine the final velocity for object 2 as it impacts object 1. I know that object 1 has a gravitational acceleration that depends on the distance from it at 1/rІ but I am unsure how to put this formula together as both that number and the velocity are changing as r shrinks – the velocity continues to climb as it gets closer due to the gravitational acceleration of the sun getting larger

ANSWER:

This is fairly easy using potential energy. The gravitational potential
energy for a mass m whose center is a distance r from
another mass M is U (r )=-MmG /r where
G= 6.67x10^{-11} N·m^{2} /kg^{2} . The
potential energy has been chosen so that U (0)=0 for r=∞. From the
example you give, we can assume that M>>m and therefore M
will remain at rest. You also need to know the radii of m and
M , call them R _{M} and R _{m} . So, the
initial energy is E _{1} =-MmG /r _{1 }
and the final energy is E _{2} =-MmG /r _{2} +Ѕmv ^{2}
where r _{1} =10^{11} m and r _{2} =R _{M} +R _{m} .
You can do all the arithmetic if you like, but I am betting that an
excellent approximation would be to take E _{1} ≈0 and R _{m} <<R _{M} ,
so 0≈-MG /R _{M} +Ѕv ^{2} or v ≈√ (2MG /R _{M} ).
If you put in M and R _{M} of the sun you find v ≈6x10^{5}
m/s. This is also the escape velocity from the survace of the sun.

QUESTION:
If a Flea was dropped from the top of the eiffel tower what would happen to it?

ANSWER:

A flea has a very small terminal velocity which means that air drag force
up on him will equal his own weight after he has achieved a very small
speed. If the air were calm, he would drift slowly to the ground and
arrive unhurt. If there were a wind, he might land blocks or miles from
the base of the tower. Think of dropping a feather—same idea.

QUESTION:
If everyone on Earth were to begin running due east at an appointed time, would the Earth slow in its rotation, even slightly? If so, would that reduction in speed be reversed once everyone ceased running (transferring their momentum back to the globe), or would the day have permanently become longer?

ANSWER:

Let's have a look at the numbers. Suppose that there are about 7 billion
people with an average mass of 80 kg. The radius of the earth is about
6.4x10^{6} m. So their moment of inertia is about MR ^{2} =7x10^{9} x80x(6.4x10^{6} )^{2} =6.5x10^{24}
kg·m^{2} . It would actually be quite a bit less than this
because most of the earth's population resides at a distance less than
6.4x10^{6 } m from the axis of rotation, but let's just suppose
everyone went down to the equator for this little game. If we
approximate the earth to be a uniform sphere with mass of 6x10^{24}
kg, its moment of inertia would be about 2MR ^{2} /5=2x6x10^{24} x(6.4x10^{6} )^{2} /5=9.8x10^{37}
kg·m^{2} . The disparities are enormous and there would be no way
you could ever notice any change. Let's look at an extreme case where
everyone is running just as fast as the whole earth is moving and in the
opposite direction (they would be running west). So, the initial angular
momentum is the total moment of inertia 6.5x10^{24} +9.8x10^{37}
kg·m^{2} times the initial angular velocity, 1 revolution/day.
The final angular momentum would just be moment of inertia of the earth
(all the people are no longer rotating) times the new angular velocity,
call it ω. Then conserving angular momentum, (6.5x10^{24} +9.8x10^{37} )x1=9.8x10^{37} ω
or ω= 1+6.5x10^{24} /9.8x10^{37} =(1+6.6x10^{-14} )
revolution/day. This means that the day would be longer by
0.0000000000066%. And, yes, in an idealized world where the people and
the earth were the only things in the universe, the earth would speed
back up again when everybody took a rest.

QUESTION:
Could an 8x12-inch sheet metal sign mounted on a 5-ft tall metal stanchion set in a 5-gal pail of concrete weighing 75 pounds be knocked over by a 30-mph gust of wind?

ANSWER:

Well, that would depend on things you have not told me, the geometry of the
pail and also on the mass of the sign+stanchion. I can estimate the
force on the sign due to the wind, call that F . This will result
in a torque of 5F ft∙lb. I will work in SI units and then convert
back to imperial units; A =96 in^{2} =0.062 m^{2} ,
v = 30 mph=13.4 m/s. The maximum force on the sign would be
approximately F ≈јAv ^{2} =јx0.062x13.4^{2} =2.78
N=0.625 lb. So now you can do a little experiment: push on the sign with
a force of about 2/3 lb and see if it tips over. I suspect, if the base
is broad enough, it will not tip over.

QUESTION:
I would like to know the measurements of each dimension of
a boat if the total weight of the boat were to carry 300 pounds. Also, we plan to make a canoe like shape so if you can help us know the measurements for that shape we would be very thankful.

ANSWER:

This is too vague. Lots of possible dimensions would do the trick. You also need to specify the weight of the boat, not just the load. The important thing
is that the buoyant force must be equal to the weight of the boat plus
load. The buoyant force is the weight of the water displaced by the
floating boat. For example, if the boat itself weighs 100 lb, the boat
must have the volume which 400 lb of water would occupy; the density of
water is about 62.4 lb/ft^{3} , so the volume of the boat would
need to be 400/62.4=6.4 ft^{3} . For example, suppose you model
your canoe as a triangular prism as shown to the left where h=b
and l =6 ft; its volume is Ѕhbl =3h ^{2} =6.4,
so h =1.5 ft. Of course, this would not be a good design since the
surface of the water would be right at the gunnels of the canoe, just
about to spill in; you would want to build in a safety factor by making
the volume of the canoe quite a bit bigger than 6.4 ft^{3} .

QUESTION:
My young son would like to know how fast something would have to move in order to defy gravity. Thanks! He's 10 and would like to become a physicist.

ANSWER:

The phrase "to defy gravity" does not really have any meaning. If you mean
that if something goes fast enough you can make gravity go away, that
does not happen. More likely you mean how fast does something have to be
going to completely leave the earth and never come back, as in
"everything which goes up must come down (except when it doesn't!)" This
speed is called the escape velocity. The mathematical expression for
escape velocity is
v _{e} =√(2 MG / R )
where
M
and
R
are the mass and the radius of the planet, respectively, and and
G
is
Newton's universal constant of gravitation . This comes out to be
about 7 m/s=25,000 mph for the earth. For a more detailed discussion
(probably over the head of a 10 year old), see an
earlier answer .

QUESTION:
Why does angular velocity act only the in case of circular motion?
I mean that if I see a car moving in a straight line(while I am in rest) still the angle subtended by the car at my eye changes (as the car approaches me) since theta is changing with time then there should be angular displacement, and accordingly angular velocity, shouldn't be?

ANSWER:

You are certainly right. You may describe the motion of any particle by
specifying angular velocity relative to some axis; you must also express
the velocity the object has radially away from (or toward) that axis to
fully describe the situation. In the figure to the right, an axis has
been chosen and the line from the axis to the car has a length r
at this instant; the velocity v of the car has been
resolved into radial (v _{r} ) and tangential (v _{θ} )
components. The rate at which r is changing is what v _{r}
is. The rate at which θ is changing is the angular velocity,
ω=rv _{θ} . If the car happened to be moving in a circle
around the axis, v _{r} =0.

QUESTION:
I have recently started flying quadcopters (QC) and on a forum I frequent, someone mentioned that by placing the battery of their QC on top of the frame rather than below it they felt that the QC quote "went from Buick to Porsche in responsiveness" Since I found this an interesting assertion I looked at the physics of the situation with my high school physics mentality and came up with this possible explanation. Note it is over 40 years since I left high school by the way, :-) and am just wondering what you think about my hypothesis and if you would care to venture your thoughts on the physics of this situation. My tongue in cheek explanation that accompanied the diagram read as "Correct me if I am wrong, it wouldn't be the first time. The quad pivots around point x and with the battery at the bottom in posn B the distance B is greater than the distance A which is the CG of a battery in posn A. Therefore to overcome the inertia of the posn B battery requires more force than overcoming the inertia of a battery in posn A. This makes the quad more responsive with the battery in posn A. Anyway that is my hypothesis but if I start believing my own BS then let me know"

ANSWER:

This is not quite as simple as I though it was going to be! I will give you
the full-blown "physics talk" explanation. First, it will not have any
effect on linear acceleration because all that determines that is the
total mass of your QC and the force you can get from the propellers.
However, moving the center of mass (COM) will have an effect on angular
acceleration, the rate at which you can cause it to turn. The
appropriate equation to understand this is Newton's second law in
rotational form, torque τ equals the moment of inertia about
the COM I _{cm} times the angular acceleration α ,
τ =I _{cm} α or α=τ /I _{cm} .
The torque due to each propeller is the force it exerts times the
distance from C/L to the propeller shaft. The moment of inertia about
the COM of the whole thing is what you need because the whole
QC in flight will rotate about its COM. Everything depends on
I _{cm} ,_{ } and the smaller you can make it, the
greater angular acceleration you can get for a given torque. The farther
away the battery is from the COM of the QC, the greater I _{cm}
will be. So to have the most responsive handling, you want to find the
COM of the QC without its battery and put the COM
of the battery right there. I am guessing that battery position A is
closer to the COM of the QC without battery than battery
position B is. If the mass of the battery is much less than the mass of
the QC, the "Porsche…responsiveness" is an illusion.

QUERY:
It would be helpful if you could give me the relative masses of the QC without battery and the battery.
And also where the COM of the QC without the battery is located. I assume it is on the C/L, but how far from the airframe labeled on your diagram?

REPLY:
The relative mass of the QC is 20 grams without the battery and 28 grams with the battery. The
COM of the QC without the battery is located on the C/L, but as to how far from the airframe labeled on my diagram I cannot answer. My whole hypothesis, for want of a better word is based on my assumption that if the left hand propeller is exerting an upward thrust and the right hand propeller a downwards thrust simultaneously, the the QC will rotate around point x.

ANSWER:
Your assumption, as I explained in the first answer, that it rotates
about x is not correct; it rotates about the COM wherever that is. It
may appear that it is rotating about another point because the overall
motion may be described as the sum of translational motion (along
a straight line) of the COM and rotational motion around
the COM. The mass you state cannot be correct since 20 gm is less than
an ounce; but this does not matter because to answer your question all I
need are the relative masses of the QC and the battery. As explained
above, the battery being a significant amount of the total mass (nearly
1/3), its placement will be important. So the whole key is that the
moment of inertia about the COM of the whole vehicle is lowest when the
two centers of mass are as close together as possible; this results in
the maximum angular acceleration for a given net torque.

QUESTION:
I'm a writer working on a nonfiction book that concerns climate and landscape, and am seeking help understanding a phenomenon I am hoping to explain to lay audiences in this book.
A hot air balloon pilot I interviewed described a near-crash. On a calm morning, this pilot took a family on a short flight over some meadows, not far from a geothermal plant. While preparing to land, the pilot's balloon was drawn uncontrollably towards the geothermal plant, where it was trapped in the column of hot air rising over the plant, like a ping pong ball over a hair drier. I am trying to understand what might have caused the balloon to be drawn towards this rising column of air. The pilot did not accidentally pass over the geothermal plant, but rather the balloon was uncontrollably sucked into the column from the side.

ANSWER:

This is a manifestation, I believe, of the Venturi effect which says, in
essence, that the pressure in a fluid decreases as the velocity
increases. It is just a special case of the Bernoulli equation, P+ Ѕρv ^{2} +ρgy =constant.
In cases like this balloon, Bernoulli's equation will not give you
precise results because it is valid only for incompressible ideal fluids
which air certainly isn't. However, it is excellent for qualitatively
understanding many things which happen in moving fluids. Examples of the
Venturi effect abound. An airplane wing generates lift because air
passes over the top faster than over the bottom resulting in a net
upward pressure. A spinning baseball curves because the pressure on one
side is greater than the other. A sheet of paper rises when you blow
across the top (see picture). Although you do not see this so much any
more, if smokers inside a moving car crack the window, the smoke will be
drawn out by the lower pressure of the outside air rushing past. In the
case of the balloon, the column of hot air is rising whereas the air
around it is still, so objects near it will feel a force drawing them
into the column.

QUESTION:
I came across a question in the back of the chapter that asked: "Because a smaller mass results in greater air resistance effects on a ski jumper, all other things being equal a lighter ski jumper flies farther than a heavier one."
According to the book the answer is true. My question is why?

ANSWER:

There are two forces on a ski jumper, his weight W and the
force of air drag on him F . The weight is always a force
which points vertically down and is proportional to the mass m .
The air drag always points opposite the skier's velocity and depends
only on the velocity magnitude, the properties of the air, and the shape
of the skier, but not his mass. Shown in the figure are the force
diagrams for the heavy (#1, leftmost) and light (#2) skiers when they
are going up. (I have chosen an extreme case, the heavy skier about
twice the mass of the light skier, to emphasize the difference.
Everything else about the two is identical.) Each
force results in an acceleration (shown as the green-dashed vectors)
which result in net accelarations shown by the full-drawn green
vectors. The lighter skier has a larger drag acceleration because the
acceleration is F /m and both have accelerations due to
their weights which are equal. But looking at the horizontal and
vertical of the net acceleration, both are larger for the light skier—he
is losing forward speed faster and is accelerating toward the ground
faster. Also shown in the figure are the
force diagrams for the heavy (#3) and light (#4, rightmost) skier when they are
going down. Now, the lighter skier still has a larger horizontal
component of acceleration, he is slowing down faster, and his vertical
component is perhaps just slightly smaller. I would conclude that your
book is wrong, the heavy skier jumps farther. Consider the following
experiment to demonstrate that the heavier will have a larger range
using two balls of the same size, say a baseball and a
baseball-sized nerf ball. Throw them each as hard as you can and
projected at about the same angle; which would you expect to go farther?
In fact, there is an analytical solution (approximation) for the range
R of a projectile: R= [v _{0} ^{2} sin2θ /g )[1-(4v _{0} sinθ /(3g ))(C /m )]
where v _{0} is the speed at launch, θ is the
launch angle, and C is a constant which depends only on the size and
shape of the projectile. Note that R gets larger as m gets
larger.

QUESTION:
I caught a fly in an empty jar and sealed it with lid. Container weighs X. Fly weighs Y. When the fly alights on the bottom, I assume the container now weighs X + Y. If the fly takes flight does the same hold true.

ANSWER:

You forgot about the air in the jar which weighs Z. I have answered
this kind of questions many times, usually we have birds in a box. It
all boils down to Newton's third law. Assume the jar is on a scale, fly
on the bottom. The scale reads X+Y+Z . Now suppose the fly
hovers or flies with constant velocity. To do that the air must exert an
upward force of Y on the fly; but that means the fly exerts a
downward force of Y on the air; but that would have to mean that
the air now exerts a downward force of Y+Z on the jar. So, again,
the scale reads X+Y+Z . Now suppose the fly has an upward acceleration a . To
do that, the air must exert an upward force of Y +(Y /g )a=Y (1+(a /g ))
where g is the acceleration due to gravity; it then follows that
the scale will read X+Y (1+(a /g ))+Z. Similarly,
if the fly has a downward acceleration of a , the scale will read
X+Y (1-(a /g ))+Z .
If the fly is in free fall in the jar, the scale will read X+Z .
My answer ignores buoyancy of the fly and, when he is in free fall, air
drag. If we consider air drag, when the fly in free fall has reached
terminal velocity the scale will, again, read X+Y+Z.

QUESTION:
Hello, I am a junior in highschool, and I am currently studying about momentum in physics. I am a bit confused with the concept though. I get it is related to newton's laws and it is mostly used for collisions. I also learned this week about impulse and collisions and conservation of momentum.
So now my teacher explained that for momentum to be conserved the net force acting on an object must be 0. That's because Ft=deltap (so if f is 0 then change in momentum is 0 so it is conserved) Now we then took about inelastic collisions and that K.E is not conserved because some is changed to hear or other forms of energy. So I've also learned previously that deltaKE=Wnet so if there is change in KE there is work and therefore a force so how then is there a net force acting in inelastic collisions but still momentum is conserved ??

ANSWER:

We often think of putty balls colliding when we think of perfectly
inelastic collisions. The balls collide, squish together, and then stick
together. Think about squishing a putty ball—does it take any work? Of
course it does. Can you get that work back? Not a chance. What happened
to it? the ball got a little warmer. Or think of a colliding ball as a
spring. That makes sense because a rubber ball will compress and then,
when you quit pushing on it, go back to its original shape. So, you put
some energy into it. If it were an ideal ball-spring and you just let go
of it, it would oscillate from squished to stretched back and forth
forever. Does it do that? Of course not because it is not an ideal
ball-spring and that energy you put into it eventually (actually very
quickly) all disappears (becomes a bit of thermal energy in the ball).
So, even balls which do not stick together usually have some
inelasticity when they collide. But, when balls collide, what is the
force which does the squishing work? It is the force which one exerts on
the other and the other exerts on the one; Newton's third law tells you
that these have to be equal and opposite, so momentum is conserved even
if energy is not. Billiard balls are very good (and very stiff) springs
and collide almost elastically.

QUESTION:
I just saw the question in your page about super powered people telekinetically lifting objects and how Newton's laws have something to say about that.
So that rose me a question about the thing. Lifting the submarine should require enormous amounts of energy, and this would naturally mean that a super powered individual who uses his or hers powers to lift an object such as a submarine would need to have some borderline magical energy production methods far surpassing anything biology gives to any creature.
So if we assume that the sub marine weighs ten thousand tons, how much energy would Magneto need to lift it at the speed of let's say ten meters per second? And how does that relate to energy consumption of a normal human, and would even a nuclear reactor be enough to produce the necessary energy for lifting the submarine?

ANSWER:

It would make more sense for your question to ask about either force, as
the original question did, or power, the rate
of delivering energy. If the mass is 10^{4} metric tons, the
weight would be about F =10^{8} N; that would be the force
you would need to exert to lift the submarine. If you were lifting at
the rate of v =10 m/s, the power P required would be about
P=F∙v =10^{9} W=1 GW. A gigawatt is a typical
output for a nuclear power plant.

QUESTION:
I understand that, with a roughly spherical object, like the earth, the gravitational force tends to act on objects towards the centre of the sphere.
What would the direction of the force be with an object shaped like a cylinder? Also would the gravitational pull be greater on each end of the cylinder than at some point in the middle? (In which case I would guess that, in nature, any massive cylinder would collapse to form a sphere?)

ANSWER:

The gravitational field of a cylinder is pretty easy to calculate on its
axis and very difficult to calculate elsewhere. At the center of each
end of a cylinder of length L and radius R , the field g can
be shown to be g _{end} =[GM /(RL )][(2L /R )+Ѕ-Ѕ√(R ^{2} +L ^{2} )/R ].
Let me first provide a qualitative argument that the field at the
"poles"
will be larger than at the "equator". At the equator, the contribution
to the field from each piece of mass in one half will have a
corresponding piece on the other side and their axial components will
cancel out, leaving only the radial components. As long as
L>R ,
there will be much less cancellation for fields at the "poles";
therefore, if the cylinder is not rigid, it will collapse to a sphere
axially (from the poles). If L<R , this argument will work in the
opposite way and you would expect the collapse to be radially in from
the equator. Next, here is an approximate analytical
solution for the case L>>R . The end fields can be
approximated as
g _{end} ≈3GM /(2R ^{2} ). At the equator, Gauss's
law may be used to
show that g _{equator} ≈2GM /(LR ).
So
g _{end} /g _{equator} ≈3L /(2R )>>1. So, your expectation was right, but
only
if R<L . On the other hand, if R>>L , the
field at the pole
approaches the field at the center of a uniform disk which is zero by
symmetry. So, whatever the field is at the equator, the force tends to collapse the
cylinder radially (inward from the equator). The details of the
calculations here are given in a
separate page
where I have also shown a
rough sketch
of the field for the L>R case.

QUESTION:
Built my kid a marshmallow shooter for a science project. Simple design. Main air chamber is 2" pvc
with a total length (including bend) of 32". It is then directed into a 3/4" inch
pvc pipe with 2 valves. The first valve (1) is the main shut off valve and inch or two from the 2" chamber. Then there is another
21" of 3/4" pvc into the 2nd valve (2). The barrel is 1/2" pvc and 24" long. We can fill the gun with 40 psi with both valves closed. We then open valve 1.
Pressure should drop a little, but not much. We then close 1 to preserve pressure and shoot marshmallows with almost 40 psi. They will go 40'. When we fill the tank with 40 psi with valve 1 open and valve 2 closed and shoot it, it'll shoot the marshmallow 100' or more. My question is:
why is releasing all the air at once shooting the marshmallows further? It's
driving me nuts. Is it the volume of gasses? How can I mathematically solve
this mystery?

ANSWER:
It is pretty easy to understand this qualitatively. As the
marshmallow moves down the barrel, the volume of the gas behind it increases
so the pressure decreases. Your first case (valve 1 closed) there is a
pretty big fractional increase in volume so, since PV is constant, a
pretty big decrease in the final pressure; in the second case (valve 1
open) there is a much smaller fractional increase in volume so there will be
a much smaller decrease in the final pressure. The average force felt by the
marshmallow over the length of the barrel will be bigger for the second
case.

Now, let's do it analytically. I will ignore the couple of inches between
the main chamber and the first valve. The operative principle is that if
temperature and amount of gas are unchanged, the product of the pressure and
volume is a constant or, equivalently, V _{initial} /V _{final} =P _{final} /P _{initial} .
An important thing to keep in mind is that 40 psi is
the gauge pressure, the pressure above atmospheric pressure which is about
15 psi; so P _{initial} =55 psi. The volume of each section is
π (d /2)^{2} L where d is the inner
diameter of the pipe (a 3/4" pipe, e.g ., specifies the diameter). So V _{barrel} =4.7
in^{3} , V _{primary} =100 in^{3} , V _{secondary} =9.3
in^{3} . In the first case V _{initial} =V _{secondary} =9.3
in^{3} and V _{final} =V _{secondary} +V _{barrel} =14
in^{3} . Therefore V _{initial} /V _{final} =0.66=P _{final} /P _{initial}
and, taking P _{initial} =55 psi, P _{final} =36
psi and the corresponding gauge pressure is 21 psi; so the average gauge
pressure during firing was (40+21)/2=30 psi. In the second case, V _{final} =V _{secondary} +V _{barrel} +V _{primary} =114
in^{3} and V _{initial} =V _{secondary} +V _{primary} =109.2
in^{3} . Going through the same procedure as the first case, P _{final} =53
psi and the corresponding gauge pressure is 38 psi; so the average gauge
pressure during firing was (38+40)/2=39 psi. This means the average force on
the marshmallow was 81% greater for the second case; this means that the
speed of the marshmallow is almost twice as great, so it is roughly in
agreement with your measurements of a distance of 100' compared to 40'. I
have not considered air drag during the flight after leaving the gun which
will be fairly important for a marshmallow. Also, I have ignored friction
between the marshmallow and the barrel; because of this and neglect of air
drag, don't expect real good quantitative predictions of range.

ADDED
NOTE:
In the first case where you pressurize to 40 psi and then open
valve 1, the pressure will drop more than just a little as you expect. V _{initial} /V _{final} =100/109.2=0.92,
so P _{final} =55x0.92=51 psi, so the final gauge pressure will
be about 36 psi, about a 10% drop.

QUESTION:
There's a weight standing equally on 2 legs, each of which is on a scale.
Each scale shows half the weight. If we disappear one of the legs (assume
it' stable and won't fall over.), the entire weight should transfer to the
remaining leg. Does it transfer instantly, or is there a time lag? If
there's a time lag, why? I say there's a time lag (but I can't explain why)
and my friend says it's instant.

ANSWER:
Generally, nature does not like discontinuities so you should always assume
first that "instantaneous" is not possible. That, indeed, is true for the
situation you describe. The two-leg scenario has some problems which
complicate things (like the center of mass is not directly over either scale
so removing a leg will result in a torque trying to tip the object over).
So, I will use a simpler problem, equivalent for your purposes. Imagine that
a weight sits on a scale. Now you place a second identical weight atop the
first. The information that the second weight has been added gets
transmitted at the speed of sound (in the first weight) to the scale. This
will still be a pretty short time, but not zero. I have ignored the time it
takes the scale to respond.

QUESTION:
Why is it that when my daughter spun the lazy Susan on our table, the salt shaker
went flying off?

ANSWER:
An object which is going in a circle requires a force to keep it
going in a circle. Imagine that you are spinning a stone attached to a
string in a horizontal circle. The force which keeps it going in a circle is
the string pulling on it. As you make it go around faster and faster, the
string has to pull harder and harder. Eventually the string can no longer
pull hard enough and breaks and the stone goes flying away. When the lazy
Susan is spinning the salt shaker is moving in a circle and therefore needs
a force to keep it going in a circle. The force which keeps it going in a
circle is friction between the lazy Susan and the shaker; if the lazy Susan
were very slippery, it would not be of much use. As it spins faster and
faster, you will need more and more friction but there is a limit to how
much friction you can get; think of trying to push a heavy box on the
floor—you push harder and harder and eventually it will start moving. So, at
a high speed the shaker will fly off. Tell your daughter to not spin it so
fast! Or maybe she just likes to make it fly off. Good thing it wasn't a
glass of milk.

QUESTION:
When a gas is pressurized (in favorable conditions) it turns into liquid.
Further pressurizing it turns into solid. Now I exert larger pressure on
a uniform solid (form all directions). What will happen?
Also what will happen if I subject a fundamental particle to very large
pressure?

ANSWER:
Very large pressures can change the properties of the material
and new phases may occur (like water and ice are two different phases of H_{2} O).
A particularly good example is water itself as shown in the figure to the
right; pressures up to about 10,000,000 atmospheres are shown in the graph
and numerous phases beyond ice/water/steam are seen at high pressures or
very low temperatures. For higher yet pressures, you need to look to stars.
If a star is massive enough it will, after going through a supernova stage,
collapse under the pressure of its own gravity to where electrons are pushed
into the protons and the whole star becomes a neutron star, essentially a
gigantic nucleus. If heavy enough, it will continue collapsing into a black
hole. What individual particles do depends on the pressure and they
essentially will retain their identities or undergo reactions with other
particles (as in the neutron star formation) or lose their identities (in a
black hole).

QUESTION:
An astronaut rotates in space an observes the universe rotating about them. How does the astronaut know whether the universe is rotating, or if it is their own motion causing that visual spin? Centripetal forces, of course, which will force blood into the fingertips which will surely be detectable in the astronauts frame of reference. Perhaps this suggests that there exists a universal state of rotational rest - detectable by measuring existence or absence of centripetal forces. Is there also a more general universal state of rest? A state that can be also measured and agreed on by any stationary or moving frame of reference? What do we already know about that has a velocity that can be measured and agreed on by all reference frames, even those with different relative velocity? What do we know of that, like zero, has an ultimate limit that cannot be exceeded? Could the speed of light actually be that universal rest frame?

ANSWER:
First of all, it is centrifugal force, not centripetal, which
causes blood to be pushed to extremities. But that is not important here.
There are two kinds of reference frames, inertial and noninertial. If you
are in an inertial frame of reference, Newton's laws of physics are true. A
noninertial frame is any frame which accelerates relative to any inertial
frame. Because Newton's laws are not valid for your rotating astronaut,
blood flows out to the fingertips even though there is nothing pushing it
out; that is why you will see centrifugal force referred to as a "fictitious"
force.
There is no preferred reference frame, no "absolute at rest."

QUESTION:
I am reading a scifi novel and in it gravity is simulated by constant thrust from the engines. If there is no inertia in space would not the spacecraft continue at that speed and therefore the gravity remain constant until the ship used thrusters to slow it down? In the novel, if the thrust is cut off the gravity is reduced. I can't get my head around it?

ANSWER:
When you are in a car which is accelerating you feel as if you
are being pushed back against the back of your seat; this is like a
horizontal "artificial gravity" and is the effect which causes the
artificial gravity when your spacecraft is burning its engines. When the car
stops accelerating and moves with constant velocity in a straight line, you
no longer feel that you are being pushed back; similarly in the spaceship
when you cut the engines. I have no idea what you mean that there is no
inertia in space. Inertia means resistant to change in velocity if acted on
by a force, so if the engines are not burning you feel weightless; gravity
is not "reduced", it disappears. It sounds like you think that you need to
push on the spacecraft to keep it going with a constant speed and nothing
could be further from the truth. Newton's first law (the law of inertia)
says that an object on which no net force acts will move in a straight line
with constant speed forever.

QUESTION:
If a gun is fired in space, how far/long will the bullet travel and how long will it spin? Why? Assumptions: 1) gun’s position is fixed relative to the earth; 2) gun’s barrel is rifled, thus the spin; 3) bullet travels in unobstructed straight line and it avoids being attracted to another body due to that body’s gravity.
In one of your earlier answers regarding the relative motion of a gun and bullet in space you say that the bullet would never stop traveling. I suppose that, in my question above, this also means that the bullet would spin forever as well?
This confuses me as it would seem to me that motion requires energy, which would eventually be depleted. What am I misunderstanding?

ANSWER:
I will first answer your last question. What you are
misunderstanding is Newton's first law which states that an object which
experiences no net force will move in a straight line with a constant speed;
also, a rigid body which experiences no net torque about its axis of
rotation through its center of mass will continue rotating with the same
rotational speed about that axis. With these you should understand the
answer to my previous answer and also why (your addition) it will spin
forever. The only proviso is that there is nowhere in the universe where
there is a perfect vacuum, always a few atoms around, and so there is always
a tiny amount of friction which would stop your bullet after maybe billions
of years. (Also, your supposition that "…it avoids being attracted to
another body due to that body’s gravity" cannot be literally true because
gravity is everywhere.

QUESTION:
This link is a video of anti gravity and
I want to know on how it really works?

ANSWER:
This one is pretty easy to debunk, I believe. The guy and all
the apparatus are upside down; the camera taking the video is also upside
down.

QUESTION:
If you have 2 plastic balls with the same outer surface but one is filled with water and the other one is empty and you released them, from rest, down the same slope, the ball with the water would accelerate more due to viscous flow right? If that is the case then would a ball that is fairly (but not completely) filled with sand accelerate faster than the empty ball and why? would there be a similar effect?

ANSWER:
There is no clear cut simple answer to this question. I prefer
to use energy conservation to look at problems like this. Let's first just
consider a hollow sphere and a solid sphere. If we have an object with mass
m , radius R , and moment of inertia (about center of
mass) I , it is easy to show that the speed of the object when it has
rolled down so that its vertical drop is h is given by

v =√[2mgh /(m+ (I /R ^{2} ))].

The moments of inertia are
I_{solid} =2mR ^{2} /5 and I _{hollow} =2mR ^{2} /3.
First note that the mass will cancel out, so
v_{solid} =√(10gh /7) and v _{hollow} =√(6gh /5)<v _{solid}
and so the solid sphere is the winner.

Now, suppose a hollow sphere, mass m ,
radius R , moment of inertial I =2mR ^{2} /3 is
filled with a fluid of mass M . (I assume that the spherical shell is
very thin compared to the radius of the ball.) There are two extremes of
what can happen, either the fluid does not rotate at all as the ball rolls
(a superfluid) or else it begins rotating with the ball almost immediately
(maybe molasses). In the first case, when the ball reaches the bottom its
rotational kinetic energy will be all due to the rotating of the ball, its
translational kinetic energy will be due to the both the ball and the fluid,
and the potential energy at the beginning will be due to both:

Ѕ(2mR ^{2} /3)(v ^{2} /R ^{2} )+Ѕ(M+m )v ^{2} =(M+m )gh .

Solving, I find
v =√ [2gh /(M+ (5m /3))];
notice that if M =0, v =v _{hollow} as it should,
so any nonzero M will result in losing to the hollow sphere. At the
other extreme, the fluid rotating rigidly the whole way, the energy
conservation equation will be

Ѕ(2MR ^{2} /5)(v ^{2} /R ^{2} )+Ѕ(2mR ^{2} /3)(v ^{2} /R ^{2} )+Ѕ(M+m )v ^{2} =(M+m )gh.

Solving, I find
v =√ [30(M+m )gh /(24M +25m )].
This can be compared with v _{hollow} :
v /v _{hollow} =√[(1+(M /m ))/(1+(24M /25m ))]
which is always greater than 1 (the denominator in the square root is
smaller than the numerator). Again, note that for M =0, v=v _{hollow} .
Therefore the filled ball will always win if the fluid rotates.

Now, to your question. Water will be at
neither of the extremes above. It will begin by not rotating, so in a race
with a hollow ball, the hollow ball will jump off to a lead. After some
elapsed time, all the water will be rotating with the ball containing it, so
it will be now be gaining speed faster than the hollow ball and be catching
up. Eventually, the water ball will catch up and pass the hollow ball; so,
you see, which ball wins will depend on how long the race is. The way you
state the sand part of the question is problematical "……fairly (but not
completely) full…" is not very quantitative. Let's just say that it
essentially fills the ball but is loosely enough packed so that it can slide
over itself. In that case, the sand will act pretty much like a very viscous
fluid and will get fully rotating with the ball more quickly than the water
did. In both cases (water and sand), trying to do a analytical solution
while the fluid is in the transition stage from not rotating to rotating is
just about impossible; because of the effects of friction (viscosity, if you
like), energy is not conserved during that time.

QUESTION:
I f you were standing on a flat platform that was free falling from the sky let's say 12000 feet up and right before impact within a certain point, or say within 12 feet, and you jump vertically off the platform, what would happen to you? Would it decrease your speed and your impact or would it just delay you hitting the ground?
This question was asked to me by my friend and we both are very curious. We have a few thoughts on the outcome but aren't sure and we really want to know for sure. Hopefully you will be able to answer this for us thank you.

ANSWER:
I answered a similar question in an old
answer , jumping in a falling elevator. There is no way that jumping can
save you because the amount of upward speed you could give yourself is tiny
compared to your downward speed. What would happen depends on the mass of
the platform. If the platform had much less mass than you, most of your
effort would be to speed up the platform rather than slow you. If it were
much heavier than you, you would still only be able to reduce your downward
speed by about 3-4 m/s. Of course, if the platform were big enough it would
act like a parachute and you would not have to worry about a thing!

QUESTION:
I am building a customizable "cat highway" of wooden shelves in my living room. The issue I am having is figuring out how much holding force I need, rather than how much weight the actual shelf can support. I know the shelves are plenty strong enough. Now I need to have them fastened strongly enough to the wall. It would be easy, if all I had to consider was the maximum weight a shelf has to take, if all three cats were to lie on it at once. However, I also need to how much weight I have to budget for for impact force - both from a descending and an ascending cat.
As there isn't room for multiple cats to jump up or come down at one time, so only the weight of the heaviest cat (3.26587 kg) needs to be used for the calculations. Also, there will not be a height of more than 0.5 m from one shelf to another, nor a distance of more than 0.5 m between one shelf and another. I can't guarantee that they won't skip a shelf, so it might be safer to double the maximum distances just to be safe.

ANSWER:
Normally I answer such questions by "you cannot tell how much
force results if an object with some velocity drops onto something". The
reason is that the force depends on how quickly the object stops; that is
why it hurts more to drop on the floor than to drop on a mattress. In your
case, however, we can estimate the time the cat takes to stop because we can
estimate the length of its legs which is the distance over which it will
stop. You are probably not interested in the details, so I will give you the
final answer: assuming constant acceleration during the stopping period, the
force F necessary to stop a cat of mass m , falling from a
height h , and having legs of length ℓ may be approximated as F=mg (1+(h /ℓ)).
E.g ., if h ≈0.5 m and ℓ≈0.1 m, F ≈6mg , six times
the weight of the cat.

FOLLOWUP QUESTION:
Thank you for your answer!
I thought you might want to know that, unlike a lot of people, I AM actually interested in the details.

ANSWER:
OK, here goes: I will use a coordinate system with increasing
y vertically upward and y =0 at the landing shelf. The cat will
have acquired some velocity -v when his feet hit the shelf. Assuming
he stops after going a distance ℓ and accelerates uniformly, we have the two
kinematic equations 0=ℓ-vt +Ѕat ^{2 } and 0=-v+at .
From the second equation t=v /a ; so, from the first equation,
0=ℓ-v (v /a )+Ѕa (v /a )^{2} =ℓ-Ѕv ^{2} /a
so a =Ѕv ^{2} /ℓ. Now, as the cat is landing there are
two forces on him, his own weight mg down and the force F of
the shelf up, -mg+F and this must be equal, by Newton's second law,
to ma , so F=m (g +Ѕv ^{2} /ℓ). This, by
Newton's third law, is the force which the cat exerts down on the shelf.
Finally, the speed if dropped from a height h is v =√(2gh ),
so F=mg (1+(h /ℓ)).

QUESTION:
Imagine the inside of a spacecraft, in orbit, so astronauts experience weightlessness and things float with no friction etc (assume there is no atmosphere in the craft ...). A pencil is floating in mid-air inside the craft. An astronaut pushes (say flicks with a finger) the pencil on one of its very ends. This push will impart linear, rotational, or both types of motion to the previously stationary floating pencil? If the pencil should rotate, about what axis does it rotate and how is angular momentum conserved?

ANSWER:
There will be very small effects due to the fact that this is
not true "weightlessness" in that the spacecraft is accelerating in a
gravitational field and not of zero size. I take it that this kind of detail
is not what you are interested in, that I can simply answer your question
for the pencil being in empty space, initially at rest. I will suppose that
the pencil has mass m , moment of inertial I about its center
of mass which is a distance d from the end where a constant force
F is applied, perpendicular to d , for a very short time t .
Newton's second law for translational motion says that the impulse equals
the change in linear momentum, Ft=mv ; after the impulse, the center
of mass will move in the direction that F was applied with speed
v=Ft/m . Newton's second law for rotational motion says that the angular
impulse equals the change in angular momentum, τt=Ftd=Iω ; after the
impulse, the pencil will rotate around the center of mass with an angular
velocity ω=Ftd /I . A relationship between v and
ω can be written as ω=mvd /I . You can also write the total
energy of the pencil, the sum of translational and rotational kinetic
energies, as E =Ѕmv ^{2} +ЅIω ^{2} .^{
} For example, if the pencil is modeled as a uniform stick of length
L , I=mL ^{2} /12 and d=L /2, so ω =mv (L /2)/(mL ^{2} /12)=6v /L
and E =3mv ^{2} /2; in this case, 2/3 of the kinetic
energy is due to the rotation. Also, the speed u of the end of the
pencil with respect to its center is u=Lω /2=3v . Regarding
angular momentum, Iω , it is conserved after the initial "flick"
because there are no torques on the pencil.

QUESTION:
In light of the recent deflated football scandal, is there a way to mathematically calculate the change in pressure as the temperature inside the ball changes? Would you assume the volume of the bladder inside the ball to change very little?

ANSWER:
Yes, the ideal gas law, PV /(NT )=constant where
P is pressure (not gauge pressure), V is the volume,
N is the amount of gas, and T is the absolute temperature.
Assuming that V and N remain constant, P /T =constant.
Let's do an example. Suppose that the ball is filled to a gauge pressure of
13 psi when the temperature is 70^{0} F. The absolute pressure is 13
plus atmospheric pressure 14.7 psi, P _{1} =13+14.7=27.7 psi.
The temperature in kelvins (absolute) is 70^{0} F=294 K. Now suppose
we cool the football to 10^{0} F=261 K. Then, 27.7/294=P _{2} /261,
P _{2} =24.6 psi, and the resulting gauge pressure is
24.6-14.7=9.9 psi. I guess it is important to fill the ball at the
temperature at which it will be played.

ADDED
NOTE:
An article in the January 30
New York Times came to essentially the same conclusion as I did here. My
answer was posted on January 26. I was astounded to read in that article,
though, that initial calculations by physicists had applied the ideal
gas law using gauge pressure rather than absolute pressure! Shame on them!

QUESTION:
If a train is traveling at 60 m/s and a person runs and jumps off the rear of said train (opposite direction of travel) at 3 m/s. Which direction would the person travel?
Would they continue to move in the opposite direction of the train, move with the train at relative speed, or drop straight downward?

ANSWER:
Variations on this question are probably the most frequent
question I answer. I always say that you always have to specify velocity
relative to what . In your question, I take it that the velocity of the
train relative to the ground is 60 m/s and the velocity of the person
relative to the train is 3 m/s in the opposite direction as the train's
velocity. The velocity of the person relative to the ground is 57 m/s in the
direction of the train's velocity. An observer on the ground would see the
person with a horizontal velocity of 57 m/s regardless of whether he was on
the train or had jumped off the back. Once he is off the train he will begin
falling and hit the ground with both horizontal and vertical components of
his velocity. For example, if he starts at a height of 2 m above the ground,
his vertical component when he hits the ground will be about 6.3 m/s so his
total speed will be √(57^{2} +6.3^{2} )=57.3 m/s.

QUESTION:
If aluminium and copper pipes are of same length and diameter ... same magnet is dropped through them ...in copper it takes more time to come out of other end, i myself have done this... please answer why is it so?

ANSWER:
Aluminum has an electrical conductivity of about 3.5x10^{8}
Ω^{-1} m^{-1} and copper has a value of about 6x10^{8
} Ω^{-1} m^{-1} . Therefore, at any speed, the magnet will
induce a larger current in the pipe for the more conductive copper.

QUESTION:
Newton's third law of motion states that, "To every action there is equal but opposite reaction". That means if i throw a ball on a wall it will bounce back but then what happens to mud, if i throw mud on a wall then
it does not it bounce back?

ANSWER:
The mud and wall still exert forces on each other during the
collision. The force which the wall exerts on the mud causes an acceleration
of the mud which has the effect of stopping it as per Newton's second law.

QUESTION:
Is there is any situation which fulfills the first condition of equilibrium but not second condition of equilibrium?

ANSWER:
Of course. Consider a uniform stick of which has opposite forces
on its ends, both perpendicular to the stick

QUESTION:
I understand the bullet leaving the gun at high velocity and bullet dropped from end of barrel at zero horizontal velocity hit the ground at the same time (assuming level ground etc.) but I have a different one. Two bullets leave the same barrel and we do not figure in anything other than level. One has twice as much mass as the other (so it needed more energy to get to the same velocity). So assuming they have the same shape and aerodynamics I think the air friction will act the same and they both hit the ground at the same time. But, others in my group think it doesn't seem right. What is the right answer?

ANSWER:
I do not understand your phrasing "…leave the same barrel and we
do not figure in anything other than level…" I will just assume that the two
bullets begin simultaneously at the same height with the same speed and in
the same direction. The force of air friction is determined by the shape of
the projectile, so each will experience the same force at any particular
speed. But the force F on the heavier bullet will result in a smaller
acceleration a than for the lighter bullet because of Newton's second
law, a=F /m . Therefore the lighter bullet will slow down faster
and hit the ground earlier. You can understand this intuitively with an
extreme example. Imagine a spherical balloon and a spherical cannonball of
the same size both projected horizontally at 100 mph. Which do you think
will hit the ground first?

QUESTION:
I doing some work with model rockets and I am trying to understand more about the center of mass. Specifically what happens to the center of mass as more weight is added. I understand that the center of mass begins to move nearer to where the weight was added, but as I continue to add more weigh does the movement become larger or smaller? My observation is that it becomes larger, but I've read another document that suggest it should be come smaller. I'm having a difficult time finding other comments on the topic. I would appreciate any help you can provide. I'd like to understand if my observation is incorrect and if so - maybe understand why.

ANSWER:
Suppose the mass of the rocket before you start adding mass is
M and that the center of mass of the rocket is a distance L
away from where you will adding mass. If you start adding mass m , the
center of mass will move away from M and toward m such that it
will be a distance d from M . Then the equation for the center
of mass is d /L =(m /M )/[1+(m /M )].
The graph to the left is for no mass added (obviously, d =0) up to
m=M (d=L /2). To answer your question, note that as m
increases, the slope of the curve decreases, that is the rate of change of
position of the center of mass decreases. As m gets bigger and
bigger, the center of mass approaches d=L , and so clearly it will the
rate at which the position changes gets smaller. The graph on the right is
plotted up to m =10M so you can see this happen.

QUESTION:
I'm doing an assignment on helicopter flight, and I'm a little confused about the Bernoulli principle. He said that if a pipe is bigger at the beginning and smaller at the end, the fluid traveling through the end of the pipe would have a lower pressure.
This seems counter-intuitive. I would have thought that there would be more pressure on the fluid that's "squeezed in together". I don't think I fully understand the concept of pressure.

ANSWER:
Let's first write Bernoulli's equation, P+ Ѕρv ^{2} +ρgy =constant.
At any point in the fluid P is the pressure, ρ is the density,
v is the speed, g is acceleration due to gravity, and y
is the distance relative to some chosen reference point (above, y >0,
below y <0). Maybe it would help you to accept this if I tell you
that Bernoulli's equation is simply conservation of energy. Since To answer your question we can ignore the y part,
it is a horizontal pipe, P+ Ѕρv ^{2} =constant. You
should also understand that this equation is exactly true only under
idealized conditions:

An incompressible
fluid (water is a very good approximation)

Laminar flow
which means flowing smoothly, no turbulence

Irrotational
which means there is never any local circulation like whirlpools.

There must be no
viscosity or other kind or friction

What goes on around a
helicopter blade satisfies none of these conditions! Nevertheless,
Bernoulli's equation can be a very useful approximation to understand what
is going on even if it is not exactly correct. The most important one for
aerodynamics is that which you cite, that when v increases, P
increases. I have struggled to come up with a way you could understand this
intuitively, but don't seem to find a simple explanation. You should
convince yourself that your intuition is wrong by doing some experiments.
For example, notice that inside a moving car the smoke from a smoker is
drawn through a cracked-open window because of the lower pressure outside
where the velocity is higher. Or, do the old blowing on the piece of paper
demonstration.

QUESTION:
I'm working on a story and I'm trying to give a correct scientific explanation I can as justification for the character's powers.
My questions are regarding speed and its various effects.
Considering a human of mass 60kg can move at the speed of 200 m/s to 300 m/s, and can accelerate and decelerate at the same rate as the acceleration of bullet fired at muzzle velocity (that is almost instantly)

What is the effect of drag at that speed on a human?

How much energy is required to gain this speed with the above mentioned acceleration and deceleration rate and weight moving for 50 meters?

What is the effect of G-force on the human?

How much kinetic energy will the human produce when moving at such speed?

Will the human be clearly visible (or observable) to another normal human, moving at such speed?

ANSWER:
This is really a ridiculously impossible scenario! I will assume
that v =250. I will not include all the details of the calculations.

You can estimate
drag force as F ≈јAv ^{2} where A is the
cross sectional area. If we take A ≈2 m^{2} , F ≈3x10^{4}
N. With this force acting, the power (rate of energy loss) is about
P=Fv =8x10^{6} W=8000 kW. This would cause him to burn up, I
would guess. I estimate that, if there were not some force to keep him
moving, he would lose 90% of his speed in about 4 s.

He has an energy
of Ѕmv ^{2} =2x10^{6} J. The acceleration and
distance are irrelevant.

I took an M16
rifle as an example. The muzzle velocity is about 1000 m/s and the
barrel length is about 50 cm, so I calculated that the acceleration
would be about 10^{6} m/s^{2} , about 100,000 g. The
maximum which the human body can withstand for short times is about 10
g. The time to accelerate the bullet would be about 10^{-3} s.

This makes no
sense. The body has kinetic energy but does not produce it. Because of
the drag, it loses kinetic energy to heat.

Your speeds are
about the same as a commercial airliner, 500-600 mph, and there is
certainly no problem seeing them.

QUESTION:
I am looking to find out at what speed an object (object 1) was traveling while hitting an other object (object 2) and pushing it straight ahead.
Object 1 was traveling straight, rubber wheels on asphalt (no skid marks), had a weight of 4500 lbs, and hit the object 2 with a weight of 3000 lbs at a 90 degree angle (side impact, rubber
sliding on dry asphalt). Object 2 was pushed 50 feet. I am looking for the formula to calculate
the speed of object 1 at impact. I know there are several factors involved and I do not need a exact number,
just an approximate but fairly close value.

QUERY:
I take it that car 2 was at rest initially and that the two
remained in contact the whole time (since you said "pushed"). Also, that car 1 did not apply brakes.

REPLY:
Yes that is correct - object 2 was at rest and object 1 did not apply
brakes.

ANSWER:
I prefer to work in SI units, so I will transform back to mph in
the end. The masses of the cars are about m _{1} =4500 lb≈2040 kg and
m _{2} =3000 lb≈1360 kg. The coefficient of kinetic friction for rubber
on dry asphalt is in the range 0.5-0.8, so I used μ =0.65. The
acceleration due to gravity is g =9.8 m/s^{2} . The distance
pushed is d =50 ft≈15
m. The frictional force F acting on the two cars as they slide is due
only to the friction between the wheels of car 2 sliding on the
asphalt is F=μm _{2} g =0.65x1360x9.8=8660 N. The work
done by this friction was W=-Fd =-8660x15=-130,000 J. This took away
the kinetic energy K the two vehicles had just after the collision,
so K =Ѕ(m _{1} +m _{2} )v _{2,1} ^{2} =1700v _{2,1} ^{2} =130,000,
so v _{2,1} =8.7 m/s≈20 mph; this is the speed the two cars had
immediately after impact. Finally, use momentum conservation to get the
speed of car 1 before the collision: m _{1} v _{1} =(m _{1} +m _{2} )v _{2,1} =2040v _{1} =3400x8.7=29,600
k∙m/s, and so v _{1} =14.5 m/s≈32 mph.

QUESTION:
A motorcyclist strikes an automoble that turns directly in front of him. The motorcyclist had locked his brakes and was traveling at @ 30 to 35 mph when he struck the front left fender of the automobile and went over the top of his widnshield at approx. 5 feet. It the motorcylist weighs 210 lbs. how far would he have traveled before he struck the ground?

ANSWER:
There is no way to calculate this with the information you have
given me. All I can tell you is the absolute maximum distance he would go if
launched with a speed of about 30-35 mph at 45^{0} angle to the
horizontal, about 25 meters. It would certainly be less than this since he
would lose some kinetic energy in the collision and would not be launched
with a speed as great as he came in with and he would likely be launched at
another angle.

FOLLOWUP QUESTION:
I'm actually thinking that for me to clear a 5 foot windshield directly in front of me from a sitting position about 15 inches behind that windshield and from a sitting position my launch angle was very close if not more than 45 degrees. If that is the case, loss of kinetic energy is the only limiting factor to the distance I traveled before landing on my shoulder on the other side of the car. Is the loss of kinetic energy linear to the distance traveled? In other words if I lost 1/3 of my kinetic energy in the crash would the distance traveled be equal to 1/3 less distance? 25 meters is 82 feet minus 28 feet equals 54.

ANSWER:
OK, here is the full equation if you want to play around with
it: R=v ^{2} sin(2θ )/g where R is the
horizontal range in m, v the speed of launch in m/s, θ is the
launch angle, and g=9.8 m/s^{2} is the acceleration due to gravity;
this is for the projectile landing at the same level where it was launched
from, not really exactly true but of minimal consequence since we are only
doing a rough calculation. 30 mph=13.4 m/s, so you can put in any speed as
v _{m/s} =0.45v _{mph} .
For θ= 45^{0} , R=v ^{2} /g . To answer your
question, the distance is not proportional to speed, rather to the square of
the speed. So, if you lost 1/3 of your speed (30 mph) in the collision your speed
in m/s would be v =20x0.45=9 m/s, so R =81/9.8=8.3 m≈27 ft. This is
more realistic, I believe, than 25 m which would be predicted for 35 mph.

QUESTION:
Most physics texts that I read state that a net torque tends to produce rotation - certainly true in case of a sphere rolling down an incline (friction produces the torque and this results in rotation). However, consider a particle of mass m in the XY reference frame, acted upon by a force F in a direction along the X axis producing an acceleration a = F/m. Let us assume that the mass m is located at (x= 0, y = y0) from the origin. Then there IS a net torque on the mass m (seen about an axis perpendicular to the XY plane passing through the origin) given by (y0) multiplied by F (angle being 90 deg). But mass m undergoes NO rotation! It simply translates in the direction of F. What's confusing for me - is there a definite criterion defining the situation when a net torque would produce rotation and when it would not.

ANSWER:
The example you stated is the simplest to understand because
this force is perpendicular to its moment arm and so the torque is τ=Fy _{0} .
First of all, a point mass cannot, by definition, rotate about an axis
through itself. If you calculated the torque about an axis ay x =0,
y=y _{0} , you would get zero torque and zero "rotation". I think
you are confusing yourself by saying that a torque causes a rotation. In
fact, a torque causes an angular acceleration about the axis, so you need to
ask whether your applied force causes an angular accleration. Angular
accleration a is related to the acceleration a by
α=a _{t} /R where R is the moment arm (y _{0}
in your example)and a _{t} is the tangential acceleration, the
component of the acceleration a perpendicular to the moment arm. In your
notation, the acceleration is all tangential and so a _{t} =F /m.
Therefore, there is an angular acceleration α=F /(my _{0} )
about the z axis at the instant you start your problem. Even if there
were no force and the mass were just moving with some velocity v in
the x direction right now, it would be considered to be rotating
about the z axis with an angular velocity ω=v /y _{0} .
If you exerted some force which kept it moving in a circle of radius y _{0}
with speed v (that would be a force which had no tangential
component, pointed toward the origin, and had a magnitude mv ^{2} /y _{0} ),
you would surely say that it was rotating around the origin, wouldn't you?
In general, whenever the velocity has a component perpendicular to the line
drawn to some axis, it has an angular velocity around that axis.

QUESTION:
From both my high school and university (albeit, introductory at university) physics courses, we've encountered the concept of escape velocity, as the velocity with which an unpowered mass must be accelerated to to escape the gravitational pull of a body.
We've also been given an extra piece of information about this, which is that the escape velocity is the speed at which an object will be accelerated to when it hits the surface of a body, falling from an infinitely far away point.
So here's my question - it seems then that if I sent an unpowered mass at the escape velocity of a body, then (assuming there were no other gravitational factors, etc), then when it reached the 'infinity far away point' it's velocity would equal zero. So, even though you could never reach that point - is that logic valid?
If so, doesn't that mean then that escape velocity is more of a measure of the velocity needed for an unpowered mass to reach x point far away? So then shouldn't it then, if I were to reach a point closer than one infinity far away, why wouldn't I be able to escape with a lesser velocity?
I suspect a flaw in my reasoning above, but I don't know what it is - any and all response would be greatly appreciated!!

ANSWER:
I am afraid I do not really get your point. But, escape velocity
is a concept somewhat divorced from reality because the universe is not
infinitely large and there are other things in it besides the earth and your
projectile. It is easy to see how the computation of (ideal) escape velocity
can be done. With nothing in the universe but the earth, the total energy of
the projectile at the earth's surface is Ѕmv (r=R _{earth} )^{2} -GM _{earth} m /R _{earth}
where I have chosen potential energy to be zero at r =∞. Anywhere else
the energy would be Ѕmv (r )^{2} -GM _{earth} m /r .
If we choose r =∞ and use energy conservation, Ѕmv (R _{earth} )^{2} -GM _{earth} m /R _{earth} =Ѕmv (∞)^{2}
and the smallest velocity v (R _{earth} ) could be is
called the escape velocity and corresponds to v (∞)=0, so v _{escape} =√[2GM _{earth} /R _{earth} ].
If there were indeed nothing else in the universe and the universe were
infinitly large, this is the speed you would have to give something for it
to never come back. In the real world, interaction with other objects would
affect the speed necessary for the object to never come back, so you should
not think of escape velocity as that speed because escape velocity is well
defined but the speed to escape the real earth is not the same thing.

QUESTION:
We want to know if a a gun could and was traveling at the speed of light and it was fired would the bullet simply fire as normal or would the bullet refuse to leave the gun?
We have tryed to find a definitive answer elsewhere but can't seem to find one

ANSWER:
First of all, it is physically impossible for the gun to move
with the speed of light; see my faq
page to find out why. But, I think I can get to the crux of your question by
having your gun move at 99.9% the speed of light, v =0.999c .
Let us suppose the muzzle velocity of the gun were u =0.002c
(which is, incidentally, much faster than any real bullet would travel).
Now, classical physics would have a bystander see the bullet going with a
speed v+u =1.001c , faster than the speed of light. But we know
that this is not possible, so classical physics must be wrong. The correct
formula for velocity addition in
special relativity is (v+u )/[1+(uv /c ^{2} )]=1.001c/(1+0.999x0.002)=0.999004c ,
just slightly faster than the gun. If you were moving along with the gun,
you would simply see the bullet go forward with a speed 0.002c . All
the above is if the bullet is fired in the direction the gun is traveling.
If you fire the gun backwards from its direction of travel, replace u
by -u and find that (v-u )/[1-(uv /c ^{2} )]=0.998996c ,
slightly slower than the gun but moving in the same direction. If you were
moving along with the gun, you would simply see the bullet go backward with
a speed 0.002c .

QUESTION:
I am curious as to how a spoiler on a car somehow provides less wind resistance when driving. One would think that the more surface area against wind would cause more resistance which would slow down the car at a steady RPM.
I don't understand how adding another element that increases surface area would somehow decrease resistance. When I think about it, the less surface area against air flow, the less resistance there would be.
So my question is, do spoilers on cars really work? If they do, how?
I have always been curious about this and have never gotten a straight answer out of somebody that really understands how they work.

ANSWER:
Well, one might say that "nobody really understands how they
work"! I have "sort of"
answered this question before, but mostly dodged it. Here I will do a
little better, I hope! I can give you a good qualitative explanation but in
terms of being able to design one by just sitting down with some fundamental
equation like Bernoulli's equation, forget it. Fluid dynamics can be
deceptively simple or amazingly complex. To do serious aerodynamic design
requires extremely powerful computers and trial and error wind tunnel
experiments. The things we do to reduce air drag are often counterintuitive
compared to our expectations. With that prologue, let me give you a couple
of simple examples. Our expectations are that an object should be very
smooth and this is often the case, particularly if velocities are not too
large. Your expectation that the increasing area presented to the onrushing
air causes greater drag is reasonable and you will find many examples here
on AskThePhysicist.com where I approximate the drag force as being
proportional to the cross sectional area, e.g. F _{drag} ≈јAv ^{2} .
But there are many situations, particularly at high speeds, where this
expectation breaks down and the culprit is turbulence in the air. To
illustrate how turbulence affects drag and how smooth is sometimes not good,
consider a golf ball. Have you ever noticed that a golf ball has dimples?
The purpose of these dimples is to reduce air drag. As shown on the
left above, a smooth ball at a high velocity has a long turbulent volume
behind it; the pressure in this turbulent volume is lower than the pressure
on the front of the ball and this contributes to there being a large net
drag force. If golf balls were nice and somooth, they would die and fall
very much sooner than a dimpled golf ball does. The ball on the right shows
the effect of the dimples; the rough surface induces a layer of turbulence
which actually makes the ball "slipperier" which causes the flow around the
ball to come back together and reduces the volume of turbulence contributing
to drag. (The picture on the right, with most of the turbulence erased,
would be what the smooth ball would look like at low speeds.) The hairs on a
tennis ball serve the same purpose. One other
example is the net you sometimes see replacing the tailgate of a pickup
to reduce the drag the tailgate causes. This, it turns out, is a complete fraud. With
the tailgate closed a bubble of still air forms in the bed of the truck
which deflects the air smoothly over the rear of the truck. So, in your
case, the purpose of the spoiler, very similar to the golf ball dimples, is
to disrupt the smooth flow in such a way that the net effect is less
turbulence behind the vehicle. Incidentally, there is something called the
Reynolds number
which allows you to estimate whether or not turbulence is important.

QUESTION:
In many fictional works involving superpowered individuals there are people who can lift objects with their minds. Like an example is Magneto lifting a submarine with the magnetic fields he generates in the movie X-Men First Class. From what I remember from high school physics is that because of Newton's third law, all actions produce an equal counteraction. So wouldn't Magneto lifting a submarine with his powers lead to him being heavily attracted by the submarine and flying straight towards it like a bullet and then getting himself completely flattened into a pancake when he crashes with the submarine?

ANSWER:
You are absolutely right, if he exerts a force on the submarine,
the submarine exerts an equal and opposite force on him. Now, he is on some
kind of aircraft and holding on, so he would have to be superstrong to be
able to hold on since the force pulling him is greater than the weight of a
submarine. And whatever he is holding on to would have to be strong enough
that it could hold more than the weight of the submarine. And the aircraft
had better be able to carry the weight of a submarine. But, maybe his power
is in his left arm so the force would just rip his arm off. Pretty
preposterous, isn't it? Good thinking on your part.

QUESTION:
Imagine a tank full of water on top of a house with a drain pipe down to the ground. The higher the tank, the greater the velocity and force of water coming out of the end of the pipe. But does the tank drain any faster with greater height? You can't pull on water, so the water leaving the tank shouldn't care how high up the tank is, so the answer must be no. On the other hand, it does seem from actual experience that a bucket will be filled faster from the higher tank. I've even measured this! (Imagine this in a vacuum to bypass any air-pressure effects)

ANSWER:
The operative physical principle is Bernoulli's equation, P+ Ѕρv ^{2} +ρgy =constant.
Here, ρ is the density of the fluid, P is the pressure, v
is the speed of the fluid, and y is the elevation relative to some
y =0, all at any point in the fluid; g =9.8 m/s^{2}
is the acceleration due to gravity. In your case, P is atmospheric
pressure both at the top and at the bottom (assuming the tank is not
sealed), I will assume y =0 at the bottom and y=h at the top
surface, the speeds at the bottom and top are v _{bottom} and
v _{top} . Putting these into Bernoulli's equation and solving,
v _{bottom} =√[2gh +v _{top} ^{2} ].
Now, if the tank is much bigger than the pipe, usually the case, you can
approximate v _{top} ≈0 and so v _{bottom} ≈√[2gh ].
An interesting fact: this is exactly the speed the water would have if you
simply dropped it from a height h . (You do not have to imagine a
vacuum since the pressure is easily included and, as here, often the same
everywhere. If the tank were pressurized to a pressure greater than
atmospheric, the water would drain faster.)

FOLLOWUP QUESTION:
I see the math, but I remain puzzled for this reason: the water flow (in gpm, say) into the pipe (where it connects to the tank) is, obviously, the same as the flow at the bottom end of the pipe. Having constant gpm flow through a constant-diameter pipe necessarily means the water is passing through the entire pipe at the same velocity. So the velocity of the water that’s just entered the pipe must be the same as the velocity of the water that’s about to exit. But Bernoulli’s equation in the form you’ve given implies that the water is accelerating as it flows down the pipe. That would require more gpm out than what went in!

ANSWER:
No, I am assuming that the tank has water in it and the point I call
y=h in my answer is the top surface of the water in the tank which I subsequently approximate as a much larger area than the pipe.
You could, I guess, let y=h be the top of the pipe but then the
pressure there would not be atmospheric. Then you, as you suggest, would
have to say v _{top} =v _{bottom} ≡v ; the
subscript top now means the top of the pipe. So, let's call the depth of the
water in the tank to be d and apply Bernoulli's equation to the tank
only: P _{top} +Ѕρv ^{2} +ρgh ≈P _{A} +ρg (d+h )
where I have again approximated the top surface of the tank to have zero
velocity. Then the pressure at the top of the pipe is P _{top} ≈P _{A} +ρgd -Ѕρv ^{2} .
Now apply Bernoulli's equation to the pipe: P _{A} +ρgd -Ѕρv ^{2} +Ѕρv ^{2} +ρgh ≈P _{A} +Ѕρv ^{2}
or Ѕv ^{2} ≈g (d+h ) which you will see is the same
answer as above except h now means the bottom of the water in the
tank rather than the top of the water in the tank.

ADDED
NOTES:
Bernoulli's equation is only applicable for incompressible,
nonviscous fluids having laminar (nonturbulent) flow. It is a fairly good
approximation for gases with velocities small compared to the speed of
sound. It is a statement of conservation of energy.

QUESTION:
In a rotating frame of reference, do the Coriolis force and the centrifugal force act opposite to each other?

ANSWER:
Certainly not. The centrifugal force always points perpendicular
to the rotation axis. The centrifugal force always points in the direction
of v xω where ω is the angular
velocity and v is the velocity of the particle. It is possible
for the two to point in opposite directions, but not usual. An example where
they do point opposite is shown to the right. You might be interested in a
recent answer .

QUESTION:
I was trying to calculate the ideal performance of a bow. I wanted to relate the draw weight of a bow to the speed of the arrow after releasing it. And I couldn't do it. The things I have are arrow weight(300 gramms) and the draw weight peak(that means 27kg at 72cm). The draw weight works like this. At 72cm it's 27kg and at 0cm it's 0kg. It is a linear decrease in kg. 27kg at 72cm, 22.5kg at 60cm, 18.75kg at 50cm and so on. And now I wanted to calculate how fast the arrow will fly after shooting it. That means I would like to know how to calculate the speed of an object with a mass of 300g after 72cm of accelerating with a force of 27kg that linearly decreases over 72cm to 0kg. Could you please tell me the formula I need to use to calculate that? I realy would love the learn the way how I have to do it.

ANSWER:
If the force increases linearly with distance, it is behaving
like an ideal spring. The force F necessary to stretch a spring with
spring constant k a distance x is F=kx . Now, 27 kg is
not a force, it is a mass; but, if you mean that the force is the weight of
a 27 kg mass (which you doubtless must), then F =27x9.8=264.6 N. So,
since x =0.72 m, the spring constant of your bow is k =264.6/0.72=367.5
N/m. The potential energy of a spring stretched by x is Ѕkx ^{2} =Ѕ(367.5)(0.72)^{2} =95.3
J. The kinetic energy of a mass m with a speed v is Ѕmv ^{2} =Ѕ(0.3)v ^{2} =0.15v ^{2} .
Ideally, all the potential energy of the bow is given to the kinetic energy
of the arrow, so v =√(95.3/0.15)=25.2 m/s≈56 mph. If the arrow is not
launched horizontally, there would be a small correction for change in
gravitational potential energy. I am not sure how good a model this is for a
real bow.

FOLLOWUP QUESTION:
The reason why I was asking is because I build bows and wanted to find out how good the efficiency of my bows are. And with your answer I found out that the weight of the arrows I mentioned to you and I was using in my own calculatoins was far too high. In the world of archery the arrows are measured in grains instead of grams but I forgot that and used 300 grams instead of 300 grains. So my calculations of the arrow speed were so low that I thought that I made a mistake in calculating. So I asked you. As the speed you calculated was very low too I started asking myself why and found out what mistake I made.

ANSWER:
300 gr=0.0194 kg, so Ѕmv ^{2} =Ѕ(0.0194)v ^{2} =0.0097v ^{2} ,
so v =√(95.3/0.0097)=99.1
m/s≈125 mph. Incidentally, for easy unit conversion I recommend a free
little program called
Convert .

FOLLOWUP QUESTION:
I understood your answer and started calculating bows to understand how efficient they work. I was very surprised when I calculated the efficiency of a compound bow(http://en.wikipedia.org/wiki/Compound_bow ). The bow I used to make my calculations is a 2014 PSE Full Throttle from the Pro Series of PSE Archery (http://pse-archery.com/c/pro-series-compound-bows_full-throttle_full-throttle-black ). The IBO-Bow Speed is measured with a speed chronograph with a bow with 70 pounds draw weight, 30 inches draw length and 350 grain arrow (http://www.archeryexchange.com/shopcontent.asp?type=amoibo ). The speed measured is 362-370 fps.
The astonishing thing I found out that this compound bow has an efficiency of 110-120% compared to a perfect spring. How is this possible? Is it because of the special mechanical function of the cams(wheels) and the cables? Do these cams and cables use a special leverage provided by the cams to gain more power?
Here is a link to a pdf of a product review of several compound bows:
http://www.arrowtrademagazine.com/articles/july_14/July2014-FlagshipBowReport.pdf . On page 7 there is a draw force curve. And in the article they are talking about efficiency as well. But I don't realy understand what it means (dynamic efficiency and draw cycle efficiency).

ANSWER:
Actually, I have
previously answered a question about a compound bow which you might want
to look at to get a brief overview of its advantages. As you can see from
the draw force curve which you refer to (see left), the biggest advantage is
that you end up with a relatively small force even though you stored a large
amount of energy; this allows for conditions for more relaxed aiming and
easier steadying of the bow. You still have to put all the energy in which
you expect to get out. The important thing to understand is that the energy
stored in the bow is the area under the the force vs . pull curve. For
a simple spring, the force, kx , is a straight line and so the area
under it is the area of a triangle Ѕ∙height∙base where the height is kx
and the base is x . The efficiency normally means the ratio of
energy out to energy in times 100. As best as I can tell, the draw cycle
efficiency is calculated by measuring force vs . distance pulling and
comparing with force vs . distance if you then slowly let it back
down. It seems to me that dynamical efficiency should compare measured
kinetic energies of the launched arrows which is the total energy extracted
from the bow, but that does not seem to be the case; I cannot find a clear
definition of dynamic efficiency. These differ only because of energy
losses, mainly due to friction. Suppose we had a linear bow with the same
draw (about 21"=1.75 ft) and which stored the same energy (78.3 ft∙lb); the
spring constant would 51.1 lb/ft and the force at maximum draw would be 89.5
lb. I have added the yellow curve to the graph from the paper you referred
to which is shown in the figure. Now each arrow from these two bows travels
the same distance while rubbing on the bow, so those rubbing frictional
forces are the same assuming similar materials and friction between the bow
and arrow. But the linear bow acquires a high speed much sooner than the
compound bow which means it is going much faster on average even though the
two end up at the same speed. But, the air drag is proportional to the
square of the velocity so the linear bow will likely lose more energy to air
drag during the launch than the compound bow. At least, that is my guess.
There can be miriad other reasons why the efficiencies differ due to details
like friction in the pulleys, different materials and designs, etc . I
must admit that I do not understand what actual and effective letoff are.

QUESTION:
What happens when you compress gas (for example carbondioxide) into a gas cylinder? I am thinking that as pressure increases, so does temperature (because of the increased kinetic energy of the gas molecules). But the 0th law of thermodynamics say that heat energy goes from an object of high temperature to an object of low temperature - so I am thinking that after a while the temperature of the gas will be the same as the room temperature?! Since pressure is constant in the cylinder, isn't the kinetic energy of the molecules? What happens with the gas as energy is transferred to the room? Am i misunderstanding something? I just can't seem to Get my head around this, so I am really hoping you can help med understand!

ANSWER:
When you say "…compress gas…into a gas cylinder…" I assume you
mean that you are adding gas to a cylinder with constant volume. All you
really need to know is the ideal gas law, PV=NRT where P is
pressure, V is volume, N is some measure of how much gas you
have, T is the absolute temperature, and R is just a constant
of proportionality. You need to make some approximations for a real-world
situation. If you add the gas really slowly and/or the cylinder has very
thin conductive walls, T will remain constant for the reason you
state—because the whole system will keep in thermal equilibrium with its
environment; this is called an isothermal process and as N increases,
P increases. The other extreme is that you take care to insulate the
cylinder and/or add the gas very quickly so no heat enters or leaves. In
this case, increasing N will increase the ratio P /T ;
however, you do not have enough information to determine how P and
T independently change. If you are using some sort of pump, you will be
doing work on the system which will increase its temperature and therefore
the pressure has to increase also to keep P /T from decreasing.
Another way to add gas to the tank is to take another tank with much more
gas in it and connect the two of them together. Since the volume of the two
tanks and the total amount of gas in them is constant, the temperature of
the whole system will decrease. Again, pressure in your tank must change in
such a way that P /T increases and that could mean that P
would increase, decrease, or stay the same, depending on initial situations.

FOLLOWUP QUESTION:
Thanks for the quick reply! I am still wondering though, is it possible for a compressed gas in a cylinder to have higher temperature than the room it is stored in over time? I am thinking comparing to coffee in a thermos after some hours will be the same temperature as its surroundings. Is the gas cylinder any different due to pressure inside the tank?

ANSWER:
Eventually the cylinder, room, and gas will all be at the same
temperature. If (see above) the temperature right after filling were higher
than the room temperature, it would cool down and the pressure would
decrease because now both V and N would be constant while the
gas cooled. Here is a numerical example: suppose that the room temperature
is 20^{0} C=293 K and the gas temperature is 40^{0} C=333 K
and the hot gas has a pressure of 4.0 Atm. Then, P /T =constant=4/333=P _{final} /293
or P _{final} =3.5
Atm.

QUESTION:
What is the effect of earth rotation on the human upright posture?
Please visit my website . I would
like to corroborate my hypothesis from a physics point of view.

ANSWER:
My original response, via email, was "The forces are negligibly small. I am quite certain that they are not
a factor in back pain." However, the questioner persisted.

FOLLOWUP QUESTION:
Thank you for responding to my question regarding the effect of rotation on the upright posture. As you stated the effect is indeed subtle but it is there. The point of my work is that the human spine is much more sensitive to these forces than researches realize.
Did you solve the question of the effect of rotation on the standing posture and what was the result percentage please. Did you fiqure a formula for this and was it compatible with my rudimentry estimate of 30% percent anterior left torque and spinal differential findings of 8%?

ANSWER:
I cannot resist responding in some detail to this since it is so
wrong! What we need to understand is how to do physics in a rotating
(accelerating) frame of reference such as the earth. In such a frame, there
are real forces and fictitous forces. As the questioner correctly notes,
even though we call the fictitious forces fictitious, they can certainly be
felt by our bodies if they are large enough. The best known fictitious force is the centrifugal
force; this is what we feel trying to throw us off a rotating
merry-go-round. A person at rest on the earth's surface experiences a real
force called the weight which points toward the the center of the earth and
has a magnitude W=mg ; a fictitious centrifugal force pointing
perpendicular to and away from the axis of rotation and has a magnitude F _{cen} =mrω ^{2}
where r is the distance from the axis and ω =7.3x10^{-5}
radians/s is the angular velocity of the earth's rotation*; and whatever
force the floor exerts on the object to maintain equilibrium. If the person
is moving along the earth's surface, there is an additional fictitious force
called the Coriolis force which I will discuss later. I start with the
simplest situation which is someone standing on the equator. The first
figure is a view from the north pole; the earth rotates counterclockwise;
the man has a mass m =100 kg so his weight (red) is W =980 N;
the radius is r=R _{earth} =6.4x10^{6} m, so the
centrifugal force (green and not to scale ) is F _{cen} = 100x6.4x10^{6} x(7.3x10^{-5} )^{2} =3.4
N; finally the force from the floor (black) is vertically upward and of
magnitude 980-3.4=976.6 N. Note that there is no net torque for a person at
the equator and the only effect is that there is approximately a 0.35%
reduction in apparent weight. Next let us examine a person at rest but at
some latitude, say 45^{0} north as shown in the second figure above;
the color coding of the forces is the same as in the first figure. The
weight is the same, 980 N; the radius is r=R _{earth} sin45^{0} =4.5x10^{6}
and so F _{cen} = 100x4.5x10^{6} x(7.3x10^{-5} )^{2} =2.4
N; F _{cen} has a radial component 2.4sin45^{0} =1.7
N which reduces the apparent weight (radial component of the floor force) by
that amount; F _{cen} has a tangential component
2.4cos45^{0} =1.7 N which results in a torque and is also the
magnitude of tangential component of the floor force. The torque tries to
rotate the person about his feet in a southerly direction (would be
northerly in the southern hemisphere) and has a magnitude of 1.7 N∙m,
estimating the center of gravity to be 1 m above the feet. To put this in
perspective, if the earth were not rotating the person would have to lean at
an angle of tan^{-1} (2.4/980)=0.14^{0} to have a comparable
torque about his feet; this angle is far smaller than leaning we do in
normal activities. Furthermore, since the direction we happen to be facing
is random, the average effect over time of these very tiny torques would be
zero.

Finally, there is another (fictitious) force
which acts on objects on the rotating earth, the
Coriolis force . This is the force responsible for the cyclonic motions
of weather patterns. However, the object has to be moving and the direction
of the force depends both on location and direction of the velocity. Its
maximum possible magnitude is F _{Cor} =2mωv where v
is the speed; its direction is always perpendicular to the velocity
vector. The fastest the average person is likely to be moving is the speed
of a passenger jet, about 500 mph≈220 m/s, so F _{Cor} =2x100x7.3x10^{-5} x220=3.2
N. Again, this force is so small compared to other forces acting that its
effect on physiololgy would be essentially zero.

*ω= (1 revolution/day)(2π
radians/revolution)(1 day/24 hr)(1 hr/3600 s)=7.3x10^{-5} s^{-1}

QUESTION:
If I was throwing a ball around with a friend on a torus-shaped space station which was rotating in order to simulate gravity, would I have to lead her? Alternatively: if I "dropped" the ball (i.e. let go of it) from, say, waist-high, would the ball land at my feet, or somewhere else?
( Assume that the torus is roughly 50m in diameter, neither
accelerating in space nor in terms of its rotation, and that we're using a
standard baseball.)

ANSWER:
My first inclination was to say that, assuming that the station
was big enough, it would be pretty much like playing catch on earth. The
reason for this guess was that I had worked out a similar problem for an
earlier answer . The question
there was if you were to jump straight up what would happen; this is
equivalent to throwing a ball straight up with some speed v . The
answer was that for sufficiently large R and small v , in
particular R>>v ^{2} /g , the time the ball was in the
air, the "height" it rose, and the time it took to "return" would all be
about the same as if you did the experiment on earth; also, the ball would
"land" back where you were. I believe that answers your question about
whether a dropped ball would land at your feet—it would (approximately). You
should be sure you understand that earlier answer. The leftmost figure above
reproduces the original figure for the ball thrown straight up. The other
two do not have all the vectors labeled, but you can tell by comparison what
they are. The components of v are shown by light blue vectors
and the red is just to aid me in adding the vectors v and
Rω . The second figure is if your partner is ahead of you (ahead
meaning in the direction of rotation). The third figure shows how you would
throw it if your partner were behind you. It is not at all clear to me where
you and your partner would be located when the ball "came back down" to the
ring (shown by green dots). Also, trying to do the problem analytically in
the nonrotating frame gets very messy. I am just interested, as are you, in
your situation—thrown baseballs in a 50 m ring rotating such that Rω ^{2} =g
where ω is the angular velocity in radians per second. So, I am going
to do the calculations in the rotating frame (where you and your partner are
at rest). This gets sort of tricky since we will have to introduce
fictitious forces to describe the motion of the ball; once you throw the
ball, there are no real forces on it. Now, maybe you throw the baseball
(mass about 0.15 kg) with a speed of 10 m/s≈22 mph. There are two fictitious
forces: the centrifugal force F _{r} =mRω ^{2} =mg
which points down for you and the Coriolis force F _{Cor} =2mωv= 2mv √(g /R )
and points in a direction perpendicular to the vector v as
shown in the figure to the left. Now, let's calculate the magnitudes of
these forces. F _{Cor} =2x0.15x10x√(9.8/50)=1.3 N and F _{r} =0.15x9.8=1.5
N. The Coriolis force is not small compared to the centrifugal force and
therefore the ball will not behave at all as it would in a true uniform
gravitational field. You would go nuts trying to play catch on this space
station! The problem is that R =50 m is not much greater than v ^{2} /g= 10.2
m. See the added thought below for a discussion of the dropped ball.

ADDED
THOUGHT: When the ball is simply dropped from say h =1 m, the
Coriolis force is relatively small because the velocity is small for most of
the time of the fall. So, the deflection should be modest. I will try to
estimate the amount of deflection. The centrifugal force is mg and is
always radially out, choosing +y radially out, a _{y} =-g+a _{y} ^{Cor} ;
I will argue that the ball does not acquire enough speed for the Coriolis
force to have a significant radial component, so a _{y} ^{Cor} ≈0,
and y ≈h- Ѕgt ^{2} , v _{y} ≈-gt .
So, the time to fall is approximately t ≈√(2h /g )=0.45 s
and v _{y} ≈4.4 m/s. The Coriolis acceleration, if the velocity
is purely radial, points in the backward direction (to the left in my figure
above) and would have a maximum magnitude of about 2v √(g /R )≈3.9
m/s^{2} . If the ball drops almost vertically the Coriolis
acceleration would be approximately a _{x} ≈2m √[(v _{x} ^{2} +v _{y} ^{2} )(g /R )]≈2mv _{y} √(g /R )=2mgt √(g /R )=8.68t =dv _{x} /dt ;
therefore, integrating, v _{x} ≈4.34t ^{2} and
x ≈1.45t ^{3} .
So, at t =0.45 s, v _{x} ≈0.88 m/s and x ≈0.13 m=13 cm.

QUESTION:
How does a GPS station work? I just learned about them and my teacher was a little unclear so I'm depending on you to hopefully give me a direct answer.

ANSWER:
Suppose that you were on a line and you wanted to know where you
were. If there were some point on that line and you knew where you were
relative to that point, then you would know where you are. Now suppose that
you were on a flat plane. Now you would need two other points the positions
of which you knew to figure out where you were relative to these points
(this is often called triangulation ). Now suppose that you are just
somewhere in a three dimensional volume; it would follow that you would need
three points whose positions you knew to find out where you were in that
space. So, if you want to know where you are if you are at rest, you would
need to know the positions three points of reference at rest relative to you
and simple geometry would tell you where you were. The GPS system is a bunch
of satellites which are zooming around in their orbits and therefore,
because both they and possibly you also are moving, you also need to
synchronize all the clocks on the satellites with your clock; this
synchronization means that you need one more point because you now have four
unknowns you need to solve for, your three spacial positions and the data
necessary to find the exact time. The GPS receiver in your iphone or
whatever receives the data beamed from the four satellites and computes your
position to remarkably good accuracy.

QUESTION:
According to Newton's law of gravitation, gravitational force is inversely proportional to the distance between the center of mass of the bodies. So if i place my hands together, the distance between them is very less so the gravitational force should be very high but it is easier to separate them. How?

ANSWER:
First of all, the force is inversely proportional to the
square of the distance between the centers of mass. Because gravity is
nature's weakest force, it is tiny unless there is a very large amount of
mass. Let's estimate that the center of mass of your hands are separated by
1 cm=0.01 m and that the mass of each hand is about 0.25 kg (about Ѕ lb).
The force each hand feels can be approximated as F=Gm _{1} m _{2} /r ^{2} =6.67x10^{-11} x0.25x0.25/0.1^{2} =4.2x10^{-10}
N≈10^{-10} lb. As a comparison, the gravitational force on each hand
by the earth (whose center is much farther away) is about 2.5 N.

QUESTION:
My son's friend asks: "why does a rolling object balance better than a stationary one?"

ANSWER:
I assume the friend is thinking about a bicycle which is, as we
all know, easier to stay up on if it is moving. The physics of a bicycle is
quite complicated, so I am not giving you the whole picture but rather one
aspect which will be understandable to a youngster. Any object which is
spinning has what is called angular momentum. Shown to the left is a
spinning wheel with its angular momentum vector shown. If the wheel were not
spinning, this vector would vanish. One of the most important laws in
physics is conservation of angluar momentum: if there are no torques on the
rotating object, its angular momentum will not change and that includes the
direction of the vector. So, if you are riding a bike straight down the
street, each wheel has a horizontal angular momentum pointing to the left.
If the bike starts to tip over, the angular momentum vector would try to
remain horizontal, keeping it from tipping. If you lean hard, there is now a
torque on the system and so the angular momentum will change but not
resulting in a fall but rather in a turn. You might want to get a toy
gyroscope to show the kids how angular momentum works.

QUESTION:
A tall tree cracks and falls. Can the resultant linear acceleration exceed the acceleration due to gravity?

ANSWER:
As the tree falls, it rotates about the point of contact with the ground
with some angular acceleration. Therefore, each point along the length of
the tree has a different acceleration which is perpendicular to the tree.
The falling tree is shown in the figure to the left. The weight mg
acts at the center of gravity a distance D from the rotation axis.
The tree has height L . An arbritrary distance up the tree is denoted
by x . The angle the tree makes with the vertical is θ . The net
torque τ on the tree is τ=mgD sinθ=Iα=Ia /x where
I is the moment of inertia of the tree about its pivot point, α
is the angular acceleration of the tree, and a is the
tangential acceleration of the point x. Therefore, a =mgxD sinθ /I.
Therefore, if mxD sinθ /I >1, a>g . Consider a
simple example modeling the tree as a long thin stick; then I=mL ^{2} /3
and D=L /2, so a =3gx sinθ /(2L ). Take a
particular example, θ= 60^{0} ; then a =1.3gx /L
or any point with x>L /1.3 will have a>g .

ADDED
NOTE:
The discussion above refers to the tangential acceleration of a point
on the tree which is what I thought you were referring to. However, each
point at position x also has a centripetal acceleration, call it a _{c} =v ^{2} /x =xω ^{2} .
You can get ω from energy conservation, mgD =ЅIω ^{2} +mgD cosθ
and therefore a _{c} =2mgxD (1-cosθ )/I .
So, the total acceleration is √(a ^{2} +a _{c} ^{2} );
I will let you work that out. Modeling the tree as a uniform stick, I have
calculated the total acceleration for several values of x shown in
the graph to the right. For x=L /4, all angles have acceleration less
than g . For x=L/2, the center of mass, a _{total} >g
for angles greater than about 60^{0} . For x=L , a _{total} >g
for angles greater than about 35^{0} .

QUESTION:
I'm writing a fantasy novel and I just want to check my science.
In it there's a tower which is so tall that it is higher than the troposphere. There is a scene where someone smashes a window. My question is would the air all rush out due to unequal air pressure and when the air pressure settles would everyone feel the effects of explosive decompression or at least find it hard to breathe?

ANSWER:
Presumably, all your windows are sealed such that the inside is at
atmospheric pressure. "Higher than the troposphere" would imply the
stratosphere which is where commercial jets fly. As you doubtless know, the
pressure there is very low. The figure to the right shows that just above
the troposphere the pressure is only about 200 millibars, about 1/5 of
atmospheric pressure. There is not enough air there to keep you alive. You
may recall the crash of the private jet in 1999 which killed golfer Payne
Stewart. The plane lost pressure at an altitude of 11.9 km and everyone lost
consciousness; it flew on autopilot until it ran out of fuel and crashed.
Most certainly the air would be drawn (violently) out of your broken window.
Another thing you should think about is that you cannot simply pressurize
your entire tower to atmospheric pressure; you have to pressurize in layers
of a few hundred meters and isolated from each other. Otherwise you will
have the same strong pressure gradient with altitude as the atmosphere has!

QUESTION:
If I stick a pea on the out side of a cylinder that is rotating on a fixed axis at a constant speed, what are the forces being applied to the pea?
Is the pea under going acceleration? Even at a constant speed due to the angular changes.

ANSWER:
The pea is experiencing what is called uniform circular motion.
Its velocity is a vector tangent to the cylinder's surface. Although the
magnitude of the velocity (speed) is not changing, the direction of the
velocity is constantly changing. Therefore, since acceleration is rate of
change of velocity, there is an acceleration. This acceleration, called
centripetal acceleration, is a vector, points toward the axis of the
cylinder, and has a magnitude v ^{2} /R where v
is the speed of the surface of the cylinder (and therefore of the pea) and
R is the radius. A force must be applied toward the axis of magnitude
mv ^{2} /R where m is the mass of the pea. This
force is provided by whatever adhesive you used to stick the pea there. If
the adhesive is not strong enough, the pea will fly off. Of course there is
also the pea's weight pointing vertically down and, assuming the axis of
rotation is vertical, the adhesive must also exert a force upward to hold
the pea from falling due to its weight.

QUESTION:
What would happen if you were to take a satellite that's in orbit, put a giant ball of string on it, and slowly unwind the string letting the free end fall
to earth? Would it snap? What if it was made of a material that is extremely strong such as carbon nano tubes?

ANSWER:
Think about it. The ball of string is orbiting with you, has
your same speed, and seems to be "weightless" just like you. So you grab
onto the loose end and release the ball to fall and nothing happens, it just
orbits along with you! Sort of related to your question is the
space elevator
which might interest you.

QUESTION:
Why does mass not alter the acceleration in a vacuum. If a hammer falls to the earth faster than hammer falls to the moon, then mass does matter. Consider the opposite perspective. Earth falls to the hammer faster than the moon falls to the hammer. Am I wrong in my reasoning? I've asked a few of my physics teachers, but they said that my logic was flawed and you can't consider it that way without explaining why.

ANSWER:
You are completely misunderstanding the constance of
acceleration due to gravity, independent of mass. The statement assumes all
masses experience the same gravitational field (see
FAQ page). Obviously, because of Newton's
second law, the acceleration a due to any force F
depends on mass m , a=F /m . The "falling" of the moon or
earth to the hammer is negligibly small for all practical purposes. However,
you could calculate the initial acceleration of each. The force on the earth
or moon is F=MmG /R ^{2} where M is the mass of
the earth (moon) and R is the radius of the earth (moon) (assuming
the hammer is close to the surface). So the initial acceleration of each
would be a=F/M=mG /R ^{2} and therefore a _{moon} /a _{earth} =(R _{earth} /R _{moon} )^{2} .
So, the moon actually accelerates toward the hammer faster but it has
nothing to do with the mass as you can see, it is because the center of the
moon is much closer to the hammer than the center of the earth is. If you
were to place the hammer such that it was a distance R _{earth}
from the center of the moon, the earth and moon would have equal
accelerations. The earth and moon are responding to the gravitational field
of the hammer. So, you are wrong on all your statements and, guess what,
your physics teachers are right!

QUESTION:
Is it true that if you have two objects and nothing else you can't tell
which object is moving and which object is standing still? Does this mean that we don't know how fast earth is moving?
And there is no experiment we can conduct to see if we are moving or not?

ANSWER:
First, you have to understand the difference between inertial
and noninertial frames of reference. If Newton's first law is true where you
are, you are in an inertial frame of reference. (Newton's first law says
that if an object is at rest in your frame the sum of all forces on it is
zero.) Any other frame which moves with constant speed in a straight line
relative to your frame is also an inertial frame. Noninertial frames are any
frames which have an acceleration relative to an inertial frame. The
principle of relativity says that the laws of physics are exactly the same
in all inertial frames of reference. Since the laws of physics are what will
determine the result of any experiment you can do, it makes no sense to
refer to an object's being at rest unless with reference to something else.
Does this answer your question? Also, we do not know how fast the earth is
moving because that is a meaningless question unless we say, for example,
how fast is it moving relative to the sun, or how fast is it moving relative
to the center of the galaxy, etc .

QUESTION:
Why is a hard ball more likely than soft ball of equal mass and volume to break a glass window if they are thrown at the same speed?

ANSWER:
The average force F which an object exerts during a
collision is the change in momentum (Δp=m Δv ) divided by the
time the collision took Δt . The soft ball has two things going for
it. First, because it is "squishy", the collision lasts longer, so Δt _{soft} >Δt _{hard} ;
second, the soft ball is probably less elastic than the hard ball so that it
loses more energy in the collision and therefore bounces back with less
momentum resulting in Δp _{soft} <Δp _{hard} .
Therefore Δp _{soft} /Δt _{soft} <Δp _{hard} /Δt _{hard
} or F _{soft} <F _{hard} .

QUESTION:
I have been attempting to teach myself chaos theory, however I have had trouble understanding it and how it is involved with different levels of quantum physics as well as relativity. I am also having trouble understanding the "three body problem", which seems to occur in many different physical systems.
I was hoping that you could help me to understand at least some of Chaos theory and how it connects to both quantum physics and relativity, and what exactly the "three body problem" is.

ANSWER:
I am sorry, but your question is too technical and too unfocused for the
purposes of this site . I can tell you something about the 3-body problem, though. If two bodies interact only with each other, for example the earth and the moon, you can write the orbits in simple analytic closed form. However, if there are three interacting bodies there is, in general, no closed-form solution. There are special cases where the 3-body problem can be solved, for example if one of the bodies is held fixed, but not in general.
If you google "three body problem" you will find more information. A
particularly interesting (and newsworthy) discussion may be read in
AAAS Science News . All these special cases are not chaotic because
they repeat in a periodic way. More general cases are not periodic and are
extraordinarily sensitive to initial conditions.

QUESTION:
I recently saw the movie Elysium. The most memorable thing was the space station. The station was a completely open system. Would the station's rotation keep its own atmosphere contained? How would a spacecraft land on a rotating ring like this? Would radiation levels just fry everyone and everything on it?

ANSWER:
The rotation causes an artificial gravity. If there were side
walls, you could certainly keep an atmosphere in there. For a detailed
discussion of how living there would be, see an
earlier answer . The way the
gravity works is that it is a centrifugal force (a
fictitious force ) and you want the acceleration at the outer rim to be
equal to g =9.8 m/s^{2} ; the acceleration at the surface is
v ^{2} /R where v is the tangential speed at the
surface and R is the radius of the ring. So, v =√(g /R ).
If we take, just as an example, R =1 kg=1000 m, then v ≈0.1 m/s;
since the circumference is 2πR =6283 m, this means that the ring would
rotate about once every 63,000 s=1.7 days. I think you would agree that this
speed is very small, so a spacecraft would have no trouble landing on it
assuming that the spacecraft had gotten up to the same orbital speed. I do
not believe that the radiation level would necessarily "fry" everyone, but
it would be a concern for the long term and some kind of shielding would
have to be employed.

QUESTION:
according to newton law of motion "a body will move if a net force act on it" then why the earth moves around its axis?

ANSWER:
Whoa! You have Newton's first law all wrong. A body will move
with constant velocity if there are no forces on it. For
translational motion, this means that the body moves with constant speed in
a straight line; if you exert a force on the body, its velocity will change.
For rotational motion, this means that the body spins with constant speed in
a straight line; if you exert a torque on the body, its spinning speed will
change. The earth rotates on its axis with constant speed because there are
no torques on it.

QUESTION:
These are non-academic, practical questions regarding the physics of balance and weight shifting pertaining to a moveable kitchen island.
Q1: Will an 11 inch countertop overhang alone cause the island to tip?
Q2: If not, what amount of weight (lbs.) can be safely placed on the overhang before tipping?
Q3: Is there a formula I can use to calculate this?
Description:
I have a moveable kitchen island fabricated with locking casters. I wanted to place a quartz countertop over the island base with an additional 11 inch countertop overhang supported by steel beams extending from the island base. The total length of the countertop is 37 inches (26 inches on the island base plus 11 inches as the overhang).
Proposed dimensions of the island base with its portion of the quartz countertop:
W = 52.5 inches
L = 26 inches
H = 36.0 inches (34.5" base + 1.5" countertop which includes a 5/8 inch plywood sub-counter)
Total Weight of the Island Base with Its 26 Inch Portion of the Countertop = 298 lbs.
Proposed dimensions of the 11 inch overhang:
W = 52.5 inches
L = 11 inches length
H = 1.5 inches (2cm quartz plus 5/8 inch plywood sub-counter)
Total Weight of the 11 Inch Overhang Portion of the Countertop = 50 lbs.
The overhang is a key feature of the kitchen island for our household. The island is intended to serve multiple, mundane purposes in our very compact home: food prep, dining, and working on work projects / homework.

ANSWER:
I will assume that the center of gravity of the island without
the overhang is at the geometric center of the base and that the casters are
at the corners; also, that the center of gravity of the overhang is at its
geometrical center. The red vectors in the figure to the right are pertinent
forces for this problem, the 298 lb weight of the island acting at the
center of gravity (star), the 50 lb weight of the overhang acting at the
center of gravity (5.5" out), the force of the floor on the front casters (N _{2} ),
and the force of the floor acting on the rear wheels (N _{1} ).
(Ignore the force F for now.) Newton's first law stipulates
that the sum of all the forces must be zero, so N _{1} +N _{2} =348.
Also required for equilibrium is that the sum of torques about any axis must
be zero; choosing to sum the torques about the front casters, 26N _{1} -298x13+50x5.5=0=26N _{1} -3596
or N _{1} =138 lb and so N _{2} =210 lb. Now,
let's think about this answer: it tells you that this (unladen) island will
not tip over because there is still a lot of weight on the rear wheels. Now,
if you start adding weight to the overhang, eventually when you have added
enough weight, the force N _{1 } will equal zero when it is
just about to tip over. So, add the force F at the outermost
edge of the overhang and find F when the island is just about to tip:
again summing torques about the front casters, -298x13+50x5.5+11F =0
or F =327 lb. This is the extreme situation—you would have to put
twice this amount of weight, for example, halfway out the overhang to tip it
over. It looks to me that this will be safe for everyday use.

FOLLOWUP QUESTION:
The questioner sent an extremely lengthy recalculation of distances and weights. The only substantive changes were: weight of island without overhang,
W =414 lb; weight of overhang w =58 lb; distances of casters from sides
d =5.5"; distance of center of gravity from back side D =14".

ANSWER:
The relevant equations to redo the calculations for the unladen
island in the original
answer are
N_{1} +N _{2} =W+w and
(26-2d )N _{1} -(26-D -d )W +(d +5.5)w =0.
I find N _{1} =137 lb, so the unladen island will not
tip. For the second part of the calculation, adding the force F
such that N _{1} =0, the relevant equation is
(11+d )F -(26-D -d )W +(d +5.5)w =0.

I find F =124 lb at the outer edge to tip the island. Again, a force
of 248 lb in the center of the overhang would tip it. These forces are
considerably smaller than the original calculation mainly because of the
relocation of the casters closer to the center of gravity and, to a lesser
extent, the moving forward of the center of gravity. Incidentally, it may
seem that the vertical positon of the center of gravity would matter. It
does not matter if the island does not tip; if it does begin to tip, it will
tip faster if the center of gravity is higher, but not sooner. Since there
is no chance that the island will tip sideways, the location of the center
of gravity along the long edge is not relevant. Finally, if you are
uncomfortable with the amount of weight to tip, I presume that you could add
weight inside and near the back-bottom edge as a counter balance. For
example, putting 100 lb at the back edge would increase F to 249 lb.
(Anything you do which shifts the overall center of gravity closer to the
back side will help stabilize against tipping.) You should also be aware
that if you move it by pushing forward at the top of the back side, the
resulting torque could tip it over, so be careful when moving it. I believe
that if it were mine, I would look for a design modification which would
move the front casters closer to the front.

ADDED NOTE:
The questioner added some plans, one of which is shown to the left. Note
that there is a kick plate recessed by 2" all around. Presumably this plate
hides the casters and is very close to the floor. Therefore, as soon as the
island would start to tip, these would become the pivot point rather than
the caster itself which would make make it less vulnerable to tipping.
Essentially, in the calculations above you would reduce d by 2" to
d =3.5". This would result in a value of F =207 lb, quite an
improvement. You can see why I indicated above that moving the casters in by
5.5" was the main culprit in increasing "tippiness". (I am curious how, if
the kick plates hide the casters, you are able to lock them.)

QUESTION::
I'm trying to compare 2 measurements that denote impact. The
first measurement is as follows : a glass sheet can withstand a 25mm (diameter)
steel ball fired at 80 km's per hour. The second measurement is that another
sheet of glass can withstand a 277 gram steel ball dropped from 1 metre in
height with a back wind of 60 meters per second. I am trying to bring the second measurement to a
measurement comparable to the first measurement.

ANSWER:
The thing which will matter is the linear momentum (p=mv )
each ball brings to the glass. The reason is that you are interested in how
much force each glass can withstand and the force is the rate of change of
momentum. I assume that each ball will spend about the same amount of time
during the collision
and will exert its force over the same area (very nearly a point for a
sphere on a plane), so whichever ball has the most momentum when it hits the
glass will indicate the glass with the greatest strength. I took the mass
density ρ of steel as 8000 kg/m^{3} . The volume of a sphere
is 4πR ^{3} /3, so m _{1} =4ρπR _{1} ^{3} /3=4∙8000∙π ∙0.0125^{3} /3=0.0654
kg. Since the speed is given, v _{1} =80 km/hr=22 m/s, we can
immediately write the momentum for #1, p _{1} =1.44 kg∙m/s.
Finding the speed of the second ball is a much more difficult problem. If it
were just falling, it would be trivial. But it is being pushed by the
downward wind which will make it speed up faster than just falling; so, it
is necessary to understand a little about air drag forces. For spheres of
normal speeds in air the force of friction is excellently approximated by
f =0.22D ^{2} u ^{2} where D is the
diameter and u is the speed of the ball relative to the air .
From the density and mass, I find the radius to be 0.0434 m so D =0.0868
m. So, when the ball is first dropped, it has two forces pointing down, its
own weight mg =0.277∙9.8=2.71 N and the wind force 0.22∙0.0868^{2} 60^{2} =5.97
N; the wind is more than twice the force as the weight. As it falls, it
speeds up and so the effect of the wind gets smaller. Newton's second law
for v _{2} , if you care, is now of the form dv _{2} /dt =g [1+((v _{w} -v_{2} )/v _{t} )^{2} ]
where v _{t} =√(mg /(0.22D _{2} ^{2} ))
and v _{w} is the speed of the wind. The solution to this
equation is v _{2} =v _{w} -v _{t} tan[tan^{-1} (v _{w} /v _{t} )-(gt /v _{t} )].
Putting in the numbers for this situation (v _{w} =60 m/s, v _{t} =40.5
m/s), the graph for the first 15 s, with and without wind, is shown to the
left. The behavior of this graph is interesting. The ball takes about 4 s to
reach a speed of 60 m/s; at that point the ball is at rest relative to the
air and so there is no air drag and the slope of the curve (which is
acceleration) is the same as the curve for no wind at all, as expected. Now,
if there were no wind it would take the ball about 0.45 s to fall 1 m and
the wind will surely get the ball there in a shorter time; therefore we are
really only interested in the first half second of the fall and this is
shown to the right. The acceleration over this time is very nearly uniform,
about a _{2} =11/0.4=27.5 m/s^{2} ; since this is just
an estimate, I will do the calculation assuming uniform acceleration rather
than doing the exact calculation to find the velocity after 1 m. I find that
the time to reach 1 m is t =√(2/27.5)=0.27 s and so v _{2} =at =7.4
m/s. Therefore, the momentum for #2 is p _{2} =m _{2} v _{2} =0.277∙7.4=2.1
kg∙m/s. Ball #2 is the winner!

ADDED
THOUGHTS:
In retrospect, I could have saved myself some work if I had
thought about the fact that in a time less than 0.45 s the acceleration
would be essentially constant and simply written 0.277a =2.71+5.97 —>
a =31.3 m/s^{2} which would give t =0.25 s and v _{2} =7.9
m/s and p _{2} =2.2 kg∙m/s. Secondly, I noticed that my solution for v _{2}
was incorrect for speeds greater than 60 m/s, so I deleted that part of the
graph. To satisfy my own curiosity, I solved the problem for speeds greater
than 4 s. Because at later times the speed of the ball is greater than the
speed of the wind, the air drag force switches to up rather down so the
analytical solution to the problem becomes different, v _{2} =v _{w} +v _{t} tanh(g (t- 4)/v _{t} );
this corrected calculation is shown in blue.
The complete solution is now graphed to the left. At large times, the
solution approaches v _{t} +v _{w} as expected
since v _{t} is the terminal velocity in still air.

QUESTION:
I have a question about the concept of angular momentum. Suppose a disc is attached to a pole and they're stationary in deep space. If we spin the disc, it's gonna have an angular momentum. Using the right thumb rule to determine the direction of which, does this mean the unit will start moving in said direction? Also, how does the spinning action produce a momentum perpendicular to the force we exerted to cause the spin?

ANSWER:
I am assuming the pole is attached axially, i.e . at the
center and perpendicular to the disk. Angular momentum is not a force and
therefore it will not cause the system to start moving. Also, it was not the
force you exerted which caused it to spin, it was the torque (see left), and
the direction of the torque τ is also given by the right hand rule such that the
angular momentum L points in the direction as the torque (see
right).

QUESTION:
I'm a young aeronautical student and I'm doing a project on small, basic fixed-wing aircraft. I need to include some explanation of how lift is generated. My teacher has vaguely mentioned the Bernoulli theory once or twice. However after some research, I found that the Bernoulli explanation is outdated. I've found other more accepted theories, only which are too complex for both my understanding and academic level. Could you provide a more accurate but relatively simplified explanation of lift?

ANSWER:
I wouldn't say "the Bernoulli explanation is outdated," it just
isn't the whole explanation. Basically, Newton's third law is responsible
for much of the lift in flight—the air coming off the trailing edge of the
wing is deflected down which means the wing exerted a downward force on it
which means it exerted an upward force on the wing. Books about flying
usually refer to this as "angle of attack." A more complete explanation may
be found in an
earlier answer . I can also recommend a book, Stick and Rudder by
Wolfgang Langewiesche.

QUESTION:
If it was possible for every vehicle in the world to point East, and at perfect timing, all accelerate at the same time, surely the torque that is put down on the Earth would affect it's spinning energy?

ANSWER:
Yet another chance to demonstrate what a tiny speck we are in
this universe! There are about a billion, 10^{9} , vehicles in the
world. Suppose the average mass is 1000 kg (about 3000 lb) and the average
acceleration is about 5 m/s^{2} (half of g); so the average force
per vehicle is about 5000 N and the total force on the earth is therefore
5x10^{12} N. Suppose the average vehicle is at about 45^{0}
latitude, so the net torque is about τ =5x10^{12} R /√2=2.3x10^{19}
N∙m where R =6.4x10^{6} m is the radius of the earth. The
angular acceleration is then given by α =τ /I where I
is the
moment of inertia of the earth , about 8x10^{37} kg∙m/s^{2} .
So, α =2.3x10^{19} /8x10^{37} =3x10^{-19}
radians/s^{2} =5x10^{-18} revolutions/s^{2} =4x10^{-8}
revolutions/day^{2} . If all your vehicles could maintain this
acceleration for a full minute, (certainly not possible), the length
of the day would shorten by 5x10^{-12} day=4x10^{-7} s.

QUESTION:
Can you explain the effect of rifling of the muzzle in rifles ?? I just know that the muzzling is done to impart spin to the bullet when it is fired ?? So, exactly how does the spin imparted to the bullet improves it's aiming accuracy or whatever it does ?

ANSWER:
An object, like the bullet, which is spinning about an axis
along its length has angular momentum. The angular momentum vector points
along the direction the bullet is flying. An important law in physics is the
conservation of angular momentum which says that the angular momentum of an
object never changes if there is no torque acting on it. So the angular
momentum will continue pointing along the path which means that the bullet
will not "tumble". It is the same principle which governs the spiral forward
pass in American football. Without spiraling, the tiniest asymetry in the
shape of the bullet will cause it to immediately start tumbling when it
leaves the barrel.

QUESTION:
With all the talk of new weapons using kinetic energy as the destructive force instead of traditional explosives; exactly how minimal could a warheads' mass be and what would its velocity need to be to create an "explosive" force of 5,000 lbs of high explosive?

ANSWER:
Funny, I have not heard "all the talk"! I did get one other
recent question similar to yours. Anyhow, I guess you want to compare the
kinetic energy of a projectile with the same energy as 5000 lb of TNT. 5000
lb is about 2.3 metric tons and the energy content of that amount of TNT is
about 10^{10} J. Since the kinetic energy of the projectile is Ѕmv ^{2} ,
there is no clear answer to your question because every mass would have a
different velocity to have the requisite amount of energy. The speed of a
near-earth satellite is about 8000 m/s and maybe such a satellite would be
used to launch your weapon, so let's use v =8000 m/s. Then, 10^{10} =Ѕm (8000)^{2}
or m =313 kg=690 lb.

QUESTION:
I'm helping a middle school student with his physics homework. One of the concepts was mass and inertia and he told me that his teacher said they are equal and are synonyms. I was under the impression that they do not mean the same thing, that they are proportional. Could you clarify?

ANSWER:
Here is what the teacher was talking about: There are two kinds
of mass, inertial mass and gravitational mass. Inertial mass quantifies how
much an object resists changing its motion if a force acts on it.
Gravitational mass quantifies how much a gravitational force affects it and
how much it gravitationally affects other gravitational masses. It turns out
that the two are identical and there is really only one mass. This
equivalence is a prediction of the theory of general relativity and is also
the reason that all objects have the same gravitational acceleration.
Inertia is usually a qualitative term which describes how resistant an
object is to being accelerated, which is what inertial mass does
quantitatively. However, it is not so unusual to use inertia synonymously
with mass or inertial mass.

QUESTION:
My friend and I are having a discussion about centrifugal force and whether or not it exists, is real, and/or if it is present in daily life. Can you expand our knowledge and settle this dispute? We want to know everything about centrifugal force and if you can help us out that would be much appreciated.

ANSWER:
The important concept to understand here is that Newton's laws,
which describe motion, are not always true. For example, suppose that you
hang a simple pendulum from the roof of a car. When you are standing still
or driving down a straight road with constant speed, the pendulum hangs
straight down. The forces on the pendulum are its own weight, which points
straight down, and the tension in the string which pulls straight up.
This pendulum is in equilibrium here in the car and so Newton's first law
tells you that the sum of all the forces must add up to zero and so the
tension in the string must be equal to the weight of the pendulum bob. Now
suppose that you smoothly accelerate; you will find that the pendulum swings
toward the rear and hangs at some angle rather than straight down. You will
look at that pendulum and say, it is just hanging there at rest in the car
and so it must be in "equilibrium". On the other hand, there is no way the
(not parallel) forces of the weight and the tension can add up to zero.
Newton's first law is a false law in this car! If the first law is true, you
are in an inertial frame of reference. If it is false, you are in a
noninertial frame of reference. Generally, it is easy to identify a
noninertial frame—it accelerates relative to an inertial frame. A frame of
reference which is rotating is a noninertial frame because an object moving
on a curved path has an acceleration even if its speed is constant because
the direction of its velocity is always changing. It is shown in any
elementary physics text that the magnitude of this acceleration is a=mv ^{2} /R ,
where m is its mass, v is its speed, and R is the
radius of the circle it is going around; the direction of the acceleration
vector is toward the center of the circle. The acceleration is called the
centripetal acceleration from the Latin verb peto which means
"I seek". The question is, is there any trick we can pull to force Newton's
laws to be true in a noninertial frame. The answer is yes; if you make up a
fictious force on any object of mass m which is F _{fictitious} =-ma
in the opposite direction as the acceleration, Newton's laws will work!
The centrifugal force is a fictitious force added so that you can
apply Newton's laws in rotating systems. (The Latin verb fugo means "I
flee".) To see examples of how fictitious forces work, see an earlier
question about
falling down in a bus which will link you to a question about a
car
rounding a curve and that will link to a
bicycle
leaning into a curve.

QUESTION:
Suppose a container is partially filled with a liquid. A small sphere made of a material whose density is less than the liquid is in equilibrium inside the liquid with the help of a thread such that one end of the thread is tied to the sphere and the other end to the bottom of the container. the whole apparatus is kept on a weighing machine. if the thread is cut, then will the reading of the weighing machine change?
Since the tension in the thread is an internal force, the reading should not change.
However the free body diagram suggests that the reading should change.

ANSWER:
When you refer to "…the free body diagram…", you must specify
the body. Solving problems like this are most often easiest if you make a
clever choice of body. I would choose the body as the
container+liquid+ball+thread (taken as weightless, probably); in that case
the only downward force is the weight of all three and the only upward force
is the scale, so the scale reads the total weight with no reference to
whether the thread is connected to the ball or not. You can make this
problem difficult by focusing on a different body, maybe the container, but if you
draw all your free-body diagrams correctly and apply Newton's third law, you
still get the same answer that the the scale reads the total weight. To the
left I have shown all the forces on each of the bodies: red is the tension in
the string which is also the force the string exerts on the sphere and the
on the container; light blue is the force the container and fluid exert on
each other; black represents the weight of each; green is the buoyant force
which is the force of the fluid on the sphere and the force of the sphere on
the fluid; purple is the force the scale exerts on the container.

FOLLOWUP QUESTION:
Suppose the sphere is accelerating upwards inside the fluid, under the influence of
the buoyant force. Now its acceleration is dependent upon the buoyant force, while the buoyant force is in turn dependent upon its acceleration. How do we precisely calculate these two quantities, at a particular instant of time?

ANSWER:
First of all, the buoyant force is not dependent on the
acceleration. As long as the sphere is fully submerged, the buoyant force (B )
is equal the the weight of the displaced fluid. But, you must also include
the drag force (D ) which the fluid exerts on the sphere and that is
not simple to include. But, most fluids have sufficiently large viscosity
that the sphere will quickly come to its terminal velocity and move upward
with constant speed. When that happens B-D-W =0 where W is the
weight of the sphere. If you really want to pursue this farther, the
simplest approximation for the drag is that it is proportional to the speed
of the sphere, Stokes's
law . In that case, a =(B-W-Cv )/m where C is a
constant . If you were
to neglect drag altogether, which would be a poor approximation for any real
fluid, the acceleration would be uniform, a =(B-W )/m.

QUESTION:
I recently saw an experiment on the popular show Mythbusters. The
experiment was: a truck moves in one direction at a constant velocity, carrying a canon facing the opposite direction. The canon then fires a football such that it will travel at the same speed at which the truck is going. Successfully performed by the Mythbusters team, the ball dropped straight down to the ground the moment it left the barrel, as theorized, to a remote observer. I wonder, since Earth rotates at about 470metres/s eastward at the equator, if a bullet is fired westward at the same speed but westward, will it drop like the football did? Although I highly doubt this is the case, my calculations so far prove it's theoretically plausible. Am I missing something? Please also elaborate if this is different from the truck-canon experiment.

ANSWER:
In the first example, both the canon and the truck were moving
relative to the ground and it was the ground relative to which the final
velocity was observed. So, the experiments are not equivalent unless the
rifle in the second example were moving east with a speed of 470 m/s. If the
rifle was at rest relative to the ground and the observer was not on the
ground but looking from space, she would observe the bullet drop straight
toward the center of the earth as the earth spun under it.

QUESTION:
I was discussing ballistics with someone the other day, and a thought came up about a detail that I am not 100% sure on.
With bullets designed to expand, generally speaking materials hardness and velocity determine that expansion. What I am wondering, is does the bullet's energy upon striking the target cause the expansion? Or is the expansion caused by the opposite forces imposed on the bullet by what it is striking? (Or would one say it is a combination thereof?)

ANSWER:
The thing to appreciate is that energy and force are not two
separate things. Everything you need to know about collisions is contained
in Newton's three laws. The idea of energy often makes problems easier to
solve or understand, but the first step in developing the formalism of
energy, called the work-energy theorem, is just Newton's second law "in
disguise". So, I will discuss the bullet collision both ways:

The bullet hits a wall and stops. What stops it? The force which the
wall exerts on the bullet stops it. It begins expanding when it first
touches the wall; a point off axis will start moving perpendicular to
the direction of the bullet's velocity which has to mean that it is
feeling forces from other parts of the bullet since that point is not
touching the wall.

The bullet has
kinetic energy when it hits the wall. After it stops, that energy is
gone. Where did it go? Part of the energy went into doing the work
necessary to "squish" the bullet and part was lost to internal friction
intrinsic in squishing something soft but not elastic and ends up as
thermal energy—the squished bullet is hot.

QUESTION:
I am trying to help my son understand velocity, but find myself confused.

Person A drives in a circle. A physics website tells me this represents acceleration, a change in velocity, because his direction is constantly changing even though his speed may not be. Fair enough.

Person B takes a step forward and then a step backward to his original position. A different physics website tells me that this represents zero velocity, no acceleration, because he has not changed position.

But these answers seem contradictory, because person A, driving in a circle, will arrive at his original position at some point. In this respect, he is no different than person B, and could be considered zero velocity.
Can't both of these examples be considered changes in velocity? I suppose it depends on the timeframe you use to measure the change in position (?). So the person driving in a continuous circle can be considered to not be accelerating in some cases? This makes no sense.

ANSWER:
The first example refers to the instantaneous velocity
and the instantaneous acceleration of person A; instantaneous refers
to an instant in time and both acceleration and velocity are continuously
changing. The second example refers to average velocity and
average acceleration. Average refers to the value of the quantity
averaged over some time period and the time period here is the time from
when he first stepped forward until he finished stepping back. Suppose that
was 10 s and the length of his step was 1 m; then the average velocity is
distance traveled divided by the time, 0/10=0 m/s and the average
acceleration was the change in velocity divided by the time, 0/10=0 m/s^{2} .
Person A's average velocity and acceleration over exactly one time around
would also be zero.

QUESTION:
If two neutrons (just for the sake of ignoring charge) were separated from each other 1 light year, how long would it take for them to "touch" each other based on their gravitational attraction only? They are also in complete isolation from the rest of the universe.

ANSWER:
This is a very strange question. I have answered a nearly
identical question before but with much larger masses and smaller distances,
but the method is identical so I refer you
there . For your masses I calculate about 14x10^{18} years, about
a billion times the age of the universe. (I would also like to add that I do
not believe that this should be taken too seriously because no theory of
gravity has been accurately tested for either such large distances or such
small masses.)

QUESTION:
I am a fabricator and I currently have a task where I am attempting to create a braking system for downhill rapid propulsion (downhill racing).
Although the product exists, it is primitive and not fit for extreme measures, reliabilty, or convenience. There are many variables besides weight, drag coefficient, mass and gravitational acceleration. I would greatly appreciate your professional advice on creating a formula in which
I could create, change or gauge different systems for different masses. I have done much research and am increasingly frustrated yet interested.
I have come too some conclusions and have run many tests. I am using a polyester blend material for the canopy that is expansive and durable yet retractable, however. My cable system and my rapid cut down on drag are a problem. So my question is if
I weigh 160 pounds I am traveling at a speed between 30-60 miles per hour (we will say 45mph) at a down grade of 45% and i would like a slowing to 10 miles per hour between 40-50 feet from deployment. What would my initial area of my canopy be and what would the tensil strength of my cable need to be set at?
There are 3 points of contact for the cable system two high and one low.

ANSWER:
First, all air drag calculations are approximate and without
extremely complex computer simulations you can only do order-of-magnitude
calculations. Since you do not mention any sliding or rolling friction of
whatever is going down the incline, I will assume they are negligible. There
are, therefore, two forces on the mass, the gravitational force down the
incline mg sinθ and the drag of your "canopy" up the incline
which I will take as c _{2} v ^{2} ; here v
is the speed, m the mass, g the acceleration due to gravity,
θ the angle relative to the horizontal, and c _{2} is a
constant determined by the geometry of your canopy. A reasonable
approximation for c _{2} is c _{2} ≈јA
where A is the area presented to the direction of motion of the
canopy (only valid in SI units). Since I am a scientist, I will work
entirely in SI units here. Newton's second law, which governs the motion of
this system, is m dv /dt =-mg sinθ +јAv ^{2}
or dv /dt=g sinθ [1-(v ^{2} /v _{t} ^{2} )]
where v _{t} ^{2} =mg sinθ /c _{2} =4mg sinθ /A ;
v _{t} is called the terminal velocity, the speed to which the
mass will slow as it goes forward. Solving the differential equation (this
is worked out in any intermediate-level classical mechanics book), the
following equation is found: v ^{2} =v _{t} ^{2} (1-exp(-2gx sinθ /v _{t} ^{2} )+v _{0} ^{2} exp(-2gx sinθ /v _{t} ^{2} )
where x is the distance traveled and v _{0} was the
speed where x =0. That is everything you need since you know
everything except A . And you may want to use a fancier value for c _{2
} more tailored to the details of your canopy. As an example, I will use
your numbers: m =160 lb=73 kg, v _{0} =45 mph=20 m/s,
v =10 mph=4.5 m/s, x =45 ft=13.7 m, θ= 45^{0} .
Putting these in, I find 20≈(2000/A )(1-exp(-0.094A ))+400∙exp(-0.094A )
or A (0.05-exp(-0.094∙A ))=5(1-exp(-0.094∙A )). Someone
more clever than I could probably solve this analytically for A , but
I will just solve it numerically by plotting the left and right sides of the
equation and finding the intersection (see inset figure on the left). I find
that A ≈100 m^{2} ≈1000 ft^{2} ; this would be a square
about 30 ft on a side. (Since A is so large, one could have easily
solved this by simply neglecting the exponential functions, 0.05A ≈5.)
Regarding the strength of the cables, since I do not have any details about
the design of the canopy, the best I can do is tell you the maximum force
the canopy would exert on the mass via the three cables. Since the
acceleration is -g sinθ +јAv ^{2} /m , the
greatest acceleration is when the velocity is greatest, when the braking
initiates. So, the force the cables must exert is -mg sinθ +јAv _{0} ^{2
} or -mg sinθ +c _{2} v ^{2} . For
your specific example, this force would be about 9500 N≈2100 lb or roughly
700 lb/cable.

The figure on the right shows the solution I have come up with. Indeed it
begins at 45 mph and drops to 10 mph at 45 ft. However, one might just as well
say that the speed is also just about 10 mph at 25 ft, just not exactly.
That is the problem with analytical solutions sometimes—they demand
exactness. To me this graph says that you could get away with a
significantly smaller canopy and still qualitatively achieve your goal. I
think I have done enough here setting stuff up and you could proceed and
investigate how much things would change if you changed your speed at 45 ft
to be 11 mph, e.g. And, don't forget, these are approximate solutions
to be used as a rough guide.

QUESTION:
We can use the work-energy theorem in any inertial frame of reference.
When no external force is applied, and there is no change of height, the change in KE = -(Change in PE(spr)).
However, KE change depends upon the frame of reference and the extension of a spring does not depend upon frame's choice! HOW IS THIS POSSIBLE?

ANSWER:
The extension of the spring is the same in both frames, but the
work done by the spring is not because the same force acts over a different
distance in the moving frame. Forget potential energy and simply write ΔK=W
where W is the work done by any conservative external force, W =_{0} ∫ ^{X} F (x )dx.
I have chosen the starting position as x =0; to keep the algebra
simple, I will also choose the starting velocity in this (x ) frame to
be 0 and the final velocity to be V , so ΔK= ЅMV ^{2} =W .
Suppose that it takes time t to reach the position X . Now,
suppose there is another reference frame (x' ) which has a speed U
in the +x direction and x' =0 at t =0 also. Then
x'=x-Ut and the initial velocity is U and the final velocity is
U-V in this frame, so ΔK'= ЅM (V-U )^{2} -ЅMU ^{2} =ЅMV ^{2} -MUV.
Finally, calculate the work done in the moving frame: W' =_{0} ∫ ^{X} ^{'} F (x' )dx' =_{0} ∫ ^{X-Ut} F (x-Ut )(dx -U dt )=W =_{0} ∫ ^{X} F (x )dx-U _{0} ∫ ^{t} F (t )dt.
The second integral is the impulse which is the change in momentum,
MV , so W '=_{0} ∫ ^{X} F (x )dx-MUV=W-MUV.
Putting it all together, ΔK'=W'= ЅMV ^{2} -MUV =W-MUV
or ЅMV ^{2} =W.

QUESTION:
When Cavendish calculated the value of universal
gravitational constant he used mass of lead balls as reference. But how did
he know mass of the lead ball if he doest know the value of G?

ANSWER:
He could simply weigh them because we know W=mg . Since
you can also write W=mM _{earth} G /R _{earth} ^{2} ,
you can identify g =M _{earth} G /R _{earth} ^{2} ,
and g is easy to measure even if G is not.

FOLLWOUP QUESTION:
How did he build a weighing scale.
When weight is determined using a scale it should have been built w.r.t certain standard mass. But with out knowing value of G there is no way
they had a standard mass. He could have weighed a 10kg mass as 1kg mass. So, how did he weight exact mass.

ANSWER:
The kilogram was officially defined in 1795 as the mass of 1
liter of water. The Cavendish experiment was performed in 1797-98. Even if
the kilogram were not defined, there were other mass definitions which
Cavendish could have used. As an example, let me consider the oldest
standard weight I could find reference to, the beqa (b) (shown in the figure) defined
about 5000 years ago as the mass of 200 grains of barley corn which is about
6.1 grams=6.1x10^{-3} kg. So if Cavendish had used the b as his mass
standard, he would have found G =6.67x10^{-11} kg^{2} /(N∙m^{2} )=6.67x10^{-11}
[kg∙s^{2} /m^{3} ]x[1 b/6.1x10^{-3} kg)=4.07x10^{-13}
b∙s^{2} /m^{3} ] (assuming that he used seconds and meters for
time and length). It is a different number but means exactly the same thing.

QUESTION:
A ball thrown upward has zero velocity at its highest point i.e no acceleration. The resultant of applied force and weight/gravity is also zero. Thus the body is at rest. Is it in equilibrium too? This is not a homework question. I and my friend are too confused.

ANSWER:
Right off the bat, your first sentence is incorrect.
Acceleration of something at rest is not necessarily zero. You cannot
determine whether an object is accelerating by knowing only its velocity
because acceleration is the way the velocity is changing which cannot be
known by simply knowing what the velocity is right now. In your example, the
ball is at rest right now but was moving upwards just before now and will be
moving down just after now. The ball has an acceleration which happens to be
9.8 m/s^{2} ; this means that one second before now it was moving
upwards with velocity 9.8 m/s and one second after now it will be moving
downward with velocity 9.8 m/s. It is not in equilibrium because Newton's
second law tells you that any object with a net force on it is not in
equilibrium; the only force (ignoring air drag) on the ball is its own
weight and this force is the source of its acceleration.

QUESTION:
Does mass affect kinetic energy?

ANSWER:
Of course. Classically the kinetic energy K =Ѕmv ^{2 } where
m is mass and v is speed. Relativistically, K=E-m _{0} c ^{2}
where c is the speed of light, m _{0} is the mass of
the object when not moving, the total energy E =√(p ^{2} c ^{2} +m _{0} ^{2} c ^{4} ),
and the linear momentum p =m _{0} v /√(1-v ^{2} /c ^{2} ).
So, you see, mass appears all over the place. The momentum of a massless
particle, the photon, is p=E /c ; its energy is all kinetic, and
K=E=pc
since a photon has momentum even though it has no mass.

QUESTION:
I am currently doing a research paper on the perfect free-kick, could you
find an equation that suits the following variables? The soccer ball is
kicked from the origin of a coordinate system with an unknown velocity such
that it passes through the points (x,y )=(9.15
m, 2.25 m) and (x,y )=(22.3 m, 2.22 m). How can I find the magnitude
and direction of the initial velocity? Just having an equation to help me work with would be very nice.

ANSWER:
The equations of motion for a projectile which has an initial
velocity with magnitude v _{0} and angle relative to the
horizontal θ are x=v _{0x} t and y=v _{0y} t -Ѕgt ^{2}
where v _{0x} =v _{0} cosθ , v _{0y} =v _{0} sinθ ,
t is the time, and g =9.8 m/s^{2} . Solving the x -equation
for t , t=x /v _{0x} ; putting t into the
y -equation, y =(v _{0y} /v _{0x} )x -Ѕg (x /v _{0x} )^{2} .
Since you have two (x,y ) data points, you have two equations with two
unknowns, (v _{0x} ,v _{0y} ). The algebra is
tedious, but the result is that v _{0x} =21.0 m/s and v _{0y} =7.30
m/s; v _{0} =22.2 m/s, θ =19.2^{0} . To check my
answer I drew the graph shown to the right (note the different x and y
scales); it looks like my solution passes pretty close to the data points.

QUESTION:
As when sudden brakes are applied during riding a fast moving bike the back wheel leaves the ground. Why? Also what must be the conditions for the wheel
to not leave ground and when bike semicirculy revolves?

ANSWER:
To the right is the "free-body diagram" showing the pertinent
forces if the back wheel has not left the ground. The weight mg acts
at the center of gravity of the bike+rider and each wheel has a normal force
(N ) and frictional force (f ) from the ground. The bike has an
acceleration a in the direction of the frictional forces, so f _{1} +f _{2} =ma .
The system is in equilibrium in the vertical direction, so N _{1} +N _{2} -mg= 0.
The bike is also in rotational equilibrium so all the torques about any axis
must be zero; summing torques about the front axle, Rf _{1} +Rf _{2} +DN _{2} -dmg =0
where R is the radius of the wheel, D is the distance between
axles, and d is the horizontal distance between the front axle and
the center of gravity. Now, suppose that the rear wheel is just about to
leave the ground; then N _{2} =f _{2} =0.
The three equations then become N _{1} -mg= 0, f _{1} =ma , and Rf _{1} -dmg =0.
Putting the second equation into the third and solving for the acceleration,
a=g (d /R ); if you are slowing down any faster than this,
your rear wheel will lift off the ground and the bike will no longer be in
rotational equilibrium. If a is really big, you will keep rotating
until your center of gravity is forward of the front axle; then you will not
be able to stop the bike from rotating all the way over and crashing you on
the ground. This actually happened to me once when I was mountain biking
with my son and I broke a couple of ribs! I do not understand your second
question.

QUESTION:
My husband is employed as a school bus driver. He repeatedly tells the middle and HS students not to stand in the aisle until the bus comes to a complete stop. He tries to explain to them (without knowing the exact physics behind it) that if they're standing while the bus is moving at say 40 mph, and he has to stop short for some reason, they will not be able to stop their bodies from propelling forward. They never seem to "get it". How can he explain this from a physics point of view that they might better "get"?

ANSWER:
Are you kidding? You actually think a bunch of rowdy kids is
going to listen to a physics lecture on the bus? If you want a detailed
description of the physics of this exact situation, I have given it in an
earlier answer . Here is a
suggestion which will maybe work: take a tall box with little width, maybe 5
feet tall and 1x1 feet at the base. Have everyone in the bus sit down and
have one student stand the box in the center of the aisle when the bus is
moving at about 40 mph. Have your husband then stop abruptly (not so quickly
that the sitting students might hit their heads) and watch what happens to
the box. Better yet, get a mannequin to stand in the aisle. You could also
put a little cart with wheels in the aisle which would go zipping to the
front of the bus when it stopped. The physical principle is inertia: an
object in motion tends to stay in motion unless acted on by some force (like
if you were holding on to something).

QUESTION:
A person is standing on a platform. Below him is another person, not wearing a hard hat. The person on the platform drops a standard sized marble. Approximately how high would the platform have to be in order for a marble to cause trauma/injury?

ANSWER:
This is a little tricky because there is no such thing as a
"standard sized marble". I figured an average marble would have a diameter
of about 1 cm (R =0.5 cm=0.5x10^{-2} m) and a mass of about
M =1 g=10^{-3} kg. The second tricky part is that I do not know
how fast a marble needs to be going in order to penetrate the skull which is
how I would judge whether serious injury resulted. I do know that a bullet
which has roughly 10 times the mass of a marble will penetrate the skull at
a speed of about 60 m/s (about 130 mph). So, I would conclude that a marble
would have to be going quite a bit faster to penetrate the skull since it
would bring in a lot less momentum than a bullet with the same speed. So the
first question I will ask is what will be the speed if I drop it from a
height of h =5 miles (about 8000 m). Now, if you were just going to do
a simple introductory physics problem you would say "neglecting air
friction" the time to fall would be about t =√(2h /g )=√(16,000/10)=40
s and the speed at the ground would be about v=gt =10x40=400 m/s
(about 900 mph); I would guess that would do some serious damage! But wait!
It would be a really big mistake to say that air friction was negligible for
something going 900 mph. The drag on a falling object in air depends on how
fast it is going and can be approximated as F ≈јAv ^{2 }
(all quantities must be in SI units) where A=πR ^{2} ≈2x10^{-5}
m^{2} is the cross sectional area. When the drag is equal to the
weight, F=mg ≈10^{-2} N, the marble will stop accelerating and
continue falling with the terminal velocity, v _{t} =√(4mg /A )≈45
m/s (about 100 mph). So, since 45 m/s is much less than the 60 m/s necessary
for a bullet to penetrate the skull, I am guessing that no matter how high
you drop the marble from, it will not cause truly serious injury—it's going
to hurt though!

QUESTION:
Here is the scenario. I have a A volt motor that supplies B watt
to work. I connect the motor to a circular disc of R radius that is of M
mass and has a density of 1 unit. I want to use the motor to move a conveyor
belt of mass N. I will place an object of mass K on the conveyor belt. Is it
possible to find the velocity of which the object will be moving at given
the voltage? Or perhaps can we find the minimum about of voltage that is
needed to move the object. Assume that the efficiency of all the motor as
well as the whole thing is 100% and that frictional force and air resistance
can be neglected. You can add othervariables in either.

ANSWER:
This is an engineering problem, not physics.
However, it sounds to me like there is no answer if you have no frictional
losses because the motor will just keep adding energy to the system and
there is no loss.

FOLLOWUP QUESTION:
However, it should be able to work since the motor is attached to a conveyor belt, yes it will keep going but it will kep moving the conveyor belt only.

ANSWER:
If the motor is adding energy (B watts would mean B Joules of energy per second are being added) and there are no losses, where is that energy going? It will constantly accelerate the conveyer belt.

FOLLOWUP QUESTION:
Is it possible to find the acceleration of the conveyor belt?

ANSWER:
Yes! Refer to the picture above. Suppose the mass and the
conveyor belt move with speed V ; then the angular velocity of the
disc is ω=V /R . The moment of inertia of the disk is I =ЅMR ^{2} .
The kinetic energy of the belt plus mass is KE _{1} =Ѕ(K+N )V ^{2}
and of the disc is KE _{2} =ЅIω ^{2} =јMV ^{2} ,
so the total kinetic energy is KE =Ѕ(ЅM+K+N )V ^{2} .
The rate of change of the kinetic energy is equal to the power B ,
B =(ЅM+K+N )V (dV /dt )=(ЅM+K+N )Va
where a is the acceleration. Solving, a =B /[(ЅM+K+N )V ]≡C /V .
Note that when V =0, a =∞; this just means that the motor cannot
deliver power unless it is actually spinning with some rate. But, as soon as
it gets moving it will have a very large acceleration which decreases as the
speed gets bigger and bigger. So, if you are really interested in such a
system, you will have to determine energy losses and probably also have a
motor with an adjustable power output. (Also, note that the voltage has
nothing to do with it, all you need is the power output.)

QUESTION:
If you have two weights, where one is 100 grams and the other is 200 grams and you use the same spring to create a pendulum with them (one at a time) why isn't the amplitude the double for the heavier weight?
(This is the result of an exercise we did, my personal curiosity, and my teacher's unwillingness to explain
because he says it's too complicated for our level.)

ANSWER:
I think you must surely mean a mass on a spring oscillating
vertically, not a pendulum. That is what I will assume. You have probably
not studied energy yet which is why your teacher did not want to get into
it. Energy methods are, by far, the easiest way to answer your question
which is what I will do. Two things you need to know: potential energy of a
spring which is stretched by an amount y is Ѕky ^{2}
and gravitational potential energy of something a distance y above
where y =0 is mgy . Imagine taking a mass m and attaching
it to a spring with spring constant k which is unstretched and
holding it there for a moment. Where it is right now I will define to be
y =0 and so, since the spring is unstretched, the total energy of the
system is Ѕk∙ 0^{2} +mg∙ 0=0;
the energy never changes. As it falls it speeds up for a while and then
slows down for a while (acquiring and then losing what is called kinetic
energy , energy by virtue of motion), finally stopping and going back up.
If it has fallen some distance A (for amplitude) before turning
around, the energy is now 0=ЅkA ^{2} +mg (-A )
and so we find two solutions (it is a quadratic equation), A =0 (we
already knew it was at rest there) and A =√(2mg /k ). So,
you see, the amplitude is proportional to the square root of the mass, not
the mass. (Incidentally, most folks would call the amplitude of this
oscillation to be ЅA . I thought it would be clearer this way.)

QUESTION:
Is it possible to derive the formula for Kinetic Energy without using work? Or are they linked by definition?

ANSWER:
Well, I can tell you that you never have to utter the phrase
"work is defined as…" The work-energy theorem is merely the integral form of
Newton's second law. For simplicity, I will do this simple derivation in one
dimension. F=ma=m (dv /dt )=m (dv /dt )(dx /dx )=m (dv /dx )(dx /dt )=mv dv /dx .
Rearrange: F dx =mv dv . Integrate:
∫ F dx =m ∫ v dv =Ѕmv _{2} ^{2} -Ѕmv _{1} ^{2}
where the integral on the left is from x _{1} to x _{2}
and is usually called the work W . Generalizing to three dimensions,
∫ F ∙dr =Ѕmv _{2} ^{2} -Ѕmv _{1} ^{2}
where the integral on the left is from r _{1} to
r _{2} .

QUESTION:
I came across your site while looking for the answer of a physics problem I would like to program in a Smartphone app.
A toy car is dragged from position A to B giving it an initial velocity (v=d/t). At point B the car is released where it travels in a straight line until the frictional force of the ground stops the toy car completely (position C). I would like to find a formula that relates the distance from B to C with the initial velocity provided from A to B.

ANSWER:
Of course, you have to know the frictional force f and
the mass m of the car in general. Assuming that the force is constant
all along the path B to C, the distance s can be written s =Ѕmv ^{2} /f.
In most cases f is proportional to m so that you do not
need to know the mass. If you are on level ground, f=μmg where μ is called
the coefficient of friction and g =9.8 m/s^{2} is the
acceleration due to gravity (mg is the weight of the car). The
coefficient just is a parameter which is small if there is little friction.
So, finally you have s=v ^{2} /(2μg ). If the car is on a
slope making an angle θ with the horizontal, s=v ^{2} /(2μg∙ cosθ ).

QUESTION:
Okay I have wondered this question for many years, and more over have
wondered how to ask it! So I think a scenario is best; when an object is
propelled forward or in any direction very rapidly, let's say for this case
a bullet out of a gun, does it advance through every speed in between it's
current speed, zero, and it's maximum velocity or does it simply "jump" from
zero to it's maximum velocity? And also the same for when it hits its
target, let's say a thick steel plate does it go from its current velocity
straight to zero, or is there some sort of slowing down in which it goes
through every speed in between?

ANSWER:
Fundamental to classical physics is Newton's second law. That
which changes the speed v of something is a force F , a push or
a pull. Further, the bigger the rate of change of speed (acceleration a ),
the bigger the force—double the force and you double the acceleration. This
is often written as F=ma where m is the mass of the object. Suppose
that you shoot a bullet from a gun. The bullet starts with a speed v =0
and ends with speed v and this happens in some time t and so
the acceleration can be written as a=v/t . If, as you suggest, the
bullet "jumps" instantaneously to v , then t =0. But for t =0
the acceleration would be infinitely large which would imply that you had to
push on it with an infinite force. The same reasoning can be applied to
stopping the bullet. You know that the force propelling or stopping a bullet
in the real world is finite. It is a pretty good rule-of-thumb for everyday
occurances that there are no infinities or discontinuities (instantaneous
changes) in the universe.

QUESTION:
Please help a work related dispute. Can the weight of a patient in a wheelchair be calculated using f=Ma? Even if it's a guesstimate?
I thought, if m= F/a
Where a = V_{2} -V_{1} / t Where V_{2} is the velocity of the chair (and porter?) and V_{1} is coming to a stop.
Would F be the mass of the porter multiplied by the common acceleration of chair and porter?
I'm guessing if I clock the speed of the porter I can calculate a.

ANSWER:
I guess that "porter" means the guy pushing the chair. You have
it all wrong, I am afraid. But, it is worth talking about for a bit. There
are three masses involved here, M _{patient} , M _{chair} ,
and M _{porter} . There is an acceleration a which we
can agree could be roughly measured by measuring times and distances.
Suppose we first look at all three and call the sum of their masses M .
Then M=F _{all} /a where F _{all} is the
force which is causing the collective mass to stop. If you neglect the
friction which would eventually stop the wheelchair with no porter, F _{all}
is the frictional force between the porter's feet and the floor; you do not
know that force. Suppose you focus your attention on the patient. M _{patient} =F _{patient} /a
where F _{patient}
is the force responsible for stopping the patient. This would be the
frictional force the seat of the chair exerts on the patient's butt; you do
not know that force. Suppose you focus your attention on the
porter. M _{porter} =F _{porter} /a where
F _{porter} is the force responsible for stopping the porter.
If we call the force with which the chair pulls on the porter F _{c-p} ,
then F _{porter} =F _{all} -F _{c-p} ;
you do not know either of these forces. I could go on and focus on the chair
alone next, but you can see that you do not know any of the forces which are
responsible for stopping any or all of the masses, so you cannot infer the
mass of any of them. Just knowing the acceleration, you cannot infer the
mass.

QUESTION:
I am writing a couple of SF novels that take place in significant part on a beanstalk station, i.e. a space elevator station located in geosynchronous orbit, and my question concerns the design. In my stories, the station is about a half mile wide with a shape similar to a hockey puck. The center third of the puck is attached to the elevator ribbon (both up to the counterweight and down to the planet of course) and fixed, therefore essentially in freefall.
My question concerns the outer section, which in my story spins in order to provide some g-force: is this a viable model? Some of my concerns are that the spin will create a gyroscopic effect as the station orbits around the Earth that will twist it off the ribbon or perhaps exert other forces on the hub that I'm not accounting for. One possible solution I thought of would be to split the outer section into two rings rotating in opposite directions to cancel out those effects.
Assuming some clever engineer works out the minor details of moving between the hub and the rotating section(s) is this design plausible? I like the science in my science fiction to be accurate and reasonable.

ANSWER:
I believe you are right to worry about the rotation causing
problems. If you just forget about the elevator and have a rotating "puck
station" in orbit, its axis will always point in the same direction in space
(conservation of angular momentum); so, if it were in a geosychronous orbit,
from earth it would appear to do a 360^{0} flip every day. I think that having a
counter-rotating ring would be a very good idea, the ring having the same
angular momentum (moment of inertia times angular velocity) as the main
station. If the station has a radius of about a quarter of a mile, about 400
m, the angular velocity ω needed to have an earth-like artificial
gravity (g =9.8 m/s^{2} ) at the edge would be ω= √(g /R )=0.16
radians/s=0.025 revolutions/s≈1.5 rpm.

QUESTION:
I need a very brief explanation for Kindergarteners. (I just
want to use the right words; they can learn the details of what they mean
later.) In an amusement park ride in which long swings are spun around a
central pole, why do the swings rise up as the pole spins faster? (I assume
it's the same reason that a skirt rises up when the person wearing it
twirls.) Is it something to do with centrifugal force? Centripetal force?
Nothing I saw on the web about these forces seems to explain why the objects
rise, only why they move to the inside or outside of the orbit.

ANSWER:
For Kindergarteners, use of centrifugal force is probably best.
I guess the merry-go-rounds we had on playgrounds when I was a kid have been
deemed unsafe, but get your students to appreciate that the faster they are
spun on something akin to this, the harder it is to hang on. This is because
the centrifugal force which tries to push them off gets bigger as the
merry-go-round spins faster. Then you can just take a pendulum and
demonstrate that the harder you pull, the higher it rises. The little
diagram to the right shows the effect of the the centrifugal force F which will lift the pendulum bob higher as
it pulls harder.

QUESTION:
I have a question that been thats been unanswered from quite some time now. I even tried asking that to a person in NASA houston and he couldn't even understand. Hope you can help.
We know that space dust and debris keep on falling on earth. And I have read that its many tons a day. Considering mass of earth is increasing every day for millions of years now, how does it affect its revolution, speed and axis. Or does it even affect that.

ANSWER:
One estimate is that the earth gains 40,000 metric tons (4x10^{7}
kg) per year. So, in a million years that would be a gain of 4x10^{13}
kg. The mass of the earth is 6x10^{24} kg. The moment of inertia
will be proportional to the mass times the square of the radius, so,
assuming the change in radius is negligible, the fractional change in moment
of inertia over a million years would be approximately 4x10^{13} /6x10^{24} ≈10^{-9} =10^{-7}
%. Since angular momentum (moment of inertia times angular velocity) is
conserved, this would mean that the angular velocity would decrease by about
10^{-7} %, an increase in the length of a day of about (24 hr)(3600
s/hr)x10^{-9} ≈10^{-3}
seconds in a million years!

QUESTION:
I was explaining conservation of energy to my daughter when she was spinning on our office chair, and her rotational velocity increased as her moment of inertia decreased when she pulled her arms and legs toward the rotational axis. Is there an analogous example for linear motion, where the linear velocity increases as the mass decreases? I can't think of a real-world example of a body whose mass decreases (or increases).

ANSWER:
Actually, you were demonstrating conservation of angular
momentum, the product of the angular velocity and moment of inertia remains
constant for an isolated system. Energy is not conserved because rotational
energy is proportional to the product of the square of the angular velocity and moment of inertia.
But, you explained it correctly for your example. So, your question is if it
is possible for conservation linear momentum, the product of mass times
velocity, to result in a changed velocity due to a change of mass. You do
not usually see as many examples of this as for angular momentum because
rotating things change shape frequently whereas moving objects usually do
not have significant change of mass. The classic example is a conveyer belt
onto which mass is being dropped from a hopper. If you had a very long
frictionless conveyer belt with mass on it, it would have a certain linear
momentum. Now, if you start dropping mass on it, it will slow down.
Similarly, if you let mass drop off the end of the conveyer belt without
replacing it, it will speed up.

QUESTION:
How long does it take to stop a 7000 lbs vehicle at 45 mph? Time and distance please.

ANSWER:
You have not given me enough information, in particular what are
the wheels and the surface made of; surely you realize that a truck stopping
on ice will go much farther than a truck stopping on a dry road. I will work
it in general and then calculate it for a typical example. The quickest stop
you can affect is to apply the brakes hard enough that the wheels are just
about to start skidding; that is what anti-skid braking systems do.
Therefore the force F which is stopping you (on a level road) is the static
friction between the road and the wheels, F=μ _{s} W
where μ _{s} is the coefficient
of static friction between the wheels and the road, W=Mg is the
weight of the vehicle, M is the mass, and g= 32 ft/s^{2}
is the acceleration due to gravity. But, Newton's second law tells us that
also F=Ma where a is the acceleration of the vehicle.
Therefore, the acceleration is independent of the mass of the vehicle and
the acceleration is a =μ _{s} g.
Now that you have the acceleration you can write the equations of motion
for position x and velocity v as functions of time t :
x=v _{0} t -Ѕat ^{2} and v=v _{0} -at
where v _{0} =45 mph=66 ft/s. Now, you need to specify
what μ _{s} is . For example, μ _{s} ≈ 0.9 for rubber on
dry asphalt, so a≈ 0.9x32=28.8 ft/s^{2} and I find t≈ 66/28.8=2.3
s and x =66x2.3-Ѕx28.8x2.3^{2} ≈76 ft.

FOLLOWUP QUESTION:
I'm trying to figure out if a traffic light at an intersection is timed too short for a heavier vehicle to stop in time. The speed limit is 45mph, the vehicle weighs ~7000 lbs (7200 empty w/driver) , there is a downward slope of which I'm trying to find out. Yes to dry pavement. What would you need to figure surface area of rubber/tire? And I need to time the light still. With all of the above information would you be able to calculate that?

ANSWER:
As I showed above, the weight is irrelevant. Of course, this is
an approximation as all friction calculations are, but a quite good one for
this situation. It is also important that my calculation is the shortest
time and minimum distance, what you would get by flooring the brake pedal
with anti-skid braking operating. If you do not have anti-skid braking and
you lock your wheels, it will take longer and go farther. Also, there is the
possibility that μ _{s}
could be different from 0.9 depending on local conditions (temperature,
surface condition, etc .). It is important to include the slope in the
calculation. If the slope is down as you say, the acceleration (I am
assuming you are not interested in the details) is a =32(μ _{s} cosθ- sinθ )
where θ is the angle of the
slope; for example, if θ= 20^{0}
and μ _{s} = 0.9, a =32(0.9x0.94-0.34)=16.2 ft/s^{2} ,
quite a bit smaller than the value of 28.8 ft/s^{2}
for a level road. Once you calculate the
acceleration, the expressions you can use for time and distance are t =66/a
and x =2178/a , respectively. So, for a 20^{0} slope,
t =4.1 s and x =134 ft. Again, this is the lower limit. I think an
engineer would build in a factor of 2 safety factor. You do not need to know
what the surface area of the contact between the rubber and the road is.

QUESTION::
Nothing can accelerate itself by applying force on itself. Besides, isolated forces do not exist and they exist in an action reaction pair.
These are the essence of Newton's third law. When an engine exerts force on a bike how can it accelerate the bike taking bike as a system in this argument? After all, a part is exerting force on another part , right?

ANSWER:
The engine exerts a torque on a wheel trying to make it spin. If
the bike were on ice, the wheel would spin, there would be no acceleration
and therefore there must have been zero net force on the bike+engine as you
surmise. However, if there is friction between the wheel and the ground, the
ground exerts a force on the wheel which is forward; this is the force which
drives the bike forward. Note that the wheel exerts a force backward on the
ground (Newton's third law) but the ground does not move because it is, effectively, infinitely
massive.

QUESTION:
W hen a bullet hits a door and gets embeded in it, no external force acts on the system of door and bullet but why is linear momentum not conserved and the angular momentum conserved? Can you give some examples where angular momentum is conserved but not linear? Can it be possible to have both linear and angular momentum conserved?

ANSWER:
I presume you are alluding to the classic introductory physics
problem of a bullet hitting a door mounted on frictionless hinges. Angular
momentum is conserved if there are no external torques, and since hinges
cannot exert a torque, it is. Linear momentum is conserved if there are no
external forces but the hinges exert a force on the door during the
collision time and so it is not. If the hinges were not there, the door and
the bullet would move forward and rotate about their center of mass
conserving both linear and angular momentum.

QUESTION:
What is the change in velocity of the earth's rotation if a person (myself) who weighs 60 kg were to stand on something about a foot tall.
Ps this is not a homework question, I'm just a curious teen who's never taken physics. Also it's the middle of the summer for me in
New Orleans.

ANSWER:
The glib answer to this question would be, for all intents and
purposes, the change in rotation would be zero. It is a good opportunity
to talk about the physics involved and to estimate how small small is here. The moment of inertia of the earth is about I _{e} =8x10^{37}
kg∙m^{2} . Your moment of inertia if you are on the earth's
surface is about I _{y} =60x(6.4x10^{6} )^{2} =2.5x10^{15} kg∙m^{2} . The moment of the
earth plus you is I =I _{e} +I _{y} ≈I _{e} =8x10^{37} .
If you increase your distance by the amount 0.3 m, about 1 ft, your moment
of inertia increases to I _{y} +ΔI= 60x(6.4x10^{6} +0.3)^{2} =I _{y} (1+4.7x10^{-8} )^{2} ≈2.5x10^{15} (1+2x4.7x10^{-8} )=2.5x10^{15} +ΔI
and so ΔI= 9.4x10^{-8} kg∙m^{2} . So you and the earth
start with I= 8x10^{37}
and end with I +ΔI= 8x10^{37} +9.4x10^{-8} .
The operative physical principle here is conservation of angular
momentum, the product of moment of inertia and angular frequency ω= 2π /T
where T is the period, 24 hours: Iω= (I +ΔI )(ω+ Δω )=Iω+I Δω+ ΔIω+ ΔI Δω.
Neglecting ΔI Δω , Δω /ω=- ΔI /I=- 1.2x10^{-45} ;
note that since Δω /ω< 0, the frequency decreases, the rotation
slows down. Now, it is pretty easy to show that ΔT /T≈- Δω /ω= 1.2x10^{-45
} or the day gets longer by 24x1.2x10^{-45} =2.8x10^{-44}
hours! Your contribution to the earth's moment of inertia is so tiny that
anything you do to change your own moment of inertia will have no measurable
effect on the rotation of the earth.

QUESTION:
If an astronaut caught a ball in space, the ball would cause the astronaut to move backwards with the force the ball was moving at correct? then, as the astronaut is still moving backwards and throws the ball back where it came from, would the astronaut move even faster with the force of throwing the ball? or would their speed remain the same?

ANSWER:
When she catches the ball, she exerts a force on it to
stop it (relative to her); the ball exerts an equal and opposite force on
her causing her (and the caught ball) to move in the direction the ball was
originally moving. When she throws it back, she must exert a force on it
opposite the direction she is moving; the ball exerts an equal and opposite force on her
causing her to move even faster in the direction she was moving. These are
examples of Newton's third law.

QUESTION:
What is mass?

ANSWER:
There two kinds of mass. Inertial mass is the property an
object has which resists acceleration when a force is applied; the harder it
is to accelerate something, the more inertial mass it has. Gravitational
mass is the property an object has which allows it to feel and create
gravitational forces; for example, the more gravitational mass an object has
the greater the force it will feel due to the earth's gravity—the more it
will weigh. It turns out that the two masses are actually identical; this
fact is one of the cornerstones of the theory of general relativity.

QUESTION:
My question is for a legal nature. I was recently in a head on collision. The crime scene analysis could not determine the speed of my vehicle but police reports indicated the other vehicle was doing 45+- mph. We were on a dirt road on impact. The vehicle going 45+- continued to travel 20 ft past impact pushing my vehicle back 40 ft. His vehicle is a 1 1/2 ton dodge ram and my vehicle was a 1 3/4 ton dodge caravan. What speed was my vehicle? This will be instrumental in a lawsuit currently being filed. Your help would be greatly appreciated.

ANSWER:
I am afraid that the speed cannot be determined from this
information for several reasons:

I do not know how much energy
was lost in the collision.

I do not know whether the
wheels were locked (brakes applied) or not.

I do not know the frictional
forces which ultimately brought the vehicles to rest.

I believe the weight of the
Dodge Ram is wrong. The Dodge Ram 1500 is referred to as 1Ѕ
tons because this is the maximum recommended load. An unladen Ram 1500
weighs more than 6000 lb; I found that the weight of the Caravan is
about 1.75 tons, though.

QUESTION:
Friction opposes the relative motion between two surfaces. when a car travels on a circular path , how can the friction act sideways to provide necessary centripetal force. The friction should act backward relative to the motion of the car. the car doesn't tend to go sideways outward, then, how does friction act sideways?

ANSWER:
There are two important classes of friction, kinetic
friction which occurs when two surfaces are sliding on each other and static
friction when they are not sliding. Kinetic friction is the one which
usually (but not always) acts opposite the direction of motion; an example
of kinetic friction acting in the direction of motion is a car which is
accelerating from rest and spinning its wheels—the friction force on the
spinning wheels by the road is forward. Static friction can point in any
direction, depending on the situation. If a box is sitting at rest on an
incline, the frictional force points up the incline to keep it from sliding
down. If a car is moving but not skidding, the appropriate friction to think
about is the static friction between the wheels and the road. Think
about a very icy road; to drive around a curve at high speed is impossible
because there is no static friction and the car simply continues going
straight regardless of whether you turn the steering wheel or not.

QUESTION:
Gravitational potential energy is the term
that means the work done by the gravitational force to take an object to the gravitational field. Here the displacement is towards the force.So,it (Gravitational potential
energy) should be positive.But it is negative.Why?

ANSWER:
You need to be a little more careful in how you define
potential energy. And, what is actually defined is the potential energy
difference between two points in space. The definition is ΔU =U (r' )-U (r )=- _{r} ∫^{r'} F ∙ dr ,
where F is the force of gravity on m due to the
presence of M . Now, if I choose increasing r to be in the
upward direction,
F ∙ dr =-(MmG /r ^{2} )dr.
So, -_{r} ∫^{r'} F ∙ dr .=-_{r} ∫^{r'} [-(MmG /r ^{2} )]dr =-MmG [(1/r' )-(1/r )]=U (r' )-U (r ).
This is completely general. It is customary to choose U (∞)=0, so if
r' =∞, U (r )=-MmG /r. (You have to be very
careful of all these minus signs!) So, you see that the potential energy is
determined by where you choose it to be zero and the choice of coordinate
system; if we had chosen r to increase in the downward direction and
U =0 infinitely far away (at r' =-∞), U would have been
everywhere positive.

QUESTION:
ok, two stones. Both spherical and same mass and
density evenly spread in each stone. Set each about 1/2 the distance to the
moon. One leading the earth's orbit and one following the earth's orbit.
Both not moving relative to the earth, yet the same speed as the earth as it
moves around the sun. No tangent or orbital speed, the stones are starting
in freefall. Which gets to earth's surface first, neglecting
air drag.

ANSWER:
Since you stipulate that the stones are not
orbiting, the stones are at rest with respect to the sun and the earth is
not. Therefore, the stone on the leading side of the orbiting earth will win
the race because the earth is moving toward it and away from the other when
the stones start dropping.

BETTER
ANSWER:
I see that I misread this question. I guess you meant there is no orbital
speed around the earth. To make this manageable at all I will neglect the
influence of the moon and assume a
spherically-symmetric mass distribution of the earth . In the figure
above, I show your two stones and the forces (blue arrows) on them. The
down-pointing forces are the from the sun (keeping them in orbit) and the
horizontal forces are the weights making them want to fall toward the earth.
The distances and forces are not drawn to scale; when the stones are about
30 earth radii away from the earth center (the moon is about 60 earth radii
away), the weight forces are about 2 times larger than the sun forces. The
big blue arrow shows the direction everyone is orbiting the sun. So, the
leading stone slows down its orbital speed and so will slightly fall toward
the sun as it falls toward the earth; the trailing
stone increases its orbital speed and so will slightly fall away from the
sun as it falls toward earth. These deflections are shown (probably quite
exaggerated) by the red arrows. Given the symmetry of the situation, I
would expect the two to be at the same distance from the center of the earth
at any given time —which is the
crux of your question, I think. To actually do this more quantitatively,
though, would be very hard because as the stones were deflected the weight
force would change direction now having a vertical component in the figure.

QUESTION:
I have a 10' long trailer I am using to haul a 4,000# symetrical object that is 6 ft in length. I was told to move it forward of the axle on the trailer to put some weight on the tongue for safety and better hauling. The maximum wt my pick up truck can hold on its hitch is 650#. I have tried to find equations for this on-line as it would be very useful for me to know how to adjust other loads as well. There are plenty of equations on-line that deal with finding the CG, but I can't find one that discusses how the weight on the tongue changes as the load is moved fore or aft of the axle. Trailer/towing experts just wing it. Do you know of an equation that would help me?

ANSWER:
I will assume that the unladen trailer will have
approximately zero force on the hitch (which would mean that its center of
gravity (COG) is at the axle). You need to know where the COG of the load is; in your
specific case, you know that it is at the geometrical center (you said it is
symmetrical), 3' from either end. For loads not symmetrical, you need to
find it. I will call the distance between the axle and the hitch L .
Suppose that the COG is a distance x from the hitch. Then, the sum of
the torques about the axle must be zero, so Hx=W (L-x ) where
H is the force (up) by the hitch and W is the weight (down) of the load.
Solving this, x =[W /(H+W )])L . For your case, with
H being 650 lb (although I cannot see why the maximum would be the
optimal) x =(4000/4650)L =0.86L . If your
trailer has more than one axle or its COG is not over the axel, I would need
more information like the geometry and weight of the trailer.

QUESTION:
Here's a question that I have been pondering. If a truck is driving on the freeway and a car pulls in closely behind the moving truck to take advantage of the draft created by the truck is there an energy cost to the truck or is having a car in it's wake energy neutral?
My gut feeling is that there would be a slight energy cost to the truck due to it's turbulence wake being interfered with.

ANSWER:
This, I discovered, is not a trivial question. For a
lengthy discussion, see
The Naked Scientists . Here is my take on it. There is no question that
the trailing car consumes less gas. The reason for this is not so much that
the truck is pulling the car but that the car experiences a much lower air
drag when drafting; the drag is approximately proportional to the square of
the velocity and the truck's wake is moving forward with the truck. What
seems to be controversial is the crux of your question —is
there a cost to the truck? Some argue that the composite truck-car system
has less total air drag, others that there is a net cost to the truck which
need not (and almost certainly will not) equal the gain by the car. There is
certainly no conservation principle here because the new system has
different forces on it than the separate systems. My feeling it that there
is at least a small cost to the truck and I base this on an observation from
nature. Why do geese fly in a V? There is less overall air drag than if the
flock all flew individually. But periodically, the leader drops back and
another goose takes a turn at the front; must be because the leader has to
do more work.

QUESTION:
A
friend asked me this and we disagreed with the answer.
If we put 25 kg of weight on top of 25 kg person, how much force would he
feel?

ANSWER:
Technically, a kilogram is not a weight but a mass. But,
since so many countries use it as a weight, I will do that for this problem.
The person feels the downward force of her own weight, 25 kg; the downward
force of the object pushing down on her, 25 kg; and the upward force of the
floor pushing up on her, 50 kg. The net force is zero because she is in
equilibrium.

QUESTION:
If an ultra high energy cosmic ray with energy of 10^{20}
eV were to strike an astronaut will that kill an astronaut?

ANSWER:
This is only about 6 J of energy. That is the energy
needed to lift 1 kg about 60 cm. And, it would probably not leave all its
energy in the astronaut. Certainly would not kill her.

QUESTION:
If an ultra high energy cosmic ray with energy of 10^{20}
eV were to strike an astronaut will that kill an astronaut?

ANSWER:
This is only about 6 J of energy. That is the energy
needed to lift 1 kg about 60 cm. And, it would probably not leave all its
energy in the astronaut. Certainly would not kill her.

QUESTION:
I have a question releated to weight/mass placement on a bar. My friend and I are weight lifters. We got into a discussion about the center of gravity on the bar. Here is the question. If we are using a 45 pound plate on each side and also have a 5 and 10 on each side. Each taking up the same space and the end of the bar is the same distance from the last weight and will not change. Does it change anything if the weights are not in the same order, from one side to the other? My friend says the side with the 45 pound plate close to the end is slightly heavier becuase the ratio has changed. I say nothing has changed becuase the weights on the bar are still taking up the same space. I believe it would only change if the distance to the end of the bar is changed, which it is not. I hope I explained this
well enough.

ANSWER:
Assuming that the bar itself is uniform (has its center
of gravity (COG) at its geometrical center), the COG of the total barbell
depends on the location of the weights. Relative to the center of the bar,
the position of the center of gravity may be written as COG=(45x _{1} +10x _{2} +5x _{3} -45x _{4} -10x _{5} -5x _{6} )/120
where the x _{i} s are the distances of weights from the
center. Suppose that the weights are placed symmetrically (x _{1} =x _{4} , x _{2} =x _{5} , x _{3} =x _{6} );
then COG=0, the center of the bar. Now, suppose we interchange two of the
weights, exchange the 45 lb with the 10 lb on one side: COG=(45x _{2} +10x _{1} +5x _{3} -45x _{4} -10x _{5} -5x _{6} )/120=(45x _{1} +10x _{2} -45x _{2} -10x _{1} )/120=(35/120)(x _{1} -x _{2} );
since x _{1} ≠x _{2} ,
COG≠0, the barbell is no longer balanced. If that explanation is too
mathematical for you, try a more qualitative argument. Each weight W
a distance D from the center exerts a torque about the center and the
magnitude of that torque is WD. The net torque due to all weights
must be zero if the bar is to balance at its center. This means that the sum
of all the WD s on one side must be precisely equal to those on the
other if the barbell is to be balanced about its center. If you change the
D s on only one side, the bar will not be balanced at its center.
(This qualitative argument is just the mathematical argument in words.) What
certainly does not change is the total weight.

QUESTION:
After watching the Bond classic YOU ONLY LIVE TWICE, I read that the scene where a craft in space overtakes a capsule ahead in the same orbit in order to "swallow" it, but would be impossible because it would have to be in a separate orbit. Then when it catches up, turn vertical and move upward. Why is it not possible for an object to accelerate in the same orbit as a slower object?

ANSWER:
If the two satellites were in the same orbit, they would
maintain the same separation. If they were in different but crossing orbits,
you could have them come together if properly synchronized; if you were to
observe this, say from the perspective of the "chased" satellite, it would
appear that the other satellite was coming at you from slightly above or
below. Finally, if the "chasing" satellite had rockets which he could point
in any direction with any thrust, he could move exactly on the same path as
the "chased" satellite but with a different speed.

FOLLOWUP QUESTION:
I don't grasp the physical law that would prevent the "chasing" capsule to catch up in the same orbit (as we see in the movie) if its thrusters accelerate it.

ANSWER:
It is easiest if we just think about circular orbits;
near-earth orbits are nearly circular and I will consider only orbits whose
altitude is very small compared to the radius of the earth. With each orbit
there is one special speed v for an orbiting satellite where the
centripetal force equals the weight, mv ^{2} /R=mg or v= √(gR )
where g is the acceleration due to gravity, R is the radius of
the orbit (approximately R _{earth} ), and m is the mass
of the satellite. If you are going faster or slower than that you will not
be in that circular orbit but some elliptical orbit which happens to cross
the circular orbit. But, let us just suppose that you are going a little
faster than v , say v+u where u<<v ; you got there
by briefly firing your rockets out the rear tangent to the orbit. If you do
nothing else, you leave that orbit. However, the force necessary F to
keep you in that orbit would be m (v+u )^{2} /R =F .
But part of F is the weight, so you can write F=mg+f where
f is what your rockets have to do. Therefore f =m (v+u )^{2} /R-mg=m (v ^{2} + 2vu+u ^{2} )/R-mg.
Now, mv ^{2} /R =mg from above and you can
neglect u ^{2} because it will be very small compared to 2uv ,
so f ≈2muv √(g /R ); you would have to point your
rockets away from the center of the earth so that this force would be down
but you would keep on that circular orbit going faster than other satellites
in that same orbit.

QUESTION:
For a statement to be a law it must be based on observations and
experiments. Newton, certainly didn't perform experiments to verify his universal law of gravitation.
Was it correct then to state it as a law?

ANSWER:
Newton may not have done the experiments, but his law was
the result of experiments done by others. Most important were
Kepler's three laws which were empirical summaries of a large body of
data on the motions of the planets. His law of gravitation, F=-MmG /r ^{2} ,
provided a complete explanation of Kepler's laws. However, since the mass of
the sun was not known, only the product MG could be determined from
the data. A good
measurement of G was not done until more than 70 years after
Newton's death. Because gravity is such a weak force, this is a very
difficult measurement to make on a laboratory scale.

QUESTION:
This question is in regards to flowing water and buoyancy. Lets say I have two connected reservoirs at different heights, therefore creating a pressure difference and fluid flow between them. There is a pump that refills the higher reservoir so the flow is constant, and there is a section of tubing that is vertical. If I were to put a buoyant object like a balloon in the vertical section of tubing with flow, could I keep it from floating to the top with enough flow? Basically can fluid flow in the opposite direction of the buoyancy force keep it from floating. I feel like it can but I am having trouble understanding why (seems like the only factor is density/displacement, maybe fluid flow increases drag?), just curious because my friend and I got into a random debate about it.

ANSWER:
Usually when we
think of buoyant forces we are thinking about fluid statics, all fluid at
rest. Your balloon in the tube will experience a buoyant force up and a
force down from its weight just as it would in a nonmoving fluid. If the
water is moving down, the balloon will also feel a downward force due to the
drag it experiences. What this drag force is will depend on the size of the
balloon, the size of the tube, and the speed of the water. It would be very
complicated to calculate, but I am sure there would be a correct speed for
any geometrical situation where the balloon would remain stationary. If that
is enough, you can stop reading here. If not, here is an example below:

Basically, this is just a terminal velocity
problem. Suppose that we imagine just releasing a spherical balloon with
radius R , volume V= 4πR ^{3} /3, cross sectional
area A =πR ^{2} , and mass m under water. The net
force upward would be F=ρgV-mg= 4ρgπR ^{3} /3-mg
where ρ= 1000 kg/m^{3} is the density of water and g =9.8
m/s^{2} is the acceleration due to gravity. The drag force can be
approximated as f=πv ^{2} R ^{2} ρC _{d} /2
where the drag coefficient for a sphere is C _{d} =0.47 and
v is the speed of the balloon. So the net force is
F _{net} = 4ρgπR ^{3} /3-mg-πv ^{2} R ^{2} ρC _{d} /2
and this is zero when v=v _{t} , the terminal velocity, v _{t} =√[(8gR /(3C _{d} ))-2mg /(πR ^{2} ρC _{d} ))];
the second term in the square root is much smaller than the first because the mass of the
balloon is very small (about 0.04 kg if R =0.1 m and the air is at
atmospheric pressure) compared to the buoyant force. This is how fast a
balloon would rise in still water. So, that would be the speed the water
would have to be moving down for the balloon to stay in place. I did a rough
calculation for R =0.1 m and found v _{t} ≈2.4 m/s. These
estimates are all for the size of the pipe much greater than the size of the
balloon. Things get much more complicated if that is not the case, but you
would still be able to use the water flow to keep the balloon in place.

QUESTION:
Using real-world estimates for the coefficient of friction between his feet and the ground, how fast could the Flash run a quarter-mile? Assume that the limiting factor for his acceleration is the force parallel to the ground that his feet can apply.

ANSWER:
Suppose he is running on a dry asphalt road with
rubber-sole shoes. Then the coefficient of static friction is approximately
μ ≈0.8. The maximum force of friction on level ground would be f _{max} ≈μN=μmg ≈8m
where m is his mass. So, his acceleration would be a =f _{max} /m =8
m/s^{2} . A quarter mile is about 400 m, so assuming uniform
acceleration the appropriate kinematic equation would be 400=Ѕat ^{2} =4t ^{2} ,
so t =10 s.

QUESTION:
I have a doubt about static friction and number of wheels.
As for elementary physics principles
1) static friction depends is mass times the coefficient of static friction
2) static friction does not depend on surface static friction is independent by the number of wheels.
... but it is hard to accept to me!
Let's suppose to design a cart to be pushed by a worker. The total weight (cart + content) is about 1000 kg.
The question is: as for the static friction it is better to use 4 or 6 wheels?

ANSWER:
If the cart is to be "pushed by a worker" it is not
static but rather kinetic friction which is in play unless all the wheels
are locked. And this is not friction due to the contact between the wheels
and the ground but friction due to the axles rubbing on the wheels. But,
let's talk about friction anyway because you seem to have a serious
misconception. First of all, the friction is proportional to the normal
force which presses the wheel to the road, not the mass. If there were one
wheel, the maximum static frictional force you could get before the cart
started slipping (call that f _{max} ) would be the weight W times the
coefficient of static friction μ _{s} (on level ground), f _{max} =μ _{s} W .
If you had two wheels, each wheel would hold up half the weight so the
maximum static frictional force you could get from each wheel would
be μ _{s} W /2; but the total
force is still μ _{s} W. Things are more
complicated on a slope, but the conclusion is still that you do not gain an
advantage regarding traction by having more wheels. The reason big trucks,
for example, have many wheels is so that each wheel does not need to support
so much weight, not to get more traction.

QUESTION:
I understand that acceleration due to gravity decreases with distance, specifically by the inverse square law. That being said, what is the maximum distance for which one can use 9.81 m/s^{2} as g for Earth?

ANSWER:
That depends entirely on how accurate you want to be,
there is technically no place other than the surface of the earth where this
is the acceleration. Furthermore, the number 9.81 is simply an average
value; it varies over the surface of the earth due to local density
variations, rotation of the earth, influences of the moon's gravity,
altitude variation, etc . You need to ask something like "at what
altitude h from the surface is the value of g changed by X %?"
Then
X /100=((1/R )^{2} -(1/(R+h )^{2} ))/(1/R )^{2}
where R is the radius of the earth. Provided that h is small
compared to R , you can solve this equation approximately as h≈XR /200.
For example, g will be reduced by 2% when h≈R /100. Another
example: the International Space Station is at an altitude of about 230
miles, about 6% of the earth's radius. Then X _{ISS} ≈ 200x0.06≈ 12%
smaller than 9.81 m/s^{2} .

QUESTION:
If I am driving my car with a bowling ball in the trunk, does it take the same energy to accelerate the vehicle to a given speed at a given time if the ball is free to roll around as it would if it were fixed to the vehicle? I assume that the net energy use would be the same in both situations (same total vehicle mass), but the acceleration rates would be different - ie: the fixed ball would result in a constant acceleration to speed, while the rolling ball would result in a non-constant acceleration. If this is true, could I harness the energy of the ball's movement relative to the vehicle (using some sort of linear generator) without causing parasitic energy loss to the vehicle?

ANSWER:
As long as the ball and the car end up going the same
speed, the total energy to get them there is the same (neglecting frictional
and air drag forces). If you devise some way to take enegy away from the
ball, that energy ultimately must come from the engine.

QUESTION:
Perhaps you can help solve a disagreement we have at work. The question being "Does a person's initial velocity during a jump equal their final velocity once the land?" My contention is "no" in that the jumper could theoretically produce any velocity on the way up, but downward would be limited to terminal velocity. Who's right?

ANSWER:
Technically, you are correct. If air drag is present,
energy is lost which results in the landing speed being less than the launch
speed. In practice, however, for a person jumping into the air the height
acquired is not high enough for this to be a measurable effect; that is,
this is an example where we can say, as we often do in an elementary physics
course, that air drag is negligible. A typical terminal velocity for a human
is about 120 mph≈54 m/s. If you jumped with this speed you go over 100 m
high, obviously not in the cards. I did a rough estimate assuming the
maximum height you could jump would be about 2 m; if the person drops from 2
m his speed at the ground would be about 6.32 m/s without drag, 6.30 m/s
with drag, a 0.3% difference. For comparison, dropping from 100 m the speeds
would be roughly 44.1 m/s and 37.1 m/s for no drag and drag,
respectively. It is good to be precisely correct as you are, but it is also
good to be able to make reasonable estimates in real-world situations.

QUESTION:
When a ball is thrown vertically upwards ignoring air resistance, and another ball is also thrown upwards with air resistance, the time taken is less for the ball with air resistance to reach max height. Why is this "because average acceleration/force is greater"? Wouldn't there be less acceleration/force because the air resistance cancels some out?

ANSWER:
The reason is that the ball with air resistance does not
go as high. The force on the ball without resistance is the weight of the
ball pointing in the downward direction; but the downward force is greater
for the ball with air resistance because the drag force is also pointing
down. Therefore the ball with resistance slows down faster so it stops more
quickly. Think of an extreme example: if you throw the ball upwards in honey
which has very great resistance, it stops almost immediately.

QUESTION:
Whenever we roll a ball or spin a quarter it will slow down and eventually stop, since energy cannot just dissapear where does it go?

ANSWER:
The kinetic energy is being taken away from the ball or
coin by friction. That energy shows up as thermal energy, the
ball/coin-table-air all get a little bit warmer. Also, since you can hear
the ball rolling and the coin spinning, some of the energy must be lost to
sound.

QUESTION:
Something
ridiculous I thought of, if the Moon suddenly stopped moving and began to
fall toward the Earth, how long would it take to impact? I'm stumped as to
how to calculate this, as the force on the Moon gradually increases as it
falls, and the Moon also pulls the Earth toward it, and the radius of each
object would have to be included.

ANSWER:
I guess I am going to have to put
questions like this one on the
FAQ page. You should read the details of these
earlier questions since I do
not want to go over all the details again. It is tedious and uninstructive
to try to do this kind of problem precisely. I, being a great advocate of
"back of the envelope" estimates, use Kepler's laws to solve this kind of
problem; I have found that a very excellent approximation to fall time can
be found this way. I note that the mass of the moon is only about 1% of the
mass of the earth, the period of the moon is about 28 days, and the moon's
orbit is very nearly circular. The trick here is to use Kepler's third law
and recognize that a vertical fall is equivalent to the very special orbit
of a straight line which is an ellipse of semimajor axis half the length of
the line. Kepler's third law tells us that (T _{2} /T _{1} )^{2} =(R _{2} /R _{1} )^{3}
where T _{i} is the period of orbit i and R _{i}
is the semimajor axis of orbit i. Now, T _{1} =28 and R _{2} =R _{1} /2
and so T _{2} =T _{1} /√8=9.9 days. But this is
the time for this very eccentric orbit to complete a complete orbit, go back
out to where it was dropped from; so, the time we want is half that time,
4.9 days.But this is not what you
really wanted since I have treated the earth and the moon as point masses.
What you really want is when the two point masses are separated by a
distance of the sum of the earth and moon radii, 6.4x10^{6} +1.7x10^{4} ≈6.4x10^{6
} m. To see how much error this causes, I can use the equation for the
velocity v at the position r =6.4x10^{6} m if dropped
from r=R _{moon-orbit} =3.85x10^{8} m which I derived
in one
of the earlier answers: v =√[2GM (1/6.4x10^{6} -1/3.85x10^{8} )]=1.1x10^{4}
m/s. It would continue speeding up if the collision did not happen, but even
if it went with constant speed the time required would be about t=R /v =6.4x10^{6} /1.1x10^{4} =580
s=9.6 min. This is extremely small compared to the 4.9 day total time, so,
to at least two significant figures, 4.9 days is the answer to your
question.

An important part of doing physics, or any
science, is knowing when to eliminate things which are of negligible
importance!

QUESTION:
I'm a Science Olympiad coach trying to optimize the performance of our "Scrambler", a car which must be accelerated by only a falling mass.
Most competitors simply tie a weight to a string and route that string over a set of pulleys (using no mechanical advantage to convert the vertical falling acceleration horizontal.
…Read a whole lot more !

ANSWER:
Sorry, but if you read site groundrules you will see that "concise, well-focused questions" are required.

FOLLOWUP QUESTION:
I was hoping you'd like the challenge of a motion/force problem that must span across several formula -- PE, KE, PEspring, velocity solved by
acceleration and distance only, etc. Something to sink your teeth into...

ANSWER:
It is really not that interesting to work the whole thing
out, but on second thought it is interesting to talk qualitatively and
generally about the questioner's proposal; so I will do that. I will
summarize the situation since I am sure none of you loyal readers will want
to read the whole original question. By using a falling mass M
attached to a car of mass m , it is wished to maximize the speed v
of the car for M having fallen through some some distance H .
The car moves only horizontally. The simplest thing to do is to have the two
simply attached by a string over a pulley. Then, using energy conservation,
0=Ѕ(M+m )v ^{2} -MgH or v =√(2MgH /(M+m )).
What the questioner proposes is to hold the car at rest and insert a rubber
band in the string so that the falling weight stretches the rubber band
which has been carefully chosen to be just right that, when M has
fallen H , it has just come to rest and is held there. Now,
presumably, the rubber band has a potential energy of MgH . If the car
is now released, the rubber band will presumably contract back to its
original length giving its potential energy to the car, MgH =Ѕmv ^{2
} or v =√(2MgH /m ), a considerable improvement. My
suggestion would be to use a spring rather than a rubber band since a rubber
band has much more damping (energy loss due to internal friction) and
hysteresis (will not return to its original length). Since the rules fix
M and H , one obviously wants m to be as small as possible.

QUESTION:
Does acceleration have momentum?
In other words if you fire a rifle, does the highest velocity of the bullet occur as it exits the barrel or does the acceleration increase after it leaves the barrel?

ANSWER:
In terms of physics nomenclature, your first question has no
meaning. But your second question seems to clarify what you mean: if
something has an acceleration does it keep accelerating even if there are no
forces on it? The answer is an unequivocal no. The only thing which causes
acceleration is force and when the bullet exits the barrel of the gun the
force which was accelerating it disappears. If there were no new forces on
it, it would continue with the same velocity it had when it exited. There
are, however, two important forces on the bullet when it is outside the
gun—gravity and air drag. Gravity causes it to accelerate toward the ground
and air drag causes it to slow down.

QUESTION:
We have a metal ruler 1 yard in length. If held in exact balance center then "pinged" it causes vibrations. nothing new there BUT at exact equal distances to either side there is a point at which it appears that the vibrations stop (a calm spot) for about 1 inch then the vibrations start again and continue to the end. We've tested and the vibrations don't stop, they are just a much smaller wave length. What is is this "calm spot" phenomenon? What causes it? Does it happen with earthquakes too?
Really geeked out about this! Way cool!

ANSWER:
You are exciting standing waves when you "ping" the stick. These are waves
which bounce back and forth from the ends of the stick and, for special
wavelengths, are just right to to resonate like a guitar string or an organ
pipe. The various wavelengths for which resonance occurs are called the
modes of oscillation. For a stick clamped at the middle, the lowest mode,
called the fundamental, has approximately 1/4 of a wavelength on either side
of the center as shown by the upper part of the figure above. A point with
zero amplitude, the center for the fundamental, is called a node and points
with maximum amplitude, the ends for the fundamental, are called antinodes.
What you are seeing is the next mode, called the first overtone, which has
approximately 3/4 of a wavelength on either side; this mode has three nodes
and four antinodes. To the right are animations for a stick clamped
at the end but they are exactly what your stick is doing on one half. Here
the nodes are near the darkest blue. Earthquakes are traveling waves and
therefore do not have nodes.

QUESTION:
I've been reading about rotational space habitats for a while now and haven't found an answer to this question by googling. So how would a space habitat be rotated? By cogs? By propulsion systems?

ANSWER:
If you want to read more about the details of such habitats,
see my earlier answers (1 and
2 ). In answer to your question,
you would need thrusters to get the ring spinning and then to occasionally
correct minor changes, but once it was spinning, it would continue to spin
just the same forever if there were no external torques on it as would be
the case in empty space.

QUESTION:
The moon's gravity is one sixth that of the earth. Thus if you kicked a box with a force of 60 N across a frictionless floor on earth, the box would travel the same distance in 1 second as when the same box was placed on a frictionless floor on the moon and kicked with a force of 10 N. Am I wrong?

ANSWER:
You are wrong on both your conclusion and on your
"…kicked…with a force of…" premise. First the premise: you need to go the
FAQ and read the link
from the question about how much force does it take to make something move
with some speed. Just knowing the force you cannot know the resulting speed;
you need either how long the force was applied or over what distance it was
applied. Now, your question implies that you think it will be easier to get
the box on the moon moving with the same speed as a box on earth with that
speed. But, in fact, the box has the same mass on both the earth and moon
and you are not lifting it against gravity, so it is equally easy to move a
box horizontally on earth or the moon. It is six time harder to lift a box
on earth as on the moon. If there were friction, however, it would be harder
to move the box on earth than on the moon because the frictional force on
the moon would be six times smaller.

QUESTION:
If you have two permanent cylindrical magnets (the kind with a hole in the center) and you stack them with poles opposite on a pencil, the top magnet will "float" above the bottom magnet. Energy is being expended to keep the magnet "up" the pencil. Where is the energy coming from? The bottom magnet will be pushing down with an equal but opposite force, but that does not cancel the energy needed to float the top magnet as far as I can see.

ANSWER:
I am afraid you do not understand energy. The lower magnet
exerts a force on the upper magnet. The force holds it there in equilibrium,
it does not require energy to hold it there. It is no different from saying
that if one of the magnets were hanging from a string, where does the energy
to hold it there come from? Or, if one of the magnets were sitting on a
table top, where does the energy to hold it there come from?

QUESTION:
I read of "gravity assist" swingy-bys of Jupiter to speed a spacecraft up to reach the outer planets. As the spacecraft approaches Jupiter, it speeds up. But it retreats from Jupiter on a symmetric path (a hyperbola I think) and Jupiter will therefore slow the spacecraft down by the same amount on the outbound path. It appears there should be no net increase in speed, just a bending of the spacecraft path. But bending a spacecraft path also takes energy. So Jupiter is providing the energy in some manner though it is unclear to me how.

ANSWER:
The trick is that the planet, with much greater mass than the
spacecraft, is moving in its orbit and the boost comes from using the speed
of the planet to speed up the spacecraft. The figure shows the idealized
one-dimensional interaction with planet; because the mass of the spacecraft
is much less than the mass of the planet, the spacecraft picks up twice the
speed of the planet. For those who have studied elementary physics, this
should look vaguely familiar: a perfectly elastic collision between a BB at
rest and a bowling ball with speed U results in the BB going with
speed 2U and the bowling ball still going U . (Of course, this
is only approximately true if the mass of the BB is much smaller than
that of the bowling ball; the bowling ball will actually lose a very tiny
amount of its original speed.)

QUESTION:
My son is in 5th grade and he is supposed to be doing a Science Fair project and presenting in front of judges, some of which I have heard work at IBM. His teacher helped him set up an experiment where he rolled a car down an inclined plane 5 times adding more weight each time. Then he is to see if the car goes faster with more weight added. And then explain that due to research. I have read your other answers but I am so confused. Is gravity the only force working on the car to get down the incline? and if so does that mean that technically the car should always travel the same speed? Any help on where I can learn about this, or any information you may be able to help me with would be so greatly appreciated.

ANSWER:
The best place on my site to read is
this earlier answer .
There you will see that gravity is certainly not the only force acting on the car. Friction is very important. The normal kinds of friction, in axles, wheels,
etc . should, according to the simplest approximation, not affect the speed of the car—all masses should take the same time. That, as I explain in that post, is because all these forces are approximately proportional to the weight of the car and, of course, gravity is also proportional to the weight. The other kind of friction is air drag. The air drag is proportional to the speed of the car squared, so the faster the car goes the more air drag it has. You know that this is true because you have a very different experience if you stick your hand outside the window of a car going 10 mph and 80 mph. If air drag is important, the heavier car wins. I have said before that this is not really a very good experiment for a science fair because there are too many variables and approximations. Friction can be a very sticky thing in physics (pun intended!).

If he did not find the heaviest car the fastest, I do not know what to tell you except it is difficult to understand.
If he did find the heaviest the fastest, here is how he could justify it: Two cars, each going the same speed down the ramp, have masses 1 kg and 2 kg. Since they are identical in all respects except mass, each experiences the same air drag frictional force, let’s call that
F _{drag} . Newton’s second law says that the acceleration a _{drag
} is equal to the force F _{drag} divided by the mass m and so a _{drag} =F _{drag} /1 for the 1 kg mass and
a _{drag} =F _{drag} /2 for the 2 kg mass. Here the acceleration is the rate of slowing down due to air drag (you might call that deceleration) and you see that the heavier car slows down
less than the lighter car. All the other forces result in the same acceleration for each car since those forces are all proportional to the mass; maybe I should elaborate on that a little: Forces due to friction and gravity can be written
F _{other} =Cm where C is some constant. And so
a _{other} =F _{other} /m=C which means that all accelerations should be the same, so only air drag gives advantage to the heavier car.

Finally one more example to illustrate: Drop two balls exactly the same size, one is a nerf ball (very light) and one is a lead ball (very heavy). If there were no air, both would fall to the ground in the same time. But you can easily show that the heavier one wins if there is air.

FOLLOWUP QUESTION:
I'm still a little confused because his teacher is telling me it is not gravity that is pulling the truck down because it is a ramp and not falling and she says it is force and momentum moving the truck down the ramp. From what I have researched it looks to me like momentum is not really a force and only forces actually move objects? Also I think I understand that gravity is not the only force acting on the toy truck but it seems to be the only one pulling or pushing it downward and the frictional forces push upwards, opposite gravity?

ANSWER: Your son's teacher has
this seriously wrong if she really said that it is not gravity pulling it
down the ramp. It most certainly is falling, it is just not falling straight
down. And you are right, momentum has nothing to do with its going down the
ramp; it has momentum but that is not what is causing it to go down the
ramp. So, I guess I will have to give you a complete primer on the motion of
an object on a ramp. I will draw some figures and use some equations but I
will try to give you as many words as I can to qualitatively explain; then
you can help your son create an explanation on his level. The basics are
that an object which has no net force on it moves with a constant velocity
(Newton's first law). To change the velocity (accelerate or decelerate) you
must push or pull on it (exert a force) (Newton's second law). In equation
form, Newton's second law may be stated F=ma where this says
force=mass times acceleration .

Let us first talk
about the ideal case where there is no friction at all. The first
picture shows the weight (gravity) as a red arrow. But you see, the
weight actually does two things as indicated by the blue
arrows—accelerate the car down the ramp (smaller blue arrow) and push
the car through the ramp (larger blue arrow). The car does accelerate
down the track but it does not go through the ramp because the ramp
pushes on the car with a force opposite the blue arrow, shown as a black
arrow. So we can agree that the reason the car accelerates down the
track is gravity. Now how does the acceleration depend on the weight of
the car? If you have two cars and one has twice the weight of the other,
the force (smaller blue arrow) down the ramp will be twice as big for
the heavier one. On the other hand, Newton's second law says that the
acceleration a of some weight (mass m ) due to a given
force F depends inversely on the weight a=F/m . So even if
F is twice as big, because m is also twice as big, they
both have the same acceleration! This means that if there is no friction
of any kind, the weight of the car should make no difference.

Next let us add friction, the kind which results
from one thing rubbing on another like wheels on axles, etc . The
friction (shown as a purple arrow in the second figure) points up the
ramp and so it tends to slow the car down. The important thing to know
about these kinds of frictional forces is that they also depend on the
weight of the car. If the car is twice as heavy, friction is twice as
big. But if the friction depends on the weight, the acceleration will be
the same just as it was for the weight force. Again, all cars should get
to the bottom in just the same time again, it will just be a shorter
time because the total force down the incline is smaller.

Finally, let us add in an air drag force (the
green arrow in the third figure). Air drag is, as I explained in the
original answer, a force which depends on two things—the shape/size of
the car and the speed of the car—but not on the weight of the car .
So air drag gives the heavier car the advantage because there is less
slowing down due to the drag force.

QUESTION:
This question should be fairly quick. Are there any know situations were momentum isn't conserved? I would say no, as momentum is always conserved if you make your system big enough. The only time momentum appears to be not conserved is when you put restrictions on the size of your system, and don't account for the momentum transferred outside your system. When you include the system " outside" your system, momentum is in fact conserved.

ANSWER:
First, what you learn in first-year physics is that linear
momentum is conserved for an isolated system; an isolated system is
one for which there are no net external forces acting. This is actually a
result of Newton's third law which essentially states that the sum of all
internal forces in a system must equal zero. This works really well until
you get to electricity and magnetism where it is easy to find examples of
moving charges exerting electric and magnetic forces on each other which are
not equal and opposite (you can see an example in an
earlier answer ). In that case you would say that such an isolated
system obeys neither Newton's third law nor momentum conservation. However,
looking deeper, we find that an electromagnetic field has energy, linear
momentum, and angular momentum content and, in the end, momentum
conservation, because of the momentum contained in the field, is still
conserved for an isolated system. Thus, Newton's third law is saved, but not
always in the simplistic "equal and opposite forces" language. Finally, if
linear momentum is p =mv , linear momentum is not
conserved in special relativity. But, physicists so revere momentum
conservation that in special relativity momentum is redefined so that it
will be conserved but still reduce to p =mv , for low
speeds: p ≡γmv =mv /√[1-(v /c )^{2} ];
this then unifies energy E and linear momentum p into a single
entity which is conserved, the energy momentum 4-vector: E ^{2} -p ^{2} c ^{2} =m ^{2} c ^{4} .
Not as quick as you expected!

QUESTION:
What is the force that causes you to fall over when a moving bus comes to an immediate stop? I'm having an argument with my teacher over what the answer is, it would be great if you could explain!

ANSWER:
When the bus is stopping, it is accelerating and so it is a
noninertial frame. That means that Newton's laws are not valid if you are
riding inside the bus. But, if we watch you from the bus stop, Newton's laws
do apply and we conclude that if you move with the bus, there must be a
force which is causing you to accelerate also. Friction provides a force
which, except under extreme circumstances, accelerates your feet along with
the bus; but, unless you are holding on to something, there is nothing to
provide a force on your upper body which therefore tends to keep going
forward without accelerating. All this says that the reason you fall forward
is not due to any force, rather it is due to lack of a force. There is,
though, another way to look at this problem. If you are in an accelerating
frame, like the bus, you can force Newton's laws to be true by adding fictitious
forces. The best known example of a fictitious force is the
centrifugal force in a rotating (and therefore
accelerating) frame. In the bus which has an acceleration a you can
invent a fictitious force F _{fictitious } on any
mass m in the bus, F _{fictitious} =-ma ;
if you do that, Newton's laws become true inside the bus and the force F _{fictitious}
may be thought of as being the force which provides your acceleration.
Note that the acceleration is opposite the direction of the bus when it is
stopping, and so the fictitious force is forward as you know if you have
fallen over in a stopping bus. When the bus is speeding up you tend to fall
backwards. Since there are two answers here, depending on how you choose to
view the problem, so maybe you and your teacher are
both right!

QUESTION:
if the earth had an orbit of 100,000 miles above the surface of the sun what would a person with a weight of 100 pounds weigh during the day and during the night and would there be any difference because of the gravitational pull from the sun.

ANSWER:
The radius of the sun is about 432,000 miles, so the radius
R
of the orbit would be R =532,000 mi=8.6x10^{8} m. Since the
diameter of the earth, about 12.8x10^{6} m, is small compared to the
radius of the orbit, there is only about a 0.1% change in the gravitational
attraction to the sun if you change the distance by one earth diameter. The
mass of the sun is M =2x10^{30} kg. The force of attraction of
the sun on the m =100 lb=45 kg person would be given by F=GMm /R ^{2} =6.67x10^{-11} x2x10^{30} x45/(8.6x10^{8} )^{2} =8117
N=1825 lb, much bigger than the 100 lb force which the earth exerts on the
person. Now, to answer your question you need to define weight. I will
assume that we mean the net sum of all forces on the person so that W _{day} =1725
lb upwards and W _{night} =1925 lb downward. Or, maybe you mean
what a scale on the floor would read (not actually what weight means) in
which case W _{night} =1925 lb and W _{day} =0 lb.
These are really only noon and midnight weights since the forces due to the
earth and sun would not be parallel at other times.

QUESTION:
If a bullet was shot through a window of a moving train and was to come out on the other side of the train through a window. Would it come out through window 2 on the same the exact opposite side as window 1 or would it look as if the bullets direction was bended?

ANSWER:
If the bullet is shot straight at the train with some speed
V and the train is moving with some speed v , an observer on
the train sees the bullet moving with a speed V toward the opposite side and, at the
same time with a speed v toward the back of the train so that the bullet would
be traveling, as measured on the train, in a straight line across and
rearward with speed √(V ^{2} +v ^{2} )

QUESTION:
I've found the formula for gravitational attraction between two objects, but I can't quite "do the math" mainly because of the metric/english conversions...I want an answer that I can relate to in pounds or ounces, not dynes, ergs, or grams (I am aware of the distinction between mass and force) Here's my question:
I am driving my 18-wheeler truck which weighs 80,000 lb. fully loaded. I am driving due west at sunset, heading straight toward the sun which is about 93 million miles away. What is the 'tidal force' of attraction between my truck and the sun?

ANSWER:
I have the feeling that you want gravitational force, not
tidal force. Tidal force is the tendency for the
truck to be stretched because the gravitational force on the front of the
truck is a tiny bit larger than on the rear. The gravitational force is
computed by F=GMm /R ^{2
} where G =6.67x10^{-11} N∙m^{2} /kg^{2} ,
M =2x10^{30} kg, m =8x10^{4} lb=3.6x10^{4}
kg, and R =93x10^{6} mi=1.5x10^{11} m. So I find F =213 N=48
lb. There is a very handy little free program called
Convert which
you can use to convert just about any units you might want to work in.

QUESTION:
Let's assume I have a magnet that can lift 100 tons. And I attach the magnet in a chain and attach the chain into roof for a system to magnetically lift items and then drop them to other places. Would the chain have to be able to take the 100 ton load or would the magnet take the 100 ton load because after all it is the one keeping the lifted item up?

ANSWER:
The magnet itself holds up the 100 ton weight. The chain
holds up the 100 ton weight plus the weight of the magnet itself. The roof
holds up the 100 ton weight plus the weight of the magnet plus the weight of
the chain.

QUESTION:
In 1973 a Physics instructor explained via math that the sidewalls of a regulation tire need not be present if the velocity of the vehicle was above a speed of 65+ mph. I tried to explain this to family members at Christmas and was scoffed at and then ridiculed. The Physic instructor had been let go from the GMC/Chevrolet plant several years before and he took the educational retraining route. His job was to change out instruments on GM cars running around a track and in excess of 100+ mph and his driver advised him that they had had a blow out and he needed to get out from under the dash and safety belted in at which time the slowed below the critical speed with the result that they did not crash but came close to it. Can you provide a link or the math to show that the speed is somewhere about 70+ miles and then the centripetal force will hold up the outer part of the tire. He did the math as part of educating us on acceleration, speed and force as it involved that part of the class curriculum.
The instructor was a good instructor in that he made the physics relevant to the real world if there is such a thing today and even at that time. Also this is why tires need sidewalls as they won't hold up in gravity and below a specific velocity.

ANSWER:
This is nonsense. If there is no air pressure to connect the tire to the axle, which would be the case if there were no sidewalls, what is going to hold up the weight of the car?

FOLLOWUP QUESTION:
No not really if you get the tire up to speed as well as providing
forward momentum the circumference and the center point about which the
tire is rotating will hold the tire up even if there is a blow out as
the forward speed or acceleration is sufficient to hold it up will
prevent deflation aka collapse of the tire above a speed. Once the speed
or acceleration drops below a key critical the tire will start to
collapse and according to the GMC aka Physics instructor all hell broke
lose on the track and only the drivers expertise kept the ensuing
deceleration from causing him and driver problems. Once you reach
velocity the outer rim of the circle/sphere need not have anything to
hold it up if the instructors explanation and the math were correct.
Key elements:

Tires inflated to recommended PSI

Vehicle an experimental test GMC product running in excess of 100+ miles an hour.

According to the LAHC Physics instructor a knowledgeable and well trained driver
at the wheel who on sensing the blow out got him out from under the dash and into the
multi-point seat belt. Instructors job was instrument technician for the test bed aka the
vehicle a GMC automobile with changeable instruments.

He was alive to prove it to the class with not sure what Physic concept/principal that escapes me.

ANSWER:
Well, maybe I misunderstand something here, but let's boil
this problem down to the simplest equivalent I can think of: imagine a tire
with sidewalls and just an axle which is supported by the sidewall, shown on
the left in my figure. Now, we would agree, I believe, that if the sidewall
suddenly disappeared, the axle would fall because there would be nothing
holding up that weight. How is that situation any different if the car is
moving? So, let's agree that "the
sidewalls…need not be present" is wrong because there has to be some
physical contact of the outer surface of the tire and the axle. So, my first
answer was a knee-jerk response to the notion that the sidewalls were not
needed.

THE ANSWER
YOU WILL LIKE:
However, there is still a way that you might have a point. When the
blowout occurs, the pressure inside the tire is lost; this pressure is
typically 30 PSI=21,000 N/m^{2} above atmospheric pressure (which is
about 100,000 N/m^{2} ). If the car is sitting still, this loss of
pressure results in the wheel collapsing because the sidewalls alone are
insufficient to hold up the weight of the car unless the force due to the
pressure pushing on the outer part of the tire holds the sidewalls taut.
Now, imagine that you are driving with some speed V and viewing a
spinning tire from its axis, you see every point on the outer surface of the
wheel accelerating with an acceleration V ^{2} /R where
R is the radius of the tire. Therefore, every little piece of the tire
with mass m experiences a (fictitious) force (called the centrifugal
force) of mV ^{2} /R. That would be equivalent to there
being a pressure P exerted on that little piece of tire of P =mV ^{2} /(aR )
where a is the area of that little piece. But, every little piece
behaves like this, so it is equivalent to a pressure of P =MV ^{2} /(AR )
acting on the outer surface where M is the mass of the tire (assuming
the sidewalls are a small fraction) and A= 2πRW is the area of
the outer surface and W is the tread width. So, if that pressure is
equal to 21,000 N/m^{2} , it will be like the blowout never happened!
I took R ≈16 in≈0.4 m, W ≈12 in≈0.3 m, and M ≈20 lb≈9 kg
and solved 21,000=MV ^{2} /(AR )=MV ^{2} /(2πR ^{2} W )
and found V= 27 m/s=60 mph. (Incidentally, the "forward momentum" has
nothing to do with it.)

QUESTION:
which team wins in a tug of war: the team that pulls harder on the rope or the team that pushes harder against the ground.Can you explain please?

ANSWER:
Focus your attention on the guy in the red shirt. There are
two horizontal forces on him, the rope pulling to the left and the ground
pushing to the right. If he is not moving, these two forces must be of equal
magnitude. To win, he must accelerate to the right and so the ground must
exert a bigger force on him than the rope exerts on him. To complete the
answer, use Newton's third law which says that the force the rope (ground)
exerts on the man must be the equal and opposite to the force that the man
exerts on the rope (ground). So the winner must push harder on the ground.

QUESTION:
If a high jumper clears the bar, is it possible that the centre of mass of the body of the jumper passes below the bar? If so can you make me visualize the scenario by a video or image illustration or a vivid description? I think that the centre of mass can be below the bar during the jump, but it has come there after travelling above the bar.

ANSWER:
You can find dozens of pictures and videos on the internet. A
nice one is shown to the right here. The path under the bar of the center of
gravity of the jumper is shown. When the body is bent the center of gravity
is outside the body. Going over with the back down is called the Fosbury
flop after Dick Fosbury, the American high jumper who won the gold medal at
the 1968 Olympics.

QUESTION:
I want to know why does torque is able to do work. I Mean Torque vector always acts perpendicular to the surface therfore meaningthat angle between torque vector and angular displacement vector is always 90 degree. Whereby meaning work done by torque is always zero but it is obviously not so. I know how to derive that work expression but still I am wondering why is it so.

ANSWER:
You are mistaken, angle and torque vectors are not always
perpendicular. Shown in the figure to the right are the vector directions
relevant to your question. The angle through which this cylinder rotates,
θ , increases as the cylinder rotates counterclockwise as seen
from above; the vector direction is seen in the black vector θ
shown in the figure. If there is a force F
acting at a distance r from the
axis, the torque vector is given by the red vector
τ shown in the figure, τ= F xr .
So, W=τ ∙θ .

QUESTION:
In a system where I have a 1600mm beam with I=1,300,000kg-mm^2 rotating about an axis (shaft) at the far left end of the beam where it's initial position is resting at 0 degrees horizontal then rotating 180 degrees counterclockwise (from 3 o'clock to 9 o'clock) about the axis (shaft) at the end of the beam described above by way of a cable wrapped around a 12" diameter disc (disc attached to beam & shaft) where the cable runs out to the right tangentially at 6 o'clock horizontally over a pully wheel then down to a weight, say 300lbs, that will fall to apply enough torque to effect the rotation of the beam and accelerate it.
How do I account for role gravity plays in the falling weight keeping up with the rotation of the beam where the beam first has to overcome some amount of gravity to get from 3 o'clock to 12 o'clock but then gravity works in it's favor from 12 to 9.
My goal is to get a point at the end of the beam to hit a plate at about 100mph at the end of the rotation (9 o'clock).

ANSWER:
I am not certain that I understand exactly what the
configuration is and some of the numbers are missing (like mass of the disc)
or not fully specified (like the axis for the moment of inertia of the
beam). I will work it out in general and you can apply it to your situation.
The situation as I understand it, before and after, is shown to the right. I
will call the length of the beam L , its mass M _{1} ,
the radius of the disc R , its mass M _{2} , the hanging
mass M _{3} . I will assume the beam and disc are uniform so
that their moments of inertia are M _{1} L ^{2} /3
and M _{2} R ^{2} /2
respectively. I assume that the mass of the pulley is negligible and that
all friction can be ignored. In the final situation the beam and disc rotate
with angular velocity ω and the hanging mass has a speed v=Rω .
Note that the hanging mass has fallen a distance half the circumference of
the disc, πR . If you are interested only in how fast everything is
going at the end, you do not need to worry at all about what is going on at
any other time because this is a classic energy conservation problem and the
final energy is equal to the initial energy. I am not going to give you all
the details, I will just give the initial equation and its solution for the
end.

Once you get ω ,
you can get the speed V of the end of the beam by calculating
V=ωL.

If I take your
numbers, M _{3} =300 lb=136 kg, L =1.6 m, M _{1} =1.52
kg (assuming the I you have given me is about an axis through the end
of the beam), M _{2} =0 (assuming it is small since you did not
specify it), R =6"=0.15 m, and g =9.8 m/s^{2}
(acceleration due to gravity), I find ω= 17 s^{-1} so V= 17x1.6=27
m/s=60 mph.

ADDITIONAL
THOUGHT:
You might think that you can increase the speed of the end of
the beam by increasing M _{3} . However, there is an upper
limit. In the limit as M _{3} —›∞,
ω —›√{2gπ /R }=20.3 s^{-1}
giving V =72 mph. Assuming that the length of the beam is fixed, you
would have to decrease the radius of the disc to reach 100 mph.

QUESTION:
Orbiting astronauts are weightless because they are essentially in free-fall and don't "feel" Earth's gravitational force. Earth is orbiting the Sun and similarly Earth should not "feel" the Sun's gravity. However, the Sun has an effect on the ocean tides therefore it follows that the ocean water "feels" the Sun's gravity. Please explain.

ANSWER:
An astronaut is not weightless but she is in free fall and
therefore feels like she is. However, she is not aware of what are called
tidal forces because they are too small because she is relatively small; I
will try to clarify this. Let's take the case of the earth-moon interaction.
Because the moon's gravitational force falls off like 1/r ^{2} ,
the forces felt on the earth are different at the side nearest to the moon,
farthest from the moon, and at the earth's center; this is shown in the
upper of the two figures to the left. The net effect, obtained by
subtracting the central gravitational force, is as shown in the lower
figure; that is why the tidal force raises the ocean level on both near and
far sides of the earth. The astronaut will also experience a tidal force but
since her size is so small compared to the size of the earth, she will not
feel the force try to stretch her; if you think about it, you will see that
the tidal force on an astronaut is greatest when she is standing on the
ground. When an object falls into a black hole, tidal forces become huge.
The above discussion contains nothing regarding how the moon and earth are
moving relative to each other so all the same arguments apply to tidal
forces the earth feels in the presence of the sun; these are, though, much
smaller since we are much farther from the sun.

QUESTION:
Conventional automobile steering produces a weight transfer from the inside wheels to those on the outside of the turn.
If, instead of deflecting the front wheels, one steers by turning the outside wheels faster than the inside wheels, will that also cause a weight transfer?

ANSWER:
There is an earlier
answer about a bicycle making a turn. It would be helpful for you to
read that first. I assume you want to understand why the weight is more
supported by the outside wheels, so I will try to show that. The easiest way
to do the problem of the car turning a curve is to introduce a fictitious
centrifugal force which I will call C , pointed
away from the center of the circle; the magnitude of this force will be
mv ^{2} /R where m is the mass of the car, v
is its speed, and R is the radius of the curve, although we do not
really need to know that to answer your question. The picture to the right
shows all the forces on the car: W is the weight and the green x is the
center of gravity; f _{1} and f _{2} are the
frictional forces exerted by the road on the inside and outside wheels
respectively; N _{1} and N _{2} are the normal
forces exerted by the road on the inside and outside wheels respectively;
the center of gravity is a distance H above the road and the wheel base is
2L (with the center of gravity halfway between the wheels). Newton's
equations yield:

f _{1} +f _{1} =C
for equilibrium of horizontal forces;

N _{1} +N _{2} =W
for equilibrium of vertical forces;

CH+L (N _{1} -N _{2} )=0
for equilibrium of torques about the red x .

If you work this out, you find the normal forces
which are indicative of the weight the wheels support: N _{1} =Ѕ(W-C (H /L ))
and N _{2} =Ѕ(W+C (H /L )). A few things to
note are:

the outer wheels support more weight,

if C =0 (you are not turning), the inner
and outer wheels each support half the weight,

at a high enough speed C will become so large
that N _{1} =0 and if you go any faster you will tip over, and

if the road cannot provide enough friction you
will skid before you will tip over.

Now we come to your question. Regardless of how you
cause the car to turn, the analysis done above will always be the same; the
centrifugal force is trying to tip the car over and that is why the weight
distribution shifts.

QUESTION:
If a gun was fired at the escape velocity of the Earth in a direction that it would not hit anything but not straightly up but instead more like in an angle of 45 degrees would the round still go to space? And if not what would the velocity needed be for a round that is fired at 45 degrees in to the sky be in order for the round to escape Earth?

ANSWER:
First, let's be clear that we are talking about an ideal
situation where we neglect the effects of air drag (which will slow the
projectile) or earth's rotation (which can add or subtract from the velocity
you give the projectile). The easiest way to get the value of the escape
velocity (v _{e} ) is to use energy conservation. Escape
velocity is that velocity which results in the projectile being at rest at
r =∞. Taking the potential energy U (r )=-GMm /r+C
to be zero at r =∞ (i.e ., choose C =0), energy
conservation gives Ѕmv _{e} ^{2} -GMm /R =0
where M and R are the mass and radius of the earth,
respectively; so, v _{e} =√(2MG /R ). You will
notice that this derivation has been done without any reference to the
direction of the velocity v _{e} , so the direction
makes no difference.

QUESTION:
This is one of the most baffling physics questions so far. Suppose a cyclist is pedalling briskly, thus accelerating forward. He exerts force on the tyres, the tyres exert force on the ground, and the ground exerts a reaction force which accelerates the cyclist. Right? But now the cyclist applies brakes. Now he starts to retards. But since the tyres are still moving in the same direction, hence friction must also be acting in the same direction. Then which force is responsible for the retardation? I suppose the brakes cannot retard the 'cyclist and the cycle' system, since it is an internal force.

ANSWER:
There is nothing "baffling" here. You are right, all internal
forces should be ignored if you want to understand the motion of the
bicycle. The only external forces on the cycle and its rider are its own
weight, the normal force up from the road on the wheels, and the frictional
force between the road and the wheels. (I am neglecting internal friction
and air drag.) The first of these two add to zero (assuming a horizontal
road) and can be ignored for our purpose. If the acceleration is forward,
the friction if forward. If the acceleration is backward, the friction if
backward. The direction of the friction is not determined by the motion of
the wheels because the wheels are not slipping on the road (you hope) so
static friction, not kinetic friction, is responsible for the acceleration.
It is friction which speeds you up or slows you down.

QUESTION:
Consider a tug of war game. is the net work the ground on the two teams is negative, positive or zero?

ANSWER:
Just after one team has won, all players are moving. They
have a kinetic energy. Therefore positive work was done on the whole system
by friction with the ground. Once everyone stops moving, the net work done
on the system would have been zero.

QUESTION:
Trying to understand physics concepts is
an interest of mine. I was hoping you could shed some light to help my
understanding? The equation for Kinetic Energy: KE = 1/2 m v^{2} .
I understand the value of velocity squared in that; the kinetic energy
of an object is proportional to its speed squared:double the speed,
quadruple the KE. But I am having trouble with why and what the 1/2
value is telling me about what is going on in the equation as to the
motion of mass and speed. It is pretty elementary stuff but if I want to
know about Energy, Work, Power and Heat I need to get the basics right.

ANSWER:
To do this on the simplest level, you need to know kinematics
for uniform acceleration [x=x _{0} +v _{0} t+Ѕat ^{2}
and v=v _{0} +at where the 0 subscripts denote the
position and velocity when t= 0] and Newton's second law [F=ma ].
Suppose that you have a constant force F that you exert on a mass
m over a distance s ; the work done is W=Fs and work
changes the energy; that is, we say that the work done equals the energy
given to m . I will assume that m begins at rest and at the
origin, so x=s=Ѕat ^{2} and a=v /t ; then
F=mv /t and s =Ѕ (v /t )t ^{2} =Ѕvt .
Finally, W=Fs =(mv /t )(Ѕvt )=Ѕmv ^{2} .
So, in the case of a constant force, the factor of Ѕ comes from the Ѕat ^{2}
part of the kinematic equation for position.

EXTRA ANSWER:
If you know calculus, this is much simpler. Write dW=F dx
and F=m (dv /dt ) so W =_{0} ∫^{s} F dx= _{0} ∫^{s} m (dv /dt )dx= _{0} ∫^{v} m (dx /dt )dv= _{0} ∫^{v} mv dv= Ѕmv ^{2} ;
and this does not require F to be constant.

QUESTION:
why do we need to have a new expression for calculating the kinetic energy of a body that is in rotational as well as translational motion?

ANSWER:
Whether you do or not depends partly on how the object is
moving. For example, if the object is rolling on a surface without slipping,
it is instantaneously rotating about the point (or line) of contact and you
can write K =ЅIω ^{2} where I is the moment of
inertia about that axis; however, you could also show (using the parallel
axis theorem) that K =ЅMv _{cm} ^{2} +ЅI _{cm} ω _{cm} ^{2}
where v _{cm } is the speed of the center of mass, ω _{cm
} is the angular velocity about the center of mass (which is the same as
ω )_{ } and I _{cm } is the moment of inertia
through the center of mass. For an object not rolling, K =ЅMv _{cm} ^{2} +ЅI _{cm} ω _{cm} ^{2
} is still the correct expression for kinetic energy at any instant but
you can no longer write the kinetic energy as pure rotation as you could for
the rolling object because there is no relationship between v _{cm
} and ω _{cm} .

QUESTION:
is work energy theorem valid in non inertial frames?

ANSWER:
The work-energy theorem says that the change in kinetic
energy of an object is equal to the work all forces do on it. Imagine that
you are in an accelerating rocket ship in empty space, a noninertial frame.
You have a ball in your hand and you let go of it. You observe this ball to
accelerate opposite the direction in which the ship is accelerating and therefore see its kinetic energy change. But, there are no
forces acting on it so no work is done. Another way you could come to this
conclusion is that the work-energy theorem is a result of Newton's laws and
Newton's laws are not valid in noninertial frames. You can, though, force
the work-energy theorem to be valid if you introduce fictitious forces, a
way to force Newton's laws to work in noninertial frames. (Centrifugal force
is an example of a fictitious force.) If you invent a
force on the objects of mass m in the accelerating (a )
rocket ship above of F _{fictitious} =-ma ,
this force will appear to do the work equal to the change in kinetic energy.

QUESTION:
First, how much energy (in the most basic sense) does a car expend driving at a moderate speed for one mile. Next, how big would a spring have to be (roughly) to store the energy equivalent of what that car expended during that mile.

ANSWER:
I can only do a rough estimate. Suppose your car has an
internal combustion engine and gets 40 mpg. The energy content of 1 gallon
of gasoline is about 120 MJ, so you would use about 120/40=3 MJ of energy
per mile. But, a typical engine has only about 20% efficiency, so the energy
supplied to the car is only about 0.2x3=0.6 MJ=600,000 J. Now, you want to
store that much energy in a spring. For a spring, the energy stored is Ѕkx ^{2}
where x is the amount by which it is compressed (or stretched) and
k is the spring constant which is determined by how stiff the spring is.
(The constant k can be measured by k=F /x where F
is the force you have to exert to stretch or compress it by x .)
Clearly, the bigger k is the less the spring will have to be
compressed to store a given amount of energy. The coil spring of a car, a
pretty stiff spring, has a typical spring constant of k ≈25,000 N/m,
so to store 600,000 J of energy, 600,000≈Ѕ∙25,000x ^{2 } or
x ≈7 m! I do not see much possibility of having a spring powered car. You
probably had in mind a spiral spring like a watch spring, but to store that
much energy in any kind of spring is going to be impractical.

FOLLOWUP QUESTIONS:
Thanks for helping me with that.
I will admit that I'm still a little unclear about the answer though.
You are correct that I am wondering about a spring powered car but I definitely would not use a spiral spring. A compression spring would probably work best.
I don't understand your final answer. I am getting lost in the spring conversion formula. What does the final answer mean in layman's terms please?

ANSWER:
The spring would have to be compressed by 7 meters. That would mean that it would need to be at least twice that long, about 40 feet! You could use a stiffer spring
(see below) to get a smaller required compression, but imagine the force you would have to exert to compress it. Keep in mind that you would have to supply the energy to the spring in the first place, a lot of energy.
I know this spring is not going to work, but how much force would it
take to hold it compressed by 7 m? F=kx =25,000x7=175,000 N≈40,000
lb. Another issue is that as the spring unstretches, the force it exerts
gets smaller, so you would need to have some kind of governor mechanism
to deliver the energy smoothly.

CONTINUED
Also, there are two things that are working in favor of the feasibility of the spring car:

I know from researching that there are many other factors that affect the gas- to engine- to -transmission to -wheels to- road, etc cycle so that the total efficiency of the entire loop would be far less than 20%.

ANSWER:
This is taken care of by the fact that I have specified the
miles per gallon for the hypothetical car. The miles per gallon you get is
determined by how well you have minimized the effects other than engine
efficiency , mainly frictional effects. You asked for an estimate of the
energy which the car would use to keep going a constant speed, and this is
the best way I can think of doing that. Here is another way to estimate the
energy consumed: suppose that I push on the car to keep it going a constant
speed. I could probably do that with a force of about 100 lb≈445 N; one mile
is about 1600 m, so the work I do is the product of the force times the
distance and the work I do is the energy I use: W =445x1600≈712,000 J.
(Actually, I am pretty impressed by how close this is to my other estimate
of 600,000 J! It's just an accident that they are so close, but good that
they are of the same order of magnitude—it increases my confidence that the
energy consumed is pretty well approximated.)

CONTINUED
Despite my very basic and incomplete understanding of the formulas that give exact measurements for potential energy in springs and the potential energy available in compressed air it is obvious to me that if a car can run 300 miles off of a bottle of compressed air and a compressed air driven motor (these cars already exist and are in production in places like holland, india, etc) then a large spring perhaps three feet long and two feet around with a wire diameter of over half an inch could at least do the same.
Getting the potential energy out of the spring in a useful way to power the car is another ball of wax altogether though.

ANSWER:
If you want a 3 ft long spring, about 1 m, the most you could
probably compress it is about a half meter. To store 600,000 J of energy,
the equation would be 600,000=Ѕkx ^{2} =Ѕk (0.5)^{2} =k /8,
and so the spring constant would be k =4,800,000 N/m. Then the force
to hold it at 0.5 m would be F=kx =2,400,000 N≈540,000
lb. If you were not faced with such enormous forces, like if you just wanted
to have a toy car go a few meters, you could certainly get the energy out of
the spring with a cleverly designed gear box; but the structural problems
you would face with forces of hundreds of thousands of pounds would be
insurmountable, I believe.

QUESTION:
In the case of a block that's dropped some vertical distance onto a spring,
it's reasonably easy to compute this value & to find the KE of the mass just before it hits the spring, hence the velocity.
What doesn't ever seem to be explained is what happens to the block's acceleration ("g" when it hits the spring). The motion of the block continues downward, but now the net force is Fnet = kx - Weight = ma, not Fnet = mg. This means that the acceleration continues to be positive as the block compresses the spring, but @ a slower rate. This also means that the velocity continues to increase until all the block's PE is converted to spring elastic energy (Eelastic). At maximum compression the block stops & motion ceases. Since the ideal spring is massless there are no losses due to friction, heat, sound, et. al.
Where does the Vmax occur? It's hard imagine for Vmax to happen @ the instant it stops? Is this process for the spring's resisting force linear?

ANSWER:
OK, suppose the speed when it hits the spring is v _{0} .
Choosing +y to be up and y =0 at the end of the uncompressed
spring, ma _{y} =-mg+k (-y ). (Note that when the
spring is compressed its force is upward and y <0 so the force is in
the +y direction; this is why I write the force of the spring as k (-y ).
Now, until y=-mg /k , the acceleration is negative (points down)
and the mass is still speeding up; below that point, the net force is
positive (points up) so the mass slows down. Therefore the greatest velocity
will be at y=-mg /k . Note that this position would be the
equilibrium position if you gently placed m on the spring. You could
also get this answer with energy conservation; using the same coordinate
system, E= Ѕmv _{0} ^{2} =Ѕmv ^{2} +Ѕky ^{2} +mgy .
This can be rearranged to give v ^{2} =-(k /m )y ^{2} -2gy +v _{0} ^{2} ;
if you differentiate this with respect to y and set equal to zero,
you find y=-mg /k for the maximum velocity position. At that
position you can solve for the speed, v= √[(mg ^{2} /k )+v _{0} ^{2} ].

The key here is that you need to be very careful
with the coordinate system and the signs of forces and potential
energies. If you do not choose +y up, the potential energy will
not be mgy (it would be -mgy ). If you do not choose the
unstretched spring as zero potential energy for the spring, Ѕky ^{2 }
will not be the potential energy.

QUESTION:
How long would it take for a 50lb solid steel sphere to sink a quarter mile through ordinary water?

ANSWER:
I will assume that the sphere quickly reaches terminal
velocity so that I can assume that it goes the whole half mile with that
constant speed. This should be an excellent approximation. The terminal
velocity may be written as v _{t} =√[2mg /(ρAC _{d} )]
where m is the mass, g =9.8 m/s^{2} is the acceleration due to
gravity, ρ is the density of water (1000 kg/m^{3} ), A is the
cross sectional area of the sphere, and C _{d} is the drag
coefficient which is 0.47 for a sphere. There is one catch here, that this
is without buoyant force and the buoyant force in water is equal to the
weight of the displaced water which is not negligible here; when it comes
time to put in the mass I will put in an effective mass of the mass of the
steel sphere minus the mass of an equal volume of water. I will work in SI
units, so 50 lb=22.7 kg and 0.5 mi=805 m. The density of steel is 7850 kg/m^{3}
so the volume occupied by 22.7 kg is V =22.7/7850=0.0289 m^{3} ;
the radius of a sphere with this volume (using V =4πR ^{3} /3)
is R =0.0884 m and so the area (using A =πR ^{2} )
is A =0.0245 m^{2} . Finally, the effective mass would be m =22.7-1000x0.0289=19.8
kg. Putting all these into the equation for v _{t} , I find
v _{t} =5.81 m/s. The time an object going this speed takes to
travel 805 m is 805/5.81=139 s, 2 minutes and 19 seconds.

QUESTION:
A plumb bob is hung from the ceiling of a train compartment. If the train moves with an acceleration 'a' along a straight horizontal track, the string supporting the bob makes an angle with the normal to the ceiling whose tangent is 'a/g'. Suppose the train moves on an inclined straight track with uniform velocity. If the tangent of angle of the incline is 'a/g', the string again makes the same angle with the normal to the ceiling. Can a person sitting inside the compartment tell by looking at the plumb line whether the train is accelerated on a horizontal straight track or it is going on an incline with uniform velocity? If yes, how? If no, is there a method to do so?

ANSWER:
First you should carefully read an earlier answer on the
accelerated pendulum . (Actually, I see that you
are the person who asked that question!) To answer your question, you can't
tell by "looking" but you certainly can tell by measurements. For example,
in the accelerated problem the tension in the string is m √(g ^{2} +a ^{2} );
for the inclined track the tension is mg . Or, in the accelerated
problem you feel youself being pushed back with a force (fictitious) Ma ,
but on the inclined track you feel yourself being pushed back with a force
(real) Mg sinθ =Mg [a /√(g ^{2} +a ^{2} )].

QUESTION:
what would happen if someone were to HOLD the gun and shoot it. The person would not be tethered to anything and would be floating freely in space. If that person is around 160Ibs and shot a High power rifle how fast would it project the wielder in the opposite direction? I know it wouldnt be as fast as the bullet because of the mass of the person but I'm curious to what extent they would be projected into space... if thats would even happen at all.

ANSWER:
First, read an earlier answer to a
question similar to yours. The concept you want to use here is momentum conservation.
Momentum is the product of mass times velocity and the total momentum of
a system must be the same before and after the gun has fired. The mass
of a 160 lb man plus his gun is about 75 kg, the mass of a typical bullet is
about 0.015 kg, and the muzzle velocity of a high-power rifle is about 250
m/s. So, since the momentum is zero before the rifle is fired, 0=75v -250x0.015=75v -3.75
or v =0.05 m/s which is about 10 ft/min.

QUESTION:
A rigid container filled with air is placed in vacuum. If a small hole is created on one side of the container, air leaks out and the container moves in the opposite direction. How would the container move if the situation were reversed, i.e. a rigid container of vacuum placed in air with a hole on one side of the container? I can't seem to apply Newton's 3rd law and momentum conservation to solve this convincingly.

ANSWER:
Let's just think of a cubical box with a hole in one face.
Any molecule which finds the hole will enter the cube, go to the opposite
face, collide with it, and rebound, thereby transferring some momentum to
the box. The box is now moving in the direction in which the molecule was
originally moving (I will call that the forward direction). The rebounding
molecule will either go back out the hole or hit inside the cube and bounce
back again. As long as the molecule stays inside the box, the net effect
will be zero but eventually it will find the hole and so the net effect will
be one collision moving the box in the forward direction. Now think of a
huge number of molecules entering the hole. At the beginning, more will be
coming in than going out so the net force on the box will be forward.
Eventually, there will be the same density of molecules inside and outside
the cube so the net force will become zero. Bottom line—the cube moves in
the same direction as if air were being released from it. An intuitive way
to see this is to note that each entering molecule carries a momentum in the
forward direction, so that is the momentum available to be transferred to
the cube.

QUESTION:
I have a question about force on a lever. I am building a set of oars
for a whitewater raft. The industry is full of opinions but very few
physicists. We all agree that the force is greatest at the oar lock
(fulcrum) but nobody has any idea how much less force is present at the
neck (the narrow part just above the blade). If you will do this
one...here are the values. Using a 10' oar the fulcrum is at the 32-36"
mark and the neck is at the 90" mark. The blade occupies the remaining
30". We taper the oars knowing they don't need to be as strong away from
the fulcrum but nobody knows how much we can taper because we don't know
how much less force they need to withstand. The taper results in the
neck having anywhere from 65% to 80% of the wood on the shaft. I suspect
80% is overkill. Any thoughts.

ANSWER:
A disclaimer: I am not an engineer and an engineer would
probably be a better person to ask this question. When you are rowing, the
water exerts a force, call it F , on the blade of the oar. I am
thinking that the thing we should be thinking about is the torque which this
force exerts about the lock compared to the torque it exerts on the neck.
This force may be thought of as being 15" from the neck, approximately in
the middle of the blade; so the torque from the water force at the neck
would be 15F . The torque at the lock would be, assuming the lock is
35 inches from the handle, about 55"; so the torque from the water force at
the lock would be 55F. So the torque on the lock would about 55/15≈3.7
times bigger than the on the neck. Now we get into some pretty complicated
materials engineering, see this
link .
It turns out that the stress σ is inversely proportional to a
quantity Z called the section modulus. For a cylindrical shape of
radius R , the section modulus is Z =0.78R ^{3} .
So, I surmise that an estimate of how much smaller the neck would be than
the lock would be R _{neck} ^{3} /R _{lock} ^{3} ≈1/3.7
or R _{neck} ≈0.65R _{lock} , about 2/3 the
thickness.

QUESTION:
If I have a ramp that is 28 feet long, fixed at the upper end (shore) and weighs 400 lbs that has a 6 foot rise, how do I calculate the weight at the lower end (dock)? I am trying to determine how much floatation I need under the water end to support the weight of the ramp at that end. That rise varies during the course of the year from zero feet to a maximum of 7 feet.

ANSWER:
I am assuming that the ramp is a uniform plank, that is, that
its center of gravity is at its geometrical center (14'). Refer to the
picture on the left. Two equations must be satisfied for equilibrium, the
sum of all forces must equal zero and the sum of all torques about any axis
must be zero. The first condition gives us that F _{1} +F _{2} -400=0
and the second condition (summing torques about the center of the ramp)
gives us that F _{1} -F _{2} =0. Solving these two
equations, F _{1} =F _{2} =200 lb. Note that the
answer, 200 lb, is independent of the rise.

QUESTION:
Two dice are suspended in outer space with no visible forces acting on them. Their center of masses are 10 cm apart, and they each have an identical mass of .0033 kg. How long would it take for the force of gravity between them to cause them to touch? (We will assume they are volumeless for more ease in calculation).

ANSWER:
This seems a very difficult problem because the gravitational
force between them changes as they get closer and so it is not a case of
uniform acceleration. However, this is really just a special case of the
Kepler problem (the paths of particles experiencing 1/r ^{2} forces) which I
have done in detail
before .
You can go over that in detail. For your case, K=Gm _{1} m _{2} =6.67x10^{-11} x(3.3x10^{-3} )^{2} =7.26x10^{-16}
N∙m^{2} /kg^{2} , the reduced mass is
μ=m _{1} m _{2} /( m _{1} + m _{2} )=0.0033/2=1.65x10^{-3}
kg, and the semimajor axis
a =2.5
cm=2.5x10^{-2} m. Now, from the earlier
answer ,
T =√(4πμa ^{3} /K)= 5.98x10^{4}
s. The time you want is T /2=2.99x10^{4} s. This is only 8.3
hours and seemed too short to me. To check if the time is reasonable, I
calculated the starting acceleration and assumed that the acceleration was
constant and each die had to go 5 cm; this time should be longer than the
correct time because the acceleration increases as the masses get closer.
The force on each die at the beginning is K /r ^{2} =7.26x10^{-16} /0.05^{2} =3.04x10^{-13}
N; so, the resulting initial acceleration is F /m =3.04x10^{-13} /3.3x10^{-3} =9.21x10^{-11}
m/s^{2} . So, assuming uniform acceleration, 0.05=Ѕat ^{2} =4.61x10^{-11} t ^{2} .
Solving, t =3.3x10^{4} s. So, the answer above is, indeed,
reasonable.

QUESTION:
I have a question concerning a magnet suspended inside a copper tube. Does the copper tube accumulate the mass of the magnet? In other words, does the copper tube now weigh more with the magnet suspended in the middle? or is it partial weight because the magnet does fall inside, albeit slowly.

ANSWER:
First, some terminology. Weight is the force which the earth
exerts on something, so the weight of the copper tube is always the weight
of the copper tube. If a magnet falls through a copper tube, it induces
currents in the copper and these currents exert a force on the magnet which
tends to slow it down. In fact, the force becomes strong enough that the
magnet quickly reaches a terminal velocity—it falls with a constant speed.
That means that the tube is exerting an upward force on the magnet equal to
the weight of the magnet. But, Newton's third law says that if the tube
exerts a force on the magnet, the magnet exerts an equal and opposite force
on the tube. Therefore, if the tube is standing on a scale, the scale will
read the weight of the tube plus the weight of the magnet, but that does not
mean that the tube got heavier. It is just the same as if you put the tube
on the scale and pushed down on it with a force equal the weight of the
magnet; you would not say that the tube got heavier because you pushed on
it. A good demonstration of this can be seen at
this link .

QUESTION:
Suppose a constant force is acting on an particle, due to which the particle is accelerated. The the velocity of the particle is increasing at a constant rate. Now if I use the relation P=F.v, i get the power delivered to the particle different at different instants since the velocity is different at different instants. But it is difficult for me to digest that though the force applied is constant, the power goes on increasing. Am I thinking right?

ANSWER:
What is power? It is the rate of change of energy. In the
example you give, a constant force in one dimension, the energy at any
instant is Ѕmv ^{2} . Just to illustrate, let's let m= 2
kg, a =1 m/s^{2} , and the mass begins at rest at t =0.
After the first second, E _{1} =Ѕx2x1^{2} =1 J; after
the second second, E _{2} =Ѕx2x2^{2} =4 J; after the
third second, E _{2} =Ѕx2x3^{2} =9 J; after the fourth
second, E _{2} =Ѕx2x4^{2} =16 J; etc . So, the
average power delivered over the first second is ΔE /Δt= (1-0)/1=1
W; the average power delivered over the second second is ΔE /Δt= (4-1)/1=3
W; the average power delivered over the third second is ΔE /Δt= (9-4)/1=5
W; the average power delivered over the fourth second is ΔE /Δt= (16-9)/1=7
W; etc . The reason you are not "thinking right" is that you deliver
more energy to a faster-moving object with a given force over a given time
because energy is proportional to the square of the speed. Another way to
look at it is that average energy delivered by a constant force F
acting over a distance Δx
is ΔE=F Δx but, in any
given time, the force acts over an ever-increasing distance adding an
ever-increasing amount of energy; of course, that is where your power
equation comes from, P =ΔE /Δt=F Δx /Δt=Fv.

QUESTION:
I've read about space habitat concepts for a while and I've ran into an interesting concept. The concept I've ran into is the McKendree Cylinder which is basically an O'Neill Cylinder made of carbon nanotubes. The O'Neill cylinder made of steel would be 32km long and 6km in diamter. The McKendree Cylinder would be 4600 km long and 460km in diameter. And the maximum length for MvKendree Cylinder is 10000km and diameter of 1000km. So McKendree one could be built a lot bigger than O'Neill one because the carbon nanotubes have greater endurance. But a habitat of thousands of km's seems to be really big when compared to what we can build from other materials. And as I recall we don't have any ways to produce Carbon Nanotubes in large quantities. Is it theoretically possible to build a habitat 10000km long and 1000 wide put of carbon nanotubes. And is the McKendree cylinder more of a theoretical design than a practical design that actually could be built?

ANSWER:
I presume that the issue is more a strength issue than
anything else. To illustrate how the strength of the material and its mass
determine the size the habitat can be, consider a rotating string of beads,
each of mass m . The rotation rate must be such that a =v ^{2} /R=g
where v is the tangential speed of each bead. Therefore each bead
must experience a force F=mg . This force can only come from the two
strings attaching each bead to its nearest neighbors and, from my drawing to
the left, F=mg =2T sinθ . But, we will imagine many, many
beads on this string and we will call the distance between them d ; so
we can make the small angle approximation that sinθ≈θ =d /R .
Solving for T , T=mgR /(2d ). Now imagine that the beads
are atoms; d will be about the same for steel or carbon, g is
just a constant, m _{steel} ≈ 5m _{carbon} ,
and the
Young's modulus of carbon nanotubes is about 5 times bigger than steel,
T _{steel} ≈T _{carbon} /5.
So, R _{carbon} /R _{steel} ≈(T _{carbon} /T _{steel} )/(m _{carbon} /m _{steel} )≈25.
Your numbers are R _{carbon} /R _{steel} ≈460/6=77;
I would have to say that my calculation is pretty good given that I have
made very rough estimates and I am not an engineer! I do not know what
considerations would limit the length of the habitat. (Of course, neither of
these models is presently practical to actually build, so call them
theoretical if you like. However, there is certainly no problem building
them if resources and manufacturing capabilities were available.)

QUESTION:
If I mounted a gun against a force gauge that measured in pounds and fired it in a vacuum and then fired it in my back yard would it read any different? How much of the recoil is actually produced by the gas pushing against the air in front of the muzzle or is it all newton thrust? There are people who claim that brakes on guns actually pull the gun forward counteracting newton force from acceleration which makes about as much sense to me as moving a boat by standing on the deck and pushing on the mast. Others who claim that part of recoil is from newton force and part is from the gas pushing against the air in front of the muzzle.

ANSWER:
This is a little tricky as are most problems which involve
air drag. First of all, look over an
earlier answer about the
recoil of an M4 carbine. To do my rough calculations, I will assume all the
data refer to a gun fired in vacuum; muzzle velocity of 940 m/s, bullet mass
m =0.004 kg, gun mass 2.77 kg, bullet diameter 45 mm. I will assume
that the acceleration along the length of the barrel (0.37 m) is uniform.
(Be sure that you realize that the muzzle velocity is the speed with
respect to the gun , not the ground.) So, the bullet starts at rest and
the equations which give its velocity and position at the end of the barrel
are 940=at and 0.37=Ѕat ^{2} where a is the
acceleration and t is the time to reach the end of the barrel.
Solving these two equations, I find that t =7.87x10^{-4} s and
a =1.2x10^{6} m/s^{2} . Using Newton's second law,
F=ma , we can now estimate the force experienced by the bullet during its
flight down the barrel as 0.004x1.2x10^{6} =4800 N=1079 lb. Now,
suppose that there is air in the barrel. As the bullet flies down the barrel
it will experience air drag which will be a force which will work against
the force propelling the bullet and therefore the muzzle velocity will be
smaller. As you will see from earlier
answers , a fairly good approximation for air drag force on an object
with cross sectional area A going with speed v is F _{drag} ≈јAv ^{2} .
For the 45 mm bullet, A ≈1.59x10^{-3} m^{2} , so taking the average speed to
be 940/2=470 m/s, I find F _{drag} ≈88 N=19.8 lb. This will be,
I believe, an underestimate because the bullet is not just plowing through
the air as it would outside the rifle but compressing the air in front of
it. Now the net force on the bullet is about 4800-88=4712 N. Now, the
average acceleration will be about 4712/0.004=1.178x10^{6} m/s^{2} ,
slightly smaller; the corresponding time and muzzle velocity are t '=7.93x10^{-4}
s and 934 m/s. In the earlier
example I found the bullet speed v and recoil velocity V
to be v =938.6 m/s and V =1.4 m/s; my (very rough) estimate for
including the effects of air are v =932.7 m/s and V =1.3 m/s.
There is slightly less recoil in air. The average force on the gun during
the firing time is the same as the force on the bullet (Newton's third law) ≈4800 N=1080 lb for vacuum, ≈4712
N=1059 lb including air. That sounds like a lot, but keep in mind that it
only lasts about 0.8 milliseconds. I have included a lot of details here, but you can
understand it qualitatively: because the bullet has to accelerate through
the air which causes drag, the muzzle velocity will be smaller so the recoil
velocity will be smaller as well.

QUESTION:
In Halo video game series there are Magnetic Accelerator Cannons on orbit around planets that can launch a 3000 ton magnetic projectile to 4% lightspeed. These cannons use the principle of the coil gun. These projectiles have a kinetic energy of 216000000000000000000 joules which translates to around 51.6 gigatons of tnt. So these cannons seem to have really unrealistic velocities for these projectiles. What would be the problems in developing these cannons to defend the human species from possible alien invaders? I know energy is one but I've heard that that if you were to accelerate a projectile to these kinds of speeds they would turn into plasma from the sheer amount of energy being transferred into them.

ANSWER:
I think you will get the picture of why this is a
preposterously impossible weapon if you read an earlier
answer . There the speed was much higher but the mass much smaller. Here
are the practical problems in a nutshell:

To accelerate it to this speed in a reasonable
distance the force required would be so large as to totally disintegrate
the projectile and the cannon for that matter.

Think about the recoil of the cannon. Unless its
mass was much bigger than 3000 tons, much of the energy expended would
be wasted, not to mention the disruption of the orbit. This would be a
good reason to have it mounted on the ground rather than orbit.

Where are you going to get the necessary energy?
I agree with your number for the kinetic energy of the projectile (Ѕmv ^{2}
works fine for this relatively low speed and a ton here is a metric
ton), ≈2.16x10^{20} J. Suppose it took one minute to get this
much energy; then the power required would be 2.16x10^{20} /60≈3.6x10^{18}
W=3.6x10^{9} GW. This is about 1,440,000 times greater than the
current total power generated on earth of about 2500 GW. (Of course,
that does not take into account the recoil energy of the cannon itself.)

Oh yeah, I almost forgot. There is no evidence
whatever for alien bad guys.

QUESTION:
If I were to drop an empty wine bottle out of an airplane flying at say 35,000 feet above the ocean at 300 mph, would the bottle hit the surface of the water hard enough to break the bottle? I read somewhere something about terminal velocity being 120 mph, so would the resistance of the atmosphere slow the wine bottle to 120 mph by the time it made impact with the ocean? And would 120 mph be enough to shatter the wine bottle, or would it depend on how choppy the seas were versus a flat water surface?

ANSWER:
When I answer questions involving
air drag and terminal velocity, I
usually use the approximation that (in SI units) the force F of air
drag is F≈ јAv ^{2 } where A is the area presented
to the wind and v is the speed. So, as something falls, the faster it
goes the greater the drag force on it so that, eventually, when the drag
equals the weight, the object will be in equilibrium and fall with constant
speed. Since the weight W is mg where m is the mass and
g =9.8 m/s^{2} , the terminal velocity can be calculated: јAv _{terminal} ^{2} ≈mg
or v _{terminal} ≈ 2√(mg /A ). So the
terminal velocity depends on the mass and size of the falling object and
your 120 mph is most likely not correct. Also, how it falls determines the
terminal velocity since it has a much bigger area falling broadside than
with the top or bottom pointing down. I figure that if it falls broadside
there will be a bigger pressure on the fat side than the neck which will
cause a net torque which will make it want to turn with its neck pointing
down; so I will assume that is how it falls . I happened to have an empty
wine bottle in my recycle bin which has a mass of about 0.5 kg and a
diameter of about 8 cm. When I calculate the terminal velocity I get v _{terminal} ≈ 63
m/s=140 mph. The 120 mph number you heard was probably a typical terminal
velocity of a human, and it is just coincidence that the wine bottle has a
terminal velocity close to that.

It is hard to say whether it would break or not. I
think probably not. Suppose that it took 1 s to stop. Then the average
force on the bottle would be F=ma =(0.5 kg x 63 m/s)/(1 s)=31.5
N≈ 7 lb which the bottle should be able to withstand. I
know that they say that at high speeds hitting the water is like hitting
a brick wall, but if the stopping time were 0.1 s the force would still
only be about 70 lb.

(Who would have thought that I would find a picture
of a falling wine bottle? You can find anything on the web!)

QUESTION:
This question is regarding Newton's cradles. Let's say a particular
one had 5 balls. We pull back 3 balls and let go. Now the other side
must rebound with 3 balls as well. This means that the middle ball must
carry on swinging. Does it ever theoretically stop for the briefest of
moments? What about in real life with compression and other factors?

ANSWER:
I presume you have read the recent
answer below . Usually, in collision problems, we ask what is going on
before the collision and after the collision, but not during the collision.
The details during the collision time depend on the details of the
interactions among the balls which we generally do not know. So in cases
like Newton's cradle, we approximate the collision to occur in zero time and
therefore approximate all accelerations to be instantaneous; this is clearly
not the case since it would require an infinite force to stop or start a
ball instantaneously. So, to be more realistic, we must, as you have done by
mentioning "compression", devise some model for the collisions. With steel
balls we could assume that each was a very stiff but perfectly elastic
spring which compressed during the collision. Then, assuming the collision
were perfectly elastic (again an approximation), the collisions would happen
in a very short but nonzero time. The two incoming balls would stop in a
short time because the middle ball would exert a large backward force on
them and the two outgoing balls initially at rest would get up to speed in a
very short time because the middle ball would exert a forward force on them.
Therefore, with this simple model, assuming all the balls are identical, the
middle ball would experience a zero net force during the collision time and
so it would proceed forward never changing its speed. I could imagine that
the middle ball might slow down slightly and then speed back up to its
original speed, but not stop. Even if the collisions were perfectly
inelastic, the middle ball (and all others) would move with speed 3/5 the
speed it came in with afterwards. Whatever happens, the middle ball would
certainly never be at rest.

QUESTION:
This one has bothered me since seeing it on TV the other evening. The device called Newton's Cradle, that executive metal ball swing toy found on many desks - I understand the basic principles of the conservation of motion, energy, etc. My mind can also wrap itself around the fact that when you raise and drop one ball, one ball on the opposite side responds by moving. What I don't understand though is why when you raise and drop two balls, TWO balls on the other side respond by moving. It would seem that to conserve energy/motion, the kinetic energy of the two balls that was raised would be transferred into the last ball and it would swing out twice as much as the two balls that started the motion. How do the two balls on the end "know" that it was two balls that started it? Can you please explain this?

ANSWER:
Be clear that the approximation to use is that the collisions
are elastic, both energy and momentum must be conserved; for steel balls,
this is a pretty good approximation. OK, let's assume that two balls come in
with speed v _{1 } and one ball goes out with speed v _{2} .
Conserving momentum, 2mv _{1} =mv _{2} or v _{2} =2v _{1} .
Now, look at energy: E _{1} =Ѕ(2mv _{1} ^{2} )=mv _{1} ^{2}
and E _{2} =Ѕmv _{2} ^{2} =Ѕm (2v _{1} )^{2} =2mv _{1} ^{2} .
Energy is not conserved, mv _{1} ^{2} ≠2mv _{1} ^{2} .
You can actually prove if that a mass M comes in, only if the mass
going out is M will both energy and momentum be conserved. Here is
the proof:

momentum: MV=mv => v =(MV /m )

energy: ЅMV ^{2} =Ѕmv ^{2} =Ѕm (MV/m )^{2
} => M=M ^{2} /m => m=M

since m=M, v=V from momentum.

Nature "knows"! You can't fool Mother Nature.

QUESTION::
I have searched for an answer to this question in vain.
A dropped large ball, with a small ball on top, will bounce and transfer the upwards momentum into the small ball, launching the small ball very high into the air. (As far as I understand this, I saw a demo).
Ignoring air resistance for now, how heavy would the large ball have to be, to launch a one kilogram weight into orbit, and what would the height of the drop have to be?
As an example, a 100 tonne ball dropped from 10 metres, with a 1kg ball on top, does a bounce, what is the velocity result for that?
How does the elasticity of the ball affect it? (eg, a large metal ball or weight might just impact the ground).

ANSWER:
This problem is fully explained and worked out at
this link . The final result, if the mass of the big ball is much bigger
than the mass of the little ball, is that the speed the little ball rebounds
is given by v ≈3√(2gh ) where h is the height from which it was
dropped. This means that the masses really do not matter. The height to
which the smaller ball bounces is 9 times the height from which it was
dropped. It is assumed that all collisions are perfectly elastic. The speed
required for a near earth orbit is about 8x10^{3} m/s, so h=v ^{2} /(18g )=3.6x10^{6}
m! To put this in perspective, the radius of the earth is about 6.4x10^{6}
m, so h is about half this. This is a really rough calculation because g
will be considerably smaller (by a factor of about 0.4) at this altitude.
Even if you did make this work, the ball goes straight up and you need it to
go horizontally to go into orbit. I think it is not a very practical idea!

QUESTION:
What role does the force of friction play in the movement of a cycle....?? I have learnt that the friction acts in the forward direction on the rear wheel and backward direction on the front wheel.How can it be...?? I think I am missing a broader link to why a wheel actually rotates.

ANSWER:
In this discussion I will ignore all friction except that
which occurs due to the contact of the wheels with the road. The two kinds
of friction we normally learn about in elementary physics courses are static
and kinetic friction. Kinetic friction is the frictional force between two
surfaces which are sliding on each other; it is kinetic friction which stops
a box sliding across the floor. Static friction is the force between two
surfaces which are not sliding on each other; it is static friction which
keeps your bike from skidding when turning a corner. A third contact
friction force is called rolling friction; this is not terribly important
for a bike but is the force which will eventually stop you if you coast on
level ground and you are most aware of it if your tires are
under-inflated—harder to pedal. Rolling friction on both wheels will always
point backwards. First think about a bike which is not skidding. If you go
in a straight path on level ground without pedaling, only rolling friction
stops you and neither kinetic nor static friction are in play. Now suppose
you start pedaling to accelerate forward. Think about what your back wheel
"wants to do"; if the road were icy, the wheel would spin and the force
which keeps it from spinning on a dry road is static friction. The road will
exert a forward force on the wheel to keep it from spinning. A force is
required to accelerate anything and this static friction force is what
accelerates your bike forward. The friction on the front wheel is just the
rolling friction backwards. Finally, suppose you brake: if both brakes are
applied gently enough that you do not skid, static friction on both front
and rear wheels will point backward; if both wheels skid during braking,
kinetic friction on both front and rear wheels will point backward. You
probably know that if you apply the brakes so that they are not quite
skidding, you will stop in a shorter distance than if you skid. When you
round a curve, both wheels have a static friction force which points
perpendicular to your direction and toward the curve's center because you
are accelerating when you move in a circle even if you are going with
constant speed. More detail on
rolling friction and on
bicycle turning can be found
in earlier answers.

QUESTION:
I had asked you a question and you had answered it. To illustrate your point, you had quoted an example that a pendulum hanging from the ceiling of a car moving with a uniform acceleration makes an angle with the vertical. Now we have been taught in our course as well. We have also learnt that tangent of the angle is equal to a/g. My question is that will that pendulum exhibit uniform oscillations if it is displaced from its normal position?

ANSWER:
I presume that by "its normal position" you mean its new
equilibrium position off the vertical. The answer to your question is that
it will exhibit oscillations about this position; the period, though, will
not be given by T ≈2π √(L /g ) but rather by T' ≈2π √[L /√ (g ^{2} +a ^{2} )].
This is most easily understood by introducing a fictitious force ma
opposite the direction of the acceleration. Now you can see that the force
exerting a torque on this pendulum has a magnitude of m√ (g ^{2} +a ^{2} )
rather than mg as in the unaccelerated pendulum. So it is just like
the usual pendulum analysis with a somewhat larger acceleration due to
gravity.

QUESTION:
Are the equations of motion v = u + at, S = ut + 0.5at^2 , and v^2 = u^2 + 2aS applicable if a particle travels at a speed close to that of light? I think that they are applicable, and only F =ma is not, but my friend says that those 3 equations are also not. Who is right?

ANSWER:
Your friend is right. These are the classical equations for
uniform acceleration, and uniform acceleration is not possible in the theory
of special relativity for a constant force. Constant force, where F =dp /dt ,
(which, incidentally, is the way Newton originally wrote his second law) can
be constant but does not result in constant acceleration. I have treated
this problem in detail in an
earlier answer ; there I find that
v =(Ft /(m ))/ √[1+(Ft /(mc ))^{2} ]
and
S =(mc ^{2} /F )( √[1+(Ft /(mc ))^{2} ]-1)
(for your
u =0,
that is, the particle at rest at
t =0)
for constant force.

QUESTION:
This is a question about tractive effort of locomotives. I am a model railroader and we have been debating this situation for a while with no clear answer. If there is a locomotive on the track that has 4 driving wheels touching the rails, and the only change you make is to the number of wheels , now 6 driving wheels on the track, will tractive effort go up, stay the same or drop. The argument for staying the same is that each wheel now supports less weight, so tractive effort of each wheel would drop. The argument for tractive effort going up would be that there are more contact points on the rail, so it would pull more.

ANSWER:
What drives your train is the static friction between the
wheels and the track. The nature of static friction is that a maximum amount
of force may be achieved before the surfaces slip on each other. This force
F _{max} is determined, to an excellent approximation in many
cases, by the nature of the surfaces (steel on steel for your case, I
presume) and how hard they are pressed together (called the normal force
N ); F _{max} =μN where μ is the coefficient
of static friction, a number determined by the materials. You will note that
this force does not depend on the area of contact. Essentially, adding
wheels adds surface area but, as you note, the force pressing the surfaces
together (namely the normal force which is determined by the weight) is
simply distributed over a larger area. So, the simplest first-order physics
answer is that the same maximum force from the wheels without slipping will
be achieved regardless of the number of wheels. Friction, though, can be a
tricky business and first-order physics does not always work for all
situations. For example, there are
instances where a car
tire can have greater "road hugging" (read increased friction) if the
road-tire contact area is increased. I believe in your case, though, since
the wheels and track are so little deformed by contact, that increasing the
number of wheels will not increase friction provided that the total
mass of the locomotive remains the same. Of course, if you are adding wheels
to an existing locomotive, you are adding mass so you will be increasing
F _{max} . There is a very easy way to test this. Put the
locomotive (wheels not free to rotate) on a piece track which is attached to
a horizontal board. Gradually increase the angle of the board until the
train slides down. The tangent of the angle at which it starts to slide will
be equal to the coefficient of static friction. If increasing the number of
wheels does not increase the slip angle, there is no advantage.

QUESTION:
If a plane, while mid-flight, had an explosion from the wing or rear area, would the debris maintain the exact same momentum as the plane and go forward alongside it, would the debris go in front of the plane, or would the debris go behind the plane when the accident first happens? Me and two friends got on the topic after watching a movie dealing with a plane crash. One friend thinks debris would go in front, the other thinks the debris would maintain precise momentum with the plane, and I'm thinking it would go behind the plane. We'd really love for a physicist to clear up this question for us, thank you so much!

ANSWER:
What anything does is determined by the forces on it. A plane
flying in a straight horizontal line with constant speed has four main
forces acting on it: gravity (its own weight); lift which is the force which
counteracts the gravity to keep it flying level, drag caused by the air, and
the forward force exerted by the engines which counteracts the drag to keep
it from slowing down. If a piece of the plane suddenly separates from the
plane, it no longer has any foward force and it no longer has any
significant lift; so, it will start dropping vertically and slowing down
horizontally, falling down from and behind the plane. The falling is
generally more prounounced than the slowing down (gravity usually a greater
force than drag) so there would be a tendency for the piece to appear to
just drop straight down as seen from the plane. (Be sure to realize that the
forward velocity of the piece is approximately maintained so that someone on
the ground sees it moving forward with about the same speed as the plane.)
This is often seen in bombs dropped from a plane. On the left photograph the
drag on the bombs is small so they keep pace with the planes as they drop.
On the right, the bombs have little parachutes to increase drag and so they
fall behind the plane.

QUESTION:
If some object, at some distance far from the surface of the Earth, but much closer to the Earth than anything else (meaning everything else in the universe is negligible) is at rest initially but then begins accelerating towards the Earth due to gravity, how long will it take to get to the Earth?
I get that the acceleration is a=GM/r^2. And then you can write a as the second derivative of r which would make it a non-linear differential equation but I have no idea how to solve those really. I have taken a differential equations course but it was 3 years ago so I don't remember if we even solved non-linear ODE's ever.

ANSWER:
I have solved variations of this problem twice before,
one very recently regarding Coulomb's law
(simply another 1/r ^{2} force) and
the other where the object
was falling into the sun instead of earth. You should read through these
first since I will skip a lot of the detail here. In the spirit of those two
answers, I will take you at your word that the object is "far from the
surface of the earth", that is, r>>R _{earth} . Hence, it is
just the Kepler problem with an orbit of eccentricity 1 and semimajor axis
a=r /2 and we need to find half the period T . Kepler's third
law states that T ^{2} =4π ^{2} a ^{3} /(GM _{earth} ).
Therefore, T /2=π√ [r ^{3} /(8GM _{earth} )].
For example, if r= 100R _{earth} (which would mean at an
altitude of 99R _{earth} ), T /2=9x10^{5}
s=250 hr.

You can always estimate how much error is made in this approximation
(assuming the earth and objects are point objects going all the way to zero
separation) by calculating the speed v the object arrives with at
earth's surface. Energy conservation gives -GM _{earth} m /r =Ѕmv ^{2} -GM _{earth} m /R _{earth}
or v =√[2GM (1/R _{earth} -1/r )]. For
the example I did, v =√[(198/100)GM _{earth} /R _{earth} ]=1.1x10^{6}
m/s. If the earth were a point mass, the object would continue speeding up.
If it kept going into the earth at a constant speed, the time it would take
to reach the center would be about 6 seconds, enormously shorter than the
time to get to the surface, so the approximation is superb for
r= 100R _{earth} .

QUESTION:
i tried to find the time it would take for two charges to collide
under electrostatic force,realizing simple kinematics won't cut it,tired
to integrate but failed,how is the question done?and how does it differ
from gravitational force??

QUERY:
You have to define what "collide" means. Since it is a 1/r ^{2} force, if you use point charges the velocity will be infinite when they collide but they will do so in a finite time. Also, what are the initial conditions (velocities, positions), masses, charges.

REPLY:
My initial question was the time it would take for a 1/r^2,to
collide,for example two bodies lets say 1g,and a charge of 1micro
coluomb,initial at rest attract each other,and collide,I can't use
simple kinematics to solve this question,what shall I do?

QUERY:
How far apart are they?

REPLY:
ok,for simplicity 1m apart,is there a general formula than can help?

ANSWER:
Whew! I finally have everything I need. This is the Kepler
problem, the same, as you suggest, as the solar system with gravity. You
may want to look at an
earlier answer similar to yours. It is very lengthy to work out the
whole problem in detail so I will refer you to a very good
lecture-note document from MIT; I will just give you some of the
necessary results to calculate what you want. First, a brief overview of
two of Kepler's laws:

Kepler's laws refer to problems where the force
is of the form F =K /r ^{2} where K is
a constant and the force is attractive. So it could refer to either two
masses or two opposite charges.

The first law states that bound planets move in
ellipses with the sun at one focus. This is really only true if the sun
is infinitely massive but the generalization still leads to an elliptical orbit
for each body, both of which move around the center of mass of the two.
Still, the semimajor axis a of the ellipse (which we will later
need) in the center of mass system can be found for any orbit from the
simple equation a=-K /(2E ) where E is the energy of
the system.

For your case, the particles move in a straight
line toward each other and then turn around and return to their original
positions. This is just the most elongated possible ellipse with an
eccentricity of 1. Of course this would never really be possible in the
real world since the particles would be going an infinite speed when
they "collide". That means we really should do the problem
relativistically which would greatly complicate the problem. Keep in
mind that you are asking an unphysical question requiring point charges
and infinite forces and velocities. But the answer below should be a
good approximation of the time if they have some finite size small
compared to their initial separation.

The third law relates the period of the orbit
T to a : T ^{2} =4πμa ^{3} /K
where μ=m _{1} m _{2} /(m _{1} +m _{2} )
is the reduced mass. In the gravitational problem, K =Gm _{1} m _{2
} and in the electrostatic problem, K =k _{e} q _{1} q _{2
} where k _{e} =9x10^{9} N·m^{2} /C^{2} .

For your case, the energy is given since the charges
are initially at rest and separated by some distance S , so E =V (S )=k _{e} q _{1} q _{2} /S
and so a =k _{e} q _{1} q _{2} /(2k _{e} q _{1} q _{2} /S )=S /2=0.5
m; the reduced mass in your case is μ=m _{1} m _{2} /(m _{1} +m _{2} )=10^{-3} x10^{-3} /(2x10^{-3} )=0.5x10^{-3}
kg; and K =kq _{1} q _{2} =9x10^{9} x10^{-6} x10^{-6} =9x10^{-3}
N·m^{2} . Finally, the time it takes for a complete "orbit" (which
would correspond to the particles returning to their original positions)
would be T =√[4πμa ^{3} /K ]=√[4π (0.5x10^{-3} )(0.5)^{3} /9x10^{-3} ]=0.3
s. But, the time you want is just half a period, T /2=0.15 s.

To help you visualize the orbits, the figure below
shows the orbits for the two charges when the eccentricity is just less than
1; imagine the orbits getting flatter yet, approaching two straight lines.

NOTE
ADDED:
I got to wondering what the limits of doing this classically
are, that is, how good an approximation my calculation above would be for
some real system. This requires that I determine how close the two charges
would approach each other before their speed v became comparable to
the speed of light c . I will use the same notation as above and write
things classically. If released a distance S apart, then when they
reach a distance r apart energy conservation gives: k _{e} q _{1} q _{2} /S=k _{e} q _{1} q _{2} /r+ Ѕμv ^{2
} which results in v =√[(2k _{e} |q _{1} q _{2} |/μ )(1/r -1/S )].
For the case in point, if I solve for r when v=c /10, a
reasonable upper limit for a classical calculation, I get r =4x10^{-8}
m, about 100 times bigger than an atom. Alternatively, we could ask what the
velocity would be for a 1 mm separation, r=S /1000: v =√[(2k _{e} |q _{1} q _{2} |/μ )(999/S )]=1.9x10^{5}
m/s=0.00063 c . In either case, I think we can conclude that the time
remaining to complete the half orbit will be extraordinarily small compared
to 0.15 s.

QUESTION:
does the terminal velocity in a parachutist's fall occurs two times?
one before opening the parachute
and one some time after opening the parachute Am i right?

ANSWER:
The terminal velocity is determined by (among other things)
the geometry of the falling object. In the simplest approximation, what
matters is the cross sectional area of the falling object. So, when the
parachute is opened, the area gets much bigger and the terminal velocity
gets much smaller. If the sky diver has achieved some larger terminal
velocity, she will slow down to the new terminal velocity. However, this
need not be the first time the terminal velocity changed. The sky diver can
orient in a ball or like a down pointing arrow and have a relatively large
terminal velocity or she can fall spread-eagle and slow down. So the
terminal velocity probably changed several times before the parachute was
opened.

QUESTION:
i am unable to understand a pulley problem

ANSWER:
This is a most peculiar problem! However, its solution ends
up being very easy to do (if not to visualize). I will assume that all
pulleys are massless and frictionless and that the string is massless and
unstretchable. I always tell my students when attacking this kind of problem
to "choose a body" upon which to focus and apply Newton's second law. There
are two obvious choices here, the left-side m which results in
-T+mg=ma _{left} and the right-side m which results in
- 2T+mg=ma _{right} . So, we have two equations and three
unknowns; the best you can do is to find the relation between the two
acclerations, 2a _{left} +a _{right} =3g .
To generate a third equation, choose the left pulley which results in 2T-T =0=T .
So, the only result which works is a _{left} =a _{right} =g ,
both masses in free fall.
I was having a lot of trouble visualizing how this happens, so you I made
a little model and made a movie of it. To see it, go to the
Ask The
Physicist facebook page. (I know, it's not going to win an academy
award, but it satisfied my curiosity!) The two masses, with equal
accelerations, will move equal distances in equal times which then made it
pretty easy for me to make a little model on the floor with strings and
paint cans. Apparently the left pulley moves twice the distance the right
pulley moves in any given time.

QUESTION:
I've been trying to write a action sci-fi screenplay, but there is one problem that I can't get my head around.
During the climax, two characters fall from a 91 story building, with the first jumping off, and the second falling approximately 10 seconds later. The second character proceeds to catch up to the first, and they then brawl.
My question is, how much time would pass before they would hit the ground?

ANSWER:
There is a good reason you can't get your head around it:
the second character will not catch up to the first! I will give you a
little basic physics tutorial on possible scenarios. For purposes of
computation, I will assume that the height of your building is about 400 m
(about 13 ft/story) and I will approximate the acceleration due to gravity
to be g =10 m/s^{2} .

First I will
assume air drag is negligible, that they fall as if they are in a
vacuum. Then the height y _{1} above the ground of
character 1 is given by y _{1} =400-5t ^{2}
where t is the time since he jumped. This tells us that when
character 2 jumps, y _{1} =400-500=-100 m; since y =0
is the ground, he has already hit the ground! The freefall time (the
time when y _{1} =0) from 400 m is about 8.9 seconds and
the speed when he hits the ground will be about 89 m/s=200 mph.

So, maybe we just
need to add some air drag. That certainly will be important for speeds
on the order of 200 mph. When air drag is taken into account, you do not
continue speeding up forever but eventually fall with some maximum
constant speed called the terminal velocity; this happens when the air
drag (up) is equal to your weight (down). If your mass is M , your
weight is Mg or, with my approximation, 10M . Let's use
M =100 kg (about 220 lb) so Mg =1000 N. A good approximation
for air drag is F =јAv ^{2} where A is the
cross sectional area of the falling object and v is the speed. I
will choose A =1 m^{2} so, if I set F =1000 and solve for
v , I find a terminal velocity of v _{t} =63 m/s=140
mph. The details of the math gets a little complicated here, so I will
give you the results and spell out the details below for anybody
interested. I find that the time that it takes to reach the ground now
is about 10.8 seconds at which time the speed is about 59 m/s. So
character 2 has less than a second to catch up to character 1, obviously
impossible.

Finally, I should
tell you that if the two characters are about identical, about the same
size and weight, character 2 will never catch up with character 1
regardless of the height of the building. In the no-air-drag situation,
both have the same acceleration and so their paths never cross. In the
air-drag situation, both have the same terminal velocity so they will
end up having that speed and separated by some constant distance. The
only way to make it happen is to greatly change the cross sectional area
of character 1, for example, give him a parachute!

I am guessing that, if
you want to have any semblance of reality, you will want to rethink your
climax!

ADDED
DETAIL:
Vertical fall with quadratic air drag is a well-known
problem worked out in any intermediate classical mechanics textbook. The two
results which I used were v=v _{t} √[1-exp(-2gh /v _{t} ^{2} )]
and t =(v _{t} /g )tanh^{-1} (v /v _{t} )
for an object dropped from height h .

QUESTION:
Was killing time on a slow day at work, and involved a coworker in the site. Some time later he ask a question, which might be up the physicist alley. Assuming we had a person seal in a vacuum with a piece of paper and a baseball, if he threw the baseball and then crumpled up the piece of paper and threw it. Would the paper fly as far as the baseball having no friction to affect it, or does gravity rule supreme and bring it to a halt just as well as it does outside a vacuum?

ANSWER:
Let's assume that you are throwing both as hard as you can
which means you exert equal forces on them. Also assume that you exert that
force over the same distance for each case which means you do the same
amount of work (force times distance) on each. That means that they have
equal kinetic energies when you release them. But, kinetic energy is ЅMV ^{2}
where M is the mass and V is the speed. So, suppose the
baseball has a mass 100 times the sheet of paper; then, if they have equal
kinetic energies, the speed of the paper will be 10 times larger than the
baseball and therefore go 10 farther. Your last sentence makes no sense
because gravity does not "bring it to a halt", it is friction which does
that.

QUESTION:
Suppose a bullet is fired parallel to the ground, due to perpendicular direction of work done with respect to gravity, no work is done against it. Then why does not the bullet fall immediately to the ground as it does when it is not in motion? I have speculated that motion in in dimension reduces the effects of forces of other dimensions. Is there really such a thing in physics? If not, then what is the correct explanation?

ANSWER:
First, do not worry about what was happening to the bullet
while the gun was firing it; that is past history and has nothing to do with
what happens to the bullet after it leaves the gun. Newton's first and
second laws tell us that the only reason something will not move in a
straight line with constant speed (or be at rest which is a constant speed
of zero) is if there is a force acting on it. Further, the change in motion
the object experiences is in the direction which the force points. Normally
in elementary physics classes, we neglect the force of air drag (which is a
pretty poor approximation for a bullet) such that the only force is that due
to gravity which we call the weight of the object. So, all the bullet can do
is change its motion in the vertical position because that is the direction
that the weight points. So even though your bullet starts with no vertical
motion, it will fall just as it would if you simply dropped it. In other
words, it will take just the same time to drop to the ground whether you
fire it horizontally or simply drop it. In an
earlier answer , you can see
a strobe photograph where a ball launched horizontally and one dropped fall
vertically the same. In the real world, air drag is not negligible and is a
force which always points in the direction opposite the velocity. This
causes the bullet to slow down in the horizontal direction but to also speed
up more slowly vertically. The figure shows an example where the conditions
allow the whole range of scenerios to play out: because of the air drag, the
horizontal motion is eventually stopped and, because of the interplay
between weight and drag, the projectile ends up falling vertically with a
constant speed called the terminal velocity.

QUESTION:
I am in my powerful rocket. If my orbital speed around the sun was zero and I was the same distance from the sun as the Earth, and my rocket was firing at the exact velocity to hold my position to keep the rocket from from falling towards the sun but no greater, what would my 200 lb Earth body weigh on the rocket ship scales? I am guessing a bit less than 800 lbs. What do you say?

QUESTION
TO QUESTIONER:
What does that mean, "my orbital speed around the sun was zero"? If your speed is zero you are not orbiting. If you are at rest a distance of the earth's orbit from the sun, I would call that hovering. Then you have to calculate the sun's gravitational force on you to know your "weight".

REPLY:
You are right, I did wish to know my weight when hovering,

ANSWER:
The force W which the sun exerts on you (aka your
"weight") is given by W=GMm /R ^{2} where G =6.67x10^{-11}
Nm^{2} /kg^{2} is the universal constant of gravitation,
M =2x10^{30} kg is the mass of the sun, R =1.5x10^{11}
m is the distance to the sun, and m =91 kg is the mass of a 200 lb
weight. If you do the arithmetic, the result is W =0.54 N=0.12 lb.
Quite a bit less than 800 lb! You might be interested in a
similar question where
the questioner wanted to know the time it would take the earth to fall
into the sun if it had no orbital speed.

QUESTION:
i have to know that why the mass of bob in a pendulum does not
affect the time of oscillation.....i think it should as the more massive
bob should have more potential energy when string is pulled for
10degrree but at the same time when its moving more air resistance would
also act on it. My book of physics says that the number of oscilllations does not affect the final answer of time period per each oscillation.
as we know that a pendulum should automatically stop moving after some time.
This means that if 100 oscillations are taken then the time per one oscillation should be decreased. I know im wrong as my physics book of GCSE says so but i want to know why im wrong and what part of my staement does not make sense

P.S : please use simple language or formulas(if u
wanna use) as i wont be able to understand complex ones.

ANSWER:
Let's get straight what we are talking about. You refer to
air resistance and you note that the pendulum will eventually stop. Of
course, the main reason it stops is air resistance. Any air resistance is
assumed to be negligible in a treatment of the pendulum at your level (9th
grade), so imagine that there is none. Second, your book should tell you
that the pendulum onl