Here
is a history of older questions and answers processed by "Ask the Physicist!".
If you like my answer, please consider making a
donation to help support this service.
that . The
americium is in a sealed container and
cannot get out. Also, the color of dust
depends on where it comes from. There is
probably some source of brown dust in your
house and you wouldn't notice its color
except when something like your smoke
detector accumulates a relatively large
amount over a long time.
2 until the cork reaches
a terminal velocity of 5 m/s. Why does the
cork reach terminal velocity?
I
can interpret everything completely
literally as written. Of course, this
cannot be physical because the air
resistance F A must
depend on the speed v (usually
F A ∝v 2 )
of the cork and it will certainly not,
in the real world, result in an
acceleration which is constant. If the
mass of the cork is m , then
F A =4m ; there is
also the force of gravity F G =-mg
where g =9.8 m/s2 . So
the net force is F net =ma net =m (4-9.8)=-5.8m ,
so a net = -5.8
m/s2 . So the cork will fall
with a downward acceleration of
magnitude 5.8 m/s2 and never
have a terminal velocity since it just
keeps speeding up.
Maybe you didn't really mean to say it
has a "uniform acceleration" but that
F A =kv 2
at the instant you release it with speed
down of 2 m/s and the acceleration due
to that force is 4 m/s2 ; so
k (2)2 =4m or
k=m . Now we have F net =ma net = (mv 2 -mg );
when v=√g= 3.13 m/s, a net = 0,
so if this is the interpretation of the
problem, the terminal must be 3.13 m/s,
not 5 m/s.
I
cannot think of any interpretation of this
problem which would be self-consistent.
The kinetic energy
E of an object with mass m
and speed v is E =½mv 2 .
If you want the energy to be expressed in
Joules, m must be in kilograms and
v must be in meters per second. If
the object takes a time of t to
achieve an energy E , the average
power expended to do so is P=E /t .
If you want the power to be in Watts, E
must be expressesed in Joules and t
must be expressed in seconds.
M , the light cart m .
There are two ways you can apply the force
F :
Push
over the same distance d for
each. In that case you give the same
kinetic energy to each, ½MV 2 =½mv 2 ,
so V=v √(m /M )
where V (v )is the
starting speed of M (m ).
Note that, as you would expect, v>V .
Push for the same time for each. In that
case you give the same linear momentum
to each, MV=mv , so V=v (m /M ).
Again, v>M but by a different
factor.
Now,
what stops the carts? Friction f
which is proportional to the weight of each
car. For the loaded cart, fM =-μMg
and for the empty cart, fm =-μmg
; here the negative sign indicates that
the force is slowing the cart down, μ
is the coefficient of kinetic friction, and
g is the acceleration due to
gravity. Because of Newton's second law, the
accelerations of the two cars are A=fM /M =-μg
and a =fm /m =-μg.
Because the accelerations of the two
cars are the same, the one which started
fasted (empty) will go farther and take a
longer time to stop.
The
force between nucleons (neutrons and
protons) is very strong and results in
the nucleus being extremely small
compared to the the atom (nucleons and
electrons). The size on an atom is on
the order of Angstroms (10-10
m) where the nucleus is on the order of
femtometers (10-15 m). If the
atom were about the size of a football
field, the nucleus would be about the
size of a golf ball.
One
of the first successful models of the
nucleus was the shell model. Because of
the average force due to all the other
nucleons, each nucleon moves in orbitals
like electrons do in atoms under the
influence of the Coulomb (electric)
force.
A
later, also successful, model is the
liquid-drop model where the particles
all move collectively. Imagine a liquid
drop which, when bumped will oscillate
in some way. Or imagine a nucleus which
is deformed like an American football.
It could rotate about its center of
mass. Many heavier nuclei have
rotational band structures. In these
cases the nucleons all move in
choreographed unison, maybe more like a
line dance than a disco dance.
Hope
this gives you the idea that structure of
nuclei is fairly well understood. The
examples I give are over simplified because,
among other things, when inside a nucleus,
nucleons lose their individuality.
G .
So you cannot just think of how things would
change if G were changed. Usually
the question is not about whether a universe
could be "born" if the fundamental constants
were different (partly because we really do
not know the mechanism of the big bang);
rather, the question is usually "how much
could we change the fundamental constants
and still have a universe where life is
possible?" Here is an interesting example I
read about of the consequences of changing
fundamental constants: If the strong nuclear
force were increased by 2%, the diproton
(two protons bound together) would be
possible. The consequence would be that
stars could not exist because the
proton-proton cycle which produces the
energy in stars would not occur if two
protons could just stick together.
QUESTION:
g -force.
The g -force is technically not a
force at all, it is an acceleration. As you
likely know, the acceleration due to gravity
is g =9.8 m/s2 . The g -force
is usually expressed in g s; for
example, because of some force you
experience an acceleration of 10 g s,
so that means that if your mass is 100 kg,
the force you experience is F=ma =100x10x9.8=9800
N=9.8 kN.
hovering helicopter is being held up by
the atmosphere which rotates with the earth.
No, you cannot travel around the earth by
staying still.
v
in a circle of radius R . It is
attached to string which passes through a
hole in the table; you are holding it in its
orbit by pulling down with just the right
force F (which happens to be
F=mv 2 /R , but we
don't need to know that now). Its angular
momentum is equal to L 1 =mvR .
Now you pull down with a bigger force so
that the new radius is R /2; as you
did this, no forces on m exerted
any torque so the angular momentum was
conserved. L 2 =mv'R /2=L 1 =mvR ,
so v' =2v ., spinning faster
than before your force pulled it in. But,
you argue, if there are many particles with
random velocities there is, surely, for each
particle a second particle which is rotating
in the opposite direction so the net
rotation is zero. But, this argument will
only work exactly if there are an infinite
number of particles. In a very large
assembly of particles even a very small
difference between a pair of particles can
cause a large angular momentum if they are
far away from the center of mass of the
assembly.
F =¼Av 2
where F is the force in Newtons,
A is the cross sectional area in m2
of the object, and v is the
velocity in m/s; you must work in SI units
because the factor ¼ contains constants like
drag coefficient, density of air, etc .
Now, F=ma= ¼mAv2
where a is the acceleration. The
hailstone has a downward force on it, its
weight mg where g =9.8 m/s2
is the acceleration due to gravity; when
v gets big enough that the drag force
(upwards) equals the the weight down, the
acceleration becomes zero. This leads to the
terminal velocity v =2√(mg /A ).
I calculated the mass to be about 0.74
kg=1.6 lb and the area about 0.0167 m2 ,
so v =48 m/s=107 mph.
Regarding the cannon you want to make, you
cannot just know the psi but the distance
over which the force from that pressure will
act. If you want to push it with a force
F over a distance d with a
pressure P , you will (ignoring any
friction) give it a speed v =√(2Fd /m )=√[2PAd /m ]
where m is the mass of the
projectile and A is the cross
sectional area of the tube. So, you can
either push it with a big pressure over a
short distance or a small pressure over a
long distance. Let's use the number I used
above, m=0.74 kg, A=0.0167 m2 ,
and use a d =2 m long tube; then
v= 48 m/s, 48=√(2xP x2x0.0167/0.76).
Solving, I find P =2.83x105
N/m2 =41 psi. Don't forget that
there is atmospheric pressure, 14.8 psi, on
the front of the ball so a rough estimate
would be 56 psi. The actual required
pressure would be quite a bit larger than
this because friction (including air drag)
would not be negligible.
discussed this question in some detail.
The bottom line is that when a bullet hits a
man the recoil is negligibly small.
P 1 V 1 =P 2 V2
(called Boyle's law) which you state is
correct provided that the temperature and
the amount of gas inside the cylinder do not
change. But then you do not apply it
correctly. I take it that you are not adding
air to the cylinder but are compressing the
air in the cylinder. I am also assuming that
the cylinder has a height of 6 in and a
radius of 1 in and that so the volume is
V 1 =6x3.14x12 =18.8
in3 . I also assume that when the
cylinder is at 18.8 in3 there is
1 atmosphere of pressure, P 1 =1
atm=14.7 psi, so P 1 V 1 =276
in·lb. We now have that V 2 =276/P 2 ;
if P 2 =40 psi, V 2 =6.90
in3 ; if P 2 =10
psi, V 2 =27.6 in3 .
Notice that for the 10 psi case the volume
is bigger than the the original volume
because the pressure is smaller than
atmospheric pressure. If you can't make the
volume any smaller than 18.8 in3 ,
the only way to reduce the pressure is to
remove some gas or cool it. If you are
interested, the most general expression for
an ideal gas is PV /(NT )=constant
where T is the absolute temperature
and N is some measure of how much
gas you have.
3 .
In that case, keeping V and T constant, the
appropriate relation would be P 1 /N 1 =P 2 /N 2
or N 2 =P 2 N 1 /P 1 .
Suppose we call the amount of gas you can
put in the cylinder at atmospheric pressure
1 gas unit. Then if you fill the tank with
10 times atmospheric pressure, you will
store 10 gas units.
2 plate of
aluminum (thermal conductivity 225.94
W/mK)that is 0.1 m thick. In the center of
the aluminum plate protrudes a long skinny
aluminum bar that is 20 m high, 0.1 m wide,
0.1 m long. This would look like a dirt
tamper. The temperature of the large plate
is at 100°C and the top of the long skinny
pole is 10°C. How can I calculate heat flow
in this system? I certainly appreciate your
help.
Is the plate maintained at 100
and the top of the bar maintained at 10 and
you want the rate of heat flowing through
the bar? Or do you want to know the final
temperature is allowed to come to
equilibrium insulated from the environment?
And if you actually want an analytic
expression for any point in the system as it
is coming to equilibrium, it would require
that I have the shape of the plate; this
problem probably cannot solved analytically
and would require a numerical solution which
I am not able to do.
Plate is maintained at 100; top heats up but
I understand that is hard to model because
the rate would decrease as the top increases
in temperature (eventually becoming zero
when the top reaches 100). I guess I'm
really interested in the heat flow if the
top is at 10 (so we could imagine it is held
at 10)
I am going to
assume that there is no heat leakage from
the sides of the bar. Initially I will
assume that the top end of the bar is kept
at a constant 10°C, so heat flows through
the bar at a constant rate. We can figure
out the heat rate as a function of the
temperature difference, ΔT =100-T .
I will neglect any edge or geometry effects
in the bar, in other words I will treat it
as a one-dimensional problem where the heat
flow vector in the bar is in the direction
of the bar and uniformly distributed across
the cross-sectional area. In that case the
equation for the rate R is
R =ΔQ /Δt=kA ΔT /L
where
A =0.01 m2 is the cross
sectional area, k =226 W/(m·K) is
the thermal conductivity, and L =20
m is the length. So R =0.113ΔT
W. For ΔT =90, R =10.2
W. Now, if the top remains at 10°C, that
means that energy at the top is being taken
away at the rate of 10.2 W; since ΔT ∝L ,
the temperature must increase linearly along
the bar when equilibrium has been achieved.
The
one-dimensional heat equation is
∂T /∂t =c 2 ∂2 T /∂x 2
where c 2 =k /(C p ρ )
where k is thermal
conductivity, C p is
specific heat at constant pressure, and
ρ is the mass density. (You may find a
derivation
here . ) The general solution of this
equation is
T (x,t )=exp(-c 2 s 2 t )[A sin(sx )+B cos(sx )]+Cx+D
where A ,
B , C , D , and
s are constants to be determined for
the specific problem. (For a derivation, go
here . Go to Section 2.2)
Suppose the boundary
conditions are that (1) the temperature of
the bar is at some constant value T 1 ,
(3) one of the bar is held at a temperature
T 2 , and that (2) the
other end of the bar (and the sides) are
insulated:
T (0,t )=T 0
∂T (x ,t )/∂x|x= L =0
T (x ,0)=T 1
These boundary
conditions lead to the following:
B exp(-c 2 s 2 t)+D=T 0 B =0, D =T 0
exp(-c 2 s 2 t )[sA cos(sL )]+C= 0C =s (exp(-c 2 s 2 t) [A cos(sL )])s ≠0 so C =0A n cos(s n L )=0⇒s n = ½nπ /L,
n odd
∫(T 0 -T 1 )sin(s m x )dx =-Σ∫{A n sin(s n x )sin(s m x )}dx =(L /2)A n δmn n =- (2/L )(T 0 -T 1 )∫ sin(½nπx /L )dx 8(T 0 -T 1 )sin2 (nπ /4)/(nπ )
So,
finally,
T (x,t )=T 0 -Σ{exp(-c 2 s n 2 t )A n sin(s n x )}
c 2 =k /(C p ρ ),
n =8(T 0 -T 1 )sin2 (nπ /4)/(nπ )=4(T 0 -T 1 )/(nπ ),
n =π/L , n
odd
We should tabulate
the constants to be used:
T (x,t )=100-Σ{(115/n)exp(-0.00206n2 t )sin(nπx /40)}
(t is in hours here.)
The plot of this
function including just the first three
terms of the series (1,3,5) is shown in the
first figure. Note that because the bar is
so long it takes the system several hundred
hours to fully equilibrate. The second
figure shows the calculation for the first
hour; this is clearly wrong since the whole
bar is at 10° at the beginning. The reason
for this is that so few terms have been
included in the infinite series. However, to
understand the long-term behavior, n>1 plays
almost no role because the n2
behavior in the exponential damps out higher
n contributions, particularly at large t .
One more
calculation, for the initial temperature
profile linearly decreasing over the length
of the bar may be seen
here .
Here is the trick you need to
understand: Newton's laws, by which we
usually do classical physics problems like
what you are describing, are not valid in
systems which are accelerating. When your
astronaut is in empty space and turns on her
engines, she thinks she feels a force
pushing her into her chair; but she cannot
understand this because she is at rest in
her frame and Newton's first law says that
if she is at rest all the forces on her
should be zero. If we look at it from the
outside we see the chair pushing her, that
being the force which is accelerating her in
accordance with Newton's second law.
However, she can do Newtonian mechanics in
her frame if she invents a force which is
equal in magnitude to her mass times the
acceleration of the system (rocket) but in
the opposite direction as the acceleration;
this is the "force" pushing her into the
chair and in physics we call it a fictitious
force. You have probably heard of a
centrifugal force,
the force which tries to throw you from a
merry-go-round; that is a fictitious force
because rotation is a kind of acceleration.
In the
following examples, the green dotted line
vector indicates the orientation of the
chair desired for most comfort for the
astronaut. Note that in each instance this
vector points exactly opposite the net
acceleration vector.
First the initial reentry, entering the atmosphere
at a steep angle. The rocket has some
velocity v a g a net .
The chair will orient with the green aligned
with the net acceleration as shown.
Next
we have the situation where the rocket is
rotating to be tail down for landing. It is
still falling with some velocity
v so there
is still an upward acceleration due to the
air drag, labeled a v
here. There is still the acceleration due to
gravity g a c experienced by the astronaut
which points toward the axis of rotation.
Add all three to
get the net acceleration a net .
Now the chair will be aligned as shown.
Finally, the landing position which is the
easiest. There is still the gravitational
acceleration g
Keep in
mind that I have not made any attempt to
have any particular relative values of the
various accelerations but they are not
unreasonable. They will vary considerably
depending on the specific conditions at
any particular time.
All
that aside, let's do the calculation
assuming that the density of the air is
independent of altitude, the same as at sea
level. The air drag is proportional to v 2
where v is the speed and I will
call the proportionality constant k ;
k is where both the density of the
air and the geometry of the falling object
are 'hiding'. Then Newton's second law may
be written as ma=mg-kv 2
where m is the mass, g is
the acceleration of gravity, and a
is the acceleration. Terminal velocity v t
is when a= 0 or v t =√(mg /k ).
Since we have stipulated that k is
the same for both A and B and is altitude
independent, only their masses would affect
v t . If they both jumped
and never opened their parachutes, B would
clearly be the winner (loser?!) since his
terminal velocity is larger than A's as
would be his speed at all times. Similarly,
if each deployed his parachute immediately,
B would get to the ground first. And,
certainly, if A deploys first he will reach
the ground after B since he was already
behind and will end up falling more slowly
than B. So, finally we come to your scenario
where A deploys after B. What will happen
depends on the altitudes where each deploy
and how far ahead B is when A deploys. If A
does not pass B during the 5 second free
fall, B will get to ground first. If A does
pass B he will end up below B but going more
slowly; if they fall for a very long time B
will pass A after some time. If that time is
less than the fall time left for A to hit
the ground, A will win.
Now,
falling from 100,000 ft will be quite
different because terminal velocities at
very high altitude will be much greater than
at atmospheric pressures. That means that
the distanced between the two will be much
greater when they have both deployed. If
they wait until they are fairly close to the
surface, it seems to me quite likely B will
hit first.
β -decay. The nucleus begins with
some mass M . It emits a β + ,
mass mβ and kinetic
energy Kβ ; and a
neutrino, mass mν and
kinetic energy Kν . The
nucleus now has mass M' and kinetic
energy K' . The energy of this
isolated system must be conserved,
Mc 2 =(M'+mβ mν c 2 +K'+Kβ +Kν .
Essentially, the energy came from the mass
of the nucleus; you would find the new
nucleus more tightly bound than the original
nucleus which you could interpret as being
why it decayed in the first place.
Therefore,
β+ -decay
does not occur if the new nucleus is less
tightly bound than the original. (By the
way, the neutrino mass and kinetic energy of
the nucleus are negligible in most cases.)
etc .) and the angular
momentum L of this system must
remain constant regardless of how you
move around. Suppose that you are at the
north pole; then the angular momentum of the
system is L 1 =Iω 1
where I is the moment of the earth
and ω 1 is its angular
velocity. Now you walk down to the equator.
The angular momentum is now L 2 = (I+mR 2) ω 2
where m is your mass and R
is the radius of the earth. But, L 2 =
L1 so if you do the algebra
you will find that ω 2 =ω 1 /(1+ (mR 2 /I )).
So the earth slows down by the tiniest
amount; I estimate (ω 1 -ω 2 )/ω 1 -29 =10-27 %!
Regarding the energy, the initial energy is
E 1 =½Iω 1 2
and the
final energy is
E 2 =½(I+mR 2 )ω 2 2 =½Iω 1 2 /(1+ (mR 2 /I ))
and so
E 2 /E 1 =(1+ (mR 2 /I ))-1 .
So, the
energy has not been conserved but is a tiny
bit smaller even though your energy has
increased by
½mR 2 ω 2 2 =½mv 2 .
If you
work it out you will find that the energy of
the earth alone is
½Iω 1 2 /(1+ (mR 2 /I ))2 .
θ , one force you will
experience is the centrifugal force shown in
the figure, C N T T f f f
sounds
dumb but I can't seem to make her believe it
i.e.
change its speed, all that is needed is for
the sum of all forces on an object to be not
zero. Even if you were on a surface with no
friction and you were in a vacuum, you could
start moving if someone behind you pushed
you. The best way to understand a rocket is
to consider the following example using the
above situation of you in a vacuum with no
friction on the floor.
You
have a ball in your hand;
you
throw the ball with some speed;
in
order to give it that speed, you need to
exert some force on it;
Newton's third law says that if you
exert a force on the ball, the ball
exerts an equal and opposite force on
you;
therefore you end up moving in the
opposite direction as the ball;
it
turns out that the speed you acquire is
smaller than the the ball's speed by the
ratio of the ball's mass to your mass;
for
example, if the ball has a mass of 1 kg
and your mass is 100 kg, your speed will
be 1/100 of the ball's speed.
The
rocket engine is essentially throwing
countless tiny "balls" (atoms, molecules,
ions) with very high speeds to propel the
rocket ship.
M 1
and M 2 of the modules
themselves are much larger than the mass
m of the person moving from one to the
other. The pair will rotate about their
center of mass which, if M 2 =M 1 ,
is at the halfway point. If the person is
very light relative to the rest of the
system, she will experience no centrifugal
force at the center as you guessed; the
reason is that the force is proportional the
square of the speed she is traveling and her
speed will be zero at the axis of rotation.
But if m is not relatively small
the location of the center of mass will vary
when she moves from one module to the other.
The
following is probably more than you want,
but I like to do it anyway. I will denote
the masses of modules as
M 2
and M 1 , including
anybody who is in them, and will treat them
as point masses separated by a distance
R ; the traveling astronaut has mass
m and is a distance d from
M 1 and will also be treated
as a point mass and I will assume the
passage has negligible mass. The center of
mass is located a distance r cm
from M 1 . Calculating the
center of mass location, relative to M 1 , is
straightforward and results in
r cm =Σ(m i r i )/Σ(m i )= [(M 2 -m )R +md ]/(M 2 +M 1 )
For
example, if M 2 =M 1 =M ,
r cm =½R -(m /M )(R-d );
and if
m<<M ,
r cm ≈½R.
Now, you
are interested in when d=r cm :
d=R (M 2 -m )/(M 2 +M 1 -m ).
And
again, for a check,
if M 2 =M 1 =M ,
d=R (M-m )/(2M -m ),
which
is, if
m<<M ,
d ≈½R.
v =5 mph=2.25 m/s, m =200
lb=90.7 kg) which is usual for physicists
and will convert back to imperial units
(pounds) at the end. I will assume that the
collision is perfectly inelastic, that is
both runners are at rest following the
collision; in that case all the kinetic
energy the two had (each has ½mv 2 =230
J) is lost in the collision.
Suppose,
first that the two have lowered their heads
much like bighorn sheep do when fighting.
Since there is not much flesh on the
forehead, they will stop in a very short
distance, let's say d=5 mm=0.2 inches. The
work done by the average force each man
feels must equal the energy he lost, Fd /2=230
J so F =460/0.005=92,000 N=20,700
lb; of course, this force will be spread out
over an area of several square centemeters.
Still it is pretty darned big.
Suppose
the two men have big beer bellies and that
is where they collide. Then the distance
over which the collision occurs will be more
like 5 cm and the resulting force more like
2000 lb. And, the force will be spread over
a much larger area.
In the
real world, such a collision would occur
over a large amount of the body so the total
energy, converted as a force, would be
spread out over maybe a square meter. places
which are hard (like your head) would be
hurt more badly than places which are softer
(like your torso). Also, if the forces look
unbelievably big, keep in mind that they
last for a very short time.
2 be
written as M=ec2 and mean the
same thing? Thanks! I’m totally ignorant
maybe there’s is an obvious reason why not
here.
E=mc 2 .
If you wanted to know how much mass could be
realized by converting an amount of energy
into mass, m=E /c 2 .
Another way to see why your M=ec 2
is an impossible equation is by doing
dimensional analysis: The units of mc 2
must be kg·m2 /s2 so
energy must have units of kg·m2 /s2 ;
but your equation would have the units of
energy being kg·m4 /s4
which is not what the units of energy are.
F=mv 2 /R .
no external forces doing
work on them. You have to be very exact when
applying conservation principles. Suppose you choose the
solar system as the system; if the solar system were in
the middle of empty space its energy would be conserved,
but it isn't an isolated system in real life and the rest
of the galaxy exerts forces on it which might change its
energy. Suppose you choose the milky way galaxy as the
system; if the galaxy were in the middle of empty space
its energy would be conserved, but it isn't an isolated
system in real life and the Andromeda galaxy, the
nearest major thing which exerts forces on it, would
change its energy. You can see where I am going with
this. Suppose the entire universe were the system; then
there are only internal forces acting on this system so
its energy will not change.
First we talk about
electrostatics where all electric fields are constant and
conservative. A conservative
field is one for which, if you move an electric
charge from one point in the field to another, the
amount of work you do is independent of the path you
choose. This means that potential difference
(voltage) between two points is a meaningful number.
You are worried because you have been told that a wire loop with
no battery in it might have a current cannot flowing in
it.
Next we talk
about magnetic fields which are caused by electric
currents which are constant and presumably caused by
potential differences. This is called
magnetostatics .
Finally we come
to the full theory of electromagnetism where the
fields are no longer constrained to be constant.
All of
electromagnetism is described by Maxwell's four
equations. It gets mathematically dense so I have
previously
qualitatively described these equations without any math.
Electric charges cause electric fields.
Electric currents cause
magnetic fields.
Changing electric fields cause magnetic
fields.
Changing magnetic fields cause electric
fields.
So the answer to your
question is that you do not need charges to create the
electric field which will drive a current; a loop of wire
will have a current running in it if you cause the
magnetic field passing through its area to change. This
is called Faraday's Law.
here .
article on gravitational lensing on Wikepedia.
However, for everyday life the bending is hardly
noticeable because it is small. Even the entire earth has
too little mass for this bending to be noticable (see a
recent answer ).
M attached to a massless
string of length L . Therefore you cannot solve
this problem using that formula. However, you could guess
that the period would be proportional to the square root
of the distance D from the suspension point to the center
of gravity of all the mass (the person, the swing, the
ropes). In that case, the period would get smaller when
the person stood up because the center of gravity would
be closer to the suspension point. The correct equation for the period
is T =2π √[I /(MgD )]
where I is the moment of inertia about the
suspension point.
m =10-30 kg, so their total mass
is about 2x10-30 kg. The energy is then
E=mc 2 =2x10-30 x(3x108 )2 =1.8x10-13
J where c =3x108 m/s is the speed of
light. To put this into perspective, this would be the
energy of a particle of dust which has a speed of about 2
inches per second. Furthermore, nearly all of the energy
(which is in the form of two x-rays) escapes without
leaving any energy in the brain.
question : "If the earth is curved how is
it you can get a laser to hit a target at
same height at sea level more then 8 km
away? How is it that it's bent around the
earth?" Part of your answer states that the
laser is perfectly straight. But spacetime
is bent (curves) within the gravity well of
a massive object like the earth. Astronomers
have shown that it's possible to see stars
that are actually behind such massive
objects because the light from the star is
bent around the massive object as it
necessarily follows (somewhat) the curvature
of spacetime around the object. So how is
the laser beam in the question you answered
perfectly straight?
θ= 4GM /(rc 2
to a beam of light tangent to the surface of
the earth at the surface of the earth, you
will find that θ≈ 0.0006"=1.67x10-7 °.
I think you will agree that this is
negligible in the context of the question I
answered!
earlier answer . So, suppose that you are standing on
the equator; there are three forces acting on you, your
own weight (gravity) which points toward the center of
the earth, the centrifugal force which points away from
the center of the earth, and the force (which points up)
which whatever you are standing on exerts to keep you at
rest. Suppose that your mass is 100 kg and the
acceleration due to gravity is approximately 10 m/s2 ;
then your weight is approximately 1000 N. The centrifugal
force is your mass times your speed (464 m/s) squared
divided by the radius of the earth (6.4x106
m), C =100x(4642 )/(6.4x106 )=3.4
N. Suppose you are standing on a scale; since you are at
rest, the force which the scale exerts up on you is
1000-3.4=996.6 N, about 0.3% smaller than your weight.
This is a long-winded answer to your question: yes,
forces due to rotation apply to anything regardless of
its mass or size. (If you are not comfortable with metric
units, 1000 N=225 lb and 3.4 N=0.76 lb. If you weighed
yourself at the poles the scale would read 1000 N if the
earth were a perfect sphere.
Here is another example: the earth is not a
perfect sphere, it bulges at the equator.
The reason is that the earth, when it was
just forming billions of years ago, was very
hot, almost molten, and therefore more
"plastic"; so the centrifugal force caused
it to flatten as it rotated. That would mean
that your weight at the poles would really
be greater than what it is at the equator
because it is closer to the center of the
earth.
answer to the "hot-water-freezes-faster" hypothesis.
As you will see, what happens depends a lot on the
conditions of any measurement or experiment you might try
to do. Your attempt to bring in "potential" is pretty
muddled, so let's review rate of flow in electrodynamics
and in thermodynamics. Materials have a property called
electrical conductivity; this property tells you how much
electric current you will get if there is a voltage
(potential difference) between two points in the material—the
larger the conductivity, the larger the current will be.
Similarly materials have a property called thermal
conductivity; this property tells you the rate at which
heat travels throught the material for a given
temperature difference—the larger the temperature
difference, the larger the energy flow will be. But, this
will have very little influence on how quickly the water
will freeze. First, the change in thermal conductivity of water
changes by only about 10% between say 20°C and 70°C;
second, water is not a very good thermal conductor; and
third conduction is not the primary way water cools
because the density of water, unlike most materials in
the molten state, gets larger as it cools, the cooled
water at the surface sinks so most of the transfer of
heat inside the water is by convection. Certainly the hot
water loses energy at the surface faster than the cold
water, but it will eventually catch up with the cold
water and then the two will be indistinguishable unless
the cold water freezes before the hot water catches up
with it. As I explained in the earlier answer,
evaporation cools the hot water more and if a significant
amount of the hot water evaporates before it catches up
to the cold water, it will win the race because there is
less of it; but that is really cheating, isn't it. Sorry,
I do not know anything about the mpemba effect.
speed of everything on the
carousel is constant. What happens if the carousel
suddenly stop. All those radial forces quickly drop to
zero. The tendency is for everything on the carousel to
continue moving with the velocity they had before, but
now in a straight line . But now, although the
horizontal radial forces are gone, each object
experiences horizontal forces opposite the direction of
their velocities. The girl would initially move forward
until she smashed into the pole; the horse would probably
be held in place by the pole although the force required
to stop the horse could very possibly bend the pole;
someone standing on the spinning floor would only have
the friction of the floor to stop her and would likely
keep moving in a tangential direction. Nothing is
"ejected", it just keeps going in the direction it was
going when the carousel stops, unless something stops it.
F F a is a=F /m where
m is your mass. The magnitude of the wall's
acceleration A is A=F /M where
M is the mass of the wall. For all intents and
purposes, the mass of the wall is infinite so its
acceleration is zero.
i.e. they have no
components.
video
of the project that I am working on.
video . Essentially, when the magnets touch the copper
wire the battery causes a current to flow in the coil
between the two batteries which causes a magnetic field
in that section of the coil; each magnet, now being in
that field, experiences a force moving it forward.
Important things:
The same pole, north or south, must point away
from the battery.
Be sure to use bare copper wire. Often copper
wire has an insulating layer on its surface
and this would not allow current to flow
Good luck on your project. It is really not as
complicated as you thought.
d2 f (x ,t )/dt 2 =v 2 d2 f (x ,t )/dx 2 .
Here, x the position on the string, t
is time, v is the velocity of the wave, and
f (x ,t )
is the solution to the equation which will describe the
shape of the wave at a time t . You may not know
calculus and this equation is goobledy-gook to you, but
all you need to know is that, mathematically, any
function f is a solution as long as it of the form
f (x-vt ). The most commonly used example of
waves is sinesoidal, for example, f=A sin(kx-ωt )
where k is called the wave number and ω is
the angular frequency of the wave. Note that ω /k=v.
Also, to touch base with quantities you might be more familiar with,
k= 2π /λ and ω= 2πf where
f is the frequency of the wave and λ is the wavelength.
Now comes the the important part: because any f
will be a solution to the wave equation, if you have a
wave traveling to the right, f right =A sin(kx-ωt ),
and an identical shaped wave traveling to the left, f left =A sin(kx+ωt ),
their sum will also be a solution to the wave equation.
As you can see from the figure, waves traveling
simultaneously right (red) and left (green) add up to a
"standing wave" (brown) which does not appear to move but
is still oscillating*. You are correct, there are nodes
which do not move but both component waves go right on by
them nevertheless, it just isn't apparent when you look
at there sum. The fact that the net motion of the medium
(string) is just the sum of all individual waves is
called the superposition principle.
*If you're handy with trigonometry, you can
calculate f right +f left
which is apparently not moving even though
you know that before you did the calculation
you could certainly see that it was two
waves.
e.g . a spinning
wheel slowing down or a box sliding to a stop on a
surface), the lost kinetic energy shows up mainly as
heat. And, if you do work on a closed system to keep it
moving forever, that is not what we mean by perpetual
motion. Proper term for what? If you mean motion that you
keep going by pushing on it, it has no particular name.
c , is the same for all observers. If you
are in the rocket and shoot a beam of light straight
across the ship, it will go straight across the ship in
some time ts and the distance it goes
will be cts ; if the width of the ship
is w , ts =w /c .
Now, someone on the ground will see the rocket moving by
with some speed v , so the point on the inside of the ship
where the light beam will have moved a distance vtg
when the light strikes it; but since this is light, the
distance it will travel is longer because the light has
to go farther. Now, to find the angle the light relative
to the perpendicular to the velocity v θ =v /c (tg
cancels out). In your question, since v=c /2, so
θ =sin-1 (0.5)=30°. This
example also demonstrates time dilation because the two
observers see different times of transit of the light. If
you work it out, tg =ts /√[1-(v 2 /c 2 )].
θ subtends a distance s for a circle of radius
r . This angle, in radians, is defined as θ =s /r .
Since the circumference of a circle is 2πr and there are 360° in a circle, there are 2π radians in 360°.
So 10°=10x(2π /360)= 0.175=s /6440,
so s =1124 km.
T N
Mg is
the force of gravity (your weight) down on you. Choosing
+y up as indicated, Newton's second law in the
y direction is, for you, N+T-Mg=Ma
where a is your acceleration and M is
your mass. The figure on the right shows the forces on
the chair: -N T
mg is the force of gravity (weight)
down on the chair. Newton's second law in the y
direction is, for the chair, -N+T-mg=ma where
a is the acceleration of the chair (the same as
yours) and m is the mass of the chair.
Let's first assume you are exerting just the right force
T on the rope so that you are at rest, so a =0.
In that case, if you solve the two equations you will
find that T =(M+m )g /2. In other
words, you need to be able to exert a force down on the
rope which is equal to half the weight of you and the
chair combined to hold the chair from falling. All you
need to do is be able to pull just a little harder to
move upwards. That answers your question but it interests
me to look at a case where a is not zero.
If you solve the two equations for a and N
you get
a =[(2T /(M+m ))-g ] and
N=T (M-m )/(M+m ).
A few observations about what these results tell us:
If you let go of
the rope, T =0, a=-g , N =0;
you and the chair are in free fall.
If m=M ,
N =0, a =(T /m )-g ;
the motion of you and the chair are identical even
though there is no interaction between you and the
chair.
You will
accelerate upwards if T >½(M+m )g ,
half the total weight. Your strength is the only
thing limiting how fast you can accelerate upwards.
If 0<T <½(M+m )g
you will accelerate downward with an
acceleration smaller in magnitude than g . Of
course T <0 is not possible for a rope.
*It is
important to note that, because of Newton's
third law, if the rope exerts a force up on
you, you exert an equal and opposite down on
the rope. T is a measure of how hard
you are pulling. How large T can be
depends only on how strong you are and how
strong the rope is.
β -decay.
The high temperatures are due to the kinetic energy of
the fission products which have speeds typically 3% the
speed of light.
(As noted below, I assume that this means the torque on the spring when
the angle is 100° and it should be gm·cm, not gm/cm.) I know the spring I need. But I only know in terms of the CGS System. I'm following an old book and I've ascertained that the spring I need has a
"CGS number" of 0.76.
So, my challenge is to find a way to convert between these two unfamiliar units of torque measurement. A
"CGS number" and gm/cm/angle.
There are plenty of calculators for converting between different units of torque. But I don't think it's as simple as that. I have two questions.
I know a bit about the CGS system and there is loads of info online. But I'm not clear what it is in the context of classifying a torsion spring. The book only says
"the CGS number indicates the restoring couple of the spring when its diameter is 1cm." So what is the CGS number exactly? Is it dyne-centimetres? If so, then that's easy to work with since that's a measure of torque I'm familiar with. So I'm half way there. If it's something else, then perhaps you can help explain.
Then there is the question of gm/cm/angle. The problem is that none of the calculators I've seen have a notion of angle. This is specific to torsion springs. The torque, in this case, is the working load when the spring bends 100°.
Will I ever find the correct spring? Sadly, I have spoken to the manufacturer, and they were unable to help me.
x when a
force F is applied to the spring. It is
experimentally determined that, to an excellent
approximation for a good steel spring, is that the
distance stretched is proportional to the force applied,
x ∝F or F=kx . The
proportionality constant k is called the spring
constant and is measured, in SI units, as N/m (Newtons per
meter) [dynes/cm in CGS units, lb/ft in Imperial units].
k indicates the stiffness of the spring
and this equation is called Hooke's law. The form of
Hooke's law for angular motion (as in a torsion spring)
is that the angular displacement θ is proportional
to the applied torque τ or τ=κθ ;
I believe the quantity you want is κ
(kappa) and is measured, in SI units, N·m/radian
[dynes·cm/radian in CGS units, lb·ft/degree in Imperial
units]*. I have noted that κ is often called, by
manufacturers and vendors, the rate
of the spring since it is the rate of change of torque
per unit angle.
I am puzzled by your reference to a constant defined as
gm/cm/degree; 1 dyne=1 gm·cm/s2 , so κ
should have units gm·cm2 /s2 /degree;
sometimes the gram is treated as a unit of force (called
a gram force where 1 gram force=980 dynes), in which case
κ
would be gm·cm/degree, not gm/cm/degree. So, there
must be somewhere in the reference you are using a
definition of what the CGS number is. Suppose that it is
0.76 dynes·cm/radian (which would be the likeliest
units if purely CGS system of units is applied) but you want it in
lb·in/degree; then 0.76 (dyne·cm/radian)x(2.25x10-6
lb/dyne)x(0.394 in/cm)/(57.3 deg/radian)=1.18x10-8
lb·in/degree. Or, suppose you wanted it in (gm
force·cm/degree); then 0.76 (dyne·cm/radian)x(gram
force/980 dyne)/(57.3 deg/radian)=1.39x10-5 (gm force·cm)/degree).There are lots of unit converters
around, my favorite is
here .
The choice of units for κ which it seems most vendors use
is lb·in/deg. The best calculator I found to get
an idea of the magnitude of these springs can be found
here .
Until you are able to deduce the units of the so-called CGS value of
0.76 you cannot know what it is.
*Sometimes
κ is implicitly given by specifying the
torque at a particular angle.
Watch Adjustment , he says that in the CGS system
for characterizing the spring in question (which I will
call N CGS ), the unit of force is the
dyne and the diameter of the spring is 1 cm. He then
uses not the restoring torque to calculate N CGS
but rather the
restoring couple associated with the force; the
restoring couple is the restoring force times the
diameter whereas the torque is the restoring force
times the radius . Most vendors use the
restoring torque to characterize springs, so ½N CGS
should be used to convert to the vendor's units.
Although the units of the angle are not given in
Jendritkzi, I have found that the more natural radians
gives values much too small when converting to vendor
units; so N CGS =(restoring couple in
dyne·cm for 1 cm diameter)/degree.
One vendor charactorizes the spring by specifying a
torque given by N gm/cm/100°; this does not
appear to be dimensionally correct unless the gram is
gram force (I will denote it as gmf) where 1
gmf=981 dynes (the weight of 1 gm in dynes) and /cm
means that again this is for a 1 cm diameter spring. So,
assuming that the vendor's torque is indeed torque
(force times half the diameter), ½N CGS =N .
For example, suppose N CGS =2; then
N =½(2) (dyne·cm/1°)(1 gmf·cm/981
dyne·cm)/(100°/1°)=0.102 gmf·cm/100°; so the conversion
from the CGS value to this vendor's value is N =0.0491N CGS .
So, in the case of N CGS =0.76
dyne·cm/1°, N =0.037 gmf·cm/100°.
[Disclaimer: I am
not a clockmaker and I am not familiar with conventions
clockmakers might use or meanings they may attach to
certain words which are not in accordance with
definitions, units, or conventions used by physicists.
In the event that I have misunderstood or that either
Jendritkzi really meant torque or the vendor really
meant couple, the conversion factor would be 0.102
instead. I have done my best to determine conversions
between two numbers representing the same thing but in
different systems of units.]
many times . In your
variation, each gear will experience a force from the
previous gear in the line and exert a force on the next
gear. The time it takes for the force to travel from the
input force location to the output force location is
determined by the speed of sound in the gear. The message
you send would travel at the speed of sound, much smaller
than the speed of light.
standing by the side of the tracks will
see you standing still; the same as running on a
treadmill in a gym with the treadmill and you moving in
opposite directions with the same speed. If you walk in
the same direction as the train is going, you will be
seen by the trackside observer as moving with a
speed of 10 km/hr.
via gravity would solve a lot of
them; all attempts to directly detect dark matter have
failed and until there is direct observation we
understand nothing. To my mind, failure to observe dark
matter may be indicative that we do not understand
gravity as well as we think we do.
You might not want to read this rather long
introductory paragraph if you are just interested in the
final answer. ) This problem was very interesting to me
because I initially thought it was pretty easy to
understand. The billiard ball starts out with much greater
speed than the bowling ball, jumps into the lead.
Neglecting air drag is usually what is done initially in
considering such problems and there has to be some
kinetic friction for the balls sliding on the ice; but
because the kinetic friction is proportional to the
weight, each ball experiences the same rate of change of
its speed as it slows. Therefore the billiard ball is the
obvious winner. When I initially solved the problem this
way using reasonable numbers, I found that the bowling and
billiard balls had initial speeds of 14 mph and 81 mph,
respectively; the bowling ball went 686 ft and the
billiard ball went 21,700 ft, a little more than 4 miles! When I
was writing the final paragraph I discussed the possible
effects of two things I had neglected, rolling of the
balls rather than sliding and the possibility of air
drag. Just for fun I estimated the effect of air drag and
discovered that at the beginning, particularly for the
billiard ball, the drag force was considerably larger than the
sliding friction. I realized that I had to redo the whole
problem, now much more difficult and mathematically
involved than originally. Because the air drag estimate I
used is valid only if you work in SI units, I will do
that throughout and convert to imperial units when needed.
The mathematics gets very complicated and I will only
sketch my calculations and whoever is interested can fill
in the gaps.
First I will tabulate quantities needed for the problem:
We need to
quantify what is meant by "…the same amount of
physical strength/athletic ability…" I will
assume that the thrower exerts a constant force F
over a distance d on both balls. Physics says
that the result is that a kinetic energy K =½mv 2
is acquired where m is the mass of the ball and
v is the speed it has when it leaves the
thrower's hand; this kinetic energy is equal the the work
W done which is W=Fd . that is, Fd =½mv 2 .
If you solve this equation for V you find the speed,
v =√(2Fd /m ). I will use
Fd =110 J (about 25 lb over 1 m) because this
value would give the billiard ball a speed of about 80 mph
which seemed reasonable to me.
The masses of the balls are
0.17 kg=6 oz for the billiard ball and 5.4 kg=12 lb for the bowling ball.
I will use
a coefficient of sliding
friction for each ball which is very small, μ =0.01.
Each ball experiences a frictional force of f=μmg=ma
(Newton's second law) where a is the
acceleration and g =9.8 m/s2 is the
acceleration due to gravity.
For the air drag
I will use the approximation D =¼Av 2 where
A
is the area presented to the oncoming air (πR 2 )
and v is the speed. The radii of the balls
are 0.0286 m and 0.108 m. As noted above, this
equation is only valid for SI units because the
factor ¼ includes things like air density, drag coefficients,
etc .
We are now ready to
start doing the physics. Start with Newton's second law
ma=m (dv /dt )=-¼A v 2 -μmg
dv /dt=-Cv 2 -μg
(C =¼A /m ).
This equation is
integrable. The result is
t =tan-1 [v √(C /(μg ))]/√(Cμg )|v(0) 0 .
Evaluating
this expression between the limits v=v (0)
to v =0 and inverting to solve for v (t )
is straightforward but messy. The essential
results are illustrated in the two graphs below.
The effect of air drag on the bowling ball is
modest but noticeable, reducing the time to stop
by about 10 s. The effect on the billiard ball,
though, is enormous; because of its high speed
at the start, it is slowed much more than due to
sliding friction alone. Still, it stayed in
motion about 20 s longer than the bowling ball
and was going faster most of the time, so
certainly went farther before stopping. Still,
it would be useful to find the position as a
function of time to determine how far each ball
went.
To find the position, x (t ), is
straightforward but again messy: integrate the
differential equation dx /dt =v (t )
where v is the function being plotted
in the graphs above. The results are shown in
the graphs below. The bowling ball stops at
t =54 s at a position x =158 m=518
ft; the billiard ball stops at t =77 s
at a position x =535 m=1755 ft. The
winner is the billiard ball which goes more than
three times farther than the bowling ball! I
believe that regardless of any details of the
friction and drag forces or how we choose the
quantity Fd , the billiard ball will
always go faster if the initial kinetic energies
are equal.
Finally, I want to discuss rolling vs .
sliding. It is really hard to get an object
rolling without slipping on a very slippery
surface. A sliding ball will not roll until it
is nearly ready to stop, and that would have
very little effect on my calculations. Also,
rolling friction is velocity-independent and
proportional to the normal force so it would
affect both balls the same.
faq
page .
E=mc 2 .) So changes of mass
will affect the details of the gravitational field in
some volume of space but if you get very far from that
volume, the gravity you see originating from that volume
will not change.
Get her to
understand Newton's first law: an object at rest or
moving with constant speed has all the forces on it
cancelling each other out.
Get her to
understand Newton's second law. If the forces are not
cancelling each other out, the object will start
moving and continue to speed up in the direction of
the uncancelled forces.
For example,
have her imagine sitting on the trike on the
incline with her feet off the pedals or the
ground. She will start moving down the incline.
What is happening is that her force, which is
still straight down, must have a little bit of
itself pushing her down the incline. (This is
basically a component of the weight but the
concept of vectors and their components is likely
too hard for a five-year old to comprehend.)
Finally, if, on
the incline, there is a piece of the weight which she
will have to compensate for in order to keep that
force from making her go down the hill: she does this
by pedaling to create a force up the incline: the
force she generates must be greater or equal to the
piece of the weight pointing down the incline. On
level ground there is no unbalanced weight so the
pedaling is much easier.
Friction, I
would opine, is an unnecessary complication.
If she were
heavier, the piece down the incline would be bigger
so it would be harder to pedal.
(This part is
for you, the little girl will not get it.) Your
question about the horsepower is incomplete because
to calculate the power you need to know the time it
takes her to go the 100 ft. Suppose she maintains a
speed of about 2 ft/s. Now, the component of a 50 lb
weight on
a 3% grade is about 50x0.03=1.5 lb so the work done
is 1.5x100=150 ft·lb. The time it takes her is
(100 ft)/(2 ft/s)=50 s, so the power is 150/50=3 ft·lb/s=0.0055
hp.
Friction has
little to do with it because the friction is not
significantly different on level ground vs .
the incline. Position on the trike plays no role
since any change would be the ease of pedaling which
would be the same in either case.
τ=Iα=I (d2 θ /dt 2 ) where
τ is the net torque about the pivot (the blue cross in the figure),
I is the moment of inertia about the pivot, α is the angular acceleration, and
θ is the angle of the system at some time
t . If the system is a pendulum with no forces
exerting torques except weights (vertically down), all torques are
proportional to sinθ where here
θ is the angle relative to the vertical.
Now our equation may be written as
(d2 θ /dt 2 )=-(|τ| /I )sinθ
where
|τ| is the torque about the pivot when
the pendulum is horizontal (θ=π /2). The
negative sign results from the fact that the angular
acceleration and the angle are always opposite, a
restoring torque. This equation is extremely difficult to
solve, but if the angle is small (much smaller than 1
radian), you can use the small angle approximation, sinθ≈θ ,
which gives us
(d2 θ /dt 2 )≈-(|τ| /I )θ
This is the simple harmonic oscillator equation for an oscillating
system for which we know the angular frequency, ω =√(|τ| /I ).
So, let's apply it to your situation as illustrated in my
figure. There are two point masses and the mass of the
rod itself which I will take as m . The rod has a center
of gravity (orange cross) a distance of L 3
from m 1 as shown; if your rod is
uniform, L 3 =L /2. (If the
mass of the rod is much smaller than both the point
masses, you can approximate m as zero.) Now,
since all weights are of the form W=mg , you can
write |τ| :
|τ|=g (m 1 L 1 -m(L 3 -L 1 )-m 2 L 2 ).
The moments of inertia for the two point masses are
I 1 =m 1 L 1 2
and I 2 =m 2 L 2 2 .
The moment of inertia for the rod is trickier, so to
avoid a lot of messy algebra I am going to assume a
uniform rod where L 3 =L /2. I
find
I rod =(m /(3L ))(L 3 -L 1 3 )
The total I is just the sum of the three moments
of inertia. So, finally,
ω =√[(g (m 1 L 1 -m (L /2-L 1 )-m 2 L 2 ))/(m 1 L 1 2 +m 2 L 2 2 +(m /(3L ))(L 3 -L 1 3 )]
And, if m can be neglected,
ω =√[g (m 1 L 1 -m 2 L 2 )/(m 1 L 1 2 +m 2 L 2 2 )]
faq page .
earlier answer . See a simulation
video .
W ) to wing area (A )
is the same as a real bat. I find that a typical value
for that ratio for bats is about R =W /A =20
N/m2 =0.42 lb/ft2 .
So, for your case, A=W /R =4500/0.42≈10,000
ft2 or one wing has an area of about 5000 ft2 .
If I model a wing to be a triangle with base B =25 ft and
a width H , 5000=½BH =12.5x25xH
or H =400 ft, a wingspan of about 800 ft.
Since it has the same R as for a real bat, it
should be able to fly as fast as a real bat which can be
as high as 100 mph.
etc . In the diagram I have shown all the
forces on the load, the tensions in the slings,
T 1 and T 2
and the weight of the weight, W T 1 =T 2
=T . Also shown are the horizontal (x )
and vertical (y ) components of the tensions;
they are T 1x =T cosθ ,
T 2x =-T cosθ ,
T 1y =T 1y =T sinθ.
So the equilibrium equation is 2T sinθ-W= 0;
therefore T=W /(2sinθ ).
So, for your example, sinθ =cosθ =1/√(2),
W =2000 lb, and T =1414 lb. Now, your
interest is in the horizontal components; each is T cosθ=W /(2tanθ )=1000
lb "crushing force" on each side. Your main error is that
you treated the vector forces as scalars, you assumed
T=Ty +Tx where in fact T =√(Ty 2 +Tx 2 ).
90 Y is the decay product of 90 Sr
which has a half life of about 29 years. 90Sr is an
abundant byproduct of the fission of uranium and
therefore available in the waste of reactors. When the
90 Y is needed it can be chemically separated
from the 90 Sr. I am sure that the hospital
cannot keep a supply on hand and will have it delivered
when it is needed for your treatment. It does not matter
how long it has been since it was separated, only that
the radiation level be correct for the prodedure. It
probably arrives with a level too high and they wait for
it to be at the needed level.
light
clock . In order that this example is convincing to
you, you must accept that the
speed of light is a
universal constant in all frames of reference; but this
is an experimentally well-verified fact and also is a
consequence of the principle of relativity—that the laws
of physics are the same in all inertial frames of
reference.
R and speed
v in air
at sea level may be approximated as F =¼πR 2 v 2 =0.79R 2 v 2
(this is correct only in SI units); the (downard) force of
the weight W on an object of mass m is
W=mg . The net force is the mass times the
acceleration a (Newton's second law), ma =-mg +0.79R 2 v 2 .
So as the object falls it goes faster and faster until
v is large enough that it falls with a constant
value speed vt (a =0), vt =[√(mg /0.79)]/R .
For a baseball, m =0.144 kg and R =0.037
m, so vt baseball =9.28 m/s.
For a bowling ball, m =8 kg and R =0.12
m, so vt bowling ball =83.0
m/s, nearly ten times greater than the baseball. So the
baseball stops accelerating sooner than the bowling ball
and loses the race to the ground.
Wikepedia article on the fate of the universe.
not matter;
it is not anything, it is a region in which light
illuminating something is blocked by some obstruction.
You might say that a shadow is the lack of anything,
light which might have otherwise illuminated where it is.
And there are degrees of "shadowness". If all the light
illuminating an area is blocked by the obstruction, it is
called an umbra ; if only part of the light is
blocked it is called a penumbra . (Umbra is Latin
for shadow.)
Wikepedia article .
Feynman Lectures .
*Actually, the moon does exert a torque on the earth
which is causing the spinning to get smaller, but the
change is so tiny that it takes millions of years to be
significant.
†The earth's orbit is
not exactly circular. When it moves closer to the sun it
speeds up, when it moves farther from the sun it slows
down by exactly the same amount, so the orbit just
continues forever.
video is of some ladies in a shop, the pressure wave hits the shop and the lady outside the shop doesn’t get thrown back but inside the other lady does. What is the explanation here??
external force. This is also true in special
relativity where the linear momentum is not p=mv but
rather p=mv /√[1-(v 2 /c 2 )]
where c is the speed of light. In the event of
chemistry going on, there is either energy gained
(endotermic) or lost (exothermic). If lost, the momentum
of the radiation has to be taken into account. If gained,
the energy will be taken from the kinetic energies of the
incident objects but momentum will be conserved.
glacial ice
melting and sending its water to the oceans,
glacial rebound
which is the raising up of the land previously
supporting the melting glacier, and
tectonic motion,
the drift of large land masses.
excellent discussion on the golf swing which, for the
first half is fairly conceptual and the second half goes
into the detailed physics analysis of the swing.
t =0.0005 s. How big would the acceleration
need to be in order to significantly change the speed
over this time? A typical speed of the club head is about
150 mph, about v= 70 m/s. Suppose that the speed increased
over the impact time by Δv =1 m/s. Then the acceleration
would have to be a =Δv /Δt= 1/0.0005=2000 m/s2 ;
this is about 200 times greater than the acceleration due
to gravity. We could estimate the acceleration by
guessing that the club took about 0.1 s to go from zero
to 70 m/s, a =700 m/s2 . So the speed
of the club might increase by 700x0.0005=0.35 m/s and the
average speed during impact would be about 70.2 m/s. My
feeling is that this would have a negligible effect on
the speed of the ball.
This article states that Force equals mass times acceleration.
It indicates that applying a force to an object over time creates momentum.
Therefore, it seems to me that a club head that is accelerating at the point of impact transmits Force and, therefore, momentum to the golf ball. And that this momentum would be in addition to the momentum of the club head.
A club head that is not accelerating will transmit no force to the golf ball.
It seems to me that a club head accelerating is going to hit the ball further, velocity at impact being equal.
F=ma
is Newton's second law and simply says that if you push
or pull on a mass it will accelerate. Indeed, if you
apply a force over a time you will accelerate it and
therefore change its momentum. But these do not
imply that the source of the force itself needs to be
accelerating. Here is what happens: when the club strikes
the ball, it exerts a force on the ball; Newton's third
law says that if the club exerts a force on the ball, the
ball exerts an equal and opposite force on the club. For
the short time that they are in contact (after which the
forces disappear), the ball will speed up and the club
will slow down if there are no other forces on it (which
may be the case*). If the average force they exert on
each other is F and the time of contact is t ,
the ball will acquire a linear momentum of p=Ft ;
the ball will experience a change in its momentum but by
how much is not so simple since other forces act on it.
Note that F=p /t=mv /t , so the
smaller t , the larger F , all else being
the same. In my answer to your original question, using
v =70 m/s and t =0.0005 and noting that
the mass of a ball is about m =0.046 kg,
F =0.046x70/0.0005=6,440 N=1448 lb.
*If the club is accelerating when it strikes the ball
there must be some force on it—ultimately that
comes from you.
2 . So
if an object changes its speed from 2 m/s to 6 m/s in a
time of 2 s, its acceleration is 2 m/s2 . Mass
is trickier to understand but it is a quantity which
measures how resistant an object is to being accelerated
if you push on it; it is harder to accelerate an object
of mass 1000 kg than it is one of 2 kg. Note that we have
not yet quantified that push (or pull) which we normally
call force. However we can certainly imagine figuring out
a way to push or pull on something twice as hard, or
three times as hard, etc . Whatever force is,
suppose we push on something with a constant force F
and vary the mass; we will find that if we double the
mass we halve the acceleration, if we halve the mass we
double the acceleration, if we make the mass 10 times
larger, the acceleration is 1/10 of its original value,
etc .
We have found that, F being constant,
acceleration a is inversely proportional to mass,
a ∝1/m .
Now suppose we vary F but keep m constant; we find that
if we double the force we double the acceleration, etc .,
i.e. a ∝F. We can now put
the two together and get a∝F /m or F∝ma .
This is what I consider to be Newton's second law; it
tells us an important connection of how quantities depend on
each other, it is a law of physics; and, you really did
not have to have defined the meter and second and
kilogram for it to be true, you just had to understand
conceptually what mass, length, and time are. If you know
even a little physics, you do not recognize this as
Newton's second law—any textbook will tell you that
it is F=ma . Now, how can one turn a
proportionality into an equation? It is simple, just
introduce a proportionality and voila! F=Cma
where C is an arbitrary constant since we
have not defined how we will measure F . Don't
think of it as a "fundamental constant" because if we had
already defined how to measure force, we would have to
measure C . For example, if we had defined a unit
of force to be a pound but measured m and a
in SI units (kg and m/s2 ), F=ma would not be a valid
statement of Newton's second law.
Now let's discuss an example of when a constant is a
fundamental constant. Newton's law of gravitation
expresses the magnitude F of the equal and
opposite forces exerted on each other for two point (or
spherical) objects of masses m 1 and
m 2 separated by a distance of r :
F ∝m 1 m 2 /r 2 .
Now, to make this an equation, we must introduce a
proportionality constant G : F=Gm 1 m 2 /r 2 . But can we choose it to be whatever we like? No, because
there is nothing in this proportionality which is
undefined; we can take two 1 kg masses, separate them by
1 m, and measure the force between them (a really hard
experiment!). In other words, we must measure the
constant G . This is a true
fundamental constant of nature, it measures the strength
of gravitation. Of course, its numerical value depends on
how we measure mass, length, and time, but it is still a
universal constant. In SI units it is 6.67x10-11 N·m2 /kg2 .
You might be interested in two earlier answers,
1 and
2 , along similar
lines as yours.
14
Hz, so you can be sure that you cannot do this with any
macroscopic magnet. One way to get magnetic radiation is
to fabricate something called a
split-ring resonator (SRR) but technical problems
limit SRRs to about 2x1014 Hz, near infrared.
Another technique uses spherical silicon nanoparticles
and these can generate visible magnetic light. The method
is too technical to discuss here, but you can read about
it at
this link .
relativistic
definition ) remains constant; then you are not
confined to low velocities. Then we will find that very
good approximations to inertial frames can be found in
nature. There is nothing wrong with basing a theory on an
idealized (fictional) concept as long as the ultimate
theory agrees with experimental results in the real
world.
F=ma . Why? If someone is
moving past you and measures the acceleration a'
of something in your frame, she will measure exactly the
same value as you. Therefore you will both conclude the
same force is being applied to m, if a=a' then
F=F' . But if relative speeds are not small,
a≠a' so Newton's second law is no longer true.
Instead we rely more on conservation principles more in
relativity theory. In particular, we want to
redefine momentum so that it is conserved in a closed
system.
electromagnetic waves . (Although
the speed of the gravity wave does not matter for the
interferometer to work, it is just as if somebody were
jiggling the mirror ends, I want to note that it is only
assumed that the speed of gravity is c ; the
speed of gravity has never been measured, we just know
that it is very fast. Since the gravitational field has
never been quantized, we can only assume that the
graviton has no mass. Note that for decades it was
assumed that neutrinos had no mass, something we now
understand is not true.)
E 2 +p 2 c 2 =m 2 c 4 and in what publication? Wikipedia says that it was Dirac in 1928, but I cannot find evidence of this in his famous 1928 paper, "The Quantum Theory of the Electron."
first paper did not mention momentum at all. In the
second paper he shows that if a particle of mass m
radiates energy L that the mass changes its
kinetic energy by L /c 2 ,
hence E=mc 2 , but does not mention
momentum. Finally in 1907 he writes an
article in which
he shows that if relativistic momentum is p=mv /√[1-(v /c )2 ],
then it is a conserved quantity as in classical Newtonian
physics. From here, and knowing E =mc 2 /√[1-(v /c )2 ],
it is
straightforward to get the energy-momentum
relation.
N is the force which the
ground must exert up on the man, W is his
weight, and F is the force which the woman
exerts down on the man. (I have drawn N as one
vector, but it is really distributed among his two feet
and two hands. Nevertheless, the size of N is
proportional to the force he must exert to lift himself.)
The man is in equilibrium, so the forces on him must add
to zero, N-W-F =0 or N=W+F . In white are
the forces on the woman: w is her weight and
f is the force which the man exerts up on the woman
which by Newton's first law equal and opposite to the
force which she exerts down on the man so F=f .
Since she is also in equilibrium, the sum of the forces
on her must add to zero, so f-w =0 or f=w=F .
So, finally, we see that the total weight the man must
cope with is the sum of his and the woman's weight,
N=W+w .
So now I put a blue brick under the woman's foot so
that it rests on the brick. This is a new force
n up on her, shown in red. Each
person's weight is the same as before, your
weight is just the force down due to gravity.
But now there are three forces on the woman
adding up to zero: n+f '-w=0 or f
'=w-n . The force f ' exerted
up on her by the man, which is the same
magnitude of the force F '
which she exerts down on the man is smaller than
before. So now,
N '-W-F ' =0
or N '=W+F '=W+w-n. He has a smaller
total force to contend with than before.
The mistake you made is that you assumed that the
force down on the man was the woman's weight.
But the woman's weight is not a force on the
man, it is a force on the woman. This is
accidentally correct in my first example because
if only the man is supporting her, it just
happens that the force she exerts down is equal
to her weight in magnitude. But in the second
example the brick is supporting part of her
weight so the man supports less. She does not
become lighter!
h and has a mass m , then the energy she
starts with is E 1 =mgh , where
g is the acceleration due to gravity. When
she hits the water she will have a smaller energy because
of air drag falling down and when she comes to rest she
will lose some energy to heat and sound and damage to her
body; I will say that a fraction f of the original energy
goes into the energy E of the wave created,
E=fE 1 . Now I will assume that all this
energy goes into a circular pulse which has a width of
w
and a length of 2πR (circumference of a
circle) where R is the radius of the circle at
the time of interest, namely when the wave hits the
shore; so R is the distance from shore where the
dragon hit the water. Now, I found an expression on
Wikepedia which relates the energy of a wave to its
height: E /A=ρgH 2 /16
where A =2πRw is the total
horizontal area of the wave, H is the height of the wave,
and ρ= 1000 kg/m3 is the density
of water. If you put it all together you will find
H =√[8fmh /(ρπRw )].
Suppose h =5000 m, m =2x106 kg (2000 metric
tons), f =½, and w =20 m; then I
find H ≈25 m. This is in the same ballpark as the
largest tsunami wave ever observed which was
about 100 ft. There is no way I could estimate
how far inland the wave would travel; it would
depend a lot on the terrain, obstacles, etc .
2 =[1x10-3
m]x[(1
yr)x(365 day/yr)x(24 hr/day)x(3600 s/hr)]-2 ≈9x10-22
m/s2 .
In general, the force is F=ke 2 /r 2 =ma ,
so r =√[ke 2 /(ma )]
where k =9x109 N·m2 /C2 ,
e= 1.6x10-9 C, and m =9x10-31 kg. I
calculate, approximately, that r ≈5x1011
m≈300,000,000 miles which is about 3 times the
distance to the sun. Of course, you could never do such
an experiment since you would need to be in a universe
which contained nothing but these two electrons.
The laws of physics are stated most elegantly as
proportionalities rather than equations. For example,
Newton's second law states that the acceleration (a )
of an object is directly proportional to the applied
force (F ) and inversely proportional to the mass
(m ), a∝F/m ; so you double the
acceleration if you push twice as hard and halve it if
you double the mass. Now, to calculate we prefer to have
equations rather than proportionalities, so we introduce
a proportionality constant, call it C , a=CF /m.
What does C mean? Since we know how to
measure a (m/s2 ) and m (kg),
your choice of C determines how we will measure
force. I would like the force to have the magnitude of 1
if a =1 m/s2 and m =1 kg, so I
choose C =1. The constant C is not a fudge
factor, rather its choice defines how we measure forces.
(We call the unit kg·m/s2 a Newton, N.)
Newton also discovered the universal law of gravity: if
two masses (point masses or spheres) m 1 and
m 2 are
separated by a distance d , they exert equal and opposite
attractive forces on each other the magnitude of which is
F ; then F ∝m 1 m 2 /d 2 .
So the equation for universal gravitation would be of the
form F=Gm 1 m 2 /d 2 .
where G is some constant. But we know how to measure mass
and distance and force, so we cannot choose G to
be any old thing we want as we did for Newton's second
law, we must measure it. It turns out to be G =6.67x10-11
N·m2 /kg2 . This is one of the
most fundamental constants in all of nature and most
certainly not a "fudge factor".
Constants are often used to quantify a particular
situation. For example, the frictional force f
of one object sliding on another is found, approximately,
to be proportional to the force which presses one object
to the other (often called the normal force, N ).
This is not a physical law, just an approximation of
experimental measurements, and it obviously involves only
the magnitudes of the forces since their directions are
perpendicular to each other. So, including a
proportionality constant μ , called the
coeficient of kinetic friction, f=μN . This
constant tells you how slippery the surfaces are; its
value for rubber sliding on ice will be much smaller than
for rubber sliding on asphalt.
Sometimes, as you suggest, constants are added to
fit data and their physical meaning is unknown,
you might label such constants as "fudge
factors".
You may be interested in learning about Planck
units discussed in an
earlier answer . where new units are
expressed by combinations of the five
fundamental constants of nature.
The best way to understand the priciples is to look at
the simplest possible generator. The figure shows a
generator consisting of a single loop rotating in a
magnetic field which is approximately uniform. There
would be no difference in the results if it were the
magnets rotating around the coil as you describe. Since
the induced EMF in the loop is proportional to the time
rate of flux through the loop and the flux is
proportional to the cosine of the angle θ between the field
and a normal to the area of the loop, the induced EMF
will be sinesoidal. The angle may be written as θ=ωt where
ω is the angular velocity of the loop (or magnets)
and t is some arbitrary time, so the
voltage is proportional to cos(ωt ). If
there is a load across the terminals, and therefore a
current through the loop, there will be a field due to
the current which does not affect the output voltage if
the source of power maintains a constant angular
velocity. If you are drawing power from the generator, it
is harder to "turn the crank" than if there is no current
flowing. If you need detailed information on "real-life"
generators, it is more an electrical engineering question
than physics.
Einstein: His Life and Universe ".
angular velocity, (revolutions per second,
e.g .) but not the same translational velocity (miles per
hour, e.g .). In general, v=rω
where v is the translational speed, r is the distance
from the axis of rotation, and ω is the
angular velocity in radians per second.
On the left of the figure I have drawn the particle
moving with constant speed v around a circle of radius R.
In some time interval Δt the particle moves through
an angle theta and has velocities v 1
and v 2 but both
speeds are equal to v , v 2 =v 2 =v .
On the right I have placed the velocities tail-to-tail
and drawn the difference vector v 2 -v 1 =Δv v /Δv =R/Δs.
If I now rearrange and divide each side by Δt ,
I find Δv /Δt =(v /R )Δs /Δt.
Now, Δv /Δt is the magnitude
of the average acceleration; to get the instantaneous
acceleration, we must take the limit as Δt approaches zero.
But, when Δt approaches zero, Δs /Δt approaches
v . So, finally, a=v 2 /R .
Planck units came from the
desire to have a system of units which are based only on
constants of nature, not on the specific units like the
meter, the second, etc . You take the five
universal constants:
the speed of
light in vacuum, c,
the universal
gravitational constant, G,
the rationalized
Planck's constant, ℏ,
the Boltzman
constant, k B , and
the Coulomb
constant, k e =1/(4πε 0 ).
You now form
combinations of these constants which have the correct
dimensions of the units you want:
length,
mass,
time,
temperature, and
electric charge
I have copied the table from
Wikipedia showing the values of these Planck units in SI
units:
It is hypothesized that, at
distances smaller than l P =1.6x10-35
m, the usual physics as we know it will not work. I think
you might say that the origin of the idea of Planck units
is quantum field theory.
MR 2 ω
where M is the mass of the top, R its radius, and ω
is the rate at which it is spinning; the direction of the
angular momentum vector is straight up if it is spinning
counterclockwise if viewed from above, straight down if
clockwise.
There is a physical law which states that if there are no
torques on a spinning object, its angular momentum never
changes. There are no torques on the top if its spin axis
is vertical and so it does never stops.
x ,y ,z );
they were named in honor of the 17th century philosopher
and mathematician
René Descartes . Legend has it that he
was once watching a fly buzzing around his room and asked
"what is the minimum number of numbers I must specify to
describe the position of the fly at any time?" He figured
it was three: the distance up from the floor, the
distance from the back of the room, and the distance from
one of the side walls; his room is said to have three
dimensions. One dimension is a line; here you simply
measure the distance from one point on the line to
wherever the particle is. The line need not be a straight
line. A circle is one-dimensional and the location of a
point on the circle is usually specified by an angle
relative to some point we call zero. Two dimensions is a
surface. The surface of the earth is two-dimensional, the
position being angles, latitude and longitude. One
surface of a flat sheet of paper is two-dimensional and
the two numbers are usually Cartesian coordinates (x ,y ).
The spatial universe in which we live is
three-dimensional, just like Decartes' room. A solid
sphere has three dimensions usually measured by the
distance from the center and two angles (longitude and
latitude).
twin paradox . How fast a clock runs and how fast it
appears to actually runs are two different things; see
earlier answer .
C r ),
W R ),
N
f
F
Note that I have
resolved F
H V
The sum of all
forces on a body is equal to the product of the mass
(m ) and the acceleration (a ) of the
center of mass of the body;
The sum of all
the torques on the body about any axis is equal to
the product of the moment of inertia (I ) of and the
angular acceleration (α ) about that axis.
You seem to
think that a force which exerts a torque about the
axis you have chosen (I will choose the axle for
the bike wheel) only can cause angular
acceleration; in fact, every force on the object
contributes to the translational acceleration.
Suppose now that you are standing on the pedal but
also applying the front brake such that the bike
remains at rest; both translational and angular
acceleration are zero. So the translational N2
yields two equations, N-W-V =0 and
C+f-H =0; the rotational N2, choosing the axle
as the axis, yields Cr-fR =0. (F ,
W , and N exert no torque about
the axle.) If you now release the brake,
H will get much smaller and
the wheel will start moving forward and rotating.
If you are
interested in the entire bike, not just the rear
wheel, the force C has no contribution to the
motion because it is an internal force; the
chain exerts a force C -C
m ) and the acceleration (a ) of the
center of mass of the body". Note that it does
not say
…the sum of all forces which are directed through the
center of mass … Let me give you a simpler
example. Suppose that you are in the middle of empty
space and you have a wheel which has a string wrapped
around its circumference. You pull on the string so that
the tension in that string is the only force on that
wheel and it is tangential . Surely you do not
think that your pulling will cause the wheel to only spin
and not accelerate toward you too. Here is another
example, a yoyo. There are two forces on it, its own
weight which passes through the center of mass and points
down and the tension of the string which you are holding
which points up and does not pass through the center of
mass. If the string does not affect the translational
acceleration, the only force on the yoyo would be its
weight so it would fall with acceleration due to gravity,
g , which it clearly does not do.
F
N-V-W =0. But in the horizontal direction there will
be an acceleration, ma=H-f . Only f F F f
Here is the take-away for you: just use N2 as I have
given it to you, always look at all forces on the object,
and do not continue insisting that a force which exerts a
torque does not affect translation.
m and speed
v suddenly loses or gains mass, what happens?
The mass either disappears without trace or appears
without a source. Suppose your musket ball has a mass
m and speed v and your cannon ball has a
mass M . Since there are no external forces on
the ball, linear momentum must be conserved, that is
mv=MV where V is the speed of the cannon
ball. So V=mv/M , much smaller than v .
However, the energy is not conserved: E 1 =½mv 2 and
E 2 =½MV 2 =½mv 2 (m /M )=(m /M )E 1 ;
a large amount of energy is lost.
m with a velocity v 1
collides with a much more massive object at rest and
rebounds with velocity v 2 , the
average force F can be estimated as F≈m (v 2 -v 1 )/t
where t is the time the collision lasts. ("Much more
massive" means the struck object does not recoil
significantly.) You seem to want to get answers in
imperial units (lb, ft, s); without going into details,
m =0.5 lb/(32 ft/s2 )=0.016 lb·s2 /ft=0.016
slug
and v 1 =90 mph=132 ft/s. Suppose that
v 2 =0, called a perfectly inelastic
collision; then F =-132x0.016/t =(-2.1
lb·s)/t . So, the smaller the time, the
larger the force. (The fact that the force is negative
means that this is the force the struck object exerts on
the ball because I have chosen the incoming velocity to
be positive. The force the ball exerts on the object is
positive and equal in magnitude.) That addresses your
second question: the ball will certainly stop in a much
shorter time when hitting a hard surface than a soft one.
Of course that should not be too surprising to you
because you know that if you fall onto a mattress it
hurts much less than if you fall on a concrete floor. So,
if t =1 s, F =2.2 lb whereas if t =0.01
s, F =220 lb. Now, if the ball hits a hard
surface it does not penetrate very far and therefore the
force is spread over a small area and pressure is large.
If the ball hits a soft surface, it penetrates deeper and
therefore the force is spread over a larger area and
pressure is smaller.
c ;
it makes the problem neither easier nor harder, just
unphysical! You will still have to do some
work! I am going to point you in the direction of solving
it yourself. First, go to an
earlier answer ; you will see that I have solved your
problem but starting at rest. If you start at some
initial velocity v 0 , the result is
that Ft=p-p 0 . Here, p is the
relativistic linear momentum, p=mv /√[1-(v /c )2 ]
where m is the rest mass, 1.68x105 kg
in your case; similarly, p 0 =mv 0 /√[1-(v 0 /c )2 ].
So now you know everything except t , so just
solve for it; this is just algebra/arithmetic. I would
advise you to convert everything to SI units (kg, m, s)
before doing your calculation. Then t will work
out to be seconds.
I do not understand what you mean by "…there
is no…time dilation…" since time dilation
is simply not of any interest here but would come into
play if the whole thing were seen by an observer in a
different frame. Also, I almost never apply the idea of
relativistic mass because I view the linear momentum as
being the quantity redefined in relativity, not mass.
Read my discussion of this in
one of the links in the earlier answer.
Atoms are mostly empty space. How can they form solid objects?
earlier answer . Although almost all the atom's mass
is in a volume much smaller than the volume of the whole
atom, the electrons fill the rest of the space in a
"cloud". Bohr's model of the atom, tiny electrons in tiny
orbits, is wrong. If you need a simpler explanation
without reference to the "electron cloud" think of all
the atoms connected to their nearest neighbors by tiny
springs. Each atom connects to its neighbors by some
interaction which can be approximated as a spring. The
atoms close enough to interact with each other will stick
together and thus form a solid object.
mg , the buoyant force
of the water B , and the drag force opposite the
velocity v
which has the form -Cv 2 where C is
some constant which depends only on the shape of the
balloon. So the net force on each balloon is -mg+B-Cv 2 =ma
where I have put this equal to the mass times the
acceleration, Newton's second law. When you release each
balloon, each will accelerate up and, as v increases,
eventually a =0 and it will have a constant speed
v t ,
called the terminal velocity, thereafter. It is easy to
show that v t =√[(B-mg )/C ]
The only difference between the two is m which
is smaller for the helium which therefore has the larger
terminal velocity. The helium balloon reaches the surface
first.
t , i.e. like y (t )=A sin(ωt )+B cos(ωt )
where A , B , and ω are
constants and y is a variable describing the motion, for
example the height above the ground in your case. If the
ball were perfectly elastic, the motion would be
harmonic, but not SHM, because y (t ) would be a train of
identical downward-opening parabolas. If it were a real
ball, it would not even be harmonic because each
subsequent bounce would be a little smaller that the
previous bounce.
recently answered this question in some detail.
c ; the electrons do not
move at that speed. So all electrons see the field turn
on almost immediately. All electrons, having
negative charge, begin accelerating in the direction
opposite the field. If the only force seen by the
electrons was due to the field, they would accelerate
until they reached the positive terminal and left the
wire. But, even though they are not bound to atoms, they
certainly see the atoms and when they hit one they are
momentarily stopped or scattered. So each electron is
constantly being accelerated, then stopped, then
accelerated and the net result is that, on average, all
the electrons participating in the current move with a
very small velocity, on the order of one millimeter per
minute; this is called the drift velocity. All the
electrons are continually gaining and then losing kinetic
energy and that lost energy shows up in the heating up of
the wire. What determines the conductivity is the number
of conduction electrons per unit volume of the material
as well as its crystaline structure and how individual
atoms scatter the electrons.
m a distance r from the center of a large
spherical mass M is, choosing zero potential
at r =∞ is V (r )=-mMG /r
where G is the universal gravitational constant.
If m has a speed v at r=R , where
R is the radius of the
earth, the total energy is ½mv 2 -mMG /R=E .
Now, suppose we want v to be big enough that m will go
all the way to r =∞ before it finally
comes to rest; then E must be zero. Therefore
v e (R )=√(2MG /R );
but, MG /R=gR , so v e (R )=√(2gR ).
That is the escape velocity at the surface of the planet
(ignoring any other forces, notably air drag) where g
is the acceleration due to gravity. Now, you want it
everywhere else. That's easy, just replace R by
r , v e (r )=√(2MG /r ).
You could use g=MGr if g now means the
acceleration due to gravity at an altitude of r-R .
(Incidentally, this does not work anywhere inside the
planet.)
https://youtu.be/hs4iv7IHvGg
does not move
in a straight line. One thing I noticed is that the toy
starts moving immediately when released; this indicates
to me that the surface it is on is not level. Just
because it is a gyroscope does not mean it will precess.
There must be an external torque acting on it; imagine an
axis running between the two legs about which you would
calculate torque. In the figure that I have clipped from
the video you can see the motor and battery and the cds
which are rotating; the toy is leaning such that I would
certainly expect the center of gravity to be beyond the
being above the axis and, were the discs not spinning the
toy would certainly fall. If you could put the center of
gravity directly above the axis, it would not precess.
You can see several instances in the video where the toy
is moving in a curved path, and in one case the path is a
pretty tight circle; these are due to precession because
of the torque due to the weight.
r 2 : F feet /F head =(r+h )2 /r 2 =[1+(h /r )]2 ≈1+(h /r )2
where r is the distance to your feet from the
center of the earth and h is your height; I have
used the binomial expansion because h /r <<1. So, F feet -F head ≈F head (h /r )2 .
Clearly, either force is going to be approximately equal
to your weight, so the tidal force is about mg (h /r )2 ;
if I take h =2 m, mg =1000 N, and r =6.4x106
m, this is about 10-10 N (2.2x10-11
lb).
The pressure force is
down on your head and slightly larger and up on your
feet; so the pressure force tends to compress you
shorter. The net force on you is simply the buoyant force
which equals the weight of the displaced fluid. If I take
your volume to be about 0.02 m3 , the buoyant
force force is about 1000 N in water and 1 N in air.
So, as you can see, the stretching due to the tidal force
is negligible to forces due to fluid pressure. So, you
will be shorter than your "real" height, but more so in
water than air. Therefore you are taller up high in the
air than down low in the water.
18
m3 and their average depth is about 3700 m
(about 12,000 ft). Now, dealing with the volume of the
atmosphere is tricky because the density of the
atmosphere gets smaller with altitude so you have to be
careful how you describe volume. The mass of the whole
atmosphere is pretty well-known, though, to be about
5.15x1018 kg; so you can calculate the volume
if the whole atmosphere were at sea level pressure where
the density is 1.225 kg/m3 , 5.15x1018 /1.225=4.2x1018
m3 . So, if the atmosphere were uniform, about
1/3 of it would flow into the emptied oceans. But, it is
not uniform so less than 1/3 would come; there would be
some base pressure 3700 m below the sea level (since
there is no sea now, when I say sea level I mean what it
was) which I would estimate to be about the same as the
pressure at sea level was because 3700 m is small
compared to the radius of the earth, 6.4x106 m
(0.06% smaller). Since the oceans occupy about 70% of
earth's surface, sea level pressure would be about what
the current pressure is at about 12,000 ft, the height of
a modestly high mountain and certainly habitable. The
highest habitable place on earth is about 16,000 ft,
so lots of places (like Denver) habitable now would not
be. And certainly nobody would be climbing Mt. Everest in
this new world!
t if you are in a region of space where there is
essentially zero gravitational field. Now imagine bringing
your clock a distance R from a spherical,
nonrotating object of mass M . The same time interval will
now be given by t'=t √[1-(MG /(Rc )]
where G =6.67x10-11 m3 kg-1 s-2
is the universal gravitational constant and c =3x108
m/s is the speed of light. For example, let the location
be on the surface of the earth, M =6x1024
kg, R =6.4x106 m; then, t'=t √[1-7x10-10 ]≈t [1-3.5x10-10 ].
So, for example, the clock in space would click off 1
second while the clock on the earth would tick off
1-3.5x10-10 seconds. To see significant
gravitational time dilation requires a larger M ,
for example a black hole; for a black hole with a
Schwartzschild radius RS , the time
dilation equation becomes t'=t √[1-RS /R ]
so time stops when R=RS . If R= 2RS ,
t' =0.7t .
puts the particle into the state
which you observe. The result of a measurement is to
observe a single state, not an admixture. So if you have
an electron whose wave function is half up, half down,
your measurement will either give one or the other, not
both; if you observed a large number of electrons, all
with the half and half wave functions, you would observe
spin up for half your measurements and spin down for the
other half.
The physics is very complicated, but there is a handy
on-line calculator where you can get at least a
qualitative estimate of how big an effect it is. I did a
calculation for water in a 1 inch diameter, 100 foot
rubber hose, and for a flow rate of 3 gallons per minute;
I assumed that the hose is horizontal and straight. The
results are shown in the figure. The calculated pressure
drop is 0.38 psi which you can compare with a typical
in-city water pressure of about 40 psi. So the drop is
only on the order of 0.1%. If the water flow were much
faster, say 20 gallons per minute, the drop would be
about 10 psi, a 25% drop.
etc . But, in
the physical world, some quantities require more
than one number to adequately describe them. The
simplest are vectors which require a
specification of both magnitude and direction.
Examples are force (e.g. 10 lb straight
up), velocity (e.g. 25 mph due north),
displacement (e.g. 5 ft straight down).
Vector quantities are denoted as bold-face, for
example F F F F + F
A·B =AB cosθ AB
where θ AB
is the angle between the vectors
A and
B A and B
are the magnitudes of the vectors. An example of
a scalar product is the work done by a force
F d W =F�d . i.e .
A·B =B�A .
The vector product or cross product is defined
as
A×B =u AB sinθ AB
where u
A and B
A·B =-B�A .
An example of a vector product is torque
τ F r
τ =r×F .
faq page
for an explanation why you can't move faster
than the speed of light. Suppose that someone
was moving with speed 99% of the speed of light
from the sun to the earth, a distance of about 8
light minutes. Because of length contraction the
distance would, for her, be shortened 8x√(1-0.992 )=1.13
light minutes and the time for her to get to
earth would be 1.1/0.99=1.14 minutes; so she
would observe the sun's demise to hit earth in a
time of 1.13 minutes, and she would arrive 0.01
minutes later. But, just because she observed
the time to be 1.13 minutes does not mean
someone on earth does; the earth-bound do not
see the distance to the sun to be shorter but
see it as 8 light minutes.
t ) but by x 0 =ct
where c is the speed of light. So the
fourth dimension has units of length just like
the original three, x 1 =x ,
x 2 =y , x 3 =z .
So, if x 0 =1 mi=t ·186,000 mi/s,
then t =1/186,000 s=5.38x10-6
s=9x10-8 minutes, approximately one
nanominute.
Before Albert Einstein developed special
relativity, it was generally assumed that time
and length are universal and independent. One
second for me is the same for anyone else in the
universe; similarly, one meter is the same
regardless of the position or motion of the
meter stick I am measuring. If someone is moving
with a speed v along my x axis and I measure the
length of his meter stick, I find it is shorter;
similarly, if I measure the rate his clock runs,
I find that it is slower than mine. We are
talking about meter sticks and clocks which, if
located in my reference frame, are identical to
mine. Mathematically this is expressed in what
we call a Lorenz transformation,
x' 0 =γ (x 0 -βx 1 )
x' 1 =γ (x 1 -βx 0 )
where the
primed coordinates are in the moving frame as
measured by the stationary frame, β=v /c ,
and γ =1/√(1-β 2 ).
So, two of the coordinates in our 4-space get
mixed up together if the other frame is moving.
Does anything analogous happen in everyday
3-space? The answer is yes; if you take a
coordinate system with three axes, (x,y,z )
and rotate it through an angle θ
about the z -axis, the new x'- and
y' -axes
both depend on what θ is and they
depend on both x and y are:
x'=x cosθ+y sinθ
y'=-x sinθ+y cosθ.
Hence the Lorenz transformation may be
interpreted as a rotation in the x-t
plane. This is the impetus of mathematically
working in a 4-dimensional space (space-time).
m =30
grams=0.03 kg. I took the acceleration to be
about g =10 m/s2 . I made a
wild guess at the terminal velocity by watching
videos on the internet and chose v =20 cm/s=0.2
m/s. I also note that many videos
seem to indicate that terminal velocity is
achieved very quickly. The resistive force is
given by F=-bv (not proportional to
v 2 as most air drag problems
are) where b is a constant to be
determined. I choose the coordinate z to
increase in the positive vertical direction and
the initial position to be at z =0 and
the initial velocity to be v =0. This is
a classic problem and I will just state the
results which result from solving the Newton's
second law equation, -mg-bv=m (dv /dt ):
v =(mg /b )[e-bt/m -1]
=0.2(e-50t -1)
z =(mg /b )[(m /b )(e-bt/m -1)-t ]
=0.2[0.02(e-50t -1)-t ]
Note that, for large bt /m , v≈-mg /b=- 0.2
which is the terminal velocity. This can be
solved for b , b =1.5 N/(m/s).
The plot above for v shows that it
takes only about 0.1 s to approximately reach
terminal velocity. The plot for z shows z (t )
in black and z (t ) if the
magnet moved with the terminal velocity the
whole time; the linear portion of the curve is
shifted up by 4 mm, possibly responsible for the
"positive z-intercept" observed by the
questioner. Of course these plots are not what
is being plotted by the questioner because each
new datum starts at rest; but every datum
plotted has the same behavior as the one I have
shown, so all the individual runs are shifted by
the same 4 mm. The shift is so small compared
with the total fall distances, the non linear
time is so small compared with the total fall
times that the experiment should be able to
extract a reasonably good measurement of the
constant b . Unfortunately, I do not
have the times of the various data points, so I
cannot be more specific; the time of fall from
the top if the terminal velocity is 20 cm/s
would be 7.5 s.
etc ., the energy is conserved because
the lost potential energy of the of the water
will show up as heat in the environment and the
water. However, this would be true regardless of
whether or not the water flowed down the
steepest path. So, I would not use the choice of
steepest path to be an example of energy
conservation.
different website and didn't get much of a response.
For some reason there doesn't seem to be a lot of
information available about this subject, or maybe I just
can't find it. Anyway, I am hoping you can help shed some
light on the subject. Please forgive the formatting -
I've posted a couple of different questions, but if you
need a single, concise question to answer, the main one I
have is this: What's going on here?
I have a bottle of water that I put in a freezer for a
period of time, until the water becomes super-cooled and
is still liquid. Now we know that when we shake or
disturb the bottle, some of it crystallizes. But it
doesn't turn into solid ice. Instead it turns into slush
- a mixture of tiny ice crystals and liquid water. Here
are my questions:
supercooled water videos and watch a few. It seems
pretty clear to me that your problem with slush is that
your water is not pure enough. Possibly not cold enough
either so that generated heat warms the water just above
zero before it freezes. This is a very touchy sort of
experiment, one of the videos has the teacher admitting
it took him a number of tries to get a good enough take
to post. By the way, I thought the answer on Reddit was
pretty good.
tidal force . Because there is friction
between the oceans and the ocean floors, the
rotation of the earth drags the bulge forward,
so it is slightly ahead of the moon as shown in
the figure (again greatly exaggerated). Now the
bulge exerts a force on the moon which causes it
to accelerate a tiny bit; that's where it gets
the added energy to increase its orbital radius.
Similarly, the moon pulls on the bulge which
causes the earth to slow down. If this were the
whole story, the energy would be conserved, the
earth losing energy and the moon gaining the
same amount. But, as noted above, there is
friction between the earth and the oceans which
takes energy away from the system, giving a net
loss of energy but the moon ends up with more
than it started with.
vice
versa ), they will not process the
information at the same rate. For simplicity's
sake, I will choose the "slow" frame B to be at
rest; in relativity, only the relative velocity
is important. Let's choose a simple concrete
example of information stream: the information
is a stream of ten pulses separated by 1 second
as seen in the reference frame B; B will process
them in 10 seconds. Now, how does A process that
same information which B processed in 10
seconds? Suppose A is moving away from B; then
the time between pulses will be longer than 1
second because he moves some distance away while
waiting for the next pulse to catch up with him—the
information is processed more slowly. You should
be able to see that if you interchanged A and B
everything else would be the same so B would
process the information more slowly. Suppose A
is moving toward B; then the time between pulses
will be shorter than 1 second because he moves
some distance toward B while waiting for the
next pulse to arrive�the information is
processed more quickly. You should be able to
see that if you interchanged A and B everything
else would be the same so B would process the
information more quickly. An
earlier answer would be of interest to you.
v ,
it hovers in still air. When it suddenly finds
itself outside it sees a headwind of v ,
not still air. This headwind will exert a force
which will tend to slow the drone down so that
eventually the drone will be hovering at rest
relative to the ground.
2 =10-4 times smaller
than on earth. And 100 earth radii is not all
that far in the whole scheme of things; the
distance to the moon is 60. Anyhow, a spacecraft
always coasts nearly all the time on a mission
somewhere in the solar system. Propulsion is
used mainly for getting up to speed and then
slowing down to orbit or land on your
destination; a much smaller use is to steer,
changing direction.
etc .
is necessary to ask your question which is,
simply: if an object is hot and cools soley
by radiation, does the object have less mass
after cooling?
First, your A2 is wrong. Photons do not have
mass even though they do have momentum and
energy. Therefore, your conclusion that the
object will be lighter because the photons carry
away mass is wrong. However, because mass is
just a form of energy, we can say that the
photons carry away energy (even though they have
no mass). But, where did that energy come from?
The object had to provide it. But there are just
as many atoms after the cooling as before, so
the object evidently changed mass into energy.
This change in mass is very difficult to observe
in the case of a red hot iron object cooling down.
Inasmuch as E=mc 2 , the change in mass would be
Δm =ΔE /c 2
where ΔE is the energy carried
off by the radiation. Suppose that ΔE
is a real big number, say a 10,000 Joules; then
Δm =104 /(3x108 )2 =10-13
kg=0.1 nanograms! This is approximately the mass
of a single yeast cell.
2 )=0.44.
Since, as you correctly stated, her clock runs
normally, it only takes her 4.4 years to get to
a star 10 light years away. It would be worth
reading my explanation of the twin paradox which
is linked to from the
faq
page.
m and its speed
v , K =½mv 2 .
It has nothing to do with its pressure or its
drag force. If there are any forces on the
object, those forces do work which might change
the kinetic energy. Because of the near
incompressibility of water, the depth will have
almost no effect in how something moves. The
effect of the pressure is simply the buoyant
force which is a force equal to the weight of
the displaced water which is nearly the same 100
ft down as 10,000 ft down. Similarly, the drag
depends only on the density, speed, and
geometry, not pressure, the drag will be about
the same at any depth.
"walking ladders" that move down slopes by themselves?
I'm thinking maybe it's the same for slinkys that 'walk' down steps. Gravity and the weight of the ladder being moved around the different legs.
I tried Googling for an answer but can't find any scientific answers.
ANSWER: This was a fun problem to figure
out. First, let's think about all the forces on the
ladder when it is slightly tipped so that two legs are
off the ground. There is the weight W N front
on the front leg and N rear
on the rear leg, and frictional forces (not shown)
f front on the front
leg and f rear on the
rear leg. The weight acts at the center of mass of the
ladder which is centered relative to the two sides, but
is closer to the front than the rear because the front is
heavier. That means that the normal force on the front
leg is greater than on the rear leg; this is important to
remember as we go further. Now, I want to examine the
torque which the weight causes. I have resolved
W into components down the incline
(W x ) and normal to the incline (W y W y W x exerts a
torque about the axis I have labelled TWIST AXIS; if no
legs slip, this torque has no effect on the motion.
However, for a couple of reasons, the rear leg may slip:
Because the normal force is smaller for the rear
leg, the maximum static frictional force will be
smaller than for the front leg.
The material contacting the ground may be
different for the rear legs than for the front,
e.g. the rear legs may have a smaller
coefficient of static friction and could slip
even if the normal forces were the same.
So, for the video, the rear legs must slip and
the ladder, with each "step" twists about a
normal axis. There are frictional losses which
would cause the ladder to lose energy and
therefore eventually stop, but since it is going
down an incline, this energy is replenished by
the work done by Wx as it
moves down the incline. It seems to me that it
would be pretty tricky to get this set up since
everything must be just right for it to work.
c it is
c squared. E=m·c·c=mc 2 .
And Einstein did not find it by trial and error;
how could he, since it was not really
appreciated at the time that mass was, in fact,
just another kind of energy? It was a natural
result of his theory of special relativity.
M exerts
on a mass m a distance r away
is F=MmG /r 2 where
G is the universal gravitational
constant. You can calculate the ratio of the
force the sun exerts on the moon to the force
the earth exerts on the moon:
F sun /F earth =(M sun /M earth )(R earth /R sun )=(2x1030 /6x1024 )(3.8x108 /1.5x1011 )2 =2.1
So the sun exerts a force more than twice the
force the earth exerts on the moon. But if you
think about it, the moon does revolve around the
sun since it moves right along with the earth;
but because of the force the earth exerts, it
also revolves around the earth.
-11 /102 =6.67x10-8
N=1.5x10-8 lb.
v where v is the
speed after the collision; equating the two and
solving for v , I find v =27.3
mph. For the second case, 65,000x50/66,500=v =48.9
mph.
Human Universe with Brian Cox. If you haven't seen it, in
part one he goes to Russian to meet with the Soyuz spacecraft. He tells his
viewers that only knowing two of Newton's equations (F=ma and the law of gravitation
F=GmM /r 2 ) he can calculate that the spacecraft only
needs to reduce velocity by 128 m/s to safely reenter orbit. I have looked
and tried to figure out how the calculations work but can't. Do you have
any clue how he would have done the math?
R =6.4x106 m
(radius of the earth)
h =4x105
m (altitude of the space station)
G is
the universal gravitational constant (will not need
it)
M is the mass of
the earth (will not need it)
m is the mass of
Soyuz (will not need it)
g =9.8
m/s2 =MG /R 2
(acceleration due to gravity at r=R )
√(MG /R )=√(gR )
(this will be useful later and is why I do not need
to know G or M )
MmG /r 2
is, as you note, the gravitational force on m a
distance r from the center of the earth.
If orbits are
circular, a=v 2 /r and you can
show that v =√(MG /r ). I
am thinking that we need to find the difference of speeds
between orbits with r=R+h and r=R :
vR -vR+h =Δv
=√(MG )[√(1/R )-√(1/(R+h ))]
=√(MG /R )[1-(1+(h /R ))-1/2 ]
≈√(MG /R )[½(h /R )]
=½h √(g /R )
=247 m/s
Note that, because
h /R= 0.0625 is small compared to 1, I have used the binomial
expansion (1+x )n ≈1+nx .
My answer is about twice what Green
got. Of course, in the real world, air drag when entering
the atmosphere would significantly reduce the required
speed change due to the additional braking it would
cause; but you cannot estimate that using only those two
equations. So, I am at a loss to understand the
difference.
h ~200 km, after
leaving the ISS, but I could find no reference to its
altitude in the video.
L =√(l (l +1))h /2π.
The answer to your question is that the
Bohr model is simply incorrect. As luck would
have it, it describes the energy levels of the
hydrogen atom well, but it is incorrect
regarding angular momentum of the levels.
etc .
Now, all the atoms in the solid are constantly
vibrating as if attached to tiny springs, and
the hotter the metal gets, the larger the
amplitudes of their vibrations becomes. This
means that they present bigger targets for the
electrons to hit and so collisions happen more
often which reduces the drift velocity and
therefore the conductivity.
f (x-vt )
where f is any function; for
example where x is the position, t
is the time, and v is the speed at
which the disturbance travels through the medium
which supports it. For your examples: dominos
falling and people in a stadium sitting and
standing are both waves, but wind is not. Wind
is an example of the entire medium moving but
that would not be described by a wave equation.
The stadium wave would not be a wave if the fans
all all just ran around together, like a
hurricane of people.
twin paradox ".
etc . appeared. To see a timeline of the
universe, click
here .
earlier answer . (You could also have found
this on my
faq page.) The answer to your second
question is that the theory of general
relativity finds that any energy density warps
spacetime, that is causes a gravitational field;
a photon has energy and therefore creates
gravity; keep in mind how small this is when
compared to any macroscopic mass.
x and
momentum p (mass times velocity). Stated
mathematically, Δx Δp ≤ℏ
where ℏ is the rationalized Planck constant
(a very tiny number and Δ denotes the uncertainy of the quantity.
So if the velocity (momentum) uncertainty is
zero (particle perfectly at rest), the
uncertainty of the position of the particle is
infinite—the particle could be anywhere in
the universe.
dark matter halos would be of interest to
you.
V , the velocity of the
car/block after the collision to be U ,
the mass of the car to be M , and the
mass of the block to be m . Conservation
of linear momentum gives MV =(M+m )U
or U=V [M /(M+m )]. Note
that U<V , so the car has lost energy
and the block has gained energy.
If you are interested you can calculate the
change in energy (ΔE=E after -E before )
for each:
ΔE car = ½M (U 2 -V 2 )=½MV 2 [
(M 2 /(M+m )2 )-1]
ΔE block = ½MU 2 =½MV 2 [M 2 /(M+m )2 ]
For example, if M=m ,
ΔE car =- (3/4)[½MV 2 ]
and ΔE block =(1/4)[½MV 2 ].
The car lost three fourths of its initial energy
and the blocked gained one fourth of the initial
energy. In this case , half of the total
energy was lost. In this case , half the
speed was also lost.
μmg cosθ . As far as I know, μ cannot be greater than 1 (irrespective of contact patch size) for a Rubber tyre on concrete road. At 45 degrees, the drag due to gradient (mg sinθ ) is equal to the Normal force (mg cosθ ). Any more than 45 degrees, and the drag force increases and Normal force decreases, thus making the vehicle's tyres slip due to loss of traction.
Now, I see videos of RC cars on
youtube
climbing gradients well over 54 degrees.
Could you explain this? Is the μ value greater than 1?
mg may be considered to act at the center
of mass of the truck (white cross) which is
located a distance h above the ground,
a distance d 1 behind the
front axle, and a distance d 2
in front of the rear axel, as shown. There are
three equations, sum of the forces along the
ramp equals zero, sum of the forces
perpendicular to the ramp equals zero, and sum
of the torques about the point where the rear
wheel touches the road equals zero:
f 1 +f 2 -mg sinθ =0
N 1 +N 2 -mg cosθ =0
(d 1 +d 2 )N 2 +hmg sinθ-d 2 mg cosθ= 0
Now, we are interested in when
the truck will start rotating backward about the rear
axle; when it is just about to do that, the normal
force N 2 =0. So, the third equation
tells us that tanθ=d 2 /h ;
therefore the maximum angle before tipping is θ tip =arctan(d 2 /h ).
For example, if d 2 =1.5 m and h =1
m, θ tip =56.30 . What
determines tipping is just where the center of mass is;
you want it as far forward and as close to the ground as
you can.
But, if you lose traction, slip,
before you rotate, you do not need to worry about
rotating. As you correctly surmised above, the truck will
start slipping when θ slip =arctan(μ )
where μ is the coefficient of static. I agree
with you that μ= 1 is a reasonable estimate
for rubber on dry concrete. However, the video you sent
is not done on a concrete ramp, rather the surface
appeared to be carpet. Since the tire tread can press
into the carpet pile, it is not surprising that μ >1
for those videos.
If you cannot find what you are looking for, it was
probably archived. You may find the most recent archived
file
here .
