Classical Mechanics

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

ANSWER:
This is sort of a classic introductory physics problem. The idea is: when does an object lose contact with the surface on which it is moving, usually taken to be a segment of a circle. I will do an approximation that the shape is a circle and that the car is a point mass. You can then generalize to your case from there or else give me more information. In the sketch above, the radius of the circle is R, the weight of the car is mg, the angle specifying the current position of the car is θ, the force which is causing the car to move with some constant speed v is F, and the normal force of the road on the car is N. Note that although the car has a constant speed, it has a centripetal acceleration a=v²/R toward the center of the circle. Applying Newton's laws, F-mgcosθ=0 and mgsinθ-N=mv²/R. The first equation tells you what force you need to keep it moving at a constant speed, F=mgcosθ which is really not of interest to you; note that the force is in the direction of v on the way up and opposite on the way down since the cosine changes sign at 90⁰. The second equation tells you what N is for any position of the car, N=mgsinθ-mv²/R; note that if N is negative it points toward the center but the road cannot pull down, only push up, so the car could not stay on the road at that speed and angle. What is really of interest is under what conditions would N=0=gsinθ-v²/R or v²=Rgsinθ; this would tell you the speed (angle) at which a car with a particular angle (speed) would leave the road. Note that it is independent of the mass. Notice also that v²/Rg<1 because the sine function cannot be larger than 1.0. For example, for what speed will the car leave the road at 45⁰ if gR=300 (m/s)² (R≈30 m)? v=√(300x0.707)=14.6 m/s=32.7 mph. Or, at what angle will a car with speed 35 mph=15.6 m/s leave the road? θ=arcsin(15.6²/300)=54.2⁰.

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

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

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

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

ANSWER:
I did not misunderstand your question. You cannot simply adjust the friction to be what you want it to be. You can, however, simulate what you want to happen by wrapping a string around the cylinder and pulling up on the string with a force mgsinθ where θ is the angle of the incline and m is the mass of the cylinder; imagine that the incline is smooth (frictionless). Now there will be a net torque about the center of mass of mgRsinθ=Iα where I is the moment of inertia and α is the angular acceleration of the cylinder. The cylinder will spin in place. Your hand will have an acceleration of a=mgR²sinθ/I; for a uniform solid cylinder with I=½mR², a=2gsinθ.

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

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

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

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

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

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

ANSWER:
There is an earlier answer about a bicycle making a turn. It would be helpful for you to read that first. The easiest way to do this problem of your trailer turning a curve is to introduce a fictitious centrifugal force which I will call C, pointed away from the center of the circle; the magnitude of this force will be mv²/R where m is the mass of the box plus trailer, v is its speed, and R is the radius of the curve. The picture to the right shows all the forces on the box plus trailer: W is the weight and the green x is the center of gravity (COG) of the box plus trailer; f₁ and f₂ are the frictional forces exerted by the road on the inside and outside wheels respectively; N₁ and N₂ are the normal forces exerted by the road on the inside and outside wheels respectively; the center of gravity of the box plus trailer is a distance H above the road and the wheel base is 2L (with the center of gravity halfway between the wheels). Newton's equations yield:

• f₁+f₁=C for equilibrium of horizontal forces;

• N₁+N₂=W for equilibrium of vertical forces;

• CH+L(N₁-N₂)=0 for equilibrium of torques about the red x.

If you work this out, you find the normal forces which are indicative of the weight the wheels support: N₁=½(W-C(H/L)) and N₂=½(W+C(H/L)). A few things to note are:

• the outer wheel supports more weight,

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

• at a high enough speed C will become so large that N₁=0 and if you go any faster you will tip over, and

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

Now we come to your question. You first want the maximum speed without tipping. Solving for v in the N₁=0 equation gives

v_max=√[RWL/(mH)]=√[RgL/H]

where g=9.8 m/s²=32 ft/s² is the acceleration due to gravity. For example, suppose that R=7 ft, L=17 in=1.42 ft, H=30 in=2.5 ft. Then v_max=√[7x32x1.42/(2.5)]=11.3 ft/s=7.7 mph.

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

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

ADDED THOUGHT:
When just about to tip, all the weight is on the outer wheel and so N₂=W and f₂=μN₂=μW, where μ is the coefficient of static friction. If you work it out, the minimum value the μ must have to keep the trailer from skidding is μ_min=L/H. For the example worked out above, μ_min=0.57. For comparison, μ for rubber on dry asphalt is about 0.9, so the trailer would not skid.

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

ANSWER:
80 m is pretty high, so you will be going very fast when you hit; therefore, neglecting air drag might introduce significant error. But air drag is pretty tricky to calculate and I will neglect it; the answer I get will be somewhat bigger than what would really happen. I will take your speed to correspond to running a 100 m dash in 15 s, about v ≈6.7 m/s. The equations of motion are x=6.7t and y=-4.9t² where x and y are the horizontal and ver tical positions relative to the edge of the cliff and t is the time after jumping. Solving the y equation for t when y=-80 m (the ground), t=√(80/4.9)=4.04 s, so x=6.7x4.04=27.1 m. That is the answer neglecting air drag. Note that it is independent of the mass m.

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

ANSWER:
This is a standard intermediate-level classical mechanics problem for accelerated reference frames. All the mathematical background is much too complicated to include here but the answer is that it lands a distance d=4ωv₀³cosλ/(3g²) west of where it was shot; here ω=7.27x10^-5 s^-1 is the angular velocity of the earth, v₀ is the initial velocity of the bullet, λ is the latitude where the gun is located, and g=9.8 m/s² is the acceleration due to gravity. This, of course, neglects any air drag or wind corrections. For example, for λ=45⁰ and v₀=390 m/s (typical 9 mm muzzle speed), d≈60 m. At the poles (λ=90⁰), it goes up and comes back down perfectly vertically.

ANSWER:
Your first question ("…will water flow through each hose in the same amount of time?") is ambiguous, so let me answer the question by finding how the speed of the delivered formula (labelled V_bottom in the figure) depends on the geometry. The operative physics principal is Bernoulli's equation, P+ЅρV²+ρgy=constant where P is pressure, ρ is fluid density, y is the height above some chosen reference level, and g=9.8 m/s² is the acceleration due to gravity. In your case, P is atmospheric pressure both at the top and at the bottom, I will choose y=0 at the bottom so y=h at the top. Therefore Bernoulli's equation becomes V_top²+2gh=V_bottom² or V_bottom=√(2gh+V_top²). There are two ways that V_bottom will be independent of the geometry: (1) if h is held constant by replenishing the formula at the top or (2) if the area of the bore is much larger than the area of the distal tip A>>a. Both of these result in V_top≈0 so V_bottom≈√(2gh). If neither of these is true, it is a much more complicated problem.

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

answer as close as you can approximate.

ANSWER:
It is ill-advised to make a donation before receiving an answer because if a question makes no sense, I cannot answer it and have no means of returning your money. In your case, there are some misconceptions and you need to understand that the key word in your question is "approximate". All air-drag questions can only have approximate solutions. Your main misconception is that there is no such thing as "the most aerodynamic object ever". A second misconception is that you will ever reach the terminal velocity; the analytical solution to the problem has the terminal velocity reached only after infinite time and I will use 99% of the terminal velocity as having reached full speed. Since I have no information at all about the detailed shape of the object, I will simply use a generic form for the air drag F which can be assumed to be well approximated as a quadratic function of velocity for this case, F=-cv² where c may be approximated as c≈јA, A being the cross sectional area presented to the wind (the negative sign indicating that F is opposite v). This approximate value of c is valid only if SI units (meters, seconds, kilograms) are used.

Now, the solutions. I will not give full details since they are fairly mathematical. Convert, first, to SI units: L=25"= 0.635 m, R=0.25"=0.00635 m, m₁=1750 grain=0.1134 kg, m₂=3500 grain=0.2268 kg, A=πR²=1.267x10^-4 m², F=(3.167x10^-5 N)v². To calculate the terminal velocity v_t, find when the drag force is equal to the weight mg (g=9.8 m/s²): mg=cv_t², v_t=√(mg/c)=√(4mg/A); I find v_t1≈187 m/s and v_t2≈265 m/s. The height h from which you would have to drop these to achieve this speed with no air drag is given by h=v_t²/(2g), so h₁^{no drag}=1780 m and h₂^{no drag}=3560 m. Now, if air drag is included in the calculations, it can be shown (see any intermediate-level classical mechanics text) that v/v_t=√[1-exp(-2gh/v_t²)]. Now, setting v/v_t=0.99 and solving for h, I find h₁=6989 m and h₂=14,034 m. I have shown a graph with the velocity relative to the terminal velocity v/v_t as a function of the height h from which it was dropped. Keep in mind that these calculations are approximations as explained above; I have tried to give enough details so that you could vary the parameters if you like, in particular the drag coefficient c and the ratio of v/v_t.

ADDED COMMENTS:
I came across a more detailed discussion of the drag forces for cylinders if you are interested. You will be referring to the picture to the left above for the following discussion. First, I need to discuss the definition of c above more carefully. In more detail, c=ЅρAC_d where ρ is the density of air which is about 1.2 kg/m³ at standard pressure and temperature and C_d is a number (called the drag coefficient) which depends only on the geometry. Note that for my calculations above the drag coefficient was C_d=јA/(Ѕx1.2xA)=0.6. The graphs above show that the drag coefficient depends on the ratio of L/d=L/(2R) which has a value of 50 for your projectile. So, if your cylinder has a blunt end, C_d≈0.81 (c≈0.6xAx0.81=0.49·A≈ЅA) and, for a rounded end, C_d≈0.4 (c≈0.6xAx0.4=0.24·A≈јA). So, the calculation above is likely a good approximation to your "… most aerodynamic object…" My whole treatment of this problem, though, assumes that c is a constant. However, as you can see above, c depends on the air density and the air density depends on altitude. Surely at values of h around 7 km or 14 km, the air density is much smaller than at sea level and therefore c is much smaller and the terminal velocity is much larger; this makes the problem considerably more difficult requiring numerical solutions, beyond the scope of this web site.

ANSWER:
First of all, it is not really clear what you mean when you say "…the gravitational attraction between them should be equal at all points…" The force felt by one little piece of mass on one board due to the other board should be the same magnitude as the other board experiences; their directions, however, will not be opposite each other except at the centers. If we approximate your boards as thin sticks (pretty good since 2"<<72" and 4"<<72"), it is fairly easy to calculate the force experienced by a very small piece of mass m in either board due to presence of the other board for your parallel configuration; it is a little messy, so I will not write it here. This force will have both an x-component and a y-component. The graph to the left shows F_x and F_y as well as the magnitude of the force, F=√(F_x²+F_y²) for sticks of mass 1 kg, length 1 m, and separated by 1 m to illustrate; I have chosen x=0 at the left end of the stick and x=1 at the right end. I have normalized the forces to the gravitational constant G. On the right you can see the forces on each stick at the ends and centers (red shows the components). The addition of other masses, like your third board, will not alter the forces between these two sticks but will certainly alter the force felt by any mass on these sticks; I guess the answer to your question is that the force which one board exerts on another is not changed by adding more boards. Still, we are normally interested in the net gravitational force on something, not just the force which one other object of many exerts on it. (Incidentally, note that m will also experience a force due to the rest of its own board.) You might be interested in an earlier answer concerning gravitational fields of cylinders

ANSWER:
The center of mass and center of gravity are coincident if the gravitational field is uniform (the same everywhere across the object). The center of mass of a uniform half disk of radius R is a distance d=4R/(3π) from the center of the straight edge diameter. If the mass density is ρ and the thickness of the disk is t, the mass is M=ρtπR²/2. So you can treat the two halves as two point masses M₁ and M₂ separated by a distance 2d. So, the distance from the center of the disk to the center of mass is (dM₁-dM₂)/(M₁+M₂)=[(dρ₁tπR²/2)+(dρ₂tπR²/2)]/[(ρ₁tπR²/2)+(ρ₂tπR²/2)]=d(ρ₁-ρ₂)/(ρ₁+ρ₂); this will always be on the more massive side. For example, if ρ₁=2ρ₂, the center of mass is a distance d/3=4R/(9π)≈0.14R from the center.

There are two forces on a ski jumper, his weight W and the force of air drag on him F. The weight is always a force which points vertically down and is proportional to the mass m. The air drag always points opposite the skier's velocity and depends only on the velocity magnitude, the properties of the air, and the shape of the skier, but not his mass. Shown in the figure are the force diagrams for the heavy (#1, leftmost) and light (#2) skiers when they are going up. (I have chosen an extreme case, the heavy skier about twice the mass of the light skier, to emphasize the difference. Everything else about the two is identical.) Each force results in an acceleration (shown as the green-dashed vectors) which result in net accelarations shown by the full-drawn green vectors. The lighter skier has a larger drag acceleration because the acceleration is F/m and both have accelerations due to their weights which are equal. But looking at the horizontal and vertical of the net acceleration, both are larger for the light skier—he is losing forward speed faster and is accelerating toward the ground faster. Also shown in the figure are the force diagrams for the heavy (#3) and light (#4, rightmost) skier when they are going down. Now, the lighter skier still has a larger horizontal component of acceleration, he is slowing down faster, and his vertical component is perhaps just slightly smaller. I would conclude that your book is wrong, the heavy skier jumps farther. Consider the following experiment to demonstrate that the heavier will have a larger range using two balls of the same size, say a baseball and a baseball-sized nerf ball. Throw them each as hard as you can and projected at about the same angle; which would you expect to go farther? In fact, there is an analytical solution (approximation) for the range R of a projectile: R=[v₀²sin2θ/g)[1-(4v₀sinθ/(3g))(C/m)] where v₀ is the speed at launch, θ is the launch angle, and C is a constant which depends only on the size and shape of the projectile. Note that R gets larger as m gets larger.

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

ANSWER:

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

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

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

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

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

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

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

ANSWER:
Very large pressures can change the properties of the material and new phases may occur (like water and ice are two different phases of H₂O). A particularly good example is water itself as shown in the figure to the right; pressures up to about 10,000,000 atmospheres are shown in the graph and numerous phases beyond ice/water/steam are seen at high pressures or very low temperatures. For higher yet pressures, you need to look to stars. If a star is massive enough it will, after going through a supernova stage, collapse under the pressure of its own gravity to where electrons are pushed into the protons and the whole star becomes a neutron star, essentially a gigantic nucleus. If heavy enough, it will continue collapsing into a black hole. What individual particles do depends on the pressure and they essentially will retain their identities or undergo reactions with other particles (as in the neutron star formation) or lose their identities (in a black hole).

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

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

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

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

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

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

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

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

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

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

     v=√[2mgh/(m+(I/R²))].

The moments of inertia are
     I_solid=2mR²/5 and I_hollow=2mR²/3.
First note that the mass will cancel out, so
     v_solid=√(10gh/7) and v_hollow=√(6gh/5)<v_solid
and so the solid sphere is the winner.

Now, suppose a hollow sphere, mass m, radius R, moment of inertial I=2mR²/3 is filled with a fluid of mass M. (I assume that the spherical shell is very thin compared to the radius of the ball.) There are two extremes of what can happen, either the fluid does not rotate at all as the ball rolls (a superfluid) or else it begins rotating with the ball almost immediately (maybe molasses). In the first case, when the ball reaches the bottom its rotational kinetic energy will be all due to the rotating of the ball, its translational kinetic energy will be due to the both the ball and the fluid, and the potential energy at the beginning will be due to both:

     Ѕ(2mR²/3)(v²/R²)+Ѕ(M+m)v²=(M+m)gh.

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

     Ѕ(2MR²/5)(v²/R²)+Ѕ(2mR²/3)(v²/R²)+Ѕ(M+m)v²=(M+m)gh.

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

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

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

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

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

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

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

ANSWER:
OK, here goes: I will use a coordinate system with increasing y vertically upward and y=0 at the landing shelf. The cat will have acquired some velocity -v when his feet hit the shelf. Assuming he stops after going a distance ℓ and accelerates uniformly, we have the two kinematic equations 0=ℓ-vt+Ѕat² and 0=-v+at. From the second equation t=v/a; so, from the first equation, 0=ℓ-v(v/a)+Ѕa(v/a)²=ℓ-Ѕv²/a so a=Ѕv²/ℓ. Now, as the cat is landing there are two forces on him, his own weight mg down and the force F of the shelf up, -mg+F and this must be equal, by Newton's second law, to ma, so F=m(g+Ѕv²/ℓ). This, by Newton's third law, is the force which the cat exerts down on the shelf. Finally, the speed if dropped from a height h is v=√(2gh), so F=mg(1+(h/ℓ)).

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

ANSWER:
There will be very small effects due to the fact that this is not true "weightlessness" in that the spacecraft is accelerating in a gravitational field and not of zero size. I take it that this kind of detail is not what you are interested in, that I can simply answer your question for the pencil being in empty space, initially at rest. I will suppose that the pencil has mass m, moment of inertial I about its center of mass which is a distance d from the end where a constant force F is applied, perpendicular to d, for a very short time t. Newton's second law for translational motion says that the impulse equals the change in linear momentum, Ft=mv; after the impulse, the center of mass will move in the direction that F was applied with speed v=Ft/m. Newton's second law for rotational motion says that the angular impulse equals the change in angular momentum, τt=Ftd=Iω; after the impulse, the pencil will rotate around the center of mass with an angular velocity  ω=Ftd/I. A relationship between v and ω can be written as ω=mvd/I. You can also write the total energy of the pencil, the sum of translational and rotational kinetic energies, as E=Ѕmv²+ЅIω². For example, if the pencil is modeled as a uniform stick of length L, I=mL²/12 and d=L/2, so ω=mv(L/2)/(mL²/12)=6v/L and E=3mv²/2; in this case, 2/3 of the kinetic energy is due to the rotation. Also, the speed u of the end of the pencil with respect to its center is u=Lω/2=3v. Regarding angular momentum, Iω, it is conserved after the initial "flick" because there are no torques on the pencil.

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

ANSWER:
Yes, the ideal gas law, PV/(NT)=constant where P is pressure (not gauge pressure), V is the volume, N is the amount of gas, and T is the absolute temperature. Assuming that V and N remain constant, P/T=constant. Let's do an example. Suppose that the ball is filled to a gauge pressure of 13 psi when the temperature is 70⁰F. The absolute pressure is 13 plus atmospheric pressure 14.7 psi, P₁=13+14.7=27.7 psi. The temperature in kelvins (absolute) is 70⁰F=294 K. Now suppose we cool the football to 10⁰F=261 K. Then, 27.7/294=P₂/261, P₂=24.6 psi, and the resulting gauge pressure is 24.6-14.7=9.9 psi. I guess it is important to fill the ball at the temperature at which it will be played.

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

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

ANSWER:
Variations on this question are probably the most frequent question I answer. I always say that you always have to specify velocity relative to what. In your question, I take it that the velocity of the train relative to the ground is 60 m/s and the velocity of the person relative to the train is 3 m/s in the opposite direction as the train's velocity. The velocity of the person relative to the ground is 57 m/s in the direction of the train's velocity. An observer on the ground would see the person with a horizontal velocity of 57 m/s regardless of whether he was on the train or had jumped off the back. Once he is off the train he will begin falling and hit the ground with both horizontal and vertical components of his velocity. For example, if he starts at a height of 2 m above the ground, his vertical component when he hits the ground will be about 6.3 m/s so his total speed will be √(57²+6.3²)=57.3 m/s.

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

ANSWER:
Aluminum has an electrical conductivity of about 3.5x10⁸ Ω^-1m^-1 and copper has a value of about 6x10⁸ Ω^-1m^-1. Therefore, at any speed, the magnet will induce a larger current in the pipe for the more conductive copper.

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

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

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

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

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

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

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

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

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

ANSWER:
Let's first write Bernoulli's equation, P+Ѕρv²+ρgy=constant. At any point in the fluid P is the pressure, ρ is the density, v is the speed, g is acceleration due to gravity, and y is the distance relative to some chosen reference point (above, y>0, below y<0). Maybe it would help you to accept this if I tell you that Bernoulli's equation is simply conservation of energy. Since To answer your question we can ignore the y part, it is a horizontal pipe, P+Ѕρv²=constant. You should also understand that this equation is exactly true only under idealized conditions:

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

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

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

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

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

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

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

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

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

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

ANSWER:
I prefer to work in SI units, so I will transform back to mph in the end. The masses of the cars are about m₁=4500 lb≈2040 kg and m₂=3000 lb≈1360 kg. The coefficient of kinetic friction for rubber on dry asphalt is in the range 0.5-0.8, so I used μ=0.65. The acceleration due to gravity is g=9.8 m/s². The distance pushed is d=50 ft≈15 m. The frictional force F acting on the two cars as they slide is due only to the friction  between the wheels of car 2 sliding on the asphalt is F=μm₂g=0.65x1360x9.8=8660 N. The work done by this friction was W=-Fd=-8660x15=-130,000 J. This took away the kinetic energy K the two vehicles had just after the collision, so K=Ѕ(m₁+m₂)v_2,1²=1700v_2,1²=130,000, so v_2,1=8.7 m/s≈20 mph; this is the speed the two cars had immediately after impact. Finally, use momentum conservation to get the speed of car 1 before the collision: m₁v₁=(m₁+m₂)v_2,1=2040v₁=3400x8.7=29,600 k∙m/s, and so v₁=14.5 m/s≈32 mph.

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

ANSWER:
There is no way to calculate this with the information you have given me. All I can tell you is the absolute maximum distance he would go if launched with a speed of about 30-35 mph at 45⁰ angle to the horizontal, about 25 meters. It would certainly be less than this since he would lose some kinetic energy in the collision and would not be launched with a speed as great as he came in with and he would likely be launched at another angle.

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

ANSWER:
OK, here is the full equation if you want to play around with it: R=v²sin(2θ)/g where R is the horizontal range in m, v the speed of launch in m/s, θ is the launch angle, and g=9.8 m/s² is the acceleration due to gravity; this is for the projectile landing at the same level where it was launched from, not really exactly true but of minimal consequence since we are only doing a rough calculation. 30 mph=13.4 m/s, so you can put in any speed as v_m/s=0.45v_mph. For θ=45⁰, R=v²/g. To answer your question, the distance is not proportional to speed, rather to the square of the speed. So, if you lost 1/3 of your speed (30 mph) in the collision your speed in m/s would be v=20x0.45=9 m/s, so R=81/9.8=8.3 m≈27 ft. This is more realistic, I believe, than 25 m which would be predicted for 35 mph.

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

ANSWER:
The example you stated is the simplest to understand because this force is perpendicular to its moment arm and so the torque is τ=Fy₀. First of all, a point mass cannot, by definition, rotate about an axis through itself. If you calculated the torque about an axis ay x=0, y=y₀, you would get zero torque and zero "rotation". I think you are confusing yourself by saying that a torque causes a rotation. In fact, a torque causes an angular acceleration about the axis, so you need to ask whether your applied force causes an angular accleration. Angular accleration a is related to the acceleration a by α=a_t/R where R is the moment arm (y₀ in your example)and a_t is the tangential acceleration, the component of the acceleration a perpendicular to the moment arm. In your notation, the acceleration is all tangential and so a_t=F/m. Therefore, there is an angular acceleration α=F/(my₀) about the z axis at the instant you start your problem. Even if there were no force and the mass were just moving with some velocity v in the x direction right now, it would be considered to be rotating about the z axis with an angular velocity ω=v/y₀. If you exerted some force which kept it moving in a circle of radius y₀ with speed v (that would be a force which had no tangential component, pointed toward the origin, and had a magnitude mv²/y₀), you would surely say that it was rotating around the origin, wouldn't you? In general, whenever the velocity has a component perpendicular to the line drawn to some axis, it has an angular velocity around that axis.

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

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

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

ANSWER:
First of all, it is physically impossible for the gun to move with the speed of light; see my faq page to find out why. But, I think I can get to the crux of your question by having your gun move at 99.9% the speed of light, v=0.999c. Let us suppose the muzzle velocity of the gun were u=0.002c (which is, incidentally, much faster than any real bullet would travel). Now, classical physics would have a bystander see the bullet going with a speed v+u=1.001c, faster than the speed of light. But we know that this is not possible, so classical physics must be wrong. The correct formula for velocity addition in special relativity is (v+u)/[1+(uv/c²)]=1.001c/(1+0.999x0.002)=0.999004c, just slightly faster than the gun. If you were moving along with the gun, you would simply see the bullet go forward with a speed 0.002c. All the above is if the bullet is fired in the direction the gun is traveling. If you fire the gun backwards from its direction of travel, replace u by -u and find that (v-u)/[1-(uv/c²)]=0.998996c, slightly slower than the gun but moving in the same direction. If you were moving along with the gun, you would simply see the bullet go backward with a speed 0.002c.

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

ANSWER:
Well, one might say that "nobody really understands how they work"! I have "sort of" answered this question before, but mostly dodged it. Here I will do a little better, I hope! I can give you a good qualitative explanation but in terms of being able to design one by just sitting down with some fundamental equation like Bernoulli's equation, forget it. Fluid dynamics can be deceptively simple or amazingly complex. To do serious aerodynamic design requires extremely powerful computers and trial and error wind tunnel experiments. The things we do to reduce air drag are often counterintuitive compared to our expectations. With that prologue, let me give you a couple of simple examples. Our expectations are that an object should be very smooth and this is often the case, particularly if velocities are not too large. Your expectation that the increasing area presented to the onrushing air causes greater drag is reasonable and you will find many examples here on AskThePhysicist.com where I approximate the drag force as being proportional to the cross sectional area, e.g. F_drag≈јAv². But there are many situations, particularly at high speeds, where this expectation breaks down and the culprit is turbulence in the air. To illustrate how turbulence affects drag and how smooth is sometimes not good, consider a golf ball. Have you ever noticed that a golf ball has dimples? The purpose of these dimples is to reduce air drag. As shown on the left above, a smooth ball at a high velocity has a long turbulent volume behind it; the pressure in this turbulent volume is lower than the pressure on the front of the ball and this contributes to there being a large net drag force. If golf balls were nice and somooth, they would die and fall very much sooner than a dimpled golf ball does. The ball on the right shows the effect of the dimples; the rough surface induces a layer of turbulence which actually makes the ball "slipperier" which causes the flow around the ball to come back together and reduces the volume of turbulence contributing to drag. (The picture on the right, with most of the turbulence erased, would be what the smooth ball would look like at low speeds.) The hairs on a tennis ball serve the same purpose. One other example is the net you sometimes see replacing the tailgate of a pickup to reduce the drag the tailgate causes. This, it turns out, is a complete fraud. With the tailgate closed a bubble of still air forms in the bed of the truck which deflects the air smoothly over the rear of the truck. So, in your case, the purpose of the spoiler, very similar to the golf ball dimples, is to disrupt the smooth flow in such a way that the net effect is less turbulence behind the vehicle. Incidentally, there is something called the Reynolds number which allows you to estimate whether or not turbulence is important.

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

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

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

ANSWER:
The operative physical principle is Bernoulli's equation, P+Ѕρv²+ρgy=constant. Here, ρ is the density of the fluid, P is the pressure, v is the speed of the fluid, and y is the elevation relative to some y=0, all at any point in the fluid; g=9.8 m/s²  is the acceleration due to gravity. In your case, P is atmospheric pressure both at the top and at the bottom (assuming the tank is not sealed), I will assume y=0 at the bottom and y=h at the top surface, the speeds at the bottom and top are v_bottom and v_top. Putting these into Bernoulli's equation and solving, v_bottom=√[2gh+v_top²]. Now, if the tank is much bigger than the pipe, usually the case, you can approximate v_top≈0 and so v_bottom≈√[2gh]. An interesting fact: this is exactly the speed the water would have if you simply dropped it from a height h. (You do not have to imagine a vacuum since the pressure is easily included and, as here, often the same everywhere. If the tank were pressurized to a pressure greater than atmospheric, the water would drain faster.)

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

ANSWER:
No, I am assuming that the tank has water in it and the point I call y=h in my answer is the top surface of the water in the tank which I subsequently approximate as a much larger area than the pipe. You could, I guess, let y=h be the top of the pipe but then the pressure there would not be atmospheric. Then you, as you suggest, would have to say v_top=v_bottom≡v; the subscript top now means the top of the pipe. So, let's call the depth of the water in the tank to be d and apply Bernoulli's equation to the tank only: P_top+Ѕρv²+ρgh≈P_A+ρg(d+h) where I have again approximated the top surface of the tank to have zero velocity. Then the pressure at the top of the pipe is P_top≈P_A+ρgd-Ѕρv². Now apply Bernoulli's equation to the pipe: P_A+ρgd-Ѕρv²+Ѕρv²+ρgh≈P_A+Ѕρv² or Ѕv²≈g(d+h) which you will see is the same answer as above except h now means the bottom of the water in the tank rather than the top of the water in the tank.

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

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

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

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

ANSWER:
If the force increases linearly with distance, it is behaving like an ideal spring. The force F necessary to stretch a spring with spring constant k a distance x is F=kx. Now, 27 kg is not a force, it is a mass; but, if you mean that the force is the weight of a 27 kg mass (which you doubtless must), then F=27x9.8=264.6 N. So, since x=0.72 m, the spring constant of your bow is k=264.6/0.72=367.5 N/m. The potential energy of a spring stretched by x is Ѕkx²=Ѕ(367.5)(0.72)²=95.3 J. The kinetic energy of a mass m with a speed v is Ѕmv²=Ѕ(0.3)v²=0.15v². Ideally, all the potential energy of the bow is given to the kinetic energy of the arrow, so v=√(95.3/0.15)=25.2 m/s≈56 mph. If the arrow is not launched horizontally, there would be a small correction for change in gravitational potential energy. I am not sure how good a model this is for a real bow.

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

ANSWER:
300 gr=0.0194 kg, so Ѕmv²=Ѕ(0.0194)v²=0.0097v², so v=√(95.3/0.0097)=99.1 m/s≈125 mph. Incidentally, for easy unit conversion I recommend a free little program called Convert.

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

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

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

ANSWER:
Eventually the cylinder, room, and gas will all be at the same temperature. If (see above) the temperature right after filling were higher than the room temperature, it would cool down and the pressure would decrease because now both V and N would be constant while the gas cooled. Here is a numerical example: suppose that the room temperature is 20⁰C=293 K and the gas temperature is 40⁰C=333 K and the hot gas has a pressure of 4.0 Atm. Then, P/T=constant=4/333=P_final/293 or P_final=3.5 Atm.

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

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

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

ANSWER:
I cannot resist responding in some detail to this since it is so wrong! What we need to understand is how to do physics in a rotating (accelerating) frame of reference such as the earth. In such a frame, there are real forces and fictitous forces. As the questioner correctly notes, even though we call the fictitious forces fictitious, they can certainly be felt by our bodies if they are large enough. The best known fictitious force is the centrifugal force; this is what we feel trying to throw us off a rotating merry-go-round. A person at rest on the earth's surface experiences a real force called the weight which points toward the the center of the earth and has a magnitude W=mg; a fictitious centrifugal force pointing perpendicular to and away from the axis of rotation and has a magnitude F_cen=mrω² where r is the distance from the axis and ω=7.3x10^-5 radians/s is the angular velocity of the earth's rotation*; and whatever force the floor exerts on the object to maintain equilibrium. If the person is moving along the earth's surface, there is an additional fictitious force called the Coriolis force which I will discuss later. I start with the simplest situation which is someone standing on the equator. The first figure is a view from the north pole; the earth rotates counterclockwise; the man has a mass m=100 kg so his weight (red) is W=980 N; the radius is r=R_earth=6.4x10⁶ m, so the centrifugal force (green and not to scale) is F_cen=100x6.4x10⁶x(7.3x10^-5)²=3.4 N; finally the force from the floor (black) is vertically upward and of magnitude 980-3.4=976.6 N. Note that there is no net torque for a person at the equator and the only effect is that there is approximately a 0.35% reduction in apparent weight. Next let us examine a person at rest but at some latitude, say 45⁰ north as shown in the second figure above; the color coding of the forces is the same as in the first figure. The weight is the same, 980 N; the radius is r=R_earthsin45⁰=4.5x10⁶ and so F_cen=100x4.5x10⁶x(7.3x10^-5)²=2.4 N; F_cen has a radial component 2.4sin45⁰=1.7 N which reduces the apparent weight (radial component of the floor force) by that amount; F_cen has a tangential component 2.4cos45⁰=1.7 N which results in a torque and is also the magnitude of tangential component of the floor force. The torque tries to rotate the person about his feet in a southerly direction (would be northerly in the southern hemisphere) and has a magnitude of 1.7 N∙m, estimating the center of gravity to be 1 m above the feet. To put this in perspective, if the earth were not rotating the person would have to lean at an angle of tan^-1(2.4/980)=0.14⁰ to have a comparable torque about his feet; this angle is far smaller than leaning we do in normal activities. Furthermore, since the direction we happen to be facing is random, the average effect over time of these very tiny torques would be zero.

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

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

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

ANSWER:
My first inclination was to say that, assuming that the station was big enough, it would be pretty much like playing catch on earth. The reason for this guess was that I had worked out a similar problem for an earlier answer. The question there was if you were to jump straight up what would happen; this is equivalent to throwing a ball straight up with some speed v. The answer was that for sufficiently large R and small v, in particular R>>v²/g, the time the ball was in the air, the "height" it rose, and the time it took to "return" would all be about the same as if you did the experiment on earth; also, the ball would "land" back where you were. I believe that answers your question about whether a dropped ball would land at your feet—it would (approximately). You should be sure you understand that earlier answer. The leftmost figure above reproduces the original figure for the ball thrown straight up. The other two do not have all the vectors labeled, but you can tell by comparison what they are. The components of v are shown by light blue vectors and the red is just to aid me in adding the vectors v and Rω. The second figure is if your partner is ahead of you (ahead meaning in the direction of rotation). The third figure shows how you would throw it if your partner were behind you. It is not at all clear to me where you and your partner would be located when the ball "came back down" to the ring (shown by green dots). Also, trying to do the problem analytically in the nonrotating frame gets very messy. I am just interested, as are you, in your situation—thrown baseballs in a 50 m ring rotating such that Rω²=g where ω is the angular velocity in radians per second. So, I am going to do the calculations in the rotating frame (where you and your partner are at rest). This gets sort of tricky since we will have to introduce fictitious forces to describe the motion of the ball; once you throw the ball, there are no real forces on it. Now, maybe you throw the baseball (mass about 0.15 kg) with a speed of 10 m/s≈22 mph. There are two fictitious forces: the centrifugal force F_r=mRω²=mg which points down for you and the Coriolis force F_Cor=2mωv=2mv√(g/R) and points in a direction perpendicular to the vector v as shown in the figure to the left. Now, let's calculate the magnitudes of these forces. F_Cor=2x0.15x10x√(9.8/50)=1.3 N and F_r=0.15x9.8=1.5 N. The Coriolis force is not small compared to the centrifugal force and therefore the ball will not behave at all as it would in a true uniform gravitational field. You would go nuts trying to play catch on this space station! The problem is that R=50 m is not much greater than v²/g=10.2 m. See the added thought below for a discussion of the dropped ball.

ADDED THOUGHT:
When the ball is simply dropped from say h=1 m, the Coriolis force is relatively small because the velocity is small for most of the time of the fall. So, the deflection should be modest. I will try to estimate the amount of deflection. The centrifugal force is mg and is always radially out, choosing +y radially out, a_y=-g+a_y^Cor; I will argue that the ball does not acquire enough speed for the Coriolis force to have a significant radial component, so a_y^Cor≈0, and  y≈h-Ѕgt², v_y≈-gt. So, the time to fall is approximately t≈√(2h/g)=0.45 s and v_y≈4.4 m/s. The Coriolis acceleration, if the velocity is purely radial, points in the backward direction (to the left in my figure above) and would have a maximum magnitude of about 2v√(g/R)≈3.9 m/s². If the ball drops almost vertically the Coriolis acceleration would be approximately a_x≈2m√[(v_x²+v_y²)(g/R)]≈2mv_y√(g/R)=2mgt√(g/R)=8.68t=dv_x/dt; therefore, integrating, v_x≈4.34t² and x≈1.45t³. So, at t=0.45 s, v_x≈0.88 m/s and x≈0.13 m=13 cm.

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

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

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

ANSWER:
First of all, the force is inversely proportional to the square of the distance between the centers of mass. Because gravity is nature's weakest force, it is tiny unless there is a very large amount of mass. Let's estimate that the center of mass of your hands are separated by 1 cm=0.01 m and that the mass of each hand is about 0.25 kg (about Ѕ lb). The force each hand feels can be approximated as F=Gm₁m₂/r²=6.67x10^-11x0.25x0.25/0.1²=4.2x10^-10 N≈10^-10 lb. As a comparison, the gravitational force on each hand by the earth (whose center is much farther away) is about 2.5 N.

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

ANSWER:
I assume the friend is thinking about a bicycle which is, as we all know, easier to stay up on if it is moving. The physics of a bicycle is quite complicated, so I am not giving you the whole picture but rather one aspect which will be understandable to a youngster. Any object which is spinning has what is called angular momentum. Shown to the left is a spinning wheel with its angular momentum vector shown. If the wheel were not spinning, this vector would vanish. One of the most important laws in physics is conservation of angluar momentum: if there are no torques on the rotating object, its angular momentum will not change and that includes the direction of the vector. So, if you are riding a bike straight down the street, each wheel has a horizontal angular momentum pointing to the left. If the bike starts to tip over, the angular momentum vector would try to remain horizontal, keeping it from tipping. If you lean hard, there is now a torque on the system and so the angular momentum will change but not resulting in a fall but rather in a turn. You might want to get a toy gyroscope to show the kids how angular momentum works.

QUESTION:
A tall tree cracks and falls. Can the resultant linear acceleration exceed the acceleration due to gravity?

ANSWER:
As the tree falls, it rotates about the point of contact with the ground with some angular acceleration. Therefore, each point along the length of the tree has a different acceleration which is perpendicular to the tree. The falling tree is shown in the figure to the left. The weight mg acts at the center of gravity a distance D from the rotation axis. The tree has height L. An arbritrary distance up the tree is denoted by x. The angle the tree makes with the vertical is θ. The net torque τ on the tree is τ=mgDsinθ=Iα=Ia/x where I is the moment of inertia of the tree about its pivot point, α is the angular acceleration of the tree, and a is the tangential acceleration of the point x. Therefore, a=mgxDsinθ/I. Therefore, if mxDsinθ/I>1, a>g. Consider a simple example modeling the tree as a long thin stick; then I=mL²/3 and D=L/2, so a=3gxsinθ/(2L). Take a particular example, θ=60⁰; then a=1.3gx/L or any point with x>L/1.3 will have a>g.

ADDED NOTE:
The discussion above refers to the tangential acceleration of a point on the tree which is what I thought you were referring to. However, each point at position x also has a centripetal acceleration, call it a_c=v²/x=xω². You can get ω from energy conservation, mgD=ЅIω²+mgDcosθ and therefore a_c=2mgxD(1-cosθ)/I. So, the total acceleration is √(a²+a_c²); I will let you work that out. Modeling the tree as a uniform stick, I have calculated the total acceleration for several values of x shown in the graph to the right. For x=L/4, all angles have acceleration less than g. For x=L/2, the center of mass, a_total>g for angles greater than about 60⁰. For x=L, a_total>g for angles greater than about 35⁰.

QUESTION:
I'm writing a fantasy novel and I just want to check my science. In it there's a tower which is so tall that it is higher than the troposphere. There is a scene where someone smashes a window. My question is would the air all rush out due to unequal air pressure and when the air pressure settles would everyone feel the effects of explosive decompression or at least find it hard to breathe?

ANSWER:
Presumably, all your windows are sealed such that the inside is at atmospheric pressure. "Higher than the troposphere" would imply the stratosphere which is where commercial jets fly. As you doubtless know, the pressure there is very low. The figure to the right shows that just above the troposphere the pressure is only about 200 millibars, about 1/5 of atmospheric pressure. There is not enough air there to keep you alive. You may recall the crash of the private jet in 1999 which killed golfer Payne Stewart. The plane lost pressure at an altitude of 11.9 km and everyone lost consciousness; it flew on autopilot until it ran out of fuel and crashed. Most certainly the air would be drawn (violently) out of your broken window. Another thing you should think about is that you cannot simply pressurize your entire tower to atmospheric pressure; you have to pressurize in layers of a few hundred meters and isolated from each other. Otherwise you will have the same strong pressure gradient with altitude as the atmosphere has!

QUESTION:
If I stick a pea on the out side of a cylinder that is rotating on a fixed axis at a constant speed, what are the forces being applied to the pea? Is the pea under going acceleration? Even at a constant speed due to the angular changes.

ANSWER:
The pea is experiencing what is called uniform circular motion. Its velocity is a vector tangent to the cylinder's surface. Although the magnitude of the velocity (speed) is not changing, the direction of the velocity is constantly changing. Therefore, since acceleration is rate of change of velocity, there is an acceleration. This acceleration, called centripetal acceleration, is a vector, points toward the axis of the cylinder, and has a magnitude v²/R where v is the speed of the surface of the cylinder (and therefore of the pea) and R is the radius. A force must be applied toward the axis of magnitude mv²/R where m is the mass of the pea. This force is provided by whatever adhesive you used to stick the pea there. If the adhesive is not strong enough, the pea will fly off. Of course there is also the pea's weight pointing vertically down and, assuming the axis of rotation is vertical, the adhesive must also exert a force upward to hold the pea from falling due to its weight.

QUESTION:
What would happen if you were to take a satellite that's in orbit, put a giant ball of string on it, and slowly unwind the string letting the free end fall to earth? Would it snap? What if it was made of a material that is extremely strong such as carbon nano tubes?

ANSWER:
Think about it. The ball of string is orbiting with you, has your same speed, and seems to be "weightless" just like you. So you grab onto the loose end and release the ball to fall and nothing happens, it just orbits along with you! Sort of related to your question is the space elevator which might interest you.

QUESTION:
Why does mass not alter the acceleration in a vacuum. If a hammer falls to the earth faster than hammer falls to the moon, then mass does matter. Consider the opposite perspective. Earth falls to the hammer faster than the moon falls to the hammer. Am I wrong in my reasoning? I've asked a few of my physics teachers, but they said that my logic was flawed and you can't consider it that way without explaining why.

ANSWER:
You are completely misunderstanding the constance of acceleration due to gravity, independent of mass. The statement assumes all masses experience the same gravitational field (see FAQ page). Obviously, because of Newton's second law, the acceleration a due to any force F depends on mass m, a=F/m. The "falling" of the moon or earth to the hammer is negligibly small for all practical purposes. However, you could calculate the initial acceleration of each. The force on the earth or moon is F=MmG/R² where M is the mass of the earth (moon) and R is the radius of the earth (moon) (assuming the hammer is close to the surface). So the initial acceleration of each would be a=F/M=mG/R² and therefore a_moon/a_earth=(R_earth/R_moon)². So, the moon actually accelerates toward the hammer faster but it has nothing to do with the mass as you can see, it is because the center of the moon is much closer to the hammer than the center of the earth is. If you were to place the hammer such that it was a distance R_earth from the center of the moon, the earth and moon would have equal accelerations. The earth and moon are responding to the gravitational field of the hammer. So, you are wrong on all your statements and, guess what, your physics teachers are right!

QUESTION:
Is it true that if you have two objects and nothing else you can't tell which object is moving and which object is standing still? Does this mean that we don't know how fast earth is moving? And there is no experiment we can conduct to see if we are moving or not?

ANSWER:
First, you have to understand the difference between inertial and noninertial frames of reference. If Newton's first law is true where you are, you are in an inertial frame of reference. (Newton's first law says that if an object is at rest in your frame the sum of all forces on it is zero.) Any other frame which moves with constant speed in a straight line relative to your frame is also an inertial frame. Noninertial frames are any frames which have an acceleration relative to an inertial frame. The principle of relativity says that the laws of physics are exactly the same in all inertial frames of reference. Since the laws of physics are what will determine the result of any experiment you can do, it makes no sense to refer to an object's being at rest unless with reference to something else. Does this answer your question? Also, we do not know how fast the earth is moving because that is a meaningless question unless we say, for example, how fast is it moving relative to the sun, or how fast is it moving relative to the center of the galaxy, etc.

QUESTION:
Why is a hard ball more likely than soft ball of equal mass and volume to break a glass window if they are thrown at the same speed?

ANSWER:
The average force F which an object exerts during a collision is the change in momentum (Δp=mΔv) divided by the time the collision took Δt. The soft ball has two things going for it. First, because it is "squishy", the collision lasts longer, so Δt_soft>Δt_hard; second, the soft ball is probably less elastic than the hard ball so that it loses more energy in the collision and therefore bounces back with less momentum resulting in Δp_soft<Δp_hard. Therefore Δp_soft/Δt_soft<Δp_hard/Δt_hard or F_soft<F_hard.

QUESTION:
I have been attempting to teach myself chaos theory, however I have had trouble understanding it and how it is involved with different levels of quantum physics as well as relativity. I am also having trouble understanding the "three body problem", which seems to occur in many different physical systems. I was hoping that you could help me to understand at least some of Chaos theory and how it connects to both quantum physics and relativity, and what exactly the "three body problem" is.

ANSWER:
I am sorry, but your question is too technical and too unfocused for the purposes of this site. I can tell you something about the 3-body problem, though. If two bodies interact only with each other, for example the earth and the moon, you can write the orbits in simple analytic closed form. However, if there are three interacting bodies there is, in general, no closed-form solution. There are special cases where the 3-body problem can be solved, for example if one of the bodies is held fixed, but not in general. If you google "three body problem" you will find more information. A particularly interesting (and newsworthy) discussion may be read in AAAS Science News. All these special cases are not chaotic because they repeat in a periodic way. More general cases are not periodic and are extraordinarily sensitive to initial conditions.

QUESTION:
I recently saw the movie Elysium. The most memorable thing was the space station. The station was a completely open system. Would the station's rotation keep its own atmosphere contained? How would a spacecraft land on a rotating ring like this? Would radiation levels just fry everyone and everything on it?

ANSWER:
The rotation causes an artificial gravity. If there were side walls, you could certainly keep an atmosphere in there. For a detailed discussion of how living there would be, see an earlier answer. The way the gravity works is that it is a centrifugal force (a fictitious force) and you want the acceleration at the outer rim to be equal to g=9.8 m/s²; the acceleration at the surface is v²/R where v is the tangential speed at the surface and R is the radius of the ring. So, v=√(g/R). If we take, just as an example, R=1 kg=1000 m, then v≈0.1 m/s; since the circumference is 2πR=6283 m, this means that the ring would rotate about once every 63,000 s=1.7 days. I think you would agree that this speed is very small, so a spacecraft would have no trouble landing on it assuming that the spacecraft had gotten up to the same orbital speed. I do not believe that the radiation level would necessarily "fry" everyone, but it would be a concern for the long term and some kind of shielding would have to be employed.

QUESTION:
according to newton law of motion "a body will move if a net force act on it" then why the earth moves around its axis?

ANSWER:
Whoa! You have Newton's first law all wrong. A body will move with constant velocity if there are no forces on it. For translational motion, this means that the body moves with constant speed in a straight line; if you exert a force on the body, its velocity will change. For rotational motion, this means that the body spins with constant speed in a straight line; if you exert a torque on the body, its spinning speed will change. The earth rotates on its axis with constant speed because there are no torques on it.

QUESTION:
These are non-academic, practical questions regarding the physics of balance and weight shifting pertaining to a moveable kitchen island.
Q1: Will an 11 inch countertop overhang alone cause the island to tip?
Q2: If not, what amount of weight (lbs.) can be safely placed on the overhang before tipping?
Q3: Is there a formula I can use to calculate this? Description: I have a moveable kitchen island fabricated with locking casters. I wanted to place a quartz countertop over the island base with an additional 11 inch countertop overhang supported by steel beams extending from the island base. The total length of the countertop is 37 inches (26 inches on the island base plus 11 inches as the overhang).

Proposed dimensions of the island base with its portion of the quartz countertop:
W = 52.5 inches
L = 26 inches
H = 36.0 inches (34.5" base + 1.5" countertop which includes a 5/8 inch plywood sub-counter)
Total Weight of the Island Base with Its 26 Inch Portion of the Countertop = 298 lbs.

Proposed dimensions of the 11 inch overhang:
W = 52.5 inches
L = 11 inches length
H = 1.5 inches (2cm quartz plus 5/8 inch plywood sub-counter)
Total Weight of the 11 Inch Overhang Portion of the Countertop = 50 lbs.

The overhang is a key feature of the kitchen island for our household. The island is intended to serve multiple, mundane purposes in our very compact home: food prep, dining, and working on work projects / homework.

ANSWER:
I will assume that the center of gravity of the island without the overhang is at the geometric center of the base and that the casters are at the corners; also, that the center of gravity of the overhang is at its geometrical center. The red vectors in the figure to the right are pertinent forces for this problem, the 298 lb weight of the island acting at the center of gravity (star), the 50 lb weight of the overhang acting at the center of gravity (5.5" out), the force of the floor on the front casters (N₂), and the force of the floor acting on the rear wheels (N₁). (Ignore the force F for now.) Newton's first law stipulates that the sum of all the forces must be zero, so N₁+N₂=348. Also required for equilibrium is that the sum of torques about any axis must be zero; choosing to sum the torques about the front casters, 26N₁-298x13+50x5.5=0=26N₁-3596 or N₁=138 lb and so N₂=210 lb. Now, let's think about this answer: it tells you that this (unladen) island will not tip over because there is still a lot of weight on the rear wheels. Now, if you start adding weight to the overhang, eventually when you have added enough weight, the force N₁ will equal zero when it is just about to tip over. So, add the force F at the outermost edge of the overhang and find F when the island is just about to tip: again summing torques about the front casters, -298x13+50x5.5+11F=0 or F=327 lb. This is the extreme situation—you would have to put twice this amount of weight, for example, halfway out the overhang to tip it over. It looks to me that this will be safe for everyday use.

FOLLOWUP QUESTION:
The questioner sent an extremely lengthy recalculation of distances and weights. The only substantive changes were: weight of island without overhang, W=414 lb; weight of overhang w=58 lb; distances of casters from sides d=5.5"; distance of center of gravity from back side D=14".

ANSWER:
The relevant equations to redo the calculations for the unladen island in the original answer are
   N₁+N₂=W+w and
   (26-2d)N₁-(26-D-d)W+(d+5.5)w=0.
I find N₁=137 lb, so the unladen island will not tip. For the second part of the calculation, adding the force F such that N₁=0, the relevant equation is
   (11+d)F-(26-D-d)W+(d+5.5)w=0.

QUESTION::
I'm trying to compare 2 measurements that denote impact. The first measurement is as follows : a glass sheet can withstand a 25mm (diameter) steel ball fired at 80 km's per hour. The second measurement is that another sheet of glass can withstand a 277 gram steel ball dropped from 1 metre in height with a back wind of 60 meters per second. I am trying to bring the second measurement to a measurement comparable to the first measurement.

ANSWER:
The thing which will matter is the linear momentum (p=mv) each ball brings to the glass. The reason is that you are interested in how much force each glass can withstand and the force is the rate of change of momentum. I assume that each ball will spend about the same amount of time during the collision and will exert its force over the same area (very nearly a point for a sphere on a plane), so whichever ball has the most momentum when it hits the glass will indicate the glass with the greatest strength. I took the mass density ρ of steel as 8000 kg/m³. The volume of a sphere is 4πR³/3, so m₁=4ρπR₁³/3=4∙8000∙π∙0.0125³/3=0.0654 kg. Since the speed is given, v₁=80 km/hr=22 m/s, we can immediately write the momentum for #1, p₁=1.44 kg∙m/s. Finding the speed of the second ball is a much more difficult problem. If it were just falling, it would be trivial. But it is being pushed by the downward wind which will make it speed up faster than just falling; so, it is necessary to understand a little about air drag forces. For spheres of normal speeds in air the force of friction is excellently approximated by f=0.22D²u² where D is the diameter and u is the speed of the ball relative to the air. From the density and mass, I find the radius to be 0.0434 m so D=0.0868 m. So, when the ball is first dropped, it has two forces pointing down, its own weight mg=0.277∙9.8=2.71 N and the wind force 0.22∙0.0868²60²=5.97 N; the wind is more than twice the force as the weight. As it falls, it speeds up and so the effect of the wind gets smaller. Newton's second law for v₂, if you care, is now of the form dv₂/dt=g[1+((v_w-v₂)/v_t)²] where v_t=√(mg/(0.22D₂²)) and v_w is the speed of the wind. The solution to this equation is v₂=v_w-v_ttan[tan^-1(v_w/v_t)-(gt/v_t)]. Putting in the numbers for this situation (v_w=60 m/s, v_t=40.5 m/s), the graph for the first 15 s, with and without wind, is shown to the left. The behavior of this graph is interesting. The ball takes about 4 s to reach a speed of 60 m/s; at that point the ball is at rest relative to the air and so there is no air drag and the slope of the curve (which is acceleration) is the same as the curve for no wind at all, as expected. Now, if there were no wind it would take the ball about 0.45 s to fall 1 m and the wind will surely get the ball there in a shorter time; therefore we are really only interested in the first half second of the fall and this is shown to the right. The acceleration over this time is very nearly uniform, about a₂=11/0.4=27.5 m/s²; since this is just an estimate, I will do the calculation assuming uniform acceleration rather than doing the exact calculation to find the velocity after 1 m. I find that the time to reach 1 m is t=√(2/27.5)=0.27 s and so v₂=at=7.4 m/s. Therefore, the momentum for #2 is p₂=m₂v₂=0.277∙7.4=2.1 kg∙m/s. Ball #2 is the winner!

ADDED THOUGHTS:
In retrospect, I could have saved myself some work if I had thought about the fact that in a time less than 0.45 s the acceleration would be essentially constant and simply written 0.277a=2.71+5.97 —> a=31.3 m/s² which would give t=0.25 s and v₂=7.9 m/s and p₂=2.2 kg∙m/s. Secondly, I noticed that my solution for v₂ was incorrect for speeds greater than 60 m/s, so I deleted that part of the graph. To satisfy my own curiosity, I solved the problem for speeds greater than 4 s. Because at later times the speed of the ball is greater than the speed of the wind, the air drag force switches to up rather down so the analytical solution to the problem becomes different, v₂=v_w+v_ttanh(g(t-4)/v_t); this corrected calculation is shown in blue. The complete solution is now graphed to the left. At large times, the solution approaches v_t+v_w as expected since v_t is the terminal velocity in still air.

QUESTION:
I have a question about the concept of angular momentum. Suppose a disc is attached to a pole and they're stationary in deep space. If we spin the disc, it's gonna have an angular momentum. Using the right thumb rule to determine the direction of which, does this mean the unit will start moving in said direction? Also, how does the spinning action produce a momentum perpendicular to the force we exerted to cause the spin?

ANSWER:
I am assuming the pole is attached axially, i.e. at the center and perpendicular to the disk. Angular momentum is not a force and therefore it will not cause the system to start moving. Also, it was not the force you exerted which caused it to spin, it was the torque (see left), and the direction of the torque τ is also given by the right hand rule such that the angular momentum L points in the direction as the torque (see right).

QUESTION:
I'm a young aeronautical student and I'm doing a project on small, basic fixed-wing aircraft. I need to include some explanation of how lift is generated. My teacher has vaguely mentioned the Bernoulli theory once or twice. However after some research, I found that the Bernoulli explanation is outdated. I've found other more accepted theories, only which are too complex for both my understanding and academic level. Could you provide a more accurate but relatively simplified explanation of lift?

ANSWER:
I wouldn't say "the Bernoulli explanation is outdated," it just isn't the whole explanation. Basically, Newton's third law is responsible for much of the lift in flight—the air coming off the trailing edge of the wing is deflected down which means the wing exerted a downward force on it which means it exerted an upward force on the wing. Books about flying usually refer to this as "angle of attack." A more complete explanation may be found in an earlier answer. I can also recommend a book, Stick and Rudder by Wolfgang Langewiesche.

QUESTION:
If it was possible for every vehicle in the world to point East, and at perfect timing, all accelerate at the same time, surely the torque that is put down on the Earth would affect it's spinning energy?

ANSWER:
Yet another chance to demonstrate what a tiny speck we are in this universe! There are about a billion, 10⁹, vehicles in the world. Suppose the average mass is 1000 kg (about 3000 lb) and the average acceleration is about 5 m/s² (half of g); so the average force per vehicle is about 5000 N and the total force on the earth is therefore 5x10¹² N. Suppose the average vehicle is at about 45⁰ latitude, so the net torque is about τ=5x10¹²R/√2=2.3x10¹⁹ N∙m where R=6.4x10⁶ m is the radius of the earth. The angular acceleration is then given by α=τ/I where I is the moment of inertia of the earth, about 8x10³⁷ kg∙m/s². So, α=2.3x10¹⁹/8x10³⁷=3x10^-19 radians/s²=5x10^-18 revolutions/s²=4x10^-8 revolutions/day². If all your vehicles could maintain this acceleration for a full minute,  (certainly not possible), the length of the day would shorten by 5x10^-12 day=4x10^-7 s.

QUESTION:
Can you explain the effect of rifling of the muzzle in rifles ?? I just know that the muzzling is done to impart spin to the bullet when it is fired ?? So, exactly how does the spin imparted to the bullet improves it's aiming accuracy or whatever it does ?

ANSWER:
An object, like the bullet, which is spinning about an axis along its length has angular momentum. The angular momentum vector points along the direction the bullet is flying. An important law in physics is the conservation of angular momentum which says that the angular momentum of an object never changes if there is no torque acting on it. So the angular momentum will continue pointing along the path which means that the bullet will not "tumble". It is the same principle which governs the spiral forward pass in American football. Without spiraling, the tiniest asymetry in the shape of the bullet will cause it to immediately start tumbling when it leaves the barrel.

QUESTION:
With all the talk of new weapons using kinetic energy as the destructive force instead of traditional explosives; exactly how minimal could a warheads' mass be and what would its velocity need to be to create an "explosive" force of 5,000 lbs of high explosive?

ANSWER:
Funny, I have not heard "all the talk"! I did get one other recent question similar to yours. Anyhow, I guess you want to compare the kinetic energy of a projectile with the same energy as 5000 lb of TNT. 5000 lb is about 2.3 metric tons and the energy content of that amount of TNT is about 10¹⁰ J. Since the kinetic energy of the projectile is Ѕmv², there is no clear answer to your question because every mass would have a different velocity to have the requisite amount of energy. The speed of a near-earth satellite is about 8000 m/s and maybe such a satellite would be used to launch your weapon, so let's use v=8000 m/s. Then, 10¹⁰=Ѕm(8000)² or m=313 kg=690 lb.

QUESTION:
I'm helping a middle school student with his physics homework. One of the concepts was mass and inertia and he told me that his teacher said they are equal and are synonyms. I was under the impression that they do not mean the same thing, that they are proportional. Could you clarify?

ANSWER:
Here is what the teacher was talking about: There are two kinds of mass, inertial mass and gravitational mass. Inertial mass quantifies how much an object resists changing its motion if a force acts on it. Gravitational mass quantifies how much a gravitational force affects it and how much it gravitationally affects other gravitational masses. It turns out that the two are identical and there is really only one mass. This equivalence is a prediction of the theory of general relativity and is also the reason that all objects have the same gravitational acceleration. Inertia is usually a qualitative term which describes how resistant an object is to being accelerated, which is what inertial mass does quantitatively. However, it is not so unusual to use inertia synonymously with mass or inertial mass.

QUESTION:
My friend and I are having a discussion about centrifugal force and whether or not it exists, is real, and/or if it is present in daily life. Can you expand our knowledge and settle this dispute? We want to know everything about centrifugal force and if you can help us out that would be much appreciated.

ANSWER:
The important concept to understand here is that Newton's laws, which describe motion, are not always true. For example, suppose that you hang a simple pendulum from the roof of a car. When you are standing still or driving down a straight road with constant speed, the pendulum hangs straight down. The forces on the pendulum are its own weight, which points straight down, and the tension in the string which pulls straight  up. This pendulum is in equilibrium here in the car and so Newton's first law tells you that the sum of all the forces must add up to zero and so the tension in the string must be equal to the weight of the pendulum bob. Now suppose that you smoothly accelerate; you will find that the pendulum swings toward the rear and hangs at some angle rather than straight down. You will look at that pendulum and say, it is just hanging there at rest in the car and so it must be in "equilibrium". On the other hand, there is no way the (not parallel) forces of the weight and the tension can add up to zero. Newton's first law is a false law in this car! If the first law is true, you are in an inertial frame of reference. If it is false, you are in a noninertial frame of reference. Generally, it is easy to identify a noninertial frame—it accelerates relative to an inertial frame. A frame of reference which is rotating is a noninertial frame because an object moving on a curved path has an acceleration even if its speed is constant because the direction of its velocity is always changing. It is shown in any elementary physics text that the magnitude of this acceleration is a=mv²/R, where m is its mass, v is its speed, and R is the radius of the circle it is going around; the direction of the acceleration vector is toward the center of the circle. The acceleration is called the centripetal acceleration from the Latin verb peto which means "I seek". The question is, is there any trick we can pull to force Newton's laws to be true in a noninertial frame. The answer is yes; if you make up a fictious force on any object of mass m which is F_fictitious=-ma in the opposite direction as the acceleration, Newton's laws will work! The centrifugal force is a fictitious force added so that you can apply Newton's laws in rotating systems. (The Latin verb fugo means "I flee".) To see examples of how fictitious forces work, see an earlier question about falling down in a bus which will link you to a question about a car rounding a curve and that will link to a bicycle leaning into a curve.

QUESTION:
Suppose a container is partially filled with a liquid. A small sphere made of a material whose density is less than the liquid is in equilibrium inside the liquid with the help of a thread such that one end of the thread is tied to the sphere and the other end to the bottom of the container. the whole apparatus is kept on a weighing machine. if the thread is cut, then will the reading of the weighing machine change? Since the tension in the thread is an internal force, the reading should not change. However the free body diagram suggests that the reading should change.

ANSWER:
When you refer to "…the free body diagram…", you must specify the body. Solving problems like this are most often easiest if you make a clever choice of body. I would choose the body as the container+liquid+ball+thread (taken as weightless, probably); in that case the only downward force is the weight of all three and the only upward force is the scale, so the scale reads the total weight with no reference to whether the thread is connected to the ball or not. You can make this problem difficult by focusing on a different body, maybe the container, but if you draw all your free-body diagrams correctly and apply Newton's third law, you still get the same answer that the the scale reads the total weight. To the left I have shown all the forces on each of the bodies: red is the tension in the string which is also the force the string exerts on the sphere and the on the container; light blue is the force the container and fluid exert on each other; black represents the weight of each; green is the buoyant force which is the force of the fluid on the sphere and the force of the sphere on the fluid; purple is the force the scale exerts on the container.

FOLLOWUP QUESTION:
Suppose the sphere is accelerating upwards inside the fluid, under the influence of the buoyant force. Now its acceleration is dependent upon the buoyant force, while the buoyant force is in turn dependent upon its acceleration. How do we precisely calculate these two quantities, at a particular instant of time?

ANSWER:
First of all, the buoyant force is not dependent on the acceleration. As long as the sphere is fully submerged, the buoyant force (B) is equal the the weight of the displaced fluid. But, you must also include the drag force (D) which the fluid exerts on the sphere and that is not simple to include. But, most fluids have sufficiently large viscosity that the sphere will quickly come to its terminal velocity and move upward with constant speed. When that happens B-D-W=0 where W is the weight of the sphere. If you really want to pursue this farther, the simplest approximation for the drag is that it is proportional to the speed of the sphere, Stokes's law. In that case, a=(B-W-Cv)/m where C is a constant. If you were to neglect drag altogether, which would be a poor approximation for any real fluid, the acceleration would be uniform, a=(B-W)/m.

QUESTION:
I recently saw an experiment on the popular show Mythbusters. The experiment was: a truck moves in one direction at a constant velocity, carrying a canon facing the opposite direction. The canon then fires a football such that it will travel at the same speed at which the truck is going. Successfully performed by the Mythbusters team, the ball dropped straight down to the ground the moment it left the barrel, as theorized, to a remote observer. I wonder, since Earth rotates at about 470metres/s eastward at the equator, if a bullet is fired westward at the same speed but westward, will it drop like the football did? Although I highly doubt this is the case, my calculations so far prove it's theoretically plausible. Am I missing something? Please also elaborate if this is different from the truck-canon experiment.

ANSWER:
In the first example, both the canon and the truck were moving relative to the ground and it was the ground relative to which the final velocity was observed. So, the experiments are not equivalent unless the rifle in the second example were moving east with a speed of 470 m/s. If the rifle was at rest relative to the ground and the observer was not on the ground but looking from space, she would observe the bullet drop straight toward the center of the earth as the earth spun under it.

QUESTION:
I was discussing ballistics with someone the other day, and a thought came up about a detail that I am not 100% sure on. With bullets designed to expand, generally speaking materials hardness and velocity determine that expansion. What I am wondering, is does the bullet's energy upon striking the target cause the expansion? Or is the expansion caused by the opposite forces imposed on the bullet by what it is striking? (Or would one say it is a combination thereof?)

ANSWER:
The thing to appreciate is that energy and force are not two separate things. Everything you need to know about collisions is contained in Newton's three laws. The idea of energy often makes problems easier to solve or understand, but the first step in developing the formalism of energy, called the work-energy theorem, is just Newton's second law "in disguise". So, I will discuss the bullet collision both ways:

QUESTION:
I am trying to help my son understand velocity, but find myself confused.

• Person A drives in a circle. A physics website tells me this represents acceleration, a change in velocity, because his direction is constantly changing even though his speed may not be. Fair enough.

• Person B takes a step forward and then a step backward to his original position. A different physics website tells me that this represents zero velocity, no acceleration, because he has not changed position.

But these answers seem contradictory, because person A, driving in a circle, will arrive at his original position at some point. In this respect, he is no different than person B, and could be considered zero velocity. Can't both of these examples be considered changes in velocity? I suppose it depends on the timeframe you use to measure the change in position (?). So the person driving in a continuous circle can be considered to not be accelerating in some cases? This makes no sense.

ANSWER:
The first example refers to the instantaneous velocity and the instantaneous acceleration of person A; instantaneous refers to an instant in time and both acceleration and velocity are continuously changing. The second example refers to average velocity and average acceleration. Average refers to the value of the quantity averaged over some time period and the time period here is the time from when he first stepped forward until he finished stepping back. Suppose that was 10 s and the length of his step was 1 m; then the average velocity is distance traveled divided by the time, 0/10=0 m/s and the average acceleration was the change in velocity divided by the time, 0/10=0 m/s². Person A's average velocity and acceleration over exactly one time around would also be zero.

QUESTION:
If two neutrons (just for the sake of ignoring charge) were separated from each other 1 light year, how long would it take for them to "touch" each other based on their gravitational attraction only? They are also in complete isolation from the rest of the universe.

ANSWER:
This is a very strange question. I have answered a nearly identical question before but with much larger masses and smaller distances, but the method is identical so I refer you there. For your masses I calculate about 14x10¹⁸ years, about a billion times the age of the universe. (I would also like to add that I do not believe that this should be taken too seriously because no theory of gravity has been accurately tested for either such large distances or such small masses.)

QUESTION:
I am a fabricator and I currently have a task where I am attempting to create a braking system for downhill rapid propulsion (downhill racing). Although the product exists, it is primitive and not fit for extreme measures, reliabilty, or convenience. There are many variables besides weight, drag coefficient, mass and gravitational acceleration. I would greatly appreciate your professional advice on creating a formula in which I could create, change or gauge different systems for different masses. I have done much research and am increasingly frustrated yet interested. I have come too some conclusions and have run many tests. I am using a polyester blend material for the canopy that is expansive and durable yet retractable, however. My cable system and my rapid cut down on drag are a problem. So my question is if I weigh 160 pounds I am traveling at a speed between 30-60 miles per hour (we will say 45mph) at a down grade of 45% and i would like a slowing to 10 miles per hour between 40-50 feet from deployment. What would my initial area of my canopy be and what would the tensil strength of my cable need to be set at? There are 3 points of contact for the cable system two high and one low.

ANSWER:
First, all air drag calculations are approximate and without extremely complex computer simulations you can only do order-of-magnitude calculations. Since you do not mention any sliding or rolling friction of whatever is going down the incline, I will assume they are negligible. There are, therefore, two forces on the mass, the gravitational force down the incline mgsinθ and the drag of your "canopy" up the incline which I will take as c₂v²; here v is the speed, m the mass, g the acceleration due to gravity, θ the angle relative to the horizontal, and c₂ is a constant determined by the geometry of your canopy. A reasonable approximation for c₂ is c₂≈јA where A is the area presented to the direction of motion of the canopy (only valid in SI units). Since I am a scientist, I will work entirely in SI units here. Newton's second law, which governs the motion of this system, is mdv/dt=-mgsinθ+јAv² or dv/dt=gsinθ[1-(v²/v_t²)] where v_t²=mgsinθ/c₂=4mgsinθ/A; v_t is called the terminal velocity, the speed to which the mass will slow as it goes forward. Solving the differential equation (this is worked out in any intermediate-level classical mechanics book), the following equation is found: v²=v_t²(1-exp(-2gxsinθ/v_t²)+v₀²exp(-2gxsinθ/v_t²) where x is the distance traveled and v₀ was the speed where x=0. That is everything you need since you know everything except A. And you may want to use a fancier value for c₂ more tailored to the details of your canopy. As an example, I will use your numbers: m=160 lb=73 kg, v₀=45 mph=20 m/s, v=10 mph=4.5 m/s, x=45 ft=13.7 m, θ=45⁰. Putting these in, I find 20≈(2000/A)(1-exp(-0.094A))+400∙exp(-0.094A) or A(0.05-exp(-0.094∙A))=5(1-exp(-0.094∙A)). Someone more clever than I could probably solve this analytically for A, but I will just solve it numerically by plotting the left and right sides of the equation and finding the intersection (see inset figure on the left). I find that A≈100 m²≈1000 ft²; this would be a square about 30 ft on a side. (Since A is so large, one could have easily solved this by simply neglecting the exponential functions, 0.05A≈5.) Regarding the strength of the cables, since I do not have any details about the design of the canopy, the best I can do is tell you the maximum force the canopy would exert on the mass via the three cables. Since the acceleration is -gsinθ+јAv²/m, the greatest acceleration is when the velocity is greatest, when the braking initiates. So, the force the cables must exert is -mgsinθ+јAv₀² or -mgsinθ+c₂v². For your specific example, this force would be about 9500 N≈2100 lb or roughly 700 lb/cable.

QUESTION:
We can use the work-energy theorem in any inertial frame of reference. When no external force is applied, and there is no change of height, the change in KE = -(Change in PE(spr)). However, KE change depends upon the frame of reference and the extension of a spring does not depend upon frame's choice! HOW IS THIS POSSIBLE?

ANSWER:
The extension of the spring is the same in both frames, but the work done by the spring is not because the same force acts over a different distance in the moving frame. Forget potential energy and simply write ΔK=W where W is the work done by any conservative external force, W=₀∫^XF(x)dx. I have chosen the starting position as x=0; to keep the algebra simple, I will also choose the starting velocity in this (x) frame to be 0 and the final velocity to be V, so ΔK=ЅMV²=W. Suppose that it takes time t to reach the position X. Now, suppose there is another reference frame (x') which has a speed U in the +x direction and x'=0 at t=0 also. Then x'=x-Ut and the initial velocity is U and the final velocity is U-V in this frame, so ΔK'=ЅM(V-U)²-ЅMU²=ЅMV²-MUV. Finally, calculate the work done in the moving frame: W'=₀∫^X'F(x')dx'=₀∫^X-UtF(x-Ut)(dx-Udt)=W=₀∫^XF(x)dx-U₀∫^tF(t)dt. The second integral is the impulse which is the change in momentum, MV, so W'=₀∫^XF(x)dx-MUV=W-MUV. Putting it all together, ΔK'=W'=ЅMV²-MUV=W-MUV or ЅMV²=W.

QUESTION:
When Cavendish calculated the value of universal gravitational constant he used mass of lead balls as reference. But how did he know mass of the lead ball if he doest know the value of G?

ANSWER:
He could simply weigh them because we know W=mg. Since you can also write W=mM_earthG/R_earth², you can identify g=M_earthG/R_earth², and g is easy to measure even if G is not.

FOLLWOUP QUESTION:
How did he build a weighing scale. When weight is determined using a scale it should have been built w.r.t certain standard mass. But with out knowing value of G there is no way they had a standard mass. He could have weighed a 10kg mass as 1kg mass. So, how did he weight exact mass.

ANSWER:
The kilogram was officially defined in 1795 as the mass of 1 liter of water. The Cavendish experiment was performed in 1797-98. Even if the kilogram were not defined, there were other mass definitions which Cavendish could have used. As an example, let me consider the oldest standard weight I could find reference to, the beqa (b) (shown in the figure) defined about 5000 years ago as the mass of 200 grains of barley corn which is about 6.1 grams=6.1x10^-3 kg. So if Cavendish had used the b as his mass standard, he would have found G=6.67x10^-11 kg²/(N∙m²)=6.67x10^-11 [kg∙s²/m³]x[1 b/6.1x10^-3 kg)=4.07x10^-13 b∙s²/m³] (assuming that he used seconds and meters for time and length). It is a different number but means exactly the same thing.

QUESTION:
A ball thrown upward has zero velocity at its highest point i.e no acceleration. The resultant of applied force and weight/gravity is also zero. Thus the body is at rest. Is it in equilibrium too? This is not a homework question. I and my friend are too confused.

ANSWER:
Right off the bat, your first sentence is incorrect. Acceleration of something at rest is not necessarily zero. You cannot determine whether an object is accelerating by knowing only its velocity because acceleration is the way the velocity is changing which cannot be known by simply knowing what the velocity is right now. In your example, the ball is at rest right now but was moving upwards just before now and will be moving down just after now. The ball has an acceleration which happens to be 9.8 m/s²; this means that one second before now it was moving upwards with velocity 9.8 m/s and one second after now it will be moving downward with velocity 9.8 m/s. It is not in equilibrium because Newton's second law tells you that any object with a net force on it is not in equilibrium; the only force (ignoring air drag) on the ball is its own weight and this force is the source of its acceleration.

QUESTION:
Does mass affect kinetic energy?

ANSWER:
Of course. Classically the kinetic energy K=Ѕmv² where m is mass and v is speed. Relativistically, K=E-m₀c² where c is the speed of light, m₀ is the mass of the object when not moving, the total energy E=√(p²c²+m₀²c⁴), and the linear momentum p=m₀v/√(1-v²/c²). So, you see, mass appears all over the place. The momentum of a massless particle, the photon, is p=E/c; its energy is all kinetic, and K=E=pc since a photon has momentum even though it has no mass.

QUESTION:
I am currently doing a research paper on the perfect free-kick, could you find an equation that suits the following variables? The soccer ball is kicked from the origin of a coordinate system with an unknown velocity such that it passes through the points (x,y)=(9.15 m, 2.25 m) and (x,y)=(22.3 m, 2.22 m). How can I find the magnitude and direction of the initial velocity? Just having an equation to help me work with would be very nice.

ANSWER:
The equations of motion for a projectile which has an initial velocity with magnitude v₀ and angle relative to the horizontal θ are x=v_0xt and y=v_0yt-Ѕgt² where v_0x=v₀cosθ, v_0y=v₀sinθ, t is the time, and g=9.8 m/s². Solving the x-equation for t, t=x/v_0x; putting t into the y-equation, y=(v_0y/v_0x)x-Ѕg(x/v_0x)². Since you have two (x,y) data points, you have two equations with two unknowns, (v_0x,v_0y). The algebra is tedious, but the result is that v_0x=21.0 m/s and v_0y=7.30 m/s; v₀=22.2 m/s, θ=19.2⁰. To check my answer I drew the graph shown to the right (note the different x and y scales); it looks like my solution passes pretty close to the data points.