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Venkatesh
11-24-2004, 11:01 PM
Just a few days ago I was playing with colored pencils in Physics and came up with an interesting problem. Unfortunately I have been unable to figure out a solution to this problem. Hopefully somebody here will have more insight into this.

Imagine a large sphere at rest, on a frictionless plane, free to both rotate and/or translate. Suddenly a small spinning and translating sphere strikes the large sphere on its horizontal circumference, in line with the center of mass, and then recoils, spinning the other way. Will the large sphere start translating or spinning?

I wasn't sure what would happen, as there were two ideas I had. My first idea was that the large sphere would not start spinning, but rather sliding, as the small sphere striking it would not be providing a torque (as it is striking in line with the center of mass). Also, linear momentum needs to be conserved, so the net change of 2mv of the small sphere needs a compensation from the larger one.

On the other hand, angular momentum also needs conservation. If the sphere is now spinning at -w and was originally spinning at w, it underwent a change in angular momentum of 2Iw. This change needs a compensation, namely the larger sphere starting to rotate.

So what actually happens? I can't think of any way to empirically test this problem, so I need the help of some wizard out there.

Thanks,

Doug G
11-24-2004, 11:26 PM
Imagine a large sphere at rest, on a frictionless plane, free to both rotate and/or translate. Suddenly a small spinning and translating sphere strikes the large sphere on its horizontal circumference, in line with the center of mass, and then recoils, spinning the other way. Will the large sphere start translating or spinning?

Assuming Friction exists between the two sphere surfaces, then the answer has to be BOTH (spinning and translating)

.. as the small sphere striking it would not be providing a torque (as it is striking in line with the center of mass).

The frictional force of the small sphere on the large sphere is applied at the surface tangentially so that there will be a torque exerted on the large sphere.

.. Also, linear momentum needs to be conserved, so the net change of 2mv of the small sphere needs a compensation from the larger one....
On the other hand, angular momentum also needs conservation. If the sphere is now spinning at -w and was originally spinning at w, it underwent a change in angular momentum of 2Iw. This change needs a compensation, namely the larger sphere starting to rotate.

Yep, both ang mom. (L) and lin mom. (p) need to be conserved in a case like this where there's no outside forces or torques on the entire system. The interesting and the tough part is determining the magnitudes of the internal frictional force and resulting torque. Need to know more about the surfaces involved.

Your's truly,

Wizard of Oz

PS: Ever played with those wooden "tops" from the '50's / '60's. You play billiards lately. Those may be a good examples of this type of scenario.

Ian Mackenzie
11-26-2004, 02:32 PM
Imagine a large sphere at rest, on a frictionless plane, free to both rotate and/or translate. Suddenly a small spinning and translating sphere strikes the large sphere on its horizontal circumference, in line with the center of mass, and then recoils, spinning the other way. Will the large sphere start translating or spinning?

Can you be a little more precise with the exact question? If you're saying what I think you're saying, the situation is actually impossible, in a couple of different ways.

First impossibility: The small ball comes in with velocity v and angular velocity w, and leaves with velocity -v and angular velocity -w. Although you could find a velocity and angular velocity for the large ball that would satisfy conservation of momentum, you would have violated conservation of energy - note that the small ball leaves with exactly the same energy as before, so any translation or rotation of the large ball would mean that energy would have to come out of nowhere (generally a bad sign).

Second impossibility: The small ball 'recoils'; I'm assuming you mean by this that the small ball leaves along the same direction that it came in on. Even if you accept my first argument and say that the ball leaves with some velocity -kv, with 0 < k < 1 (i.e. the ball reverses direction and leaves with some lesser speed, since some kinetic energy was transferred to the large ball), the situation is still impossible. The tangential force between the two balls that results in a change in angular velocity ALSO accelerates both balls sideways, so the small ball will leave in a direction some angle from the direction it came in on, and the large ball will start moving in another direction entirely.

I have attached some top-down sketches of what would happen (qualitatively); let me know if they aren't clear.

Unfortunately, I think we'd need more information to solve the problem; for instance, if you're trying to solve for the final velocities of both balls (both x and y components) and the final angular velocities of both balls, you have 6 unknowns; with conservation of momentum in x, conservation of momentum in y, conservation of angular momentum, and conservation of energy (assuming a perfectly elastic collision), you only have 4 unknowns.

You could be clever and try to solve for the two components of the impulse between the balls (from which you could easily figure out everything else); however, this implicitly guarantees conservation of momentum, so you're left with conservation of energy being your only equation, and you still can't solve. You'd need to know something else about the problem.

Of course, you could also consider slipping (kinetic friction) between the two balls if you really wanted to be more realistic, but the calculations would probably become pretty hideous.

Chris Hibner
11-28-2004, 06:21 PM
As stated before, it depends on the friction between the two spheres and the point of contact.

A very interesting problem is this: how must the sphere at rest be impacted such that it rolls along the frictionless surface without slipping? The answer is that the sphere must be impacted at its "center of percussion".

You won't learn about this until advanced dynamics courses in college, but the concept of "center of percussion" is very important in a lot of areas. In sports, the center of percussion of a baseball bat, golf club, tennis racket, etc. is referred to as "the sweet spot" in laymans terms. That is, when you hit the sweet spot, you can't even feel that you hit anything at all. This is important not only in sports equipment, but also for getting a good ride in a vehicle. Anyway, just something fun to think about.

suneel112
11-28-2004, 08:43 PM
if the situation is completely free of friction, then the second sphere will not rotate, and rotation won't even be figured in the picture. If muKn for the ball to the sphere is 0 and muKn for the ball to the other ball is infinite (it transfers energy whenever touching), you assume that all rotational energy is transferred, and it becomes a rotational energy problem. Sorry, I forgot physics (I took AP last year, and I didn't understand rotation to begin with), but I think I=2/5mR^2 (for a sphere) and tau = I alpha. Make energy considerations involving energy, rotational (1/2 * I * omega^2) and translational (K = 1/2 mv^2). If the collision is elastic and the finishing rotational energy for both spheres is constant R1 + R2 = Rinitial, it becomes an easy problem, since K1i + K2i = K1 + K2.

Al Skierkiewicz
11-28-2004, 08:59 PM
Chris,
I am not sure I can totally agree on this one. As an observer of billiards I have seen examples of all of the above situations. In your example, a ball with reverse english is capable of translating enough spin to another ball to make it spin at the precise speed that would prevent slipping on the frictionless surface would it not?
Ian, it is also possible to have two balls meet, with or without spin and have one or both, recoil along the same axis at which they met. The five balls hanging on strings desk toy is an example. Yes you can argue that the strings keep them in alignment during recoil but I think it is a little simpler than that. If it was such a precise balance that is reguired for that toy to work, you wouldn't be able to mass produce them and still make money.
Thoughts?

Chris Hibner
11-29-2004, 08:55 AM
Chris,
I am not sure I can totally agree on this one. As an observer of billiards I have seen examples of all of the above situations. In your example, a ball with reverse english is capable of translating enough spin to another ball to make it spin at the precise speed that would prevent slipping on the frictionless surface would it not?
Ian, it is also possible to have two balls meet, with or without spin and have one or both, recoil along the same axis at which they met. The five balls hanging on strings desk toy is an example. Yes you can argue that the strings keep them in alignment during recoil but I think it is a little simpler than that. If it was such a precise balance that is reguired for that toy to work, you wouldn't be able to mass produce them and still make money.
Thoughts?

Al,

In billiards, all balls are roughly the same diameter - therefore, they cannot hit each other at the center of percussion of the ball (the center of percussion of a sphere is about 17% above it's center line - the balls always contact each other at the center line). Furthermore, I'm assuming no friction between the balls.

Anyway, a better billiards example is: where should you strike the cue ball (with the cue stick) such that you have absolutely zero "english" on the cue ball? The answer: at the center of percussion of the ball. That's because you want to strike the ball such that it rolls without slipping on the surface (as I stated in the original post) which means there's no english if you hit it at the center of percussion.

-Chris

Al Skierkiewicz
11-29-2004, 10:07 AM
(the center of percussion of a sphere is about 17% above it's center line - the balls always contact each other at the center line).
-Chris

So Chris,
Your answer brings more thoughts, questions. You are implying that no billiard balls just roll on the surface, they are all slipping? (Until friction with the table surface takes over and rotates the ball.) And no ball can reflect along it's trajectory track since they do not make contact at the percussion point? So the surface must have much more to do with ball interaction than it would seem, correct?

Chris Hibner
11-29-2004, 01:34 PM
So Chris,
Your answer brings more thoughts, questions. You are implying that no billiard balls just roll on the surface, they are all slipping? (Until friction with the table surface takes over and rotates the ball.) And no ball can reflect along it's trajectory track since they do not make contact at the percussion point? So the surface must have much more to do with ball interaction than it would seem, correct?

There is some interaction due to friction between the surfaces of the balls, so that does have an impact. But, if the pool table was frictionless and you hit the cue ball with the cue stick in the dead center of the ball (the motion of the cue stick is parallel to the table), then the ball will only slide - not rotate.

In fact, the friction between two billiards balls would tend to make the target ball spin in the opposite direction of it's motion (think of two gears meshing once the balls contact each other). The friction of the table will impart a force on the target ball causing the ball roll along the surface rather than slip.

So, yes, the table surface has more to do with the motion than the friction between the two balls. Otherwise, english wouldn't be so predictable. The spin of the cue ball would impart side spin on the object ball causing it to curve. As you've probably noticed, the object ball generally goes pretty straight, even with good english on the cue ball. The friction between the balls is pretty negligible compared to the friction between the balls and the felt. I don't know if this is due to coefficient of friction or more due to the very brief contact time between the two balls.

Ian Mackenzie
11-29-2004, 06:54 PM
if the situation is completely free of friction, then the second sphere will not rotate, and rotation won't even be figured in the picture. If muKn for the ball to the sphere is 0 and muKn for the ball to the other ball is infinite (it transfers energy whenever touching), you assume that all rotational energy is transferred, and it becomes a rotational energy problem. Sorry, I forgot physics (I took AP last year, and I didn't understand rotation to begin with), but I think I=2/5mR^2 (for a sphere) and tau = I alpha. Make energy considerations involving energy, rotational (1/2 * I * omega^2) and translational (K = 1/2 mv^2). If the collision is elastic and the finishing rotational energy for both spheres is constant R1 + R2 = Rinitial, it becomes an easy problem, since K1i + K2i = K1 + K2.

Even if friction between the balls was infinite, you still would not have all rotational energy transferred. I presume you're saying that the first ball would stop spinning completely, and the second ball would start spinning in the opposite direction (with some unknown speed). However, since the spin is in the opposite direction, you have clearly violated conservation of angular momentum.

EDIT: Re-reading the post, I see from the last line that you have not in fact assumed that the small ball ends up with zero rotational kinetic energy (although now I'm a bit confused as to what you mean by "all rotational energy is transferred"). In any case, you can't separate rotational and translational kinetic energy like you can with momentum; any time you have friction transferring rotational kinetic energy between the balls, you will also necessarily accelerate the balls sideways, i.e. some of the angular kinetic energy is transferred into translational kinetic energy. Consider the case of two high-friction balls, with the small one spinning very quickly and moving very slowly, and the large one stationary and having an extremely high mass. As the two balls come in to contact, the small ball will jump quickly off to the side, clearly transferring rotational kinetic energy into translational kinetic energy. The net result is that you cannot state that R1 + R2 = R1initial + R2initial, or that K1 + K2 = K1initial + K2initial; you can only state (assuming a perfectly elastic collision) that R1 + R2 + K1 + K2 = R1initial + R2initial + K1initial + K2initial.

Al: I completely agree that it's possible to have cases with straight rebound; however, the question stated that the small ball was spinning and the large ball was not, and I assumed from the way the question was asked that there was some friction between the balls (and that each ball had finite mass and radius). In this situation, you will not have direct rebound, except perhaps in one theoretical case; if you had finite sliding friction, but the balls were so stiff that they remained in contact only instantaneously (i.e. zero time taken for the balls to deform and pop back to their original shape), the impulse delivered by friction would be zero. My answer didn't address the most general case of two balls colliding; it addressed the closest physically feasible version of the problem stated.

Al Skierkiewicz
11-30-2004, 07:18 AM
Now that I have thought about this for a day I can see that there is a lot of energy transferred to surface when the balls strike. Taking Chris's explanation of the center of percussion, when equal diameter balls collide on an infinitely strong surface, something has to give and you would expect that the balls would lift off the table momentarily. Then looking at the dynamics of two low friction surfaces meeting at relatively high pressure (the collision) then force must be transferred to the ball at rest in a downward direction due to the the rotation of the moving ball. It would interesting to run this experiment in a weightless environment. Thanks for the explanation guys.

dlavery
11-30-2004, 10:56 AM
Now that I have thought about this for a day I can see that there is a lot of energy transferred to surface when the balls strike. Taking Chris's explanation of the center of percussion, when equal diameter balls collide on an infinitely strong surface, something has to give and you would expect that the balls would lift off the table momentarily. Then looking at the dynamics of two low friction surfaces meeting at relatively high pressure (the collision) then force must be transferred to the ball at rest in a downward direction due to the the rotation of the moving ball. It would interesting to run this experiment in a weightless environment. Thanks for the explanation guys.

That sounds like a great idea for a Microgravity University (http://microgravityuniversity.jsc.nasa.gov/) student experiment! Unfortunately, it is a little late for this year's application process (http://microgravityuniversity.jsc.nasa.gov/students/theProgram/), but some one save this idea for next year!

-dave

Al Skierkiewicz
11-30-2004, 01:59 PM
Oh Oh! What have I done? Well this seems like a simple project for one of you student types. Just figure out how to lauch one billiard ball with a little spin (representing forward motion) and have it contact a stationary ball as if they were resting on a common plane. Do a motion study of the trajectory and interaction on both of the balls at the moment of collision and for two seconds following the collision. It would likely work with smaller non-resilient balls like those used for playing bocci or two larger ball bearings might fit the bill as well. Needs to be light and compact I would guess. Interesting problem...

Venkatesh
11-30-2004, 08:30 PM
Oh the wonderous answers I have unleashed,

I started playing the idea of center of percussion today, rather sleep though a physics movie =). I took a pen and hung it on a paper clip. Then I hung this paper clip-pen compound on a pencil. I marked the spot at which the paper clip started and then provided it with a force at varying spots. At any point beyond 2/3rds of the length of the pen, a force causes a forward motion of the paper clip. At any point between the center of mass of the pen (determined experimentally), an applied force would cause the paper clip to slide backwards. Very cool, since it goes against intuition and first guesses.

I guess I need to reconsider the initial problem. It was just a thought experiment, since I was bored of watching a friend play on the GBA SP. The idea was born because I was thinking about rotational collisions and conservation of angular momentum in such systems.