# Rotational motion - billiard ball problem

1. May 7, 2005

### UnSaniTiZ

rotational motion -- billiard ball problem

I've been struggling with an AP physics free resposne problem. I have what I hope to be the right answer, but I'm not sure.

Heres the question:

A billiard ball has mass M, radius R, and moment of inertia about the center of mass Ic = 2/5 * M * R^2

The ball is struck by a cue stick along a horizontal line through the ball's center of mass so that hte ball initially slides with a velocity v0. As the ball moves across the rough billard table (coefficient of sliding friction u), its motion gradually changes from pure translation through rolling with slipping to rolling without slipping.

a) develop an expression for hte linear velocity v of the center of the ball as a function of time while it is rolling with slipping.

What I got for this part was v = v0 - u * g * t

What I don't know is if the rotation of the ball needs to be factored into the acceleration. If so, please advice.

b) Develop an expression for the angular velocity w of the ball as a function of time while it is rolling with slipping.

I got w = (5*g*u*t)/(2R)

This part I am more sure about... However, if that's the answer, isn't the limit as t goes to infinity infinity, which clearly cannot be? Help me out, please :)

c) determine the time at which the ball begins to roll without slipping.

v=wr
v0 - (u*g*t) = (5*g*u*t)/(2)
solve for t: t = (2*v0)/(7*g*u)

d) when the ball is struck it acquires an angular momentum about the fixed point P on the surface of the table. During the subsequent motion the angular momentum about point P remains constant despite the friction. Explain why this is so.

I'm not so sure about this, but I was thinking because the point is on the table, and the ball is on the table, as the ball loses momentum it loses it to the table, which makes the angular momentum about point P constant.

No more for now guys... Thanks a lot, and any help on any of it is appreciated (assuming what you say is correct, as I'm sure it will be :tongue2:) I am kind of in a rush with this, so get back to me asap

Thanks again,
UnSaniTiZ

2. May 8, 2005

### Pyrrhus

I took a glance at a, b, and c. They look right to me.

As for the d question what happens when traslational and rotational movement are combined to the contact point speed?

3. May 8, 2005

### UnSaniTiZ

Well, I know it is rotating slower than the top, and in the reverse direction... Does the rotational velocity of that point go to 0 when you add the translational and rotational velocities? This one part rotates faster than the other part has always been confusing to me...

Last edited: May 8, 2005
4. May 8, 2005

### OlderDan

Your dilema in part b) is resolved by your answer in part c). The angular velocity only changes during the time interval when the ball is slipping. When the slipping stops, the angular velocity remains constant.

For d) is the point P on the path of the ball, or is it any point on the table? My first reaction is that for a point off the line of motion, there will be a torque, and a change in the angular momentum relative to that point. If the line of motion goes through point P there will be no torque from the frictional force, and so there can be no change in the angular mometum. For an arbitray point, the angular momentum will have a component parallel to the plane and a component perpendicular to the plane. Only the perpendicular component will be changed by the torque. If the point P lies on the line of motion of the ball, there is no perpendicular component to change.

The reason the angular momentum about a P on the line remains constant is that the force acting on the ball is directed along the line through point P, so there is no torque about point P. Where does the angular mometum lost to the reduced velocity go? Clearly the angular momentum of the translating center of mass is reduced by the reduced velocity, but now the top of the ball is moving twice as fast as its center, while the bottom of the ball is not moving at all. I don't know if your course gets into it, but the rotation of the ball about its center will account for the "missing" angular momentum associated with the translation of the center of mass. You don't need to know this to answer the question; the torque consideration will do. But that "missing" angular mometum has to be somewhere. If you know about the angular mometum of a rotating object being $L = I\omega$ , you might want to convince yourself that the angular momentum of rotation about the center of mass is equal to the reduction in the angular momentum of the center of mass.

5. May 8, 2005

### UnSaniTiZ

well, your answer for part d both enlightened and further confused me, but point P is on the surface of the table where the bottom of the ball rests on it (before it is struck with the cue).

good to know I got the rest of it right ;)

6. May 9, 2005

### OlderDan

It is a bit difficult to visualize the case of a P not on the line of motion. Since we now know that you need not worry about that complication, you should understand that the angular momentum is conserved because the frictional force acting in this problem has no torque about point P. If you calculate the angular mometum of the center of mass with its reduced velocity after slipping stops, plus the $L = I\omega$ angular momentum due to rotation about the center of mass you should find that their sum is the anglular momentum the ball had when it was first struck and was slipping with no rotation.