# Homework Help: Analyzing the movement of a billiard ball

1. Oct 6, 2012

### assaftolko

A billiard ball is at rest on a billiard table. The player hits the ball with the stick at an height h above the center of the ball, so Vcm just after the hit is v0. The final Vcm of the ball is 9/7 * v0.

Prove that h=4/5 * R

I have here a question and as you can see Vcm at the final stage is higher than Vcm at the initial stage. This is a bit bizzard because intuativley you'd expect that the kinetic friction will act at the opposite direction of movement and slow the ball down. But I think that here, from the moment just after the ball is hit and moves in v0, and until the final stage where it's moving in 9/7 * v0 (Which I suspect is rolling without slipping mode) - the kinetic friction acts to the left - and so it generates linear acceleration to the left (as it's the only force acting to the left after the hit from the stick) which is accountable for the increase in Vcm, and at the same time it generates clock-wise angular accelration which decreases w (via the torque from the kinetic friction). So in the initial stage wR>Vcm and this friction force to the left makes it possible for the system to achieve Rw'=V'cm and to maintain rolling without slipping mode.

Am I right? Is it the same situation when you're in a standing car and you push the gas all the way down? the tires spin very fast but the car isn't moving almost at all... until it finally does.

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2. Oct 6, 2012

### TSny

Yes, that's right. Good thinking.

3. Oct 6, 2012

### assaftolko

Tnx! Is there any chance to understand how to solve this question using the concept of angular impulse?

4. Oct 6, 2012

### TSny

Yes. The angular impulse relative to the CM of the ball can be expressed in terms of the linear impulse and h. You can also relate the angular impulse to the angular velocity immediately after the impulse. Likewise, you can relate the linear impulse to the linear velocity of the CM of the ball immediately after the impulse. Then you'll need to find a way to relate the initial angular and linear velocities to the final angular and linear velocities.

5. Oct 6, 2012

### assaftolko

Tnx but both impulse and angular impulse involve integration over dt... And i have no information about the duration of each part of the movement... Am i suppose to assume that for instance the force F from the stick is constant with time? I really don't have any experience with angular impulse..

6. Oct 6, 2012

### TSny

Let J represent the linear impulse (i.e., the integral of the force over the time). You should be able to express the angular impulse in terms of the symbol J and the distance h. Think about how to relate the linear and angular impulses to the linear and angular velocities immediately after the impulse. Just write everything in terms of symbols. See if that helps to see where to go from there.

7. Oct 6, 2012

### TSny

Angular impulse is just the integral of torque with respect to time. Express the torque in terms of the force and the distance h. When you integrate the torque over time, you should be able to see how the integral can be expressed in terms of the linear impulse and h. The net linear impulse equals the change in linear momentum. So, what do you think the net angular impulse equals?

8. Oct 6, 2012

### TSny

Instead of using energy concepts, I think it's easier to just use kinematic equations to relate the final linear and angular velocities to the initial velocities (immediately after the impulse). You won't be able to actually calculate the linear and angular acceleration values since you won't know the value of the force of friction. But if you write everything in symbols, you should be able to eliminate unknown quantities such as the friction force and the time to reach the final velocity.

9. Oct 6, 2012

### assaftolko

Change in ang momentum probably :)

10. Oct 6, 2012

### rcgldr

As TSny posted, you can assume that the impulse equals the change in linear momentum, and that the same impulse also changes the angular momentum.

Are you supposed to consider the fact that mechanical energy is converted into heat by friction during the time the ball is spinning and not rolling?

11. Oct 6, 2012

### TSny

Yes :)

12. Oct 6, 2012

### assaftolko

I guess i am because the friction will result in mechanical energy loss due to heat...

But the angular momentum also changes due to the torque from the kinetic friction (from when Vcm=v0 until Vcm=9/7 * v0) and not just due to the torque from the stick right (both the forces are supposly act in different time periods but when the stick exerts its impulse to the ball isn't there suppose to be some static friction from the table that will act in response to the force from the stick at the same time?))

13. Oct 6, 2012

### TSny

You can assume that the friction force during the impulse from the stick is much smaller than the force of the stick. So, you don't need to consider the effect of friction during the short time that the stick is in contact with the ball. This type of approximation is called an "impulse approximation".

14. Oct 6, 2012

### assaftolko

Tnx! Is it due to the short time that Fstick is exerted?

15. Oct 6, 2012

### TSny

If you use kinematic equations for constant acceleration to handle the change from initial to final velocities, you do not need to worry about how much heat is produced by friction.

16. Oct 6, 2012

### TSny

I think it's due rather to the fact that the stick's force has a much greater magnitude than the friction force and to the fact that both forces act for the same time during the impulse. Even if the time of the impulse happened to be long, the impulse due to the friction would still be very small compared to the impulse of the stick.

17. Oct 6, 2012

### assaftolko

18. Oct 6, 2012

### rcgldr

Turns out this won't matter. You can determine the energy lost afterwards, but it's not being asked for in this case.

Just focus on the linear and angular momentum; the same friction force that is slowing down the angular momentum is increasing the linear momentum until the surface speed of the ball matches the linear speed of the ball.

19. Oct 7, 2012

### assaftolko

tnx guys I've solved it without kinematics equations:

let counter clock-wise be positive direction for torque.

first stage - from rest to just after the impact from the stick:

delta L = Iw0-0 = Iw0.
linear impulse: SFstick*dt=deltaP=mv0-0=mv0 / multiply by h

SFstick*hdt = mv0h (h is constant so it can get inside the integral).

angular impulse: SFstick*R*sinq*dt = Iw0 -> (since Rsinq=h) -> SFstick*h*dt=Iw0

and so: Iw0=mv0h -> (I=2/5 * mR^2) -> w0=5v0h/2R^2

second stage - from just after impact to final stage (rolling without slipping):

delta L = I*9v0/7R - I*5v0h/2R^2
linear impulse: Sfk*dt = 9mv0/7 - mv0 / multiply by R
Sfk*R*dt = 2/7 * mv0R
angular impulse: S-fk*R*dt = delta L -> Sfk*R*dt = -deltaL

and so: 2/7 * mv0R = 2/5 * mR^2 * 5v0h/2R^2 - 2/5 * mR^2 * 9v0/7R

h = 4/5 * R

20. Oct 7, 2012

Good work!