Analyzing the movement of a billiard ball

[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.

 

[TSny]

Yes, that's right. Good thinking.

 

[assaftolko]

Yes, that's right. Good thinking.

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

 

[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.

 

[assaftolko]

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.

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..

 

[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.

 

[TSny]

I really don't have any experience with angular impulse..

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?

 

[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.

 

[assaftolko]

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?

Change in ang momentum probably :)

 

[rcgldr]

Tnx but both impulse and angular impulse involve integration over dt

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?

 

[TSny]

Change in ang momentum probably :)

Yes :)

 

[assaftolko]

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?

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?))

 

[TSny]

...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?))

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".

 

[assaftolko]

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".

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

 

[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.

 

[TSny]

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

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.

 

[assaftolko]

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.[/]

so it's just pure logic? You basiclly understand that the billiard table offers little to stop the stick from moving the ball when it hits?

And about rolling without slipping-what is the reason that when the system achieves this mode (from wR>Vcm like here or from Vcm>wR like when you have rolling with slipping) it stays in this mode? I know that when we get to Vcm=wR the kinetic friction becomes static friction with magnitude 0 but why does it happen really?

 

[rcgldr]

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?

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.

 

[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

 

[TSny]

Good work!

 

[assaftolko]

Good work!

Tnx! What about what i asked about rolling without slipping can i get your answer?

 

[TSny]

Tnx! What about what i asked about rolling without slipping can i get your answer?

I'm not sure I understand your question here. When the ball achieves rolling without slipping, then the point of the ball that is instantaneously in contact with the surface has zero velocity relative to the surface. So, kinetic friction goes to zero and there is no need for static friction to come into play either. (Actually, in reality there would still be some "rolling friction" that will slow the ball down. Somewhere I read an article long ago about rolling friction where they analyzed it in terms of considering the ball as causing a small indentation in the the surface that the ball is rolling on. As I recall, they argued something to the effect that the ball is effectively always rolling "up hill" in the small indentation.)

But, I suspect I'm not getting to the heart of your question. I guess I'm not sure what it is that puzzles you.

 

[assaftolko]

I'm not sure I understand your question here. When the ball achieves rolling without slipping, then the point of the ball that is instantaneously in contact with the surface has zero velocity relative to the surface. So, kinetic friction goes to zero and there is no need for static friction to come into play either. (Actually, in reality there would still be some "rolling friction" that will slow the ball down. Somewhere I read an article long ago about rolling friction where they analyzed it in terms of considering the ball as causing a small indentation in the the surface that the ball is rolling on. As I recall, they argued something to the effect that the ball is effectively always rolling "up hill" in the small indentation.)

But, I suspect I'm not getting to the heart of your question. I guess I'm not sure what it is that puzzles you.

Never mind i got it by now, tnx a lot!

 

[rcgldr]

Actually, in reality there would still be some "rolling friction" that will slow the ball down. Somewhere I read an article long ago about rolling friction where they analyzed it in terms of considering the ball as causing a small indentation in the the surface that the ball is rolling on. As I recall, they argued something to the effect that the ball is effectively always rolling "up hill" in the small indentation.

The term is normally "rolling resistance". It's related to the loss in energy due to compression and expansion at the point of contact between ball and surface. Both ball and surface will compress and then re-expand as the ball rolls, but the restoration path involves somewhat less force than the deformation path (hysteresis), so energy is lost. There can also be losses due to kinetic friction due to the differing amounts of deformation and restoration of the ball and surface.

In case you're curious

Using h = 4/5 R or using w1 = w0 - 2/7 m v0 R you get w0 = 2 v0 / R

Initial energy = 1/2 m v0^2 + 1/2 (2/5 m R^2) (2 v0 / R)^2 = (91/70) m v0^2

Final energy = 1/2 m (9 v0/7)^2 + 1/2 (2/5 m R^2) (9 v0 / 7 R)^2 = (81/70) m v0^2

energy loss = 1/7 m v0^2

 

[assaftolko]

Thats the first check i did after i solved the question in order to verify that the kinetic friction, although acting to the left unintuativly, still doing work that results in kinetic energy loss

 

[rcgldr]

The kinetic friction, although acting to the left unintuitively, still doing work that results in kinetic energy loss

In this case, kinetic friction decreases angular energy more than it increases linear energy.

 

[assaftolko]

In this case, kinetic friction decreases angular energy more than it increases linear energy.

One should hope so if we want to maintain the natural order in the universe :)