# Billiards problem

1. Dec 9, 2003

### zekester

Good pool players say that it is possible to strike a cue ball with the cue stick held horizontal to the table ina a way that the ball does not slide. If the cue ball has a radius R, a mass M and is struck a distance H above the centerline, find H for this circumstance to occur.

I'm not really sure where to start. Any hints?

2. Dec 9, 2003

### NateTG

The idea is that the pool ball should be rotating at a speed that makes the surface of the ball move at the same speed relative to the center of the ball as the table is moving. That way the ball will not slip.

You could start by figuring out the ratio of rotational and linear speeds as a function of H.

3. Dec 10, 2003

### zekester

that makes sense but i still can't quite figure it out.

4. Dec 10, 2003

### Staff: Mentor

Consider a force (F) exerted at some height (h) from the table. That force exerts a torque about the center of mass, causing an angular acceleration; it also causes a translational acceleration. Write the dynamical equations: Torque = I &alpha; & F = ma.

Solve those two equations subject to the condition for rolling without slipping: acm = r &alpha; .

5. Dec 10, 2003

### formulajoe

so H is the moment arm of the torque correct? but the force involved in the torque comes from the mass and acceleration of the cue stick which is not given.

i dont quite understand that equation for torque. what is the I and where does the alpha come from? i thought torque was moment arm times force.

6. Dec 10, 2003

### formulajoe

i was just playin with the symbols, and i was just curious if i stumbled onto the correct answer.
i started with F = MA, and A = alpha/R. so F = M*alpha/R. than torque = F * moment arm, which is also H. and F = torque / H. so torque / H = m*alpha/R.
solving for H, I got H = torque * R/M*alpha.

7. Dec 10, 2003

### formulajoe

i played around with torque = I * alpha and F = ma.
i made F = ma, into F = m(alpha*R) and solved for alpha. which ended up being alpha = F/m*R
i plugged that into the torque equation.
torque = I *(F/m*R). torque = F*H
so F*H = I * (F/m*R)
solving for H, i get H = I/m*R
and I = m*r^2 so H = m*r^2/mr, so h = R. according to that, the ball must be struck right on the top for the no slip condition to be present initially. am i correct?

8. Dec 10, 2003

### NateTG

I don't think so. The correct answer will be in the form:
$$H=kR$$
where k is a dimensionle ss constant.

Let's say that the momentum transfer occurs at a single point on the ball at height H.

Then we have the following:
$$L=MVH$$
from the impact, and
$$L=I\omega=I\frac{V}{R}=\frac{2}{5}MR^2\frac{V}{R}=\frac{2}{5}MRV$$
from the no slip condition.
which now we can solve for H and get:
$$H=\frac{2}{5}R$$

9. Dec 10, 2003

### formulajoe

what is L = MVH? ive never seen anything like that before.

10. Dec 10, 2003

### NateTG

Sorry, I thought it simplified nicely, but I guess it doesnt.

Let's say that
1. The pool cue has mass $$m$$ and velocity $$v$$
and
2. The pool cue is at rest after the colision.

Now:
Conservation of linear momentum gives:
$$mv=MV$$
and conservation of angular momentum gives
$$L=mvH$$
since
$$mv=MV$$
we have
$$L=MVH$$

11. Dec 10, 2003

### formulajoe

ohhh, okay. we never covered angular momentum.

12. Dec 10, 2003

### Staff: Mentor

You guys are all over the map. It's got nothing to do with conservation of momentum. (Doesn't anyone listen to me? )

Considering torque about center of mass:

F(H-R) = I &alpha; = 2/5 MR2 &alpha;

But, F = Ma
And, since a = R &alpha;, &alpha; = F/(MR)

So,
F(H-R) = 2/5 MR2 F/(MR)

H = 7/5 R

13. Dec 10, 2003

### formulajoe

why is it H-R?

14. Dec 10, 2003

### Staff: Mentor

We are finding torque about the center of the ball, which is at height R. So, the moment arm is H-R.

15. Dec 10, 2003

### NateTG

Whatever. We came up with the same answer.

16. Dec 10, 2003

### StephenPrivitera

Hi, I'm trying to understand what's going on here.
When you write F=Ma, a is the acceleration of the center of mass. When you write F=I &alpha;=I a/R, a is the linear acceleration of a point at a distance R from the axis of rotation (as if the ball were not moving at all). So these are not the same, are they? They total linear acceleration of a point in contact with the surface should be the difference between these two. No?

Doesn't the magnitude of the force matter?

Last edited: Dec 10, 2003
17. Dec 11, 2003

### jamesrc

Yes, the total linear acceleration of a point in contact with the surface is the difference between these two; it is 0. Since the ball is not slipping, it must be pivoting about the point where it contacts the floor (the relative velocity between the point of contact of the ball and the pool table is 0). The center of mass is "point at a distance R from the axis of rotation." Does that help?

18. Dec 11, 2003

### Staff: Mentor

Re: Re: Billiards problem

D'oh! You are absolutely right. (I didn't realize that H was measured from the center.) Sorry, NateTQ and formulajoe.

19. Dec 11, 2003

### StephenPrivitera

So its acceleration relative to the table is zero. And since the initial velocity of that point was zero, at all successive instants, the point in contact with the surface should be at rest relative to the table. And thus no slipping because slipping would imply movement relative to the table. Yeah, I think I get it now.