Find H for Cue Ball Strike to Not Slide: Billiards Problem

[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?

 

[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.

 

[zekester]

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

 

[Doc Al]

Originally posted by zekester
that makes sense but i still can't quite figure it out.

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 α & F = ma.

Solve those two equations subject to the condition for rolling without slipping: a_cm = r α .

 

[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.

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 á & F = ma.

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

i don't 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.

 

[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.

 

[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?

 

[NateTG]

Originally posted by formulajoe
i was just playin with the symbols, and i was just curious if i stumbled onto the correct answer...

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:
$$H=\frac{2}{5}R$$

 

[formulajoe]

what is L = MVH? I've never seen anything like that before.

 

[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
$$L=MVH$$

 

[formulajoe]

ohhh, okay. we never covered angular momentum.

 

[Doc Al]

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 α = 2/5 MR² α

But, F = Ma
And, since a = R α, α = F/(MR)

So,
F(H-R) = 2/5 MR² F/(MR)

H = 7/5 R

 

[formulajoe]

why is it H-R?

 

[Doc Al]

Originally posted by formulajoe
why is it H-R?

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

 

[NateTG]

Originally posted by 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?

Whatever. We came up with the same answer.

 

[StephenPrivitera]

Originally posted by Doc Al
Considering torque about center of mass:

F(H-R) = I α = 2/5 MR² α

But, F = Ma
And, since a = R α, α = F/(MR)

So,
F(H-R) = 2/5 MR² F/(MR)

H = 7/5 R

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 α=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?

 

[jamesrc]

Originally posted by 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 α=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?

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?

 

[Doc Al]

Originally posted by NateTG
Whatever. We came up with the same answer.

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

 

[StephenPrivitera]

Originally posted by 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?

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.