# Ball on paper

1. Jan 28, 2010

### boardbox

1. The problem statement, all variables and given/known data
I have a heavy ball on a piece of paper on the floor. The paper is grabbed and moved horizontally with acceleration a. What is the acceleration of the center of the ball? The ball is assumed to not slip with respect to the paper.

2. Relevant equations

3. The attempt at a solution
I'm not really sure how the acceleration is supposed to interact with the ball. A thought I've had is to say that the ball doesn't translate but rotates with angular acceleration a/R since it doesn't slip.

2. Jan 28, 2010

### kuruman

*** Edited after reading Doc Al's Comment ***

That's almost it. See Doc Al's comment below.

Last edited: Jan 28, 2010
3. Jan 28, 2010

### Staff: Mentor

Careful. The paper exerts a force on the ball, so the ball must translate as well as rotate. Both the ball and the surface (the paper) are accelerating, but at different rates. So you can't just assume that the angular acceleration is a/R.

Hint: Assume that the paper exerts some force F on the ball. Analyze the translational and rotational dynamics of the ball using Newton's 2nd law.

4. Jan 28, 2010

### boardbox

$$\Sigma$$Fball = Fpaper = maball
$$\Sigma$$t = tpaper = I $$\alpha$$
t = r x F
$$\alpha$$ = apaper/r

plug and chug

aball = 2apaper/5

does that get it about right?

Last edited: Jan 28, 2010
5. Jan 28, 2010

### Staff: Mentor

OK.
OK.
No. Alpha is the rotation about the center of mass--compare the acceleration of the paper with the acceleration of the center of mass.

(But you're on the right track!)

Last edited: Jan 28, 2010
6. Jan 28, 2010

### boardbox

Would it be the sum of the two accelerations over the radius?

I think rolling without slipping is $$\omega$$ = v/r so the acceleration version of that should just be the time derivative. If I have a a constant v on the paper and the ball I would expect to just sum them.

7. Jan 28, 2010

### Staff: Mentor

What if the center of mass had the same acceleration as the paper? What would be the rotational acceleration of the ball in that case?

8. Jan 28, 2010

### boardbox

Zero?
You highlight sum. I'm wondering if you want magnitude of difference?

9. Jan 29, 2010

### Staff: Mentor

Exactly. If you just added them you'd get alpha = 2a/r, which doesn't make sense.
What you need to find alpha is the relative acceleration of paper and ball divided by r.

10. Jan 29, 2010

### boardbox

I see, makes sense. Thanks for the help.