# Ball rolling on incline

1. Oct 24, 2007

### sheri1987

1. The problem statement, all variables and given/known data

A ball of mass 2.30 kg and radius 0.142 m is released from rest on a plane inclined at an angle q = 42.0° with respect to the horizontal. How fast is the ball moving (in m/s) after it has rolled a distance d=1.35 m? Assume that the ball rolls without slipping, and that its moment of inertia about its center of mass is 1.70E-2 kg·m2.

2. Relevant equations

.5mv^2+.5I(v^2/R^2) = mgdsin(angle)

3. The attempt at a solution

I plugged in the numbers to equation above and got 3.2 m/s by solving for v...am I doing something wrong, because it is not the right answer?!?

2. Oct 24, 2007

### Staff: Mentor

Your error is assuming that the only force acting on the ball parallel to the plane is the component of its weight. Not so. (What other force must act so that the ball rolls instead of slides?)

Edit: I was wrong--your equation is fine.

Last edited: Oct 24, 2007
3. Oct 24, 2007

### nrqed

I don't see the error in the OP's formula. The force you are thinking about does not do any work.

to the OP: What equation did you use for the moment of inertia? The one for a full sphere?

EDIT: never mind my last question, I had not noticed that the moment of inertia itself was given. Then it must simply be a mistake of plugging in the calculator.

Last edited: Oct 24, 2007
4. Oct 24, 2007

### sheri1987

I am not sure what other force, that is probably my problem

5. Oct 24, 2007

### Dick

The OP is just using a conservation of energy equation. How can that be wrong? I put the numbers in and got a similar number but somewhat larger. I think it's just some kind of calculator slip up.

6. Oct 24, 2007

### Dick

It looks like the moment of inertia was given in the problem statement.

7. Oct 24, 2007

### Staff: Mentor

No equation needed--the moment of inertia is given.

Yep, you're right.

8. Oct 24, 2007

### nrqed

Yes, thank you. I edited my post when I noticed that.