# A ball rolling down an incline

1. Dec 12, 2006

### rleung3

Hey,

I have been thinking about this problem for a while to no avail. Any input would be greatly appreciated. Thanks!

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

A solid sphere of mass M and radius R rolls without slipping down a 30 degree incline (see figure attached).

What is the linear acceleration, a, of the sphere down the incline?

2. Relevant equations

I=(2/5)MR^2

torque = (Ia)/R = rxF

3. The attempt at a solution

I only know of two ways to solve for this: energy methods or torque methods. To do torque, I would need to know the value of the static friction that is keeping the ball from slipping, but I don't have that information. For energy methods, I don't have a height from where the ball began to roll, so I don't see how I can use that.

Here is my attempt at torque:

torque = (-Ia)/R = -fR where f=force of static friction

=> Ia = fR^2 => substitute I=(2/5)MR^2 => (2/5)(MR^2)a = fR^2 => f=(2/5)Ma => a=(5f)/(2M)

I don't know where to go from here. The answer is (5/14)g. I must be missing some type of other alternative to solving this problem, but I can't think of what. Thanks so much! I appreciate it.

Ryan

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• ###### physics practice final no 18 mc.jpg
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2. Dec 12, 2006

### marlon

You need three essential parts :
F' is friction

1) Mgsin(30) - F' = Ma (a along incline and the positive direction is along the incline as well)

2) F' R = I $$\alpha$$

3) $$\alpha$$ = a/R

Put the F' from 1) into 2) to get rid of it. Then use 3) to solve for a

marlon

3. Dec 12, 2006

### AlephZero

If you want to use an energy method, you can assume it rolls an arbitrary distance starting from rest, find the final linear and angular velocities, then find the accleration using the formulas for motion with constant acceleration.

4. Dec 12, 2006

### rleung3

Ohh! Thank you. Yea, I am definitely rusty at this..has been a while.