# Rotation down an incline

1. Apr 13, 2009

### hejo

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

A uniform solid sphere rolls down an incline without slipping. If the linear acceleration of the center of mass of the sphere is 0.22g, then what is the angle the incline makes with the horizontal?

2. Relevant equations

a=gsin(theta)

3. The attempt at a solution

The only thing I can think of is acceleration down an incline=gsin(theta). Since we have a=0.22g, I made gsin(theta)=0.22g and thereby arcsin(0.22) should give me theta...but it doesn't...any help please??

2. Apr 13, 2009

### Staff: Mentor

That's the acceleration down an incline if the only force acting parallel to the incline is gravity, for example when a box slides down a frictionless incline. Hint: Since the sphere rolls without slipping, the incline cannot be frictionless. (Solve for the acceleration using Newton's 2nd law twice; once for translation, once for rotation.)

3. Apr 13, 2009

### hejo

Okay. I set up my equation for translational motion to be ma=mgsin(theta) - (coefficient of friction)*mgcos(theta). So a=gsin(theta) - gcos(theta)*(coefficient of friction).

Is that right? I don't understand how I can get acceleration from rotational motion.

4. Apr 13, 2009

### Staff: Mentor

No, it's not quite right. Since there's no slipping, the relevant friction is static friction. And static friction can be anything up to a maximum of μN. So you can't just set it equal to that maximum value. Just call the friction "F" and continue.
That friction force also exerts a torque on the sphere which produces an angular acceleration. Set up an equation for that. (You can relate the angular acceleration to the translational acceleration.)

Note that you have two unknowns--the friction and the acceleration--but you also have two equations.