# Rotation without slipping

1. Apr 13, 2007

### ripper9100

1. The problem statement, all variables and given/known data
A solid uniform cylinder with mass 5 kg is being pulled with a constant force of 45 Newtons up a 30 degree incline. The force is acting on the cylinders center and is parallel to the incline. What is its angular acceleration?

I have a good idea on how to do the problem by using Newtons second law for a rigid body. The only part i'm confused about is on what the moment of inertia should be. I think Its either (.5mr^2) or (.5mr^2 +mr^2)?

2. Apr 13, 2007

### e(ho0n3

3. Apr 13, 2007

### ripper9100

im also iffy on the net Torque. Im thinking that friction is the only force causing torque since the 45 N force is acting on the cylinders center.
am i right?

4. Apr 13, 2007

### e(ho0n3

Yes. I agree.

5. Apr 13, 2007

### ripper9100

thanks for the help

6. Apr 13, 2007

### Staff: Mentor

The moment of inertia depends on what you are using as the rotational axis; either expression will work, if you're careful.

Again, it depends on the axis of rotation. If you are using the center of mass as the axis, then you are correct.

Even though you have a choice, I recommend using the center of mass of the cylinder as your axis--I think it gives the best understanding of what's going on.

7. Apr 14, 2007

### ripper9100

what about the friction force is it up the incline or down the incline? Im confused about because usually the direction of friction force is opposite the direction of motion.

8. Apr 14, 2007

### Staff: Mentor

Remember that friction acts to oppose slipping between surfaces. If there were no friction, which way would the surfaces (cylinder bottom and incline) slip with respect to each other? Use that to figure which way friction must act.