# Homework Help: Angular Acceleration of a rubber wheel

1. Nov 11, 2007

### jmcmillian

A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The small wheel has a radius of 2.0 cm and accelerates at the rate of 7.2 rad/s^2, and it is in contact with the pottery wheel (radius 21.0 cm) without slipping.

(A)Calculate the angular acceleration of the pottery wheel
(B)and the time it takes the pottery wheel to reach its required speed of 65 RPM.

$$\omega$$=$$\omega$$$${0}$$+$$\alpha$$t

I don't know enough about pottery wheels to know if they are all made the same, but based on the problem I want to assume that the smaller wheel is inside the bigger wheel, that way the outer edge of the small wheel turns the inner edge of the pottery wheel. This may be where I am wrong...but that's what I'm rolling with for now.

If that is the case, then the velocity of the smaller wheel ought to be the initial velocity of the larger wheel, then there should be some sort of proportional relationship between the two based on the difference in radii. As a matter of fact, I venture to say that the acceleration for the larger wheel (what I'm looking for) will be much smaller than the 7.2 rad/s^2 given for the small wheel.

Carrying on, then $$\omega$$ (small wheel) will equal 7.2t. This would be equal to the initial $$\omega$$ of the wheel.

Beyond that, I don't know where to go from there. Any clues would be appreciated!

2. Nov 11, 2007

### rl.bhat

When the two wheels are in contact and rotating without slip, there linear velocities and acceleration are the same. Hence r1w1 = r2w2 = v and
a = r1*alpha1 = r2*alpha2. From the given data find the angular acceleration of pottary wheel.

3. Nov 11, 2007

### jmcmillian

Ahhhhhhhhhh yes. I've done so much physics tonight that I think I often lose focus of the problem and wind up going on wild goose chases when it's a lot simpler than it should be. Thanks for your help!