# Torque on potter's wheel

1. Apr 18, 2007

### chamonix

1. The problem statement, all variables and given/known data
A potter's wheel of radius 0.50 m and mass 100.0 kg is freely rotating at 50.0 rev/min. The potter can stop the wheel in 6.0 seconds by pressing a wet rag against the rim.
a. What is the angular acceleration of the wheel?
b. How much torque does the potter apply to the wheel?

2. Relevant equations
t=Ia
a=w/t

3. The attempt at a solution
b. I didnt understand this portion of the question. I tried .5^2*100*5.235988 but the answer was not 11 Nm which I know is the answer. Please help! Any help is greatly appreciated. Thank you.

2. Apr 18, 2007

### hage567

You want

torque = I * alpha

not I * omega, that is angular momentum. Check your calculation.

Also, if you are assuming the wheel is a solid cylindrical disk, then I = 0.5MR^2, so your above I value is not correct if that's the case.

Last edited: Apr 18, 2007
3. Apr 19, 2007

### chamonix

So...in this case, i should go .5*100*.5^2=12.5?
But the answer given is 11 Nm. I dont understand. :(

4. Apr 19, 2007

### hage567

You didn't multiply by alpha!! torque = I*alpha. If you do that you will get the right answer. You found the right value of alpha in part (a).

5. Apr 19, 2007

### chamonix

Oh, ok. My fault. Ok. Alpha.
so.... .5*100*.5^2*.87=10.9=11Nm!! ok. Thank you!!
Sorry, I must have overlooked that alpha part. Thank you.