Calculating Torque on a Potter's Wheel

[chamonix]

Homework Statement

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?

Homework Equations

t=Ia
a=w/t

The Attempt at a Solution

a. (50/60*2*pi)=5.235988 rad/sec
5.235988/6=.87 rad/sec^2
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.

 

[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.

 

[chamonix]

So...in this case, i should go .5*100*.5^2=12.5?
But the answer given is 11 Nm. I don't understand. :(

 

[hage567]

So...in this case, i should go .5*100*.5^2=12.5?
But the answer given is 11 Nm. I don't understand. :(

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).

 

[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.