Physics
posted by JAYZ .
A potter’s wheel of radius 53.13 cm and mass 106.4 kg is freely rotating at 47.1 rev/min. The potter can stop the wheel in 5.65 s by pressing a wet rag against the rim and exerting a radially inward force of 66.2 N.
What is the angular acceleration of the wheel?

The only information of any use in this lengthy paragraph is:
wheel stops from 47.1 rev/min in 5.65 sec
we want radians and seconds
47.1 rev/min * 2 pi rad/rev * 1 min/60 s
= 4.93 radians/second
acceleration = change in velocity/change in time
= 4.93 rad/s / 5.65 s
= .873 radians/second^2 
Oh, and the answer is negative (deacceleration)

yea i had the answer but forgot the negative, thanks man! really appreciate it!
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