Question

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?

Answer 1

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

Answer 2

Oh, and the answer is negative (deacceleration)

Answer 3

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