Calculating coefficient of kinetic friction through angular acceleration

[xregina12]

A potter's wheel is 101 kg and has radius of 0.483 meters and is freely rotating at 71.9 rev/min. the potter can stop the wheel in 6.54 s by pressing against the rim and exerting a radially inward force of 103 N. find coefficient of kinetic friction.


I found alpha=71.9 x 2 x pi/60sec/ 6.54= 1.15 rad/s^2
Torque= I x alpha = 27.126
however, i don't know where to go from here. Thanks for any help.

 

[alphysicist]

Hi xregina12,

A potter's wheel is 101 kg and has radius of 0.483 meters and is freely rotating at 71.9 rev/min. the potter can stop the wheel in 6.54 s by pressing against the rim and exerting a radially inward force of 103 N. find coefficient of kinetic friction.


I found alpha=71.9 x 2 x pi/60sec/ 6.54= 1.15 rad/s^2
Torque= I x alpha = 27.126
however, i don't know where to go from here. Thanks for any help.

What type of force is the 103N force? What is the formula for the kinetic frictional force, and what direction is it in this case? After answering those questions you can find the torque due to the frictional force and plug it into your equation.

 

[tiny-tim]

A potter's wheel is 101 kg and has radius of 0.483 meters and is freely rotating at 71.9 rev/min. the potter can stop the wheel in 6.54 s by pressing against the rim and exerting a radially inward force of 103 N. find coefficient of kinetic friction.
…
Torque= I x alpha = 27.126

Hi xregina12!

(have a alpha: α and an omega: ω )

Torque = force x distance, so you can calculate the force (that's the friction, of course) …

so you now know both the friction force and the normal force, so …