Posted by David on .
What is the angular momentum of a 1.0kg uniform cylindrical grinding wheel of radius 19cm when rotating at 1700rpm ?How much torque is required to stop it in 4.6s ?

Physic Angular momentum and torque 
Damon,
r = 0.19 m
I =.5 m r^2 =.5 (1)(.19)^2 = .0181 kg m^2
1700 rev/min * 1 min/60 s * 2 pi rad/rev
= 178 radians/second = omega
L =I * omega .0181 * 178 = 3.22 kg m^2/s
final omega = initial omega  alpha t
alpha (4.6) = 178
alpha =  38.7 radians/s^2
Torque = I alpha = .0181 (38.7)
=  0.7 kg m^2/s^2 = 0.7 Newton meters