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 ?

Answer 1

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