Time required to stop:
theta = (1/2)(w0 + w)*t
163(2π) rad = (1/2)(30(2π) rad/s +0 rad/s)*t
t ~ 10.87 s
Angular acceleration of wheel:
= (0 rad/s - 30*2π rad/s) / 10.87 s
~ -17.3 rad/s^2
Required slowing torque:
T = I * alpha
= (0.015 kg*m^2)(-17.3 rad/s^2)
~ -0.26 kg*m^2/s^2
(or -0.26 N*m)
A 0.500 kg wheel that has a moment of inertia of 0.015 kg m^2 is initially turning at 30 rev/s. it starts to rest after 163 revolution. how large is the torque that slowed it?
1 answer