What is the angular momentum of a 2.5 kg uniform cylindrical grinding wheel of radius 19 cm when rotating at 1200 rpm? And how much torque is required to stop it in 5.7 s? Need help with step-by-step explanation.

Answer 1

chose your moment of Inertia model. I think solid disk is a good model. I= 1/2 mr^2

angMomentum= I*angvelocity

where w (angular velocity)=2PI*1200/60 rad /seconds

Torque=momentinertia*angularacceleration
= 1/2 m r^2 * change in angvelocity/time
= 1/2 m r^2 * (wf-wi)/5.7sec

wf=0 ; wi is 2PI*1200/60 rad/sec