Asked by Anonymous
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.
Answers
Answered by
bobpursley
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
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
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