Posted by Taylor on Monday, November 2, 2009 at 10:43am.
What is the angular momentum of a 3.7kg uniform cylindrical grinding wheel of radius 25cm when rotating at 1500rpm?
How much torque(in magnitude) is required to stop it in 5.0s?
Could you explain this in detail, I don't really understand torque and momentum well.

U. Phys  bobpursley, Monday, November 2, 2009 at 11:00am
You need to separate linear motion from rotational.
Read in two columns
mass kg...........momentofinertia kg*m^2
displacement m...angular displacement rad
velocity m/s ....ang velocity rad/s^2
acceleration m/s^2.ang acc rad/s^2
and finally, memorize this
tangential velocity on a curve= radius*ang velocity
Now your question.
ang momentum=I*w
(that is ang momentum is momentofinertia * angular velocity)
So you need I for a solid disk, make you a convenient card with common shapes..http://en.wikipedia.org/wiki/List_of_moments_of_inertia
Using thin disk model, I= mr^2/2
ang momentum= mr^2/2 * 1500rev/min*1min/60sec*2PIrad/rev
b) Torque= I*angacceleration= I*(wfwi)/time= mr^2/2(1500*2PI/60)
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