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*(wf-wi)/time= mr^2/2(-1500*2PI/60)

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