What is the angular momentum of the disk as it leaves the plane and rolls on a level surface? The radius of the disk is 0.1 m.

Question

anonymous · Answer

it rolls down an incline? how far does it fall?

anonymous · Answer

I agree with Stan, it's not a clear statement.

anonymous · Answer

Well there is a previous problem that ties in with this problem. I just realized it. My instructor didn't make this very clear to us.
The problem..."A 500 g solid disk rolls down an inclined plane 1 m long elevated by 10 cm. What is the kinetic energy of the disk at the bottom of the plane?" "Solution: 1/2(9.8)(0.1) = 0.49 J". And then the question I asked ties in with this problem I just provided.

anonymous · Answer

I assume he wants the angular momentum about the center of the disk...$$L=I \omega$$where the angular velocity wanted in the angular velocity at the bottom of the hill. You can get that from $$K = K _{rotation} + K _{translation}$$$$0.49J=1/2I \omega^{2}+1/2mv^{2}$$$$I=1/2mR^{2}$$and assuming rolling w/o slipping $$v=r \omega$$so$$0.49J=1/2(1/2mR^{2})\omega^{2}+1/2m(R^{2}\omega^{2})$$solve for omega and substitute into the equation for ang. momentum

anonymous · Answer

the change in potential energy is already taken care of in the gain of kinetic energy (0.49J).... you don't want to do it twice :)

anonymous · Answer

Sorry ! Exact ! I would be tired. So, I will supress my comment.

anonymous · Answer

Just for my curiosity, is really the question :
"What is the kinetic energy of the disk at the bottom of the plane?" => Kinetic energy or potential energy.
I just try to find a mitigating circumstance. :)

anonymous · Answer

The previous question (the lead-in to this one) wanted to find the total KE, which you get from the potential energy, but here they want L