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can someone help me with the following?

A rigid assembly of a thin hoop (of mass m and radius R of 0.15m) positioned on top of a thin radial rod (of mass m and length L = 2.0 R). (So imagine a circle on top of a stick, whereby if you extend the stick you will hit the center of the circle). If we give the assemblly a nudge the assembly will rotate around a horizontal axis that is perpendicular to the rod and lying on the bottom of the rod. Assuming that the energy given in the nudge is negligible, what is the assembly's angular speed about the rotation axis when it passes through the upside-down (inverted position)?

This is what I did:

I thought that since mechanical energy is conserved, so change in kinetic energy + change in potential energy = 0.

Kinetic energy = 1/2*I*v^2, where v = angular speed. I = moment of inertia, and in this case I found the inertia of the hoop = m*R^2 + m*(3R)^2 = 10*m*R^2.

The inertia of the rod = 1/3*m*L^2 = 1/3*m*(2R)^2.

Add the two up I got 34/3*m*R^2.

the potential energy I can treat as concentrated in the center of mass of the assembly. the y coordinate of the center of mass I found by: (m*R + m*3R)/2*m = 2R. Then I treat the total mass of 2m as concentrated at 2R (or -2R because it is inverted)

So my equation would give me

1/2*34/3*m*r^2*v^2 = 2R*2m*g

But I don't think this is right.

Any suggestions?

Thanks!

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# Homework Help: Rotational kinetic energy problem

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