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## Homework Statement

A uniform rod (length = 2.0m) is mounted to rotate freely about a horizontal axis that is perpendicular to the rod and that passes through the rod at a point of 0.5m from one end of the rod. If the rod is released from rest in a horizontal position, what is the angular speed of the rod as it rotates through its lowest position?

## Homework Equations

I believe (although I might be wrong on some of them):

[itex]\tau = I \alpha[/itex]

[itex]K = \frac{1}{2} I \omega^{2}[/itex]

[itex]L_{z} = I \omega[/itex]

For a rod the moment of inertia is:

[itex]\frac{1}{12} M L^{2}[/itex]

And the parallel axis thereom:

[itex]I = I_{cm} + M D^{2}[/itex]

## The Attempt at a Solution

So far I have used the parallel axis theorem and the moment of inertia for a rod to get the moment of inertia to be 0.5833 for this rods rotation.

I am not sure exactly what to do after that. I can figure it out if it is the moment right after release, but not sure how to do it when the rod is at the bottom.