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I wanted to consider a second case of my homework assignment. We were asked to solve the question:

A uniform rod of length b stands vertically upright on a horizontal plane in a position of unstable equilibrium. The rod is then given a small displacement at the top and it tips over. What is the rod's angular velocity when it hits the plane, assuming the rod does not slip?

I managed to get an answer from conservation of energy. However, I wanted to also understand what difference there would be if we considered if the base of the rod could slip. How would I go about understanding that situation? Any suggestions would be much appreciated! :)

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  • $\begingroup$ 'I managed to get an answer from conservation of energy.' Please incorporate this conclusion into your question. $\endgroup$ – Gert Oct 30 '15 at 3:47
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If it doesn't slip, you can model it as rotating around the end of the rod (where it touches the ground). When it strikes the ground, this restriction means that you know both the angular speed of rotation and the speed of the center of mass.

If there is no friction, then the rod will rotate around the center of mass. This changes the moment of inertia for the rod, and it decouples the rotation and the fall. At the beginning of the fall, the two will be related. But during the fall it could happen that both ends lift from the ground. That makes determining the energy due to rotation difficult.

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