A thin rod of length 2R and mass m is standing(vertically) on a perfectly smooth floor. the state of equilibrium in which the rod is at res is unstable, and the rod falls. find the trajectories that the various points of the rod describe and the velocity with which the upper end of the rod hits the floor?

Dear Aditya Nijampurkar,

Ans:- It is clear that the trajectory is a circle iff the bottommost point is in stable condition i.e it is not sleepping.

Next, the CM falls a distance R. as there is no sleepping hence this entire energy converts into rotational K.E hence

1/3M (2R)²W²=MgR

Hence W=√(4g/3R)

Hence V=2WR=2√(4gR/3)

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