When the rod falls, the mechanical energy is conserved: mgL/2= 1/2Jw2 (square), w=V/L, J= 1/3mL2 (square) w;

Conservation of angular momentum:1/2jw =1/2jw'+mv' l, w = v'/l);

Work done by friction: umgs= 1/2mv''2 (square);

I should be able to write it myself in the future.

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