Rotational 2

A rod OA of mass m and length L hinged at end O can rotate smoothly in vertical plane about horzontal axis passing through O .It is released from a horizontal position as shown. and strikes a block of mass M kept vertically below at distance L on a smooth horizontal surface.The block after collisin moves on the smooth vertical circular track of diameter L such that the force exerted by the circular track on the block at C is one third of its wieght .If the rod after collision rotates through an angle 30 degrees frm horizontal . find the ratio of masses M/m

4 Answers

535
Aditya Agarwal ·

Mm=(√37)(√2-1)

1123
Sourish Ghosh ·

I'm getting 17

213
Nandikesh Singh ·

M/m=3/7

Anonymous ·

can someone give a hint , i know that it will solved by energy conservation but still i am not able to solve it.

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