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Question

A homogeneous rod AB of length L and mass M is hinged at the centre O in such a way that it can rotate freely in the vertical plane. The rod is initially in horizontal position.  An insect S of the same mass M falls vertically with speed V on point C, midway between the points O and B. Immediately after falling, the insect starts to move towards B such that the rod rotates with a constant angular velocity omega.

If insect reaches the end B when the rod has turned through an angle of  calculate V in terms of L

  1. 3 over 7 square root of 2 g L end root
  2. 7 over 12 square root of 2 g L end root
  3. 1 over 12 square root of g L end root
  4. 2 over 7 square root of 2 g L end root

The correct answer is: 7 over 12 square root of 2 g L end root


    A angular momentum of system of rod and insect, just after collision = initial angular momentum of insect about
    therefore open curly brackets fraction numerator M calligraphic Z squared over denominator 12 end fraction plus M open parentheses L over 4 close parentheses squared close curly brackets omega equals M v L over 4
omega equals fraction numerator 12 V over denominator 7 L end fraction
L equals I omega equals open parentheses fraction numerator M I squared over denominator 12 end fraction plus M x squared close parentheses omega

L e t space t h e space i n s e c t space f a l l space o n space t h e space r o d space a t space t i m e space t space equals space 0 comma space t h e n space a t space t i m e space t comma space i n c l i n a t i o n space o f space r o d space w i t h space h o r i z o n t a l comma space
theta equals omega t

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