Two uniform rods A and B of length 0.6 m each and of masses 0.01 kg and 0.02 kg respectively are rigidly joined end to end the combination is pivoted at the ligher end, P as shown in figure. Such that it can freely rotate about point P in a vertical plane. A small object of mass 0.05 kg, moving horizontally, hits the lower end of the combination and sticks to it. What should be the velocity of the object, so that the system could just be raised to the horizontal position?

Explanation

Moment of inertia of the system of combination of rod A and B about
`P = (0.01 (0.6)^(2))/(3) + ((0.02) (0.6)^(2))/(12) + 0.02 (0.9)^(2) = 0.018 kg - m^(2)`
Moment of inertia of the object about `P = (0.05) (1.2)^(2) = 0.072 kg-m^(2)`
Centre of mass of the system of (A + B + small object)
`= (0.01 xx 0.3 + 0.02 xx 0.9 + 0.05 xx 1.2)/(0.01 + 0.02 + 0.05) = (8.1)/(8)`m from the point P, and within the system.
Conserving angular momentum about P of (the rod + object) before and after collision, `(0.05) (1.2) (v) = (0.072 + 0.018) omega`, where `omega` is angular velocity of the system just after collision.
`rArr" "0.06 v = 0.09 omega rArr 2v = 3 omega`
Now, conservation of mechanical energy of the system, (after the collision) gives
`(1)/(2) (0.072 + 0.018) omega^(2) = 0.08 xx (8.1)/(8) xx 10`
`rArr" "omega = sqrt((0.08 xx 2 xx 8.1 xx 10)/(8 xx 0.09)) = 3 sqrt(2) rad//s`
Hence, `v = (3 omega)/(2) = (9)/(sqrt(2)) = 6.36 m//s`