A uniform rod of mass `m=5.0kg` and length `l=90cm` rests on a smooth horizontal surface. One of the ends of the rod is struck with the impulse `J=3.0N*s` in a horizontal direction perpendicular to the rod. As a result, the rod obtains the momentum `p=3.0N*s`. Find the force with which one half of the rod will act on the other in the process of motion.

Explanation

Due to hitting of the ball, the angular impulse received by the rod about the C.M. is equal to `p1/2`. If `omega` is the angular velocity acquired by the rod, we have
`(ml^2)/(12)omega=(pl)/(2)` or `omega=(6p)/(ml)` (1)
In the frame of C.M., the rod is rotating about an axis passing through its mid point with the angular velocity `omega`. Hence the force exerted by one half on the other =mass of one half x acceleration of C.M. of that part, in the frame of C.M.
`=m/2(omega^2l/4)=m(omega^2l)/(8)=(9p^2)/(2ml)=9N`