Question

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

Answer 1

Due to hitting of the ball, the angular impulse received by the rod about the C.M. is equal to $p\frac{1}{2}$.
If $\omega$ is the angular velocity acquired by the rod, we have
$\frac{m{l}^2}{12}\omega =\frac{pl}{2}$ or $\omega =\frac{6p}{ml}$...... (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 $\times$ acceleration of C.M. of that part, in the frame of C.M.
$=\frac{m}{2}\left({\omega }^2\frac{l}{4}\right)=m\frac{{\omega }^{2}l}{8}=\frac{9p^2}{2ml}=9N$.