Question

A uniform rod of maas $m$ and length $l$ rests on a smooth horizontal surface. One of the ends of the rod is struck in a horizontal direction at right angles to the rod. As a result the rod obtains velocity $v_0$ Find the force with which one-half of the rod will act on the other in the process of motion.

Answer 1

Let $J$ be the linear impulse applied at $B$ and $\omega$ the angular speed of rod.
$J=m{v}_0....\left(i\right)$
$J\left(\frac{l}{2}\right)=\frac{m{l}^2}{12}\omega .....\left(ii\right)$
Solving these two equations
$\omega =\frac{6v_0}{l}$
Linear speed of $D$ (midpoint of CB) relative to $C$
$v=\omega \left(\frac{l}{4}\right)=\frac{3}{2}{v}_0$
$\therefore$ Force exerted by upper half on the lower half,
$F=\frac{\left(\frac{m}{2}\right)v^2}{\left(\frac{l}{4}\right)}$
Substituting $v=\frac{3}{2}{v}_0$ we get
$F=\frac{9}{2}\frac{m{v}_0^2}{l}$