Question

A uniform rod $AB$ of length $2l$ and mass $2m$ is suspended freely at $A$ and hangs vertically at rest when a particle of mass $m$ is fired horizontally with speed $v$ to strike the rod at its midpoint. If the particle is brought to rest by the impact, find: (a) the impulsive reaction at $A$, (b) the initial angular speed of the rod, and (c) the maximum angle the rod makes with the vertical in the subsequent motion.
$[mv/4,3v/8l,{\cos }^{-1}(1-\frac{g/v^2}{32v_2})]$

Answer 1

A uniform AB length$=2l$

Mass$=2m$

$U=0$ (initial)

See the image

Just after impact

From conservation of angular momentum, just before and just after.

Impact about point O we have

$l_i=l_f\therefore m\times \frac{1}{2}=\left(\frac{m{l}^2}{3}\right)\omega \omega =\frac{3v}{2l}{v}_c=\frac{1}{2}\omega =\frac{3}{4}v$

Impulse J has changed the momentum of pariticle from $mv$ to $O$ Hence,

$J=mv$ for rod $J_H+Jm{v}_c=\frac{3}{4}mv$

$\therefore J_H=\frac{3}{4}mv-J=\frac{3}{4}mv-mv=\frac{mv}{4}or|J_H|=\frac{mv}{4}$

$\left(b\right)$The initial angular speed of the rod

$v=\omega l\omega =\frac{v}{l}$

$\left(c\right)$ Maximum angle of the rod makes with the vertical in the subsequent motion.

Potential energy in rod

$U=mg\frac{d}{2}(1-\cos \theta )\cos \theta =(1-\frac{g/v^2}{32v_2})$