Question

A uniform rod of mass $M$ and length $a$ lies on a smooth horizontal plane. A particle of mass $m$ moving at a speed $v$ perpendicular to the length of the rod strikes it at a distance $a/4$ from the centre and stops after the collision. Find
a. The velocity of the centre of the rod and
b. the angular velocity of the rod about its centre just after the collision.

Answer 1

R.F.F image

First we converse linear momentum

initial linear momentum = final linear momentum

mv = MV

$v=\frac{mv}{M}$

$y_{com}=\frac{-ma}{4(m+M)}$

Now we converse angles momentum about com of system,

Initial angles momentum = final angular momentum

$mv(\frac{a}{4}-|y_{com}|)=\frac{M{l}^2}{12}w-MV$

$MV\left(\frac{Ma}{4(m+M)}\right)=\frac{M{l}^2}{12}w-MV(-y_{com})$

$\frac{m{l}^2}{12}w=\frac{m{v}^a}{4}\left(\frac{M-m}{M+m}\right)$

$w=\frac{3aV}{l^2}\left(\frac{M-m}{M+m}\right)\frac{m}{M}$