Question

Two identical particles $B$ and $C$ each of mass $50g$ are connected by a light rod of length $30CM$. Another particle $A$ of same mass moving with a speed $u=60CM/s$ strikes $B$, in a direction perpendicular to $AB$, and sticks to it. The whole process takes place on a smooth horizontal plane. Find the angular velocity $\omega$ of the system about its centre of mass, immediately after the impact.

Answer 1

R.E.F image

final com of system $=\frac{m\left(\frac{l}{2}\right)+m\left(\frac{l}{2}\right)+m\left(\frac{-l}{2}\right)}{3m}$

$=\frac{l}{6}$ above center a rod

com serving angular momentum about point at final com

mu $(\frac{l}{2}-\frac{l}{6})+m\left(0\right)(\frac{l}{2}-\frac{l}{6})+m\left(0\right)(\frac{l}{6}+\frac{l}{2})$

$=IW\left(1\right)$

$I=m(\frac{l}{2}-\frac{l}{6}{)}^2+m(\frac{l}{2}-\frac{l}{6}{)}^2+m(\frac{l}{2}+\frac{l}{6}{)}^2$

$=m\frac{l^2}{9}+m\frac{l^2}{9}+\frac{4l^2}{9}$

$=\frac{2}{3}m{l}^2\left(2\right)$

from (1) and (2)

$mu\frac{l}{3}=\frac{2}{3}m{l}^{2}w$

$w=\frac{u}{2l}$