Question

A uniform rod of mass $M=4kg$ and length $L=3m$ lies on a frictionless horizontal plane. A particle of same mass $M$ moving with speed $v_0=10m/s$ perpendicular to the length of the rod strikes the end of the rod and sticks to it. Find the angular velocity (in $\text{rad}/s$) about the center of mass of (rod + particle) system

Answer 1

Draw a free body diagram of given problem

Calculate the position of the center of mass

Given,
Mass $M=4kg$
Length $L=3m$
Speed $v_0=10m/s$
Therefore, according to question, mass of particle is comparable or equal to rod, hence distance of centre of mass of system from $A$ is $\frac{L}{2}$ and , location of centre of mass after of collision,
$x_{cm}=\frac{M\times 0+M\times \frac{L}{2}}{M+M}=\frac{L}{4}$

Calculate moment of inertia about the center of mass

After that, moment of inertia of system about $C$
$I_c=M{x}_{c{m}^2}+[\frac{M{L}^2}{12}+M{(\frac{L}{2}-x_{cm})}^2]$
$=M{\left(\frac{L}{4}\right)}^2+[\frac{M{L}^2}{12}+M{\left(\frac{L}{4}\right)}^2]$
$=\frac{5}{24}M{L}^2$

Calculate the angular velocity about the center of mass

As we know, after collision velocity of center of mass and angular velocity about center of mass is $\omega$.
It means, momentum is conserved
$M{v}_0=(M+M)v_{cm}$
$v_{cm}=\frac{v_0}{2}$
Momentum is conserved about c, as c is centre of mass
$M{v}_0{x}_{cm}=I_c\omega$
$M{v}_0\frac{L}{4}=\frac{5}{24}M{L}^2\omega$
$\omega =\frac{6v_0}{5L}=\frac{6\times 10}{5\times 3}$
$=4\text{rad}/s$