Question

A thin rod of mass $m$ and length $l$ is free to rotate on a smooth horizontal plane about its one fixed end. When it is at rest, it receives a horizontal impulse $J$ at its other end, at angle of ${37}^o$ with the length. Immediately after impact,

• A. Angular momentum of the rod is $0.6Jl$
• B. Angular velocity of the rod is $\frac{1.8J}{ml}$
• C. Kinetic energy of the rod is $\frac{0.54J^2}{m}$
• D. Linear velocity of the centre of mass of the rod is $\frac{9}{10}\frac{J}{m}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Angular momentum of the rod is $0.6Jl$
B Angular velocity of the rod is $\frac{1.8J}{ml}$
C Kinetic energy of the rod is $\frac{0.54J^2}{m}$
D Linear velocity of the centre of mass of the rod is $\frac{9}{10}\frac{J}{m}$

Perpendicular component of Impulse on the rod is $J\sin {37}^0=J\times \frac{3}{5}=0.6J$
Angular momentum of the rod about the fixed end : $L=0.6J\times l=0.6Jl$
MI of the rod about the fixed end: $I=\frac{m{l}^2}{3}$
Equating angular momentum L with $I\omega$
$0.6Jl=\frac{m{l}^2}{3}\omega$
$\therefore \omega =\frac{0.6\times 3J}{ml}=\frac{1.8J}{ml}$
Kinetic energy: $K=\frac{1}{2}I{\omega }^2=\frac{1}{2}\frac{m{l}^2}{3}(\frac{1.8J}{ml}{)}^2=0.54\frac{J^2}{m}$
Linear velocity of cm of the rod, $v_{cm}=\frac{l}{2}\times 1.8\times \frac{J}{ml}=0.9\frac{J}{m}=\frac{9}{10}\frac{J}{m}$