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 ${37}^{\circ }$ with the length. Immediately after impact

• A. Angular momentum of the rod is $0.6lJ$
• B. Angular velocity of the rod is $\frac{1.8J}{ml}$
• C. Kinetic energy of the rod is $\frac{1.5J}{m}$
• D. Linear velocity of the center of mass of the rod is $\frac{2J}{5m}$

Answer 1

Answer: (A), (B)

The correct options are
A Angular momentum of the rod is $0.6lJ$
B Angular velocity of the rod is $\frac{1.8J}{ml}$

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