Question

A thin uniform rod of mass $m$ and length $l$ is free to rotate about its upper end. When it is at rest, its receives an impulse $J$ at its lowest point, normal to its length. Immediately after impact,

• A. the angular momentum of the rod is $Jl.$
• B. the angular velocity of the rod is $\frac{3J}{ml}$.
• C. the kinetic energy of the rod is $\frac{3J^2}{2m}$.
• D. the linear velocity of the midpoint of the rod is $\frac{3J}{2m}$.

Answer 1

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

The correct options are
A the angular momentum of the rod is $Jl.$
B the linear velocity of the midpoint of the rod is $\frac{3J}{2m}$.
C the angular velocity of the rod is $\frac{3J}{ml}$.
D the kinetic energy of the rod is $\frac{3J^2}{2m}$.

By definition angular momentum $L=Jl$
Now as $L=I\omega$
or
$\omega =\frac{L}{I}$
or
$\omega =\frac{Jl}{\frac{1}{3}m{l}^2}$
or
$\omega =\frac{3J}{ml}$
Now Kinetic energy
$KE=\frac{L^2}{2I}=\frac{3J^2}{2m}$
and
$v_c=\frac{\omega l}{2}=\frac{3J}{2m}$