Question

A bullet of mass $10g$ and speed $500m/s$ is fired into a door and gets embedded exactly at the center of the door. The door is $1.0m$ wide and weighs $12kg$. It is hinged at one end and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the bullet embeds into it.

Answer 1

Angular momentum imparted by bullet on the door=$mvr$

$=(10\times {10}^{-3})\times 500\times 0.5kg{m}^2/s$

Moment of inertia of the door, $I=M{L}^2/3=\frac{1}{3}\times 12\times 1^2=4kg{m}^2$

Angular momentum of the system after the bullet gets embedded$\approx I\omega$

From conservation of angular momentum about the rotation axis,

$mvr=I\omega$
$⟹\omega =0.625rad/s$