Question 34: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

Let the door be a uniform rectangular plate of width $L=1.0$ m and mass $M=12$ kg, hinged along one side. Its moment of inertia about the hinge is
$$I_{\rm door}=\frac{1}{3}ML^2=\frac{1}{3}\,(12)(1.0)^2=4.0\;\mathrm{kg\cdot m^2}.$$
The bullet of momentum $p$ embeds at the centre ($r=L/2=0.5$ m), so its initial angular momentum about the hinge is
$$L_{\rm in}=p\,r=p\,(0.5)\,. $$
By conservation of angular momentum (neglecting the bulletÃ¢ÂÂs own $I$ as $m\ll M$):
$$L_{\rm in}=I_{\rm door}\,\omega   \quad\Longrightarrow\quad   \omega=\frac{p\,(0.5)}{4.0}=\frac{p}{8}\,. $$
Answer: Option C, $\displaystyle \frac{p}{8}$.

Answer

$\frac{p}{4}$

Answer

$\frac{3 p}{5}$

Answer

$\frac{p}{8}$

Answer

$\frac{3}{8 p}$

Centre of Mass and Rotational Motion

A bullet of momentum p is fired into a door and gets embedded exactly at the center of the door. The door is 1.0 m wide and weighs 12 kg . It is hinged at one end and rotates about a vertical axis practically without friction. The angular speed of the door just after the bullet embeds into it is :

Choose the correct answer: