Question

a door 1.00m wide, of mass 15kg can rotate freely about a vertical axis through its hinges. A bullet with a mass of 10 g and a speed of 400 m/s strikes the center of the door, in a direction perpendicular to the plane of the door, and embeds itself there. Find the door's angular speed. Is kinetic energy conserved? physics final

Answer 1

Answer:

This question contains collision of rotational objects, so we should use angular momentum.

$L = \omega I$ or $L = r \times mv$

By conversation of angular momentum;

$L_1 = L_2$

$|L_1| = rmv\sin(\theta) = rmv\sin(\pi/2) = rmv$ since the bullet hits the door in a direction perpendicular to the plane of the door.

$r = 0.5m$, the distance from the center of the door to the hinges.

$L_1 = (0.5)(0.01)(400) = 2 kg*m^2/s$

After the collision the bullet embeds itself to the door. Now the door and the bullet is a combined object with mass 15,01 kg.

$I = I_{door} + I_{bullet} = \frac{1}{3}Ma^2 + mr^2$

where $a$ is the width of the door, and the bullet can be considered as a point particle.

$I = \frac{1}{3}15(1)^2 + (0.01)(0.5)^2 = 5 + 0.0025 = 5.0025 kg * m^2$

$L_1 = L_2\\2 = 5.0025 \omega\\\omega = 0.399 rad/s$

Explanation:

The calculation of moment of inertia of a rectangular door rotating around its hinges is a bit long. But the answer is in every standard textbook. It is safe to use it without deriving it ourselves.