A bullet of mass 10 g and speed 500 m/s is fired into a door and gets embedded exactly at the centre 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. Find the angular speed of the door just after the bullet embeds into it.

Given:

Mass of Bullet, m = 10g = 10^-3 kg

Speed of bullet, v = 500m/s

Width of the door, w = 1m

Distance from hinge where bullet hits, r = 0.5 m

Mass of door, M = 12 Kg

We know that angular moment is given by,

α = mvr

where,

α = angular momentum

m = mass of the body

v = velocity of the body

r = radius of the body

by putting the value, we get,

⇒ α = 10 × 10^-3 kg × 500 ms^-1 × 0.5 m

⇒ α = 2.5 Kgm²s^-1 …(i)

Moment of inertia is given by,

I = ML²/3 (For rectangle)

Where,

M = mass of door

L = width of door

I = 1/3 × 12 Kg × (1m)²

I = 4 Kgm²

We also have,

α = Iω

where is α is the angular momentum

l is the linear momentum

ω is the angular velocity

⇒ ω = α /I

By putting the values, we get,

⇒ ω = 2.5/ 4

⇒ ω = 0.625 rad s^-1

Note: It is a good practice to memorise the moment of inertia for common shapes like rectangle, circle, triangle, cylinder and sphere.