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.