Question

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.
(Hint: The moment of inertia of the door about the vertical axis at one end is ML2/3.)​

Answer 1

Answer:

Mass of the bullet, m = 10 g = 10 × 10–3 kg

Velocity of the bullet, v = 500 m/s

Thickness of the door, L = 1 m

Radius of the door, r = m / 2

Mass of the door, M = 12 kg

Angular momentum imparted by the bullet on the door:

α = mvr

= (10 × 10-3 ) × (500) × (1/2) = 2.5 kg m2 s-1 ...(i)

Moment of inertia of the door:

I = ML2 / 3

= (1/3) × 12 × 12 = 4 kgm2

But α = Iω

∴ ω = α / I

= 2.5 / 4

= 0.625 rad s-1

Answer 2

Angular momentum imparted by bullet on the door = mvr

= (10×10^-3)×500×0.05 = 2.5kgm²

Moment of inertia of the door, I = ML²/3 = 1/3×12×12 = 4kgm²

Angular momentum of the system after the bullet gets embedded ≈ Iω

From conservation of angular momentum about the rotation axis,

mvr=I•ω

ω = mvr/l

ω = 2.5/4

ω = 0.625rad/s