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Home » A bullet of mass and speed is fired into a door and gets embedded exactly at the centre of the door. The door is wide and weighs . 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.
A bullet of mass 10 \mathrm{~g} and speed \mathbf{5 0 0} \mathrm{m} / \mathrm{s} is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 \mathrm{~m} wide and weighs 12 \mathrm{~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.
CBSE | Class 11 | NCERT | Physics | System of Particles and Rotational Motion
A bullet of mass 10 \mathrm{~g} and speed \mathbf{5 0 0} \mathrm{m} / \mathrm{s} is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 \mathrm{~m} wide and weighs 12 \mathrm{~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.

Velocity is given as v = 500 m/s

Mass of bullet is given as m = 10 g or 10 × 10–3 kg

The width of the door is given as L = 1 m
The radius of the door is given as r = 1 / 2
Mass of the door is given as M = 10 kg

Angular momentum imparted by the bullet on the door can be calculated as,
α = mvr
= (10× 10-3) × (500) × (1/2)  =  2.5 kg m2 s-1    …(i)

Also, Moment of inertia of the door :
I = ML2 / 3
= (1/3) × 12× 12 = 4 kgm2
We know, α = Iω
∴ ω = α / I
= 2.5/ 4 = 0.625  rad /s

i

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