Systems of Particles and Rotational Motion

Question : 24

Total: 33

A bullet of mass 10 g and speed 500ms–1 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.)

Solution:

Here, Mass of the bullet, m=10g=0.01‌kg
Velocity of bullet,v=500ms–1
Width of door = 1 m
The distance from the axis, where the bullet gets embedded in the door,
r=

=0.5m
Mass of the door =12kg
Angular velocity of door, ω=?
Angular momentum imparted by the bullet
L=mv×r =0.01×500×0.5=2.5
I=

ML3

12×1.02

=4kg m2∵L=Iω
∴ω=

2.5

=0.625‌rad‌ ‌s−1

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