__Question 7.24__:
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

(Hint: The moment of inertia of the door about the vertical axis at one end is

*ML*^{2}/3.)

__Solution__**:**

*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 X 10

^{-3}) X (500) X (1/2) = 2.5 kg m

^{2}s

^{-1}...(i)

Moment of inertia of the door:

I = ML

^{2}/ 3

= (1/3) X 12 X 1

^{2}= 4 kgm

^{2}

^{}

^{}But α = Iω

∴ ω = α / I

= 2.5 / 4

= 0.625 rad s

^{-1}

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