System of particles and Rotational motion NCERT Solutions Class 11 Physics - Solved Exercise Question 7.24

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 ML2/3.)
Solution:
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 X 10-3 ) X (500) X (1/2)  =  2.5 kg m2 s-1    ...(i)
Moment of inertia of the door:
I = ML2 / 3
= (1/3) X 12 X 12 = 4 kgm2
But α = Iω
ω = α / I
= 2.5 / 4
= 0.625 rad s-1

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