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System of Particles and Rotational Motion

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Physics Part I

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Physics

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
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.)

Given, 

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 half

      =  2.5 kg m
2 s-1                                   ...(i) 

Moment of inertia of the door is given by, 

I thin space equals space ML squared over 3

space space equals space open parentheses 1 third close parentheses space x space 12 space x space 1 squared space

space space equals space 4 space k g space m squared space

B u t comma space

alpha space equals space I space omega space

omega equals space alpha over I
space space space equals space fraction numerator 2.5 over denominator 4 end fraction space
space space space equals space 0.625 space r a d divided by s e c  

This is the required angular speed of the door just after the bullet embeds into it. 

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