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.)

Asked by Abhisek | 1 year ago |  76

Solution :-

Given, Velocity,v= 500 m/s

Mass of bullet, m = 10 g = 10 × 10–3 kg

Width of the door, L = 1 m

Radius of the door, r = $$\dfrac{1}{2}$$

Mass of the door, M = 10 kg

Angular momentum imparted by the bullet on the door:

α = mvr

= (10× 10-3 ) × (500) × ($$\dfrac{1}{2}$$)

=  2.5 kg m2 s-1    …(i)

Now, Moment of inertia of the door :

I = ML2 / 3

= ($$\dfrac{1}{3}$$) × 12× 12 = 4 kgm2

We know, α = Iω

∴ ω = $$\dfrac{a}{I}$$

$$\dfrac{2.5}{4}$$ = 0.625  rad /s

Answered by Pragya Singh | 1 year ago

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