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Question
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 = 1/2 m`
Mass of the door, M = 12 kg
The Angular momentum imparted by the bullet on the door:
α = mvr
`=(10xx10^(-3))xx(500)xx1/2 = 2.5 kg m^2s^(-1)` ...(i)
Moment of inertia of the door:
`I = 1/3 ML^(2)`
`= 1/3 xx 12 xx (1)^2 = 4 kgm^2`
But `alpha = Iomega`
`:.omega = alpha/I`
`= 2.5/4 = 0.625 rad S^(-1)`
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I1 = M.I. of thin circular ring about its diameter,
I2 = M.I. of circular disc about an axis perpendicular to disc and going through the centre,
I3 = M.I. of solid cylinder about its axis and
I4 = M.I. of solid sphere about its diameter.
Then -
