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 and rotates about a vertical axis practically without friction. Find the angular speed of the door just after the moment bullet embeds into it. (Hint: The moment of inertia of the door about the vertical axis at one end is ML²/3).
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◆ Answer-
ω = 0.625 rad/s
◆ Explaination-
# Given-
m = 10 g = 10 × 10–3 kg
v = 500 m/s
L = 1 m
M = 12 kg
# Solution-
Angular momentum provided by the bullet on the door-
α = mvr
α = (10×10^-3)×(500)×(1/2)
α = 2.5 kgm^2/s
Moment of inertia of the door-
I = ML^2/3
I = (1/3)×12×(1)^2
I = 4 kgm^2
But α = Iω
∴ ω = α / I
ω = 2.5 / 4
ω = 0.625 rad/s
Therefore, angular speed of the door is 0.625 rad/s.
Hope that is useful...
◆ Answer-
ω = 0.625 rad/s
◆ Explaination-
# Given-
m = 10 g = 10 × 10–3 kg
v = 500 m/s
L = 1 m
M = 12 kg
# Solution-
Angular momentum provided by the bullet on the door-
α = mvr
α = (10×10^-3)×(500)×(1/2)
α = 2.5 kgm^2/s
Moment of inertia of the door-
I = ML^2/3
I = (1/3)×12×(1)^2
I = 4 kgm^2
But α = Iω
∴ ω = α / I
ω = 2.5 / 4
ω = 0.625 rad/s
Therefore, angular speed of the door is 0.625 rad/s.
Hope that is useful...
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