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Systems of Particles and Rotational Motion

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Question : 24 of 33
Marks: +1, -0
A bullet of mass 10 g and speed 500 m s−1500 \text{ m s}^{-1} 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 ML23\frac{ML^2}{3}.)
Solution:  
Here, Mass of the bullet, m=10 g=0.01 kgm = 10\,\mathrm{g} = 0.01\,\mathrm{kg}
Velocity of bullet,v=500 m s−1v = 500 \text{ m s}^{-1}
Width of door = 1 m
The distance from the axis, where the bullet gets embedded in the door,
r=12=0.5 mr = \frac{1}{2} = 0.5 \,\mathrm{m}
Mass of the door =12 kg= 12\,\mathrm{kg}
Angular velocity of door, ω=?\omega = ?
Angular momentum imparted by the bullet
L=mv×rL = m v \times r =0.01×500×0.5=2.5= 0.01 \times 500 \times 0.5 = 2.5
I=ML33I = \frac{M L^3}{3}
=12×1.023= \frac{12 \times 1.0^2}{3}
=4 kg m2∵L=Iω= 4\,\mathrm{kg}\,\mathrm{m}^2 \because L = I \omega
∴ω=LI=2.54=0.625 rad s−1\therefore \omega = \frac{L}{I} = \frac{2.5}{4} = 0.625 \,\mathrm{rad\,s}^{-1}
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