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6.System of Particles and Rotational Motion
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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 . The angular speed of the door just after the bullet embeds into it will be
A
$6.25\, rad/sec$
B
$0.625\, rad/sec$
C
$3.35\, rad/sec$
D
$0.335\, rad/sec$
(JEE MAIN-2013)
Solution
Angular momentum imparted by the bullet
$L = mv \times r $$ = \left( {10 \times {{10}^{ – 3}}} \right) \times 500 \times \frac{1}{2} = 2.5$
$I = \frac{{M{L^2}}}{3} = \frac{{12 \times {{1.0}^2}}}{3} = 4kg{m^2}$
$L = I\omega $
$\therefore \omega = \frac{L}{I} = \frac{{2.5}}{4} = 0.625\,rad/\sec $
Standard 11
Physics
