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A bullet of mass $10 \mathrm{~g}$ and speed $500 \mathrm{~m} / \mathrm{s}$ is fired into a door and gets embedded exactly at the centre of the door. The door is $1 \mathrm{~m}$ wide and weighs $12 \mathrm{~kg}$. If 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 M.I. of the door about the vertical axis at one end is $\mathbf{M L}^2 / 3$ )
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Verified Answer
Angular momentum imparted by the bullet to the door $=L$
$$
\begin{aligned}
&=m r=10 \times 10^{-3} \times 500 \times \frac{1}{2}=2.5 \mathrm{kgm}^2 / \mathrm{s} \\
&\text { M.I. }=I=\frac{\mathrm{ML}^2}{3}=\frac{12 \times 1}{3}=4 \mathrm{kgm}^2 .
\end{aligned}
$$
Also, $L=\mathrm{I} \omega \therefore \omega=\frac{\mathrm{L}}{\mathrm{I}}=\frac{2.5}{4}=0.625 \mathrm{rad} / \mathrm{s}$.
$$
\begin{aligned}
&=m r=10 \times 10^{-3} \times 500 \times \frac{1}{2}=2.5 \mathrm{kgm}^2 / \mathrm{s} \\
&\text { M.I. }=I=\frac{\mathrm{ML}^2}{3}=\frac{12 \times 1}{3}=4 \mathrm{kgm}^2 .
\end{aligned}
$$
Also, $L=\mathrm{I} \omega \therefore \omega=\frac{\mathrm{L}}{\mathrm{I}}=\frac{2.5}{4}=0.625 \mathrm{rad} / \mathrm{s}$.
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