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A rod of length $l$ is held vertically stationary with its lower end located at a point $P$, on the horizontal plane. When the rod is released to topple about $P$, the velocity of the upper end of the rod with which it hits the ground is
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Verified Answer
The correct answer is:
$\sqrt{3 g l}$
In this process potential energy of the metre stick will be converted into rotational kinetic energy. PE of metre stick $=\frac{m g l}{2}$
Because its centre of gravity lies at the middle of the rod.
Rotational kinetic energy $E=\frac{1}{2} I \omega^2$
$I=$ moment of inertia of metre stick about point
$A=\frac{m l^2}{3} \text {. }$
By the law of conservation of energy
$\begin{aligned}
m g\left(\frac{l}{2}\right) & =\frac{1}{2} I \omega^2 \\
& =\frac{1}{2} \frac{m l^2}{3}\left(\frac{v_B}{l}\right)^2
\end{aligned}$
By solving, we get $v_B=\sqrt{3 g l}$
Because its centre of gravity lies at the middle of the rod.
Rotational kinetic energy $E=\frac{1}{2} I \omega^2$
$I=$ moment of inertia of metre stick about point
$A=\frac{m l^2}{3} \text {. }$
By the law of conservation of energy
$\begin{aligned}
m g\left(\frac{l}{2}\right) & =\frac{1}{2} I \omega^2 \\
& =\frac{1}{2} \frac{m l^2}{3}\left(\frac{v_B}{l}\right)^2
\end{aligned}$
By solving, we get $v_B=\sqrt{3 g l}$
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