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VITEEE 2009 Paper

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Question : 39 of 120

Marks: +1, -0

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

$\sqrt{\frac{g}{l}}$
$\sqrt{3gl}$
$3\sqrt{\frac{g}{l}}$
$\sqrt{\frac{3g}{l}}$

Solution:

Here, potential energy of the metre stick will be converted into rotational kinetic energy.

Because centre of gravity of stick lies at the middle of the rod,

PE of metre stick =

rac{mg l}{2}

Rotational kinetic energy

E=\frac{1}{2}I\omega^{2}

I

about point

A=\frac{m l^{2}}{3}

By law of conservation of energy

mg\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}

By solving, we get

v_{B}=\sqrt{3gl}

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