A rod of mass m and length l is made to stand at an angle of $60^{o}$ with the vertical. Potential energy of the rod in this position is?

mgl

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\frac{m g l}{2}

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\frac{m g l}{3}

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\frac{m g l}{4}

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Solution

The correct option is D $\frac{m g l}{4}$
$x = \frac{l}{2} cos 60 °$
As shown in the figure
$= \frac{l}{4}$
Potential energy $= m g x$
$= \frac{m g l}{4}$

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A rod of mass m and length l is made to stand at an angle of 60o with the vertical. Potential energy of the rod in this position is?

The correct option is D mgl4x=l2cos60° As shown in the figure =l4 Potential energy =mgx=mgl4

A rod of mass m and length l is made to stand at an angle of $60^{o}$ with the vertical. Potential energy of the rod in this position is?

The correct option is D $\frac{m g l}{4}$
$x = \frac{l}{2} cos 60 °$
As shown in the figure
$= \frac{l}{4}$
Potential energy $= m g x$
$= \frac{m g l}{4}$