First slide

Potential energy

Question

, A uniform rod of mass m and length / is held inclined at an angle of $60^{0}$ with the vertical. What will-be potential energy of the rod in this position ?

Moderate

A

m g I
B

m g I/2
C

m g I/3
D

m g I /4

Solution

See fig. (12). Let the center of gravity of the rod be at a
height ft. From figure,
$h = OCsin 30^{\circ} = (l / 2) \sin 30^{\circ} = l / 4 P.E. = mgh = mgl / 4$

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