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Question

A uniform rod of mass m and length l is fixed at one end released from horizontal position as shown in the figure. The angular velocity of the rod when the rod makes an angle 60o with vertical is:
1036501_cf9f42e7f90849b6ba7366b789465f05.png

A
3g2l
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B
4g3l
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C
2gl
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D
3gl
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Solution

The correct option is A 3g2l

Since total mechanical energy is conserved

P.E.=mgl(sin2(θ2))

K.E.=12I0ω2

Moment of inertia of this road is

I0=ml23

Substituting this in the above equation we get

K.E.=ml2ω26

ml2ω26=mgl(sin2(θ2))

ω=6glsin2(30)

ω=6glsin(30)

ω=126gl

ω=3g2l


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