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Question

A uniform rod of length l=3 m free to rotate in a vertical plane about a fixed horizontal axis passing through O. The rod is allowed to rotate by releasing from it's vertical position. Find the angular velocity of the rod when rod makes an angle θ=60 with vertical. (take g=10 m/s2)


A
5 rad/sec
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B
5 rad/sec
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C
52 rad/sec
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D
52 rad/sec
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Solution

The correct option is B 5 rad/sec
Let the rod having mass m kg and moment of inertia I about the one end O.

When rod rotates through an angle θ, it's centre of mass (c) falls through the distance (h).
So, Loss in potential energy ΔPE:
ΔPE=mgh
h=l2(1cos θ)
ΔPE=mgl2(1cos θ) (1)
Rod is hinged, hence constrained to rotate only.

Gain in rotational kinetic energy of rod
KERot=(12Iω2)
By applying mechanical energy conservation:
Loss in PE=Gain in KERot
(KE)rotational=12×(13ml2)ω2 (2) [I=ml23]
mgl2(1cos θ)=16ml2ω2

mgl2×2 sin2 θ2=16ml2ω2
ω=6gl×sin2θ2
Substitute the values of l=3 m and θ=60
g=10 m/s2
ω=6×103 sin 30
ω=20×12=5 rad/sec

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