From conservation of mechanical energy

Loss in gravitational PE= gain in KE
mg2lsin30o=21Iω2
mg2lsin30o=213ml2ω2
ω=2l3g=2×50×10−23×10=30rad/s
A rod of length 50cm is pivoted at one end. It is raised such that it makes an angle of 30o from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rads−1 ) will be (g=10ms−2)

Held on 9 Jan 2019 · Verified 6 Jul 2026.
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