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Asked in JEE Main 9th Jan 2nd Shift 2019 · Rotational energy and released bodies
The CM (at /L2) drops by /L2 sin 30°=/L4 as the rod goes from 30° to horizontal.
Energy: mg/L4=1/2((mL²)/3)ω²⇒ ω²=(3g)/(2L)=(3(10))/(2(0.5))=30.
ω=√30 rad/s.
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