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A thin **rod of length $L$, mass $M$** is hinged at one end and held horizontally; released from rest. The **angular speed** when it has rotated through angle $\theta$ below horizontal is:
A$\sqrt{2g\sin\theta/L}$
B$\sqrt{3g\sin\theta/L}$
C$\sqrt{g\sin\theta/L}$
D$\sqrt{6g\sin\theta/L}$
Answer & Solution
Correct answer: B. $\sqrt{3g\sin\theta/L}$
CG drops by $(L/2)\sin\theta$. Energy: $Mg(L/2)\sin\theta = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}(ML^2/3)\omega^2$ ⇒ $\omega^2 = 3g\sin\theta/L$.
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