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Centripetal Acceleration

A rod of mass...

Question

A rod of mass M & length L is kept at a angle $60^{o}$ with vartical initially & then released find angular acceleration of rod ?

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Solution

$\begin{matrix} m g (\frac{\sqrt{3}}{2} . \frac{1}{2}) = \frac{m l^{2}}{3} \times α \Rightarrow \frac{3 \sqrt{3} g}{4 l} = α α = \frac{3 \sqrt{3} g}{4 l} \end{matrix}$

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