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A uniform rod of length '10 \mathrm{~cm}' and mass ' \1 ~\mathrm{kg}' is free to rotate in a vertical plane about one of its end. The rod is kept in horizontal position and then released, then its angular acceleration just after the release is given by _____

Option: 1

150 ~\mathrm{rad} / \mathrm{s}^2


Option: 2

50~\mathrm{rad} / \mathrm{s}^2


Option: 3

100~\mathrm{rad} / \mathrm{s}^2


Option: 4

200~\mathrm{rad} / \mathrm{s}^2


Answers (1)

best_answer

\begin{aligned} & \tau=m g\left(\frac{l}{2}\right) \\ & \alpha\left(\frac{1}{3} m l^{{2}}\right)=\frac{m g l}{2} \\ & \alpha=\frac{3 g}{2 l}=\frac{3 \times 10}{2 \times 0.1}=150 ~\mathrm{rad} / \mathrm{s}^2 \end{aligned}

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Sanket Gandhi

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