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A uniform rod of mass $20 \mathrm{~kg}$ and length $1.6 \mathrm{~m}$ is pivoted at its end and swings freely in the vertical plane. Angular acceleration of rod just after the rod is released from rest in the horizontal position is:
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The correct answer is:
$\frac{15 g}{16}$
Just after the rod is released, Torque of weight $=\mathrm{I} \alpha$
$\begin{aligned}
& \mathrm{mg} \times \frac{1}{2}=\frac{\mathrm{ml}^2}{3} \times \alpha, \\
& \Rightarrow \alpha=\frac{3 \mathrm{~g}}{2 \mathrm{l}} \\
& \Rightarrow \alpha=\frac{3 \times \mathrm{g}}{2 \times 1.6} \\
& \Rightarrow \alpha=\frac{15 \mathrm{~g}}{16}
\end{aligned}$
$\begin{aligned}
& \mathrm{mg} \times \frac{1}{2}=\frac{\mathrm{ml}^2}{3} \times \alpha, \\
& \Rightarrow \alpha=\frac{3 \mathrm{~g}}{2 \mathrm{l}} \\
& \Rightarrow \alpha=\frac{3 \times \mathrm{g}}{2 \times 1.6} \\
& \Rightarrow \alpha=\frac{15 \mathrm{~g}}{16}
\end{aligned}$
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