Rotation Ques 12
12 A rod of length $50 \mathrm{~cm}$ is pivoted at one end. It is raised such that if makes an angle of $30^{\circ}$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s ${ }^{-1}$ ) will be
(Take, $g=10 \mathrm{~ms}^{-2}$ )
(2019 Main, 09 Jan II)
(a) $\frac{\sqrt{30}}{2}$
(b) $\sqrt{30}$
(c) $\frac{\sqrt{20}}{3}$
(d) $\sqrt{\frac{30}{2}}$
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Answer:
Correct Answer: 12.( b )
Solution:
$ \therefore \triangle P E=\frac{1}{2} I \omega^2 $
(where, $I=\mathrm{MOI}$ of rod and $\omega=$ angular frequency of rod)
$ \begin{array}{cc} \Rightarrow & M g \times \frac{L}{2} \sin 30^{\circ}=\frac{1}{2} \times I \times \omega^2 \\ \Rightarrow & M g \frac{L}{2} \times \frac{1}{2}=\frac{1}{2} \times I \times \omega^2 \Rightarrow \frac{M g L \times 2}{4 \times I}=\omega^2 \\ \Rightarrow \quad & \sqrt{\frac{M g L}{4 \times \frac{M L^2}{3}} \times \frac{2}{1}}=\omega \\ \omega=\sqrt{30} \mathrm{rad} / \mathrm{s} \end{array} $
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