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  • A uniform rod of mass '  ' and length ' <img alt="1 \mathrm{~m}

A uniform rod of mass ' 1 \mathrm{~kg} ' and length ' 1 \mathrm{~m} ' is pivoted at its lower end and kept vertically as shown in the figure. If rod is displaced from its position, find its angular velocity when it makes 60^{\circ} angle with the vertical.

Option: 1

5\sqrt{3} \mathrm{rad} / \mathrm{s}


Option: 2

3\sqrt{5} \mathrm{rad} / \mathrm{s}


Option: 3

\sqrt{30} \mathrm{rad} / \mathrm{s}


Option: 4

\sqrt{15} \mathrm{rad} / \mathrm{s}


Answers (1)

best_answer

Taking torque about point A:-

\begin{gathered} \operatorname{mg}\left(\frac{l}{2}\right) \sin \theta=\frac{1}{3} m l^2 \alpha \\ \alpha=\frac{3}{2} \cdot \frac{g}{l} \sin \theta \\ \omega \frac{d \omega}{d \theta}=\frac{3}{2} \cdot \frac{g}{l} \sin \theta \\ \int_0^\omega \omega d \omega=\frac{3}{2} \cdot \frac{g}{l} \int_{\theta=0}^{\theta=\pi/3} \sin \theta \cdot d \theta \end{gathered}

\begin{aligned} \frac{1}{2} \omega^2 & =\frac{3 g}{2 l}[-\cos \theta]_0^{\pi / 3} \\ \omega^2 & =-\frac{3 g}{l}\left[\frac{1}{2}-1\right]=\frac{3 g}{l} \times \frac{1}{2} \\ \omega^2 & =\frac{3 g}{2 l} \\ \omega & =\sqrt{\frac{3 g}{2 l}}=\sqrt{\frac{3 \times 10}{2 \times 1}} \\ \omega & =\sqrt{15} \mathrm{rad} / \mathrm{s} \end{aligned}

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Ritika Jonwal

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