A uniform rod of length ' $2 L$ ' is placed with one end in contact with the earth and is then inclined at an angle $\alpha$ to the horizontal and allowed to fall without slipping at contact point. When it becomes horizontal, its angular velocity will be

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Aicharya · Accepted Answer

By conservation of energy, Loss in PE = Gain in KE $\begin{aligned} & \Rightarrow m g(L \sin \mu) \frac{1}{2} I \omega^2 \\ & \Rightarrow m g L \sin \mu=\frac{1}{2}\left[\frac{1}{3} m(2 L)^2\right] \omega^2 \\ & \therefore \omega=\sqrt{\frac{3 g \sin \alpha}{2 L}} \end{aligned}$

A uniform rod of length ' $2 L$ ' is placed with one end in contact with the earth and is then inclined at…

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