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  • A metre stick is held vertically with one end on a rough horizontal floor, If it is gently displaced from its position such that the lower end does not slip, then the angular velocity of the rod at

A metre stick is held vertically with one end on a rough horizontal floor, If it is gently displaced from its position such that the lower end does not slip, then the angular velocity of the rod at the instant when its angular displacement from the vertical is 60^{\circ}.

Option: 1

\sqrt{\frac{15}{2}} \mathrm{rad} / \mathrm{s}


Option: 2

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


Option: 3

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


Option: 4

\frac{5 \sqrt{3}}{2} \mathrm{rad} / \mathrm{s}


Answers (1)

best_answer

Taking torque about point A:

\tau=m g\left(\frac{l}{2} \cdot \sin \theta\right)

using, \tau=I \alpha

\operatorname{mg} \frac{l}{2} \sin \theta=\frac{1}{3} \operatorname{m} l^2 \alpha
\begin{aligned} & \alpha=\frac{3 g}{2 L} \sin \theta \\ & \omega \frac{d \omega}{d \theta}=\frac{3 g}{2 L} \sin \theta \end{aligned}

Integrating both sides

\begin{aligned} & \int_0^\omega d \omega=\left(\frac{3 g}{2 L}\right) \int_0^{\pi/3} \sin \theta d \theta \\ & \frac{1}{2} \omega^2=\frac{3 g}{2 L}[-\cos \theta]_0^{\pi / 3} \\ & \frac{1}{2} \omega^2=-\frac{3 g}{2 L}\left[\frac{1}{2}-1\right]=\frac{3 g}{4 L} \\ & \omega^2=\frac{3 g}{2 L} \\ & \omega=\sqrt{\frac{3 g}{2 L}}=\sqrt{\frac{3 \times 10}{3}}=\sqrt{15} \end{aligned}

 

Posted by

avinash.dongre

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