Question

A rod of length $R\sqrt{3}$ is kept vertically inside a fixed smooth spherical shell of radius R such that its both ends are contact with shell.If the rod is released then the angular speed of the rod, when it becomes horizontal will be

• A. $\sqrt{\frac{2g}{R}}$
• B. $\sqrt{\frac{3g}{R}}$
• C. $\sqrt{\frac{4g}{R}}$
• D. None of these

Answer 1

The correct option is C $\sqrt{\frac{4g}{R}}$

$\begin{array}{c}\text{Note that,}AB\text{is the rod in initial position,}\\ \text{while}BC\text{is the rod in final position.}\\ \text{Given,}AB=BC=\sqrt{3}R\\ \text{Here,}\theta ={60}^{\circ }\end{array}$

$\begin{array}{c}\text{By work-Energy theorem:}\\ \begin{array}{cc}{W}_{mg}=& \Delta K\cdot E\\ \Rightarrow mg\times \frac{R}{2}=& \frac{1}{2}{I}_{rod}\cdot {\omega }^2\\ & =\frac{1}{2}\frac{m(\sqrt{3}R{)}^2}{12}\cdot {\omega }^2\end{array}\end{array}$

$\begin{array}{c}\text{There}m\text{is mass of rod,}\omega \text{is the required final}\\ \text{angular velocity of rod.}\\ \text{Thus,}\omega =\sqrt{\frac{4g}{R}}\\ \text{So, correct option is (C).}\end{array}$