Question

A thin rod of length $L$ is vertically straight on horizontal floor. This rod falls freely to one side without slipping at its bottom. The linear velocity of centre of rod when its top end touches floor is : (Given acceleration due to gravity $=g$)

• A. $\sqrt{2gL}$
• B. $\sqrt{\frac{3gL}{2}}$
• C. $\sqrt{3gL}$
• D. $\sqrt{\frac{3gL}{7}}$

Answer 1

The correct option is D $\sqrt{\frac{3gL}{7}}$

Work Done by gravity $=\frac{mgL}{2}$

Angular velocity $\omega =\frac{2v}{L}$

$KE=\frac{m{v}^2}{2}+\frac{1}{2}\times \frac{m{L}^2}{3}\times {\omega }^2$

Using Energy Balance

$\frac{mgL}{2}=\frac{m{v}^2}{2}+\frac{1}{6}m{L}^2\times \frac{4v^2}{L^2}$

$mg\frac{L}{2}=m{v}^2(\frac{1}{2}+\frac{2}{3})$

$\sqrt{\frac{3gL}{7}}=v$