Question

A weightless rod of length $2l$ carries two equal masses ${}^{\prime }{m}^{\prime }$, one second at lower end $A$ and the other at the middle of the rod at $B$. The rod can rotate in vertical plane about a fixed horizontal axis passing through $C$. What horizontal velocity must be imparted to the mass at $A$ so that it just completes the vertical circle.

• A. $\sqrt{\frac{48}{5}gl}$
• B. $\sqrt{\frac{45}{5}gl}$
• C. $\sqrt{\frac{40}{5}gl}$
• D. $\sqrt{\frac{35}{5}gl}$

Answer 1

The correct option is A $\sqrt{\frac{48}{5}gl}$

From Law of Conservation of energy potential energy = Kinetic Energy

$\Rightarrow 4mgl+2mgl=\frac{m{v}_B^2}{2}+\frac{m{v}_A^2}{2}$

The angular velocity at point $A$ and $B$ are same

$\therefore \frac{V_a}{2L}=\frac{V_b}{L}\Rightarrow V_s=2V_b$

$\Rightarrow 6mgl=\frac{1}{2}m{v}_a^2+\frac{1}{2}m\frac{V_a^2}{4}$

$V_a=4\sqrt{\frac{3}{5}Lg}=\sqrt{\frac{48}{5}Lg}$