Question

A weightless rod of length $l$ carries tow equal masses $m$, one secured at the end and the other in the middle of the rod. The rod can revolve in a vertical plane around the point A. What horizontal velocity must be imparted to the end of the rod C to deflect it to the horizontal position? (Given acceleration due to gravity $=g$)

• A. $\sqrt{2gl}$
• B. $\sqrt{2.4gl}$
• C. $\sqrt{3gl}$
• D. $\sqrt{3.6gl}$

Answer 1

Answer: (B)

$\sqrt{2.4gl}$

when string will be horizontal the heights of masses will increased by $l$ and $\frac{l}{2}$.

both masses will have same angular velocity therefore mass at the middle of rope will have half the velocity of mass at the end.

conserving energy of system.

$\frac{1}{2}m{v}^2+\frac{1}{2}m(\frac{v}{2}{)}^2=mgl+mg\frac{l}{2}\Rightarrow \frac{5}{8}{v}^2=\frac{3}{2}gl\Rightarrow v=\sqrt{2.4gl}$