Question

A weightless rod of length $2l$ carries two equal masses ${}^{\prime }{m}^{\prime }$, one tied at lower end $A$ and the other at the middle of the rod at $B$. The rod can rotate in a vertical plane about a fixed horizontal axis passing through $C$. The rod is released from rest from the horizontal position. The speed of the mass $B$ at the instant the rod becomes vertical is:

• A. $\sqrt{\frac{3gl}{5}}$
• B. $\sqrt{\frac{4gl}{5}}$
• C. $\sqrt{\frac{6gl}{5}}$
• D. $\sqrt{\frac{7gl}{5}}$

Answer 1

The correct option is C $\sqrt{\frac{6gl}{5}}$

Since both masses are attached to the same rod, their angular velocity about hinged point $\left(C\right)$ will be same i.e $\omega$


The linear velocity of mass $A$ and $B$ are
$v_A=r_A\omega =2l\omega$
$v_B=r_B\omega =l\omega$
Let $v_A=v$. Then $v_B=2v$

Applying mechanical energy conservation:
$\text{Loss in}P.E\text{of A}+\text{Loss in}P.E\text{of B}=\text{Gain in}K.E\text{of A}+\text{Gain in}K.E\text{of B}$
$\Rightarrow mg\left(2l\right)+mgl=\frac{1}{2}m(2v{)}^2+\frac{1}{2}m{v}^2$
$\Rightarrow 3mgl=\frac{5m{v}^2}{2}$
$\therefore v=\sqrt{\frac{6gl}{5}}$