Question

Two beads of mass 2m and m, connected by a rod of length l of negligible mass are free to move in a smooth vertical circular wire frame of radius l as shown. Initially the system is held in horizontal position (as shown in the figure). The velocity that should be given to mass 2m (when rod is in horizontal position) in counter-clockwise direction so that the rod can just become vertical is :

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The triangle formed will be equilateral as all these sides will have a length l. Also the speeds of both masses should be the same as their relative velocity along the rod should be zero.

$|\overrightarrow{v_1}|cos{30}^{\circ }=|\overrightarrow{v_2}|cos{30}^{\circ }$

$\Rightarrow |\overrightarrow{v_1}|=|\overrightarrow{v_2}|$

The speeds given to 2m will also be possessed by m

$h1=cos{30}^{\circ }-cos{60}^{\circ }$

=$l\left[\frac{\sqrt{3}-1}{2}\right]$

$h2=cos{30}^{\circ }+cos{60}^{\circ }$

=$l\left[\frac{\sqrt{3}-1}{2}\right]$

Work done (2mgh1 + mgh2) = $\frac{mgl}{2}(3\sqrt{3}-1)$

Work energy theorem

$\frac{1}{2}3m{v}^2=\frac{mgl}{2}(3\sqrt{3}-1)\Rightarrow v=\sqrt{\left(\frac{3\sqrt{3}-1}{3}\right)}gl$