Question

A circular tube of mass $M$ is placed vertically on a horizontal surface as shown in the figure. Two small spheres, each of mass $m$, just fit in the tube are released from the top, as shown in the figure. If $\theta$ gives the angle between radius vector of either ball with the vertical, obtain the value of the ratio $m/M$ for which the tube breaks its contact with ground when $\theta ={60}^{\circ }$ (Ignore any friction)

Answer 1

Using conservation of energy principle, if $v$ be the speed of either ball when its radius vector makes angle $\theta$ with vertically upwards direction.
$mgR[1-\cos \theta ]=\frac{1}{2}m{v}^2$
$\Rightarrow \frac{m{v}^2}{R}=2mg[1-\cos \theta ]$
Free body diagram of ball:
$N$ is the normal reaction by tube walls on ball $B$.
$N^{\prime }$ be the normal recation force by ground on tube.
From F.B.D. (ii);
$N=mg\cos \theta -\frac{m{v}^2}{R}$
$=mg\cos \theta -2mg[1-\cos \theta ]$
From F.B.D. (ii),
$N^{\prime }=2N\cos \theta +Mg$
At the instant tube breaks its contact with ground $N^{\prime }=0$
$\Rightarrow Mg+[mg\cos \theta -2mg(1-cos\theta \left)\right]2\cos \theta =0$
for $\theta ={60}^{\circ }$, we get $m/M=2$.