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. If $\theta$ gives the angle between radius vector of either ball with the vertical, obtain the value of the ratio M/m if the tube breaks its contact with ground when $\theta ={60}^{\circ }.$ Neglect any friction.

• A. $1:2$
• B. $1:3$
• C. $2:1$
• D. $2:3$

Answer 1

Answer: (A)

$1:2$

Let the speed of sphere at $\theta ={60}^0$ be $v$.

By conservation of energy, $\frac{1}{2}m{v}^2=mg(R-Rcos{60}^0)=mgR/2$

$⟹m{v}^2=mgR$

Let the normal force by tube on a sphere be $N$, towards the center.

Balancing forces in the radial direction,

$N+mgcos{60}^0=\frac{m{v}^2}{R}=\frac{mgR}{R}=mg$

$⟹N=0.5mg-\left(1\right)$

Now, by newton's third law, equal normal force is applied by each sphere on the tube in the radial direction. Total vertical component of these forces is,

$N_{vertical}=2Ncos{60}^0=N$, in the upward direction

The tube will break contact with ground when normal force by ground on the tube is zero.

So, $N_{vertical}=Mg⟹0.5mg=Mg⟹M=0.5m$

$⟹M:m=0.5:1=1:2$