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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 $θ$ 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 $θ = 60 °$ . Neglect any friction.

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$\frac{3}{5}$

$\frac{5}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

answer is B.

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Detailed Solution

Speed of each particle at angle $θ$ is,

$v = \sqrt{2 gh}$ (from energy conservation)

where, $h = R (1 - cosθ)$

$∴ v = \sqrt{2 gR (1 - cosθ)}$

$∴ N + mgcosθ = \frac{{mv}^{2}}{R}$

or $N + mgcosθ = 2 mg (1 - cosθ)$

or $N = 2 mg - 3 mgcosθ$

The tube breaks its contact with ground when $2 Ncosθ > Mg$

Substituting, $2 Ncosθ = Mg$

or $4 mgcosθ - 6 {mgcos}^{2} θ = Mg$

Substituting, $θ = 60 °$

$2 mg - \frac{3 mg}{2} = Mg$ or $\frac{M}{m} = \frac{1}{2}$

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