A thin hoop of mass $M$ and radius $r$ is placed on a horizontal plane. At the initial instant, the hoop is at rest. A small washer of mass m with zero initial velocity slides from the upper point of the hoop along a smooth groove in the inner surface of the hoop. Determine the velocity u of the centre of the hoop at the moment when the washer is at a certain point A of the hoop, whose radius vector forms an angle $ϕ$ with the vertical. The friction between the hoop and the plane should be neglected.

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$m \cos ϕ \sqrt{\frac{2 g r (1 + \cos ϕ)}{(M + m) (M + m \sin^{2} ϕ)}}$

$2 m \cos ϕ \sqrt{\frac{2 g r (1 + \cos ϕ)}{(M + m) (M + m \sin^{2} ϕ)}}$

$m \cos ϕ \sqrt{\frac{g r (1 + \cos ϕ)}{(M + m) (M + m \sin^{2} ϕ)}}$

$m \cos ϕ \sqrt{\frac{2 (1 + \cos ϕ)}{(M + m) (M + m \sin^{2} ϕ)}}$

answer is A.

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Detailed Solution

Let $v_{r}$ be the velocity of washer relative to centre of hoop and v the velocity of centre of hoop. Applying conservation of linear momentum and mechanical energy we have,

$m (v_{r} \cos ϕ - v) = M v \dots \dots \dots \dots \dots \dots \dots \dots . (i) a n d m g r (1 - \cos ϕ) = \frac{1}{2} M v^{2} + \frac{1}{2} m (v_{r} {}^{2}+ v^{2} - 2 v v_{r} \cos ϕ) . . . . . . . . . . . . . . . (i i)$