A ring of radius 3a is fixed rigidly on a tab | vedclass

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(A) Let $R = 3a$ be the radius of the fixed ring and $r = a$ be the radius of the small rolling ring.When the small ring is at the horizontal position

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$\sqrt {2ga} $

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(A) Let $R = 3a$ be the radius of the fixed ring and $r = a$ be the radius of the small rolling ring.When the small ring is at the horizontal position $A$,the height of its center from the lowest point is $h = R - r = 3a - a = 2a$.By the law of conservation of energy,the potential energy lost by the small ring equals the kinetic energy gained at the lowest point.Potential energy lost,$\Delta U = mgh = mg(2a) = 2mga$.At the lowest point,the kinetic energy $K$ consists of translational kinetic energy and rotational kinetic energy: $K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a ring rolling without slipping,$I = mr^2$ and $\omega = v/r$.Substituting these,$K = \frac{1}{2}mv^2 + \frac{1}{2}(mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.Equating energy: $2mga = mv^2$.Therefore,$v^2 = 2ga$,which gives $v = \sqrt{2ga}$.

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