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(D) The radius of the ring is $R = 4a$. The radius of the disc is $r = a$. When the disc is at the top of its path (at the same level as the center of

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$\sqrt{4ga}$

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(D) The radius of the ring is $R = 4a$. The radius of the disc is $r = a$. When the disc is at the top of its path (at the same level as the center of the ring),the center of mass of the disc is at a height $h = R - r = 4a - a = 3a$ above the lowest point of the ring. By the principle of conservation of mechanical energy,the potential energy lost by the disc is equal to the kinetic energy gained by the disc. $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ Since the disc rolls without slipping,$v = r\omega$,so $\omega = v/r$. The moment of inertia of the disc about its center of mass is $I = \frac{1}{2}mr^2$. Substituting these into the energy equation: $mg(3a) = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2$ $3mga = \frac{1}{2}mv^2 + \frac{1}{4}mv^2$ $3mga = \frac{3}{4}mv^2$ $3ga = \frac{3}{4}v^2$ $v^2 = 4ga$ $v = \sqrt{4ga}$

$A$ ring of radius $4a$ is rigidly fixed in a vertical position on a table. $A$ small disc of mass $m$ and radius $a$ is released as shown in the figure. When the disc rolls down,without slipping,to the lowest point of the ring,then its speed will be

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