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(D) The radius of the ring is $R = 4a$ and the radius of the disc is $r = a$. When the disc is at the top (or side as shown),the center of the disc is

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$\sqrt{4ga}$

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(D) The radius of the ring is $R = 4a$ and the radius of the disc is $r = a$. When the disc is at the top (or side as shown),the center of the disc is at a distance $R - r = 4a - a = 3a$ from the center of the ring. When the disc reaches the lowest point,its center is at a distance $R - r = 3a$ below the center of the ring. Thus,the vertical displacement of the center of mass of the disc is $h = 3a$. By the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy (translational + rotational). $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Since the disc rolls without slipping,$\omega = v/r$ and $I = \frac{1}{2}mr^2$. $mg(3a) = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(v/r)^2$. $3mga = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$. $3ga = \frac{3}{4}v^2$. $v^2 = 4ga$. $v = \sqrt{4ga}$.

$A$ 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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