A small disc of mass 500 g and radius 5 cm | vedclass

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Question

(D) When a disc rolls down an inclined plane without slipping,the loss in gravitational potential energy is converted into translational and rotationa

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height of the incline and acceleration due to gravity

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(D) When a disc rolls down an inclined plane without slipping,the loss in gravitational potential energy is converted into translational and rotational kinetic energy. By the law of conservation of energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. For a disc,the moment of inertia is $I = \frac{1}{2}mr^2$ and the rolling condition is $\omega = \frac{v}{r}$. Substituting these values: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2$. $mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$. Canceling $m$ from both sides: $gh = \frac{3}{4}v^2$. Solving for $v$: $v = \sqrt{\frac{4gh}{3}}$. Thus,the velocity $v$ depends only on the height of the incline $h$ and the acceleration due to gravity $g$.

$A$ small disc of mass $500 \ g$ and radius $5 \ cm$ rolls down an inclined plane without slipping. The speed of its center of mass when it reaches the bottom of the inclined plane depends on:

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