A disc of mass m and radius r rolls down an i | vedclass

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Question

(A) For a disc rolling down an inclined plane,the total potential energy at the top is converted into translational and rotational kinetic energy at t

Vedclass · Accepted Answer

$\frac{mgh}{3}$

Vedclass · Answer

(A) For a disc rolling down an inclined plane,the total potential energy at the top is converted into translational and rotational kinetic energy at the bottom.By the law of conservation of energy: $mgh = K_{trans} + K_{rot}$.For a disc,the moment of inertia $I = \frac{1}{2}mr^2$.Since the disc rolls without slipping,$v = r\omega$,so $\omega = \frac{v}{r}$.The translational kinetic energy is $K_{trans} = \frac{1}{2}mv^2$.The rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 = \frac{1}{4}mv^2$.Total energy $mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.Thus,$\frac{1}{2}mv^2 = \frac{2}{3}mgh$.Substituting this into the expression for rotational kinetic energy: $K_{rot} = \frac{1}{4}mv^2 = \frac{1}{2}(\frac{1}{2}mv^2) = \frac{1}{2}(\frac{2}{3}mgh) = \frac{mgh}{3}$.

$A$ disc of mass $m$ and radius $r$ rolls down an inclined plane of height $h$. When it reaches the bottom of the plane,its rotational kinetic energy is ($g=$ acceleration due to gravity).

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