A cylinder of mass M and radius R rolls on an | vedclass

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Question

(C) For a cylinder rolling without slipping,the total kinetic energy $(K)$ is the sum of translational kinetic energy and rotational kinetic energy.$K

Vedclass · Accepted Answer

$\frac{3}{4}Mv^2$

Vedclass · Answer

(C) For a cylinder rolling without slipping,the total kinetic energy $(K)$ is the sum of translational kinetic energy and rotational kinetic energy.$K = K_{	ext{trans}} + K_{	ext{rot}}$$K = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$ and the rolling condition is $v = R\omega$ (or $\omega = v/R$).Substituting these into the equation:$K = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{1}{2}MR^2)(\frac{v}{R})^2$$K = \frac{1}{2}Mv^2 + \frac{1}{4}Mv^2$$K = \frac{3}{4}Mv^2$Thus,the gain in kinetic energy is $\frac{3}{4}Mv^2$.

$A$ cylinder of mass $M$ and radius $R$ rolls on an inclined plane. The gain in kinetic energy is

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