A solid cylinder of mass M and radius R rolls without slipping on a horizontal surface. If the velocity of its center of mass is v , what is the ratio of its translational kinetic energy to its rotati

Question

A solid cylinder of mass M and radius R rolls without slipping on a horizontal surface. If the velocity of its center of mass is v , what is the ratio of its translational kinetic energy to its rotational kinetic energy?

TestHub · Accepted Answer

For a solid cylinder, the moment of inertia about its central axis is $I = \frac{1}{2} M R^{2}$ .

The translational kinetic energy is $K_{t r an s} = \frac{1}{2} M v^{2}$ . For rolling without slipping, the relation between linear velocity $v$ and angular velocity $ω$ is $v = R ω$ , so $ω = \frac{v}{R}$ . The rotational kinetic energy is $K_{ro t} = \frac{1}{2} I ω^{2} = \frac{1}{2} (\frac{1}{2} M R^{2}) (\frac{v}{R})^{2} = \frac{1}{4} M R^{2} \frac{v ^{2}}{R ^{2}} = \frac{1}{4} M v^{2}$ . The ratio of translational kinetic energy to rotational kinetic energy is $\frac{K _{t r an s}}{K _{ro t}} = \frac{\frac{1}{2} M v ^{2}}{\frac{1}{4} M v ^{2}} = \frac{1/2}{1/4} = \frac{2}{1}$ .

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