A solid uniform cylinder is rolling without slipping.What fraction of its kinetic energy is rotational?
A)1/3
B)2/3
C)1/2
D)1/4
E)3/4

Question

Quizplus · Accepted Answer

For a solid uniform cylinder rolling without slipping, the total kinetic energy (K) is the sum of translational kinetic energy (K_trans) and rotational kinetic energy (K_rot). The translational kinetic energy is given by $K_{trans} = \frac{1}{2}mv^2$, where m is the mass and v is the velocity of the center of mass. The rotational kinetic energy is given by $K_{rot} = \frac{1}{2}I\omega^2$, where I is the moment of inertia of the cylinder about its axis and $\omega$ is the angular velocity. For a solid cylinder, $I = \frac{1}{2}mr^2$, and because the cylinder is rolling without slipping, $v = r\omega$. Substituting $I$ and $v = r\omega$ into the equations, we find that $K_{rot} = \frac{1}{4}mv^2$. Therefore, the fraction of the kinetic energy that is rotational is $K_{rot}/K = \frac{\frac{1}{4}mv^2}{\frac{1}{2}mv^2 + \frac{1}{4}mv^2} = \frac{1}{3}$.

A Solid Uniform Cylinder Is Rolling Without Slipping