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

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 of the cylinder and v is its velocity. 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 and $\omega$ is its angular velocity. For a solid cylinder, $I = \frac{1}{2}mr^2$, and because the cylinder is rolling without slipping, $v = r\omega$. Substituting these into the equation for $K_{rot}$, we get $K_{rot} = \frac{1}{2}(\frac{1}{2}mr^2)\omega^2 = \frac{1}{4}mv^2$. Therefore, the rotational kinetic energy is one-third of the total kinetic energy, since $K_{rot} = \frac{1}{4}mv^2$ and $K_{trans} = \frac{1}{2}mv^2$, making the total $K = \frac{3}{4}mv^2$, and thus $K_{rot} = \frac{1}{3}K$.

A Solid Uniform Cylinder Is Rolling Without Slipping