Discussion
Two disks A and B with identical mass (m) and radius (R) are initially at rest. They roll down from the top of identical inclined planes without slipping. Disk A has all of its mass concentrated at the rim, while Disk B has its mass uniformly distributed. At the bottom of the plane, the ratio of velocity of the center of disk A to the velocity of the center of disk B is
Answer: A
The ratio is \(\sqrt{3/4}\).
Rolling without slipping, energy conservation gives \(mgh = \tfrac{1}{2}mv^{2}(1 + I/mR^{2})\).
\[ \begin{aligned} \text{A (ring, } I = mR^{2}) &: v_{A}^{2} = gh \\ \text{B (disk, } I = \tfrac{1}{2}mR^{2}) &: v_{B}^{2} = \tfrac{4}{3}gh \\ \frac{v_{A}}{v_{B}} &= \sqrt{\tfrac{3}{4}} \end{aligned} \]
Disk A, with its mass at the rim, stores more of its energy as rotation and so arrives slower. \(\sqrt{3/2}\) and \(\sqrt{2}\) put A ahead, and 1 would hold only if both had the same moment of inertia.
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