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Chapter 10: Problem 5

A solid sphere and a hollow sphere of the same mass and radius are rolling along level ground. If they have the same total kinetic energy, which is moving faster?

Short Answer

Expert verified
The solid sphere is moving faster.

Step by step solution

01

Calculate kinetic energy

First, we should recall that total kinetic energy of a rolling sphere includes both translational and rotational kinetic energy. Both of these forms of energy need to be accounted for when evaluating kinetic energy. The total kinetic energy of an object rolling without slipping is given by the equation: \( K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \), where m is the mass, v is the velocity of the center of mass, I is the rotational inertia, and 蠅 is the angular speed. Because the spheres are rolling without slipping, there is a relationship between the translational velocity and the angular velocity which is \( v = wR \), where R is the radius of the spheres.
02

Calculate the ratio of velocities

Since both the hollow and solid sphere have the same kinetic energy, we can equate their kinetic energy equations, and cancel the mass and radius where applicable. By doing this, we can deduce a relation between the velocities of the hollow sphere (\(v_{h}\)) and solid sphere (\(v_{s}\)). The moment of inertia of a hollow sphere is \(I_{h} = \frac{2}{3}mR^2\) and that of solid sphere is \(I_{s} = \frac{2}{5}mR^2 \). This allows us to create a ratio \( \frac{v_{h}}{v_{s}} = \sqrt{\frac{I_{s}}{I_{h}}} \). After substituting the moment of inertia equations into this, we get \( \frac{v_{h}}{v_{s}} = \sqrt{\frac{2}{3}} \).
03

Find which one is faster

As \(\sqrt{\frac{2}{3}} < 1\), this means \(v_{h} < v_{s}\). Therefore, the solid sphere is moving faster.

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