A solid sphere and a solid cylinder of the same mass and radius are released from the top of an inclined plane at the same time. If both roll without slipping, which object will reach the bottom first

Question

A solid sphere and a solid cylinder of the same mass and radius are released from the top of an inclined plane at the same time. If both roll without slipping, which object will reach the bottom first?

TestHub · Accepted Answer

The acceleration of a body rolling down an inclined plane without slipping is given by $a = \frac{g sin θ}{1 + \frac{I}{M R ^{2}}}$ .

For a solid sphere, the moment of inertia is $I = \frac{2}{5} M R^{2}$ , so its acceleration is $a_{s p h ere} = \frac{g sin θ}{1 + \frac{2}{5}} = \frac{5}{7} g sin θ$ .

For a solid cylinder, the moment of inertia is $I = \frac{1}{2} M R^{2}$ , so its acceleration is $a_{cy l in d er} = \frac{g sin θ}{1 + \frac{1}{2}} = \frac{2}{3} g sin θ$ .

Since $\frac{5}{7} > \frac{2}{3}$ , the acceleration of the solid sphere is greater than that of the solid cylinder. Thus, the solid sphere will reach the bottom first.

Answer: The solid sphere

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