Question

A solid sphere, a hollow sphere, ring and disc are released from top of inclined plane (sufficient friction); which one will reach first?

Answer 1

Solid sphere

Answer 2

For objects rolling without slipping, the acceleration down an incline is given by

$$a=\frac{g\sin\theta}{1+\frac{I}{mr^2}},$$

where $I$ is the moment of inertia. The moments of inertia for the given objects are:

• Solid sphere: $I=\frac{2}{5}mr^2$
• Disc: $I=\frac{1}{2}mr^2$
• Hollow sphere: $I=\frac{2}{3}mr^2$
• Ring: $I=mr^2$

Since a smaller $\frac{I}{mr^2}$ means higher acceleration, the ordering of accelerations is:

$$\text{Solid sphere} > \text{Disc} > \text{Hollow sphere} > \text{Ring}.$$

Thus, the solid sphere reaches the bottom first.

The object with the smallest moment of inertia factor ($\frac{I}{mr^2}$) accelerates fastest. For the given bodies, the solid sphere has the smallest factor ($\frac{2}{5}$), so it reaches first.