Problem — Object in a spherical bowl

Problem

A small object rests in a smooth spherical bowl of radius $R$. Displaced slightly from the bottom, it oscillates. Explain why the motion is harmonic for small angles, and what the period depends on. Compare with the simple pendulum.

Solution

The object at the bottom of a bowl of radius $R$ moves along a circular arc, exactly like a mass hanging from a string of length $\ell = R$: the centre of the sphere acts as the suspension point. Displaced by an angle $\theta$, the tangential component of weight is the restoring force: $F_t = -mg\sin\theta \approx -mg\,\theta \qquad (\theta \ll 1)$ For small angles $\sin\theta \approx \theta$, so the restoring force is proportional to the displacement and the motion is harmonic. With the displacement along the arc $s = R\theta$ we obtain $m\,\ddt s = -\frac{mg}{R}\,s \quad\Rightarrow\quad \ev{\omega = \sqrt{\frac{g}{R}}}$ and so $T = 2\pi\sqrt{R/g}$. It is identical to a simple pendulum of length $\ell = R$: the period depends only on the radius and on $g$, not on the mass of the object.

Links

Topics: Oscillazioni e moto armonico Concepts: Moto armonico semplice · Pendolo semplice Skills: Approssimazione delle piccole oscillazioni Methods: Approssimazione lineare delle piccole oscillazioni