A little sand lies at the bottom of a hollow sphere of inner…

A little sand lies at the bottom of a hollow sphere of inner radius $R$. Where does the sand collect when the sphere rotates about a vertical axis with angular velocity $\omega$? Neglect friction between the sand and the wall.

Explanation of Method

Sand in a rotating hollow sphere.

Step 1: Forces in the rotating sphere

Take a grain at angle $\theta$ from the lowest point; its distance from the axis is $r = R\sin\theta$. In the rotating frame it feels gravity $mg$, the centrifugal force $m\omega^2 R\sin\theta$ (horizontal, away from the axis) and the wall's normal force $N$ (toward the centre of the sphere).

Step 2: Along the wall

A smooth wall cannot push along itself, so the components along the surface must cancel. Gravity pulls the grain down toward the bottom with $mg\sin\theta$; the centrifugal force pushes it up the wall with $m\omega^2R\sin\theta\cos\theta$:

\[ mg\sin\theta = m\omega^2 R\sin\theta\cos\theta. \]

Either $\sin\theta = 0$ (the sand stays at the bottom), or

\[\cos\theta = \frac{g}{\omega^2 R}\]

Step 3: Which one happens

If $\omega^2 R \le g$ (slow rotation), the second solution does not exist: the sand stays at the bottom.
If $\omega^2 R \gt g$, the bottom becomes unstable, and the sand moves up and collects in a ring at the angle $\theta = \arccos\dfrac{g}{\omega^2R}$, at height $R - \dfrac{g}{\omega^2}$ above the lowest point.

$\theta$ $mg$ $m\omega^2 R\sin\theta$ $N$ $\omega$ $\text{sand ring}$ $\cos\theta = \frac{g}{\omega^2 R}$

Seen in the rotating sphere. The sand rests where gravity and the centrifugal force add up to a force pointing straight into the wall, so the smooth wall can balance it. The sand forms a ring at the angle $\theta$.

Result

\[\boxed{\,\cos\theta = \frac{g}{\omega^2 R}\,}\]

Remark

Note. The original result is correct; this adds the condition $\omega \gt \sqrt{g/R}$ for it to apply. The wall's push is $N = mg\cos\theta + m\omega^2R\sin^2\theta = m\omega^2 R$. The faster the rotation, the closer the ring gets to the "equator" ($\theta \to 90^\circ$), but it never reaches it.