Bead on a Rotating Hoop: Equilibrium and Stability

A standard bead-on-a-rotating-hoop problem considers a small bead constrained to move without friction on a circular hoop of radius $a$. The hoop is vertical and rotates at constant angular speed $\omega$ about its vertical diameter. Gravity acts downward.

The interesting feature is a stability transition: as the rotation becomes fast enough, the bead no longer prefers the bottom of the hoop. Two new stable positions appear away from the bottom.

Set up the geometry

Let $\theta$ be the angle measured from the downward vertical. Thus:

• $\theta=0$ is the bottom of the hoop;
• $\theta=\pi$ is the top;
• $\theta$ and $-\theta$ describe positions on opposite sides of the vertical diameter.

The bead’s height relative to the hoop’s center is $z=-a\cos\theta$, and its distance from the rotation axis is $a\sin\theta$. Since the hoop rotates with angular speed $\omega$, the bead has two contributions to its speed:

1. motion along the hoop, with speed $a\dot{\theta}$;
2. motion around the vertical axis, with speed $a\omega\sin\theta$.

Therefore the kinetic and potential energies are

$$T=\frac12ma^2\dot{\theta}^{\,2}+\frac12ma^2\omega^2\sin^2\theta,$$ $$V=-mga\cos\theta.$$

The rotation is imposed externally, so it is best to use the Lagrangian with $\omega$ treated as a fixed parameter:

$$L=T-V =\frac12ma^2\dot{\theta}^{\,2}+\frac12ma^2\omega^2\sin^2\theta+mga\cos\theta.$$

Equation of motion

Applying the Euler–Lagrange equation to $\theta$ gives

$$\ddot{\theta}=\sin\theta\left(\omega^2\cos\theta-\frac{g}{a}\right).$$

This equation can also be obtained in the rotating frame. The component of gravity along the hoop is $-mg\sin\theta$, while the centrifugal force has tangential component $m\omega^2a\sin\theta\cos\theta$. The Coriolis force vanishes for a bead instantaneously at rest in the rotating frame, and in the full equation its effect is accounted for by the velocity-dependent rotating-frame terms.

An effective potential makes the result easier to interpret. The equation can be written as

$$ma^2\ddot{\theta}=-\frac{dU_{\mathrm{eff}}}{d\theta},$$

where

$$U_{\mathrm{eff}}(\theta)=-\frac12ma^2\omega^2\sin^2\theta-mga\cos\theta.$$

The centrifugal contribution lowers the effective potential away from the vertical rotation axis, while gravity lowers it toward the bottom.

Equilibrium positions

An equilibrium has $\dot{\theta}=0$ and $\ddot{\theta}=0$. Hence

$$\sin\theta\left(\omega^2\cos\theta-\frac{g}{a}\right)=0.$$

There are two types of solution.

Bottom and top

The equation $\sin\theta=0$ gives

$$\theta=0 \quad\text{and}\quad \theta=\pi.$$

These are the bottom and top of the hoop.

Off-center equilibria

The second factor gives

$$\cos\theta_0=\frac{g}{a\omega^2}.$$

These positions exist only if

$$\frac{g}{a\omega^2}\leq 1,$$

or equivalently,

$$\omega\geq\omega_c=\sqrt{\frac{g}{a}}.$$

For every angular speed above this critical value, there are two off-center equilibria, at $+\theta_0$ and $-\theta_0$.

Stability of the equilibria

To test stability, write $\theta=\theta_e+\eta$, where $\eta$ is a small displacement from an equilibrium. Keeping only terms linear in $\eta$ gives the small-oscillation equation.

Bottom point

Near $\theta=0$,

$$\sin\theta\approx\theta,\qquad \cos\theta\approx1.$$

Thus

$$\ddot{\theta}\approx\left(\omega^2-\frac{g}{a}\right)\theta.$$

The bottom is stable when the coefficient is negative:

$$\omega<\sqrt{\frac{g}{a}}.$$

In that case, small displacements oscillate about the bottom with angular frequency

$$\Omega=\sqrt{\frac{g}{a}-\omega^2}.$$

At $\omega=\sqrt{g/a}$, the restoring force disappears to first order. For $\omega>\sqrt{g/a}$, the bottom is unstable: a small displacement grows rather than returning to zero.

Top point

Near $\theta=\pi$, let $\theta=\pi+\eta$. Since $\sin(\pi+\eta)\approx-\eta$ and $\cos(\pi+\eta)\approx-1$,

$$\ddot{\eta}\approx\left(\omega^2+\frac{g}{a}\right)\eta.$$

The coefficient is always positive, so the top is always unstable. Rotation does not stabilize it in this setup.

Off-center points

For an off-center equilibrium, use $\omega^2\cos\theta_0=g/a$. The derivative of the right-hand side of the equation of motion at $\theta_0$ is

$$-\omega^2\sin^2\theta_0.$$

Therefore small deviations satisfy

$$\ddot{\eta}+\omega^2\sin^2\theta_0\,\eta=0.$$

The off-center equilibria are stable, with small-oscillation frequency

$$\Omega=\omega\sin\theta_0.$$

Worked example

Suppose the hoop radius is $a=0.50\ \mathrm{m}$. The critical angular speed is

$$\omega_c=\sqrt{\frac{9.81}{0.50}} =\sqrt{19.62} \approx4.43\ \mathrm{rad\,s^{-1}}.$$

If the hoop rotates at $\omega=6.00\ \mathrm{rad\,s^{-1}}$, the bottom is unstable and the off-center equilibria exist. Their angle satisfies

$$\cos\theta_0=\frac{9.81}{(0.50)(6.00)^2} =\frac{9.81}{18.00} \approx0.545.$$

Thus

$$\theta_0\approx\arccos(0.545)\approx57.0^\circ.$$

The bead can therefore rest at either $57.0^\circ$ to one side of the bottom direction or $57.0^\circ$ to the other side. The two positions have the same energy because the system is symmetric about the vertical diameter.

Common mistakes

Using only the centrifugal force. Gravity and the centrifugal force both contribute to the tangential balance. The equilibrium condition is not simply “centrifugal force equals gravity”; only their components along the hoop must balance.

Forgetting the two off-center positions. The equation determines $\cos\theta_0$, so the equilibria occur symmetrically at $+\theta_0$ and $-\theta_0$.

Assuming the bottom is always stable. It is stable only below the critical speed. Above $\sqrt{g/a}$, the centrifugal effect overcomes gravity for small sideways displacements.

Treating the top as stable at high speed. The top remains unstable. The new stable points emerge near the bottom, not near the top.

Confusing angular speed with ordinary frequency. Here $\omega$ is measured in radians per second. If the rotation frequency is $f$ cycles per second, use $\omega=2\pi f$; the critical frequency is $f_c=\frac{1}{2\pi}\sqrt{g/a}$.

The transition at $\omega=\sqrt{g/a}$ is a symmetry-breaking pitchfork bifurcation: one stable bottom equilibrium becomes unstable while two symmetric stable equilibria appear.