The Bead That Climbs When You Spin Faster

physicsmechanics/rotating-framesdifficulty 4/5#rotating-frame#effective-potential#bifurcation#false-attractor#limiting-case

A bead of mass $m$ slides without friction on a circular hoop of radius $R$ that spins about its
vertical diameter at a constant angular speed $\omega$. Let $\theta$ be the bead's angle from the
downward vertical.

(a) Working in the rotating frame, write the effective potential $U_{\mathrm{eff}}(\theta)$
(gravity plus the centrifugal term). (b) Find every equilibrium angle and the critical speed
$\omega_c$ at which their number changes. (c) The bead is nudged from the bottom. A common
claim is "the bottom is the stable rest point, as for any pendulum." Decide when that is true —
and what (1) says happens above $\omega_c$.

mm θθ ωω RR

The effective potential (dropping a constant) governs everything:

$$ U_{\mathrm{eff}}(\theta) = -mgR\cos\theta - \tfrac{1}{2}mR^2\omega^2\sin^2\theta $$(1)

Below $\omega_c$ the bottom is the only stable point; above it, $\theta=0$ turns into a hilltop and
two tilted equilibria appear — a pitchfork:

0.5 1 1.5 2 2.5 3 -1.5 -1 -0.5 0.5 1 slow: bottom stable θ Ueff
Answer & solution

Answer

(a) $ U_{\mathrm{eff}}(\theta) = -mgR\cos\theta - \tfrac{1}{2}mR^2\omega^2\sin^2\theta $

(b) $ \theta = 0,\ \pi\ \text{always};\quad \cos\theta_\pm = \frac{g}{R\omega^2}\ \text{for}\ \omega>\omega_c=\sqrt{g/R}. $

(c)

The bottom is stable only for $\omega<\omega_c=\sqrt{g/R}$. Above $\omega_c$ the sign of
$U_{\mathrm{eff}}''(0)=mR(g-R\omega^2)$ flips: $\theta=0$ becomes unstable and the tilted angles
$\theta_\pm$ take over — the pendulum intuition (the false attractor) fails.

Solution

Necessary insight. In the rotating frame the centrifugal force derives from a potential
$-\tfrac12 mR^2\omega^2\sin^2\theta$; adding gravity gives (1), and every question is
then just "where are the minima of one function."

(a) Height above the bottom is $R(1-\cos\theta)$ and the distance from the spin axis is
$R\sin\theta$, so $U_{\mathrm{eff}} = -mgR\cos\theta - \tfrac12 mR^2\omega^2\sin^2\theta$ up to a
constant.

(b) $U_{\mathrm{eff}}'(\theta) = mgR\sin\theta - mR^2\omega^2\sin\theta\cos\theta = mR\sin\theta\,(g - R\omega^2\cos\theta)$. Zeros: $\sin\theta=0$ (so $\theta=0,\pi$) and
$\cos\theta = g/(R\omega^2)$, which has a solution only once $R\omega^2 > g$, i.e.
$\omega_c = \sqrt{g/R}$.

(c) $U_{\mathrm{eff}}''(0) = mR(g - R\omega^2)$. For $\omega<\omega_c$ this is positive — the
bottom is stable, exactly the pendulum intuition. For $\omega>\omega_c$ it is negative: the bottom
becomes a local maximum (the false attractor fails) and the two symmetric tilted angles $\theta_\pm$
are the new minima. Spinning faster makes the bead climb.

Grading

  • 3 pts (a) Correct effective potential $U_{\mathrm{eff}}(\theta)$ in the rotating frame.
  • 4 pts (b) Both equilibria found; the critical angular speed identified.
  • 3 pts (c) Stability decided from $U_{\mathrm{eff}}''(\theta)$, not by intuition.

Total: 10 pts