⚛️ Physics

Problem 1 (Medium): Bead on a Rotating Hoop — A Pitchfork Bifurcation

A bead of mass $m$ slides frictionlessly on a circular wire hoop of radius $R$. The hoop rotates about its vertical diameter at a constant angular velocity $\omega$. Let $\theta$ be the angle from the bottom of the hoop to the bead.

(a) Write the Lagrangian $\mathcal{L}(\theta ,\dot{\theta })$ and derive the equation of motion.

(b) Find all equilibrium positions as a function of $\omega$. Show that a critical angular velocity ${\omega }_c=\sqrt{g/R}$ separates two regimes.

© Analyze stability: show that for $\omega <{\omega }_c$, the only stable equilibrium is $\theta =0$; for $\omega >{\omega }_c$, the bottom becomes unstable and two new symmetric stable equilibria appear at $\theta =\pm \arccos (g/{\omega }^{2}R)$. Identify this as a supercritical pitchfork bifurcation.

(d) Find the frequency of small oscillations about the off-axis equilibrium when $\omega >{\omega }_c$.

Hint: The effective potential is $U_{\text{eff}}(\theta )=-mgR\cos \theta -\frac{1}{2}m{\omega }^2{R}^2{\sin }^2\theta$. Equilibria satisfy $U_{\text{eff}}^{\prime }(\theta )=0$, and stability requires $U_{\text{eff}}^{\prime \prime }(\theta )>0$. For the oscillation frequency, expand $U_{\text{eff}}$ to second order about the equilibrium and read off the curvature.

Solution

The velocity of the bead can be decomposed into two, one is rotation around the $z$ axis, the other is sliding along the hoop. $T=\frac{1}{2}m(R^2{\dot{\theta }}^2+R^2{\omega }^2{\sin }^2\theta )$. Take the center of the hoop to be potential reference, then $V=-mgR\cos \theta$. $$\mathcal{L}=T-V=\frac{1}{2}m(R^2{\dot{\theta }}^2+R^2{\omega }^2{\sin }^2\theta )+mgR\cos \theta$$Taking Euler-Lagrange equation:$$\begin{gathered} \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\theta }}-\frac{\partial \mathcal{L}}{\partial \theta }=0 \\ R\ddot{\theta }=R{\omega }^2\sin \theta \cos \theta -g\sin \theta \end{gathered}$$

To find the equilibrium positions, we set $\ddot{\theta }=\dot{\theta }=0$, $$\sin \theta (R{\omega }^2\cos \theta -g)=0$$If $\sin \theta =0$, then $\theta =0$ bottom, or $\theta =\pi$ top.
Otherwise, $\cos \theta =\frac{g}{R{\omega }^2}$ which has solution only when $\omega \ge \sqrt{\frac{g}{R}}={\omega }_c$. If $\omega ={\omega }_c$, then $\cos \theta =1,\theta =0$ meaning bottom. If $\omega >{\omega }_c,\theta =\pm \arccos (\frac{g}{R{\omega }^2})$. Namely, when $\omega \le {\omega }_c$, the only two equilibrium positions are bottom and top; otherwise there are two more $\theta =\pm \arccos (\frac{g}{R{\omega }^2})$.

$$\mathcal{L}=T-V=\frac{1}{2}m{R}^2{\dot{\theta }}^2-(-\frac{1}{2}m{R}^2{\omega }^2{\sin }^2\theta -mgR\cos \theta )$$Taking the hoop as the reference, the effective potential is $$\begin{gathered} U_{\text{eff}}(\theta )=-mgR\cos \theta -\frac{1}{2}m{\omega }^2{R}^2{\sin }^2\theta \\ {U}_{\text{eff}}^{\prime \prime }(\theta )=mgR\cos \theta -m{\omega }^2{R}^2\cos 2\theta \end{gathered}$$
When $\theta =0$, $U_{\text{eff}}^{\prime \prime }(0)=mgR-m{\omega }^2{R}^2=mR(g-{\omega }^{2}R)$, when $\omega <{\omega }_c$, $U_{\text{eff}}^{\prime \prime }(0)>0$, stable; when $\omega >{\omega }_c$, unstable.
When $\theta =\pi$, $U_{\text{eff}}^{\prime \prime }(\pi )=-mR(g+{\omega }^{2}R)$, always negative, unstable.
When $\theta =\pm \arccos \frac{g}{R{\omega }^2}$, this only happens when $\omega >{\omega }_c$, $U_{\text{eff}}^{\prime \prime }(\theta )=m({\omega }^2{R}^2-\frac{g^2}{{\omega }^2})>0$, stable.

When $\omega$ increasing from $0$, before ${\omega }_c$, there’s only one stable equilibrium at $\theta =0$, after ${\omega }_c$, $\theta =0$ becomes unstable and bifurcates into two stable branches.

From the equation of motion, $$m{R}^2\ddot{\theta }=-U_{\text{eff}}^{\prime }(\theta )$$Taylor expansion from ${\theta }_0$ which is equilibrium angle, assume $\theta ={\theta }_0+ϵ$, we get $$\begin{gathered} m{R}^2\ddot{\theta }=-(U_{\text{eff}}^{\prime }({\theta }_0)+U_{\text{eff}}^{\prime \prime }({\theta }_0)ϵ) \\ m{R}^2\ddot{ϵ}=-U_{\text{eff}}^{\prime \prime }({\theta }_0)ϵ \\ \ddot{ϵ}=-\frac{U_{\text{eff}}^{\prime \prime }({\theta }_0)ϵ}{m{R}^2} \end{gathered}$$Angular frequency is $$\sqrt{\frac{U_{\text{eff}}^{\prime \prime }({\theta }_0)}{m{R}^2}}$$

Problem 2 (Easy): The Falling Chain

A uniform chain of total length $L$ and mass $M$ rests on a frictionless table. A small length $a$ hangs over the edge at $t=0$, and the chain is released from rest.

(a) Let $x(t)$ denote the length hanging over the edge at time $t$. By applying Newton’s second law to the entire chain, show that:

$$\ddot{x}=\frac{g}{L}x$$

(b) Solve this ODE with initial conditions $x(0)=a$, $\dot{x}(0)=0$.

© Find the speed of the chain at the instant it completely leaves the table.

(d) Verify your answer to © using energy conservation directly. Which method do you find more elegant?

Solution

(a) The table is frictionless, so the force is only the gravity of the chain hanging over the edge. $$\begin{gathered} F=M\ddot{x}=\frac{Mx}{L}g \\ \ddot{x}=\frac{g}{L}x \end{gathered}$$
(b) $x=A{e}^{\sqrt{\frac{g}{L}}t}+B{e}^{-\sqrt{\frac{g}{L}}t}$ Plug in $x(0)=a,\dot{x}(0)=0$ we get $$\begin{gathered} A+B=a \\ A\sqrt{\frac{g}{L}}-B\sqrt{\frac{g}{L}}=0 \end{gathered}$$So $A=B=\frac{a}{2}$, $x=a\cosh (\sqrt{\frac{g}{L}}t)$
© $\dot{x}=a\sqrt{\frac{g}{L}}\sinh (\sqrt{\frac{g}{L}}t)$. When the chain leaves the table, meaning $x(t^{\prime })=L,a\cosh (\sqrt{\frac{g}{L}}{t}^{\prime })=L$, we get $\sinh (\sqrt{\frac{g}{L}}{t}^{\prime })=\frac{\sqrt{L^2-a^2}}{a}$, thus $$v=\dot{x}(t^{\prime })=\sqrt{\frac{g(L^2-a^2)}{L}}$$
(d) $E_0=-\frac{Mg}{2L}{a}^2$, no kinetic energy, only potential energy. $E^{\prime }=\frac{1}{2}M{v}^2-\frac{MgL}{2}$. By conservation of the total energy: $$\begin{gathered} -\frac{Mg}{2L}{a}^2=\frac{1}{2}M{v}^2-\frac{MgL}{2} \\ v=\sqrt{\frac{g(L^2-a^2)}{L}} \end{gathered}$$ The second is more elegant, no need to solve the ODE. But the first method can tell you the full information of $x(t)$.