= Solution
In <Lagrangian mechanics>, choose independent <generalized coordinates> $q_i$ compatible with the mechanical <constraints>, write $L=T-U$, and apply the <Euler-Lagrange equation>
$$
\frac d{dt}\frac{\partial L}{\partial\dot q_i}-\frac{\partial L}{\partial q_i}=0.
$$
Equivalently, the physical path makes the <action> stationary under variations with fixed endpoints. For ideal holonomic <constraints>, their unknown reaction forces need not appear in the reduced equations. Direct use of <Newton's laws of motion> instead resolves vector <forces> and reactions and enforces the <constraints> separately. Both approaches describe the same mechanics; the scalar <Lagrangian> is often shorter in curvilinear coordinates. Nonconservative generalized forces must be added separately. Here a drive maintains the prescribed azimuthal <angular velocity> $\Omega$.
The mass has polar <speed> $a\dot\theta$ and azimuthal <speed> $a\Omega\sin\theta$, so its <kinetic energy> is $ma^2(\dot\theta^2+\Omega^2\sin^2\theta)/2$. The elastic <potential energy> is $\mu\theta^2/2$, and the height is $a\cos\theta$, giving gravitational <potential energy> $mga\cos\theta$. Therefore
$$
L=\frac12ma^2\dot\theta^2+\frac12ma^2\Omega^2\sin^2\theta
-\frac12\mu\theta^2-mga\cos\theta,
$$
and the <Euler-Lagrange equation> gives
$$
\boxed{\ddot\theta=\Omega^2\sin\theta\cos\theta
-\frac\mu{ma^2}\theta+\frac ga\sin\theta.}
$$
For $\mu=2mga$, define $q=\Omega^2a/g$ and $F(\theta)=q\sin\theta\cos\theta-2\theta+\sin\theta$, so $\ddot\theta=(g/a)F(\theta)$. Near zero,
$$
F(\theta)=(q-1)\theta-\left(\frac{2q}{3}+\frac16\right)\theta^3+O(\theta^5).
$$
When $q<1$ the upright <equilibrium> is linearly stable and the <small oscillations> have \b[$\omega_1^2=g/a-\Omega^2$].
For $q>1$, the graph of $F$ initially rises, whereas $F(\pi/2)=1-\pi<0$. There must therefore be a zero inside the interval. More strongly, for $0<\theta<\pi/2$,
$$
\frac{F(\theta)}{\sin\theta}=q\cos\theta+1-\frac{2\theta}{\sin\theta}
$$
is strictly decreasing: $\cos\theta$ decreases and $\theta/\sin\theta$ increases because $\sin\theta-\theta\cos\theta>0$. The ratio goes from $q-1>0$ to $1-\pi<0$, giving exactly one zero $\theta_0$ in that interval. At the zero, $F'(\theta_0)<0$, so it is a stable <equilibrium>. The <rotating torsion-stabilized inverted pendulum> also has a symmetric stable angle $-\theta_0$ if both signs of tilt are permitted.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2015/paper-1-rotating-pendulum.png]
{title=Acceleration curves after the upright instability and the quartic potential at the threshold}
At $\theta_0=\pi/4$, setting $F(\theta_0)=0$ gives $q/2-\pi/2+1/\sqrt2=0$, so
$$
\boxed{\Omega=\sqrt{\frac ga(\pi-\sqrt2)}.}
$$
For $\theta=\theta_0+\xi$, the linearized equation is $\ddot\xi=(g/a)F'(\theta_0)\xi$. Since $F'=q\cos2\theta-2+\cos\theta$,
$$
\boxed{\omega_2^2=\frac ga\left(2-\frac1{\sqrt2}\right).}
$$
At the threshold $q=1$ the linear restoring term vanishes, but the cubic term remains:
$$
\ddot\theta=-\frac{5g}{6a}\theta^3+O(\theta^5),\qquad
U_{\rm eff}(\theta)-U_{\rm eff}(0)=\frac{5mga}{24}\theta^4+O(\theta^6).
$$
\b[The upright state remains nonlinearly stable, with no nonzero infinitesimal harmonic frequency.] Small finite disturbances produce <quartic oscillator> motion whose <frequency> is proportional to its small angular <amplitude>; its <period> tends to infinity as that amplitude tends to zero. Zero linear stiffness is not the same as an absent restoring <torque>.
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