= Bead on a uniformly rotating wire
{title2=$(1+f'^2)\ddot r=\omega^2r-gf'-f'f''\dot r^2$}
A smooth wire has instantaneous shape $z=f(r)$ in <cylindrical coordinates> and rotates at constant <angular velocity> $\dot\theta=\omega$. The <tangent vector> at fixed time is $\mathbf e_r+f'(r)\mathbf e_z$. Its <dot product> with the <constraint force> vanishes. Projecting <Newton's second law> along this <tangent vector> gives the equation in the title.
Multiplying that equation by $\dot r$ proves conservation of
$$
\mathcal H=\frac12(1+f'(r)^2)\dot r^2+U(r),\qquad
U(r)=gf(r)-\frac12\omega^2r^2.
$$
This is a <conserved quantity> in the rotating description; the motor can exchange ordinary <mechanical energy> with the bead. A relative <equilibrium> at $a>0$ satisfies $U'(a)=gf'(a)-\omega^2a=0$. Its <linearized equation> is $(1+f'(a)^2)\ddot\epsilon=-U''(a)\epsilon$. A strict local minimum with $U''(a)>0$ gives <Lyapunov stability>, while $U''(a)<0$ gives an <unstable equilibrium>. If $U''(a)=0$, higher orders are needed.
Back to article page
