Equilibrium of a bead on a circle with a spring and gravity

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A bead $P$ of mass $m=3\ \text{kg}$ is constrained to slide on a circle of centre $O$ (the origin) and radius $3$. It is joined to the fixed point $Q(1,1)$ by a spring of zero natural length and elastic constant $k=60\ \text{N/m}$; the gravitational acceleration is $g=10\ \text{m/s}^2$. (a) Write the total potential energy $U(\alpha)$, where $\alpha$ is the angle that $OP$ makes with the $x$-axis. (b) Find the stable and unstable equilibrium positions and show that for $\alpha\approx 26^\circ$ the equilibrium is stable.

Solution

(a) The bead’s position is $P=(3\cos\alpha,\,3\sin\alpha)$. Gravitational energy: $U_{grav}=mgy=3\cdot10\cdot3\sin\alpha=90\sin\alpha$. Squared spring length: $|PQ|^2=(3\cos\alpha-1)^2+(3\sin\alpha-1)^2=9-6\cos\alpha-6\sin\alpha+2=11-6\cos\alpha-6\sin\alpha.$ Elastic energy (spring of zero natural length): $U_{spring}=\tfrac12 k|PQ|^2=30(11-6\cos\alpha-6\sin\alpha)=330-180\cos\alpha-180\sin\alpha$. Total: $U(\alpha)=330-180\cos\alpha-180\sin\alpha+90\sin\alpha=330-180\cos\alpha-90\sin\alpha.$

(b) Equilibria where $U'(\alpha)=0$: $U'(\alpha)=180\sin\alpha-90\cos\alpha=0\ \Rightarrow\ \tan\alpha=\tfrac12\ \Rightarrow\ \alpha\approx 26.57^\circ\ \text{or}\ \alpha\approx 206.57^\circ.$ Stability via the second derivative: $U''(\alpha)=180\cos\alpha+90\sin\alpha.$

• At $\alpha\approx 26.57^\circ$: $U''\approx 180(0.894)+90(0.447)\approx 201>0$ → minimum of $U$ → stable equilibrium.
• At $\alpha\approx 206.57^\circ$: $U''\approx 180(-0.894)+90(-0.447)\approx -201<0$ → maximum of $U$ → unstable equilibrium. $\boxed{\;\alpha_{stab}\approx 26.57^\circ\ (\text{stable}),\qquad \alpha_{inst}\approx 206.57^\circ\ (\text{unstable})\;}$