Bead on a spinning ring — natural frequency: 22-Mec-A2 Kinematics and Dynamics of Machines, December 2013, Q5

Question 5: Bead on a spinning ring — natural frequency (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Bead $m=1$ kg on a hoop of radius $R=0.25$ m spinning at $\Omega=10$ rad/s about a vertical diameter; $g=9.81$ m/s². Let $\theta$ be the bead angle measured from the lowest point.

Find. The natural frequency of small oscillations about the stable equilibrium.

Bead on a hoop spinning about the vertical axis; the stable equilibrium lifts off the bottom to θ₀ when the spin exceeds the critical rate.

Approach. Form the Lagrangian with $\theta$ as the single coordinate (the spin $\Omega$ enters as a prescribed rotation), find the stable equilibrium from the effective potential, and linearise the equation of motion about it.

Energies. The bead’s speed has an in-plane part $R\dot\theta$ and an out-of-plane part $\Omega R\sin\theta$ (its distance from the axis is $R\sin\theta$): $$T=\tfrac12 mR^{2}\dot\theta^{2}+\tfrac12 mR^{2}\Omega^{2}\sin^{2}\theta,\qquad V=-mgR\cos\theta.$$
Equation of motion (Lagrange). $$\ddot\theta+\Big(\frac{g}{R}-\Omega^{2}\cos\theta\Big)\sin\theta=0.$$
Stable equilibrium. Setting the bracket×$\sin\theta=0$: besides $\theta=0$, a tilted equilibrium exists when $\cos\theta_0=\dfrac{g}{R\Omega^{2}}=\dfrac{9.81}{0.25(100)}=0.392$. Since $\Omega^{2}=100>g/R=39.24$, the bottom is unstable and the stable point is $\theta_0=\cos^{-1}(0.392)=66.9^\circ$.
Linearise about $\theta_0$. Writing $\theta=\theta_0+\varphi$, the natural frequency follows from the derivative of the restoring term: $$\omega_n^{2}=\frac{g}{R}\cos\theta_0-\Omega^{2}\cos2\theta_0=39.24(0.392)-100(-0.692)=84.6\ \text{s}^{-2}.$$
Natural frequency. $$\omega_n=\sqrt{84.6}=\;$$$\omega_n=9.20$ rad/s $= 1.46$ Hz.

Quantity	Value
Critical spin ($\Omega_c=\sqrt{g/R}$)	6.26 rad/s ($\Omega=10$ is super-critical)
Stable equilibrium angle	θ₀ = 66.9° from bottom
Natural frequency	ω_n = 9.20 rad/s (1.46 Hz)

Topic. Small oscillations of a one-DOF system with a rotation-dependent (centrifugal) effective potential — a classic bifurcation problem.

Key relations. $V_{\text{eff}}(\theta)=-mgR\cos\theta-\tfrac12 mR^{2}\Omega^{2}\sin^{2}\theta$; equilibria $V_{\text{eff}}^{\prime}=0$; $\omega_n^{2}=V_{\text{eff}}^{\prime\prime}(\theta_0)/(mR^{2})=(g/R)\cos\theta_0-\Omega^{2}\cos2\theta_0$.

Why this works. In the rotating frame the spin contributes a centrifugal potential that competes with gravity. Below the critical spin $\Omega_c=\sqrt{g/R}$ the lowest point is stable; above it that point turns into a maximum and the bead settles at a tilted angle. Linearising the effective potential about the true minimum gives the small-oscillation frequency.

Common pitfall. Linearising about $\theta=0$ even when the spin is super-critical (it is then unstable, giving an imaginary “frequency”), or dropping the $\sin^2\theta$ centrifugal term. Rao, Mechanical Vibrations, ch. 2 & 9.

Question 5 of 6: Bead on a spinning ring — natural frequency

Question 5: Bead on a spinning ring — natural frequency (20 marks)