An Angular Simple Harmonic Oscillator — Question 6

Question 6

Torsion can stabilize an otherwise inverted rotor

A rigid body pivots about a horizontal axis with its center of mass distance $d$ above the axis at $\theta=0$. Its mass is $M$ and inertia about the axis is $I$. Gravity exerts the supplied destabilizing torque $+Mgd\sin\theta$. A torsion spring adds $-\kappa\theta$.

1. Derive the condition for linear stability and the small-oscillation frequency.

2. For $Mgd=.80\,\mathrm{N\,m}$, $\kappa=1.20\,\mathrm{N\,m/rad}$ and $I=.040\,\mathrm{kg\,m^2}$, find the period. What is special about the threshold $\kappa=Mgd$?

Question 6 - Solution

Torsion can stabilize an otherwise inverted rotor

(a) Derivation and calculation

$$I\ddot\theta=-\kappa\theta+Mgd\sin\theta\simeq-(\kappa-Mgd)\theta.$$$$\boxed{\kappa>Mgd},\qquad\boxed{\Omega=\sqrt{(\kappa-Mgd)/I}}.$$$$\Omega=\sqrt{(1.20-.80)/.040}=\sqrt{10},\quad\boxed{T=1.98692\,\mathrm s}.$$

(b) Result and interpretation

At $\kappa=Mgd$, expand one order further:$$\tau=Mgd(\sin\theta-\theta)=-\frac{Mgd}{6}\theta^3+O(\theta^5).$$$$U=\frac{Mgd}{24}\theta^4+O(\theta^6).$$The equilibrium remains nonlinearly restoring, but its linear frequency is zero; a finite-amplitude period cannot be inferred from SHM.

Physical check

For $\kappa<Mgd$, the linear torque has the same sign as displacement and the upright equilibrium is unstable.