An Angular Simple Harmonic Oscillator — Question 5

Question 5

Tangential springs become a torsional restoring law

A rotor of inertia $I$ is connected to two identical linear springs of constant $k$, acting tangentially at radius $r$ on opposite sides. The geometry is arranged so a small rotation $\theta$ extends one spring and compresses the other by magnitude $r\theta$, and both torques restore equilibrium.

1. Find the effective torsion constant and oscillation frequency.

2. For $k=100\,\mathrm{N/m}$, $r=.10\,\mathrm{m}$ and $I=.020\,\mathrm{kg\,m^2}$, find the period. Verify the effective stiffness by summing spring energies.

Question 5 - Solution

Tangential springs become a torsional restoring law

(a) Derivation and calculation

For small $\theta$, each spring changes length by magnitude $r\theta$:$$|F|=kr|\theta|,\qquad\tau_1=\tau_2=-kr^2\theta.$$$$I\ddot\theta=-2kr^2\theta\Rightarrow\boxed{\kappa_{\mathrm{eff}}=2kr^2},\quad\boxed{\Omega=\sqrt{2kr^2/I}}.$$

(b) Result and interpretation

$$\kappa_{\mathrm{eff}}=2(100)(.10)^2=2.00\,\mathrm{N\,m/rad},$$$$\Omega=\sqrt{2/.020}=10.0\,\mathrm{rad/s},\quad\boxed{T=2\pi/10=.62832\,\mathrm s}.$$$$U=2\left[\tfrac12k(r\theta)^2\right]=\tfrac12(2kr^2)\theta^2.$$

Physical check

Although the spring forces may point oppositely, their moments about the axis both oppose rotation.