An Angular Simple Harmonic Oscillator — Question 2

Question 2

Moving identical masses outward isolates a torsion balance inertia

A torsion balance has unknown bare inertia $I_0$ and stiffness $\kappa$. Two point masses, each $m=.050\,\mathrm{kg}$, are placed symmetrically at radius $r$. Squared periods are $T_1^2=1.00\,\mathrm{s^2}$ at $r_1=.10\,\mathrm{m}$ and $T_2^2=1.60\,\mathrm{s^2}$ at $r_2=.20\,\mathrm{m}$.

1. Infer $\kappa$ and $I_0$.

2. Predict the period without the point masses. Explain why period squared is linear in $r^2$, not in $r$.

Question 2 - Solution

Moving identical masses outward isolates a torsion balance inertia

(a) Derivation and calculation

$$I(r)=I_0+2mr^2,\qquad T^2=\frac{4\pi^2}{\kappa}(I_0+2mr^2).$$$$\Delta I=2(.050)(.20^2-.10^2)=.00300\,\mathrm{kg\,m^2}.$$$$\boxed{\kappa=\frac{4\pi^2\Delta I}{1.60-1.00}=.197392\,\mathrm{N\,m/rad}}.$$

(b) Result and interpretation

$$I_0=\frac{\kappa T_1^2}{4\pi^2}-2mr_1^2=.00500-.00100=\boxed{.00400\,\mathrm{kg\,m^2}}.$$$$T_0^2=\frac{4\pi^2 I_0}{\kappa}=.800\,\mathrm{s^2},\qquad\boxed{T_0=.89443\,\mathrm s}.$$

Physical check

The bare inertia is positive and the bare period is shorter than either loaded period.