An Angular Simple Harmonic Oscillator — Question 8

Question 8

Degrees must be converted before identifying torsional stiffness

A calibration plot reports restoring torque as $\tau=-0.020\,\theta_{\mathrm{deg}}\,\mathrm{N\,m}$, where $\theta_{\mathrm{deg}}$ is the numerical angle in degrees. The rotational inertia is $I=.010\,\mathrm{kg\,m^2}$.

1. Convert the law to radians and find the period.

2. How wrong would the period be if $0.020$ were used directly as the torsion constant in SI? Explain the square-root factor.

Question 8 - Solution

Degrees must be converted before identifying torsional stiffness

(a) Derivation and calculation

$$\theta_{\mathrm{deg}}=\frac{180}{\pi}\theta,\quad\tau=-.020\theta_{\mathrm{deg}}=-\left(.020\frac{180}{\pi}\right)\theta.$$$$\boxed{\kappa=1.145916\,\mathrm{N\,m/rad}}.$$$$\boxed{T=2\pi\sqrt{.010/1.145916}=.58696\,\mathrm s}.$$

(b) Result and interpretation

Using the degree slope as a radian stiffness would give$$T_{\mathrm{wrong}}=2\pi\sqrt{.010/.020}=4.44288\,\mathrm s.$$$$\boxed{\frac{T_{\mathrm{wrong}}}{T}=\sqrt{180/\pi}=7.56940}.$$The dynamical equation uses radians because angular acceleration and rotational kinetic energy use radian measure.

Physical check

The actual stiffness is $57.2958$ times the numerical slope per degree.