The balance wheel of a watch is a thin ring of radius \(\displaystyle 0.95 \ \text{cm}\) and oscillates with a frequency of \(\displaystyle 3.10 \ \text{Hz}\). If a torque of \(\displaystyle 1.1 \times 10^{-5} \ \text{m}\cdot\text{N}\) causes the wheel to rotate \(\displaystyle 60^{\circ}\), calculate the mass of the balance wheel.
We have a torque and simple harmonic motion (oscillation), then we can use Hook's Law for Rotation:
\(\displaystyle \tau = K\theta\)
where \(\displaystyle K\) is the effective torsional constant.
Plug in the given numbers and solve for \(\displaystyle K\).
\(\displaystyle 1.1 \times 10^{-5} = K(60^{\circ})\frac{\pi}{180^{\circ}}\)
This gives:
\(\displaystyle K = 1.0504 \times 10^{-5} \ \frac{\text{m} \cdot \text{N}}{\text{rad}}\)
We can express the period in terms of the moment of inertia \(\displaystyle I\) and the effective torsional constant \(\displaystyle K\).
\(\displaystyle T = 2\pi\sqrt{\frac{I}{K}} = \frac{1}{f}\) (I will derive this result in future Episodes)
From accumulated knowledge we know that the moment of inertia of a thin ring is \(\displaystyle I = MR^2\).
Then,
\(\displaystyle 2\pi\sqrt{\frac{I}{K}} = 2\pi\sqrt{\frac{MR^2}{K}} = \frac{1}{f}\)
Plug in numbers.
\(\displaystyle 2\pi\sqrt{\frac{M0.0095^2}{1.0504 \times 10^{-5}}} = \frac{1}{3.1}\)
This gives:
\(\displaystyle M = \textcolor{blue}{0.31 \ \text{g}}\)
