= Cavendish torsion balance
{c}
A <Cavendish torsion balance> is a <torsion balance> arranged to measure gravitational <torque> from nearby source <masses>. Its suspended horizontal rod rotates about a vertical fiber. In the linear elastic regime, a twist <angle> $\theta$, measured from the fiber's untwisted orientation, gives a restoring <torque> $\tau_{\mathrm{fiber}}=-\kappa\theta$, where $\kappa>0$ is the <torsion constant>. The minus sign is essential: the fiber resists the twist. Under an approximately constant gravitational <torque>, the rotational equation using <moment of inertia> $I$ and <angular acceleration> $\ddot\theta$ is
$$
I\ddot\theta=\tau_g-\kappa\theta,\qquad \theta_{\mathrm{eq}}=\frac{\tau_g}{\kappa}.
$$
At <equilibrium>, the two <torques> balance. The <angle> is a displacement from the fiber's reference orientation, not the largest excursion during an <oscillation>. https://www.phys.uconn.edu/~hamilton/phys258/N/cav.pdf[University of Connecticut, The Cavendish Experiment, pp. 2–3: restoring torque and equilibrium deflection]{ref}
For the toy parameters in the <Cavendish experiment>, the predicted <equilibrium> twist is
$$
\theta_{\mathrm{eq}}=\frac{5.33944\times10^{-11}}{1.0\times10^{-8}}=0.00533944\,\mathrm{rad}\simeq0.3059^\circ.
$$
Here $L|\theta_{\mathrm{eq}}|/r\simeq0.0107$, consistent with using a small geometric displacement in the toy model, but not an estimate of its total <measurement error>. Reversing the source positions ideally takes the <equilibrium> <angle> from $+\theta_{\mathrm{eq}}$ to $-\theta_{\mathrm{eq}}$: the magnitude of the change is $2|\theta_{\mathrm{eq}}|\simeq0.6119^\circ$. A stable zero offset then cancels in the difference. A real reading may track a reflected light spot; the <law of reflection> doubles a mirror's rotation in the outgoing beam, so the spot's angular change must not be confused with the rod's <angle>. https://sciencedemonstrations.fas.harvard.edu/presentations/cavendish-experiment[Harvard Natural Sciences Lecture Demonstrations, Cavendish Experiment: mirror readout and source reversal]{ref}
How can the fiber's <torsion constant> be found without already knowing the gravitational <force>? Measure the <oscillation period> and calculate the <moment of inertia>. For the same toy rod, neglecting its <mass> and treating both test <masses> as <point masses>, $I=mL^2+mL^2=2mL^2$. With the source <torque> absent or treated as constant, let $\phi=\theta-\theta_{\mathrm{eq}}$ be a small departure from <equilibrium>. Neglect <damping> and any change of gravitational <torque> with <angle>. The rotational equation becomes <simple harmonic motion>:
$$
I\ddot\phi+\kappa\phi=0,\qquad \phi(t)=A\cos(\omega t+\delta),\qquad \omega=\sqrt{\frac{\kappa}{I}},\qquad T=2\pi\sqrt{\frac I\kappa},\qquad \kappa=\frac{4\pi^2I}{T^2}.
$$
Here $A$ is the angular amplitude, $\delta$ specifies the initial phase, and $\omega$ is the <angular frequency>. A softer fiber gives a larger <equilibrium> twist but also a longer <oscillation period>: greater sensitivity costs observation time. An undamped <oscillation> does not settle at <equilibrium>; <damping> is needed for settling. https://www.phys.uconn.edu/~hamilton/phys258/N/cav.pdf[University of Connecticut, The Cavendish Experiment, p. 3, equation (7): period calibration neglecting damping]{ref}
The toy <moment of inertia> and undamped <oscillation period> are
$$
I=2(0.010)(0.10)^2=2.0\times10^{-4}\,\mathrm{kg\,m^2},\qquad T=2\pi\sqrt{\frac{2.0\times10^{-4}}{1.0\times10^{-8}}}\simeq888.5766\,\mathrm s\simeq14.81\,\mathrm{min}.
$$
Timing this <oscillation period> would recover the assumed <torsion constant>. The units check explains the result: <torque> divided by <torsion constant> is an <angle> in <radians>, and $I/\kappa$ has units of <time> squared when <radians> are treated as dimensionless in <SI>.
\Image[/torsion-balance-period-calibration.png]
{title=Undamped small torsional oscillations calibrate the restoring torque: a softer fiber has a longer period}
Combining the <equilibrium> twist with the period calibration, the two-near-pair model would infer the <gravitational constant> as
$$
G=\frac{\kappa|\theta_{\mathrm{eq}}|r^2}{2LmM}=\frac{4\pi^2Lr^2|\theta_{\mathrm{eq}}|}{MT^2}.
$$
The test <mass> cancels in the second expression because it enters both the gravitational <torque> and the toy <moment of inertia>. If the measured quantity is the magnitude $|\Delta\theta|$ of the change between opposite source positions, use $|\theta_{\mathrm{eq}}|=|\Delta\theta|/2$. Substituting the toy twist and <oscillation period> returns the input $G$; it is a consistency check, not independent experimental evidence.
The sensitivity of a <Cavendish torsion balance> also explains its practical difficulty. Air motion can exert unwanted <torque>; unequal <temperatures> can drive that motion. Cavendish discusses this effect on p. 471 and describes observing from outside a closed room to avoid disturbing the instrument. https://www.infinitoteatrodelcosmo.it/wp-content/uploads/2021/07/cavendish-1997-xxi-experiments-to-determine-the-density-of-the-earth.pdf[Cavendish (1798), original paper, p. 471: temperature-driven air currents and remote observation]{ref} Mechanical vibrations can excite the same slow <oscillations> used to calibrate the <torsion constant>, while changes in the fiber or source separations alter the inferred <force>. Precision therefore depends on resolving the <equilibrium> <angle>, calibrating the full suspended <moment of inertia> and <torsion constant>, and modeling the actual source geometry. The two-near-pair model omits opposite-source attraction, rod contributions, finite-size effects, <damping>, and the variation of gravitational <torque> as the rod turns. https://sciencedemonstrations.fas.harvard.edu/presentations/cavendish-experiment[Harvard Cavendish demonstration: sensitivity, settling, and separation-dependent measurement]{ref}
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