Newton's Law of Gravitation — Question 9

Question 9

A torsion balance measures a small gravitational force through a couple

A torsion balance has two small masses $m$ at opposite ends of a light horizontal rod, each distance $\ell$ from the pivot. Two source masses $M$ each attract one small mass perpendicularly to the rod from distance $r$, producing torques in the same sense. Neglect cross-attractions. The fiber restoring torque is $\kappa\theta$.

1. Derive an expression for $G$ from the equilibrium angle.

2. For $\kappa\theta=8.0\times10^{-9}\,\mathrm{N\,m}$, $\ell=.10\,\mathrm m$, $r=.050\,\mathrm m$, $M=10\,\mathrm{kg}$ and $m=.15\,\mathrm{kg}$, infer $G$.

Question 9 - Solution

A torsion balance measures a small gravitational force through a couple

(a) Derivation and calculation

Each source supplies $F=GMm/r^2$ perpendicular to a lever arm $\ell$. Both moments have the same sign: $$\tau_g=\ell F+\ell F=\frac{2\ell GMm}{r^2}.$$ Static torque balance requires $$\kappa\theta=\tau_g,\qquad \boxed{G=\frac{\kappa\theta r^2}{2\ell Mm}}.$$

(b) Result and interpretation

Substituting the supplied measurements, $$G=\frac{(8.0\times10^{-9})(.050)^2}{2(.10)(10)(.15)} =\boxed{6.67\times10^{-11}\ \mathrm{N\,m^2/kg^2}}.$$ The inferred force on either end is $$F=\frac{\kappa\theta}{2\ell}=4.0\times10^{-8}\ \mathrm N.$$

Physical check

The two end moments reproduce $8.0\times10^{-9}\,\mathrm{N\,m}$; neglected cross-attractions belong to the model assumptions.