# Moment of inertia of the Cavendish balance

In her paper, Henry Cavendish: The man and the measurement, Isobel Falconer uses, Newton's force equation as, $$\mu \theta = 4 G M m a / d^2$$ for the force that turns the arm 1 radian. Where does she get the factor of 4 and why $$a$$, the half-length of the pendulum enters this equation?

I copied her full analysis here. And the entire paper can be found here.

• I must be having a hard time finding it because from what I can see this equation appears nowhere in the paper, where did she make this separate analysis? Cavendish did it by measuring the period of the pendulum, and I don't see that in her analysis. Commented Jan 4, 2022 at 17:14
• She has it on page 475, in the link I provided, on a footnote. Commented Jan 4, 2022 at 17:23
• Yep I see that now, I can't really go through it on mobile, hopefully someone comes along with some clarification for you. Commented Jan 4, 2022 at 18:45

When the large spheres are moved from one side to the other, the rod swings through an angle $$\theta$$, but $$\frac {\theta}{2}$$ from the equilibrium position, line $$L$$
The torques acting on the rod to keep it in position $$A$$ were
$$\mu \frac{\theta}{2} = 2\frac{GMm}{d^2} \times a$$
where the $$2$$ on the right hand side is due to there being another pair of masses on the bottom left of the diagram, the result follows.
When one big sphere is to the right of its neighbouring little sphere, and the other big sphere is to the left of its neighbouring little sphere, a couple is exerted on the torsion bar. The torque of the couple is $$T=\text{magnitude of one force}\times \text{perpendicular separation of forces}$$ That is $$T=\frac{GMm}{d^2}2a$$ When the big spheres are on the other sides of the small spheres (but still with centres of neighbouring spheres distance $$d$$ apart), the torque is reversed. So the change in torque is $$\Delta T=\frac{GMm}{d^2}4a$$ So if the torque per unit angle of twist in the suspension is $$c$$, the angle turned through because of the reversal of gravitational torque will be given at equilibrium by $$c\theta=\frac{GMm}{d^2}4a.$$