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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.

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  • $\begingroup$ 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. $\endgroup$
    – Triatticus
    Commented Jan 4, 2022 at 17:14
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    $\begingroup$ She has it on page 475, in the link I provided, on a footnote. $\endgroup$
    – zeynel
    Commented Jan 4, 2022 at 17:23
  • $\begingroup$ Yep I see that now, I can't really go through it on mobile, hopefully someone comes along with some clarification for you. $\endgroup$
    – Triatticus
    Commented Jan 4, 2022 at 18:45

2 Answers 2

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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 $L$

enter image description here

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.

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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 $$c\theta=\frac{GMm}{d^2}4a.$$

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