Suppose a plumb bob hangs without swinging, then the string defines the effective direction of gravity. Suppose you are holding the bob on the surface of the earth at colatitude $\theta$, where $\theta$ measures the spherical angle as measured from the north pole, then the error $\epsilon$ between $\vec g_{eff}$ and $\vec g$ is given by

$$\epsilon \approx \frac{R \Omega^2 sin(\theta) cos(\theta)}{g-R \Omega^2 sin^2(\theta)}$$

We can find the maximum error, by differentiating $\epsilon$ with respect to $\theta$. I was curious just if we can answer this question conceptually.

  • $\begingroup$ It seems the denominator with $R\Omega$ has not correct units ? $\endgroup$
    – Cretin2
    Oct 20, 2022 at 4:23
  • $\begingroup$ @ Cretin2 good catch, I have corrected it. Thanks! $\endgroup$ Oct 20, 2022 at 4:25
  • $\begingroup$ The easiest way is to graph the function, no? And you only need a domain of 0 to 2pi. $\endgroup$
    – RC_23
    Oct 20, 2022 at 4:54
  • $\begingroup$ If you think that amount of calculation is tedious, just wait until you learn E&M or Quantum. $\endgroup$ Oct 20, 2022 at 11:45
  • $\begingroup$ @ Michael Seifert, experienced that in E&M already haha. I was just curious if there is a way to conceptually answer this. $\endgroup$ Oct 20, 2022 at 15:20

1 Answer 1


I will define $\alpha \equiv R\Omega^2/g$ so that we can write $$ \epsilon = \frac{\alpha\sin(\theta)\cos(\theta)}{1-\alpha\sin^2(\theta)}. $$ We can differentiate this using the quotient rule to find that $$ \frac{d\epsilon}{d\theta} = \frac{\alpha(1-\alpha\sin^2(\theta))(\cos^2(\theta)-\sin^2(\theta)) + 2\alpha^2\sin^2(\theta)\cos^2(\theta)}{(1-\alpha\sin^2(\theta))^2}; $$ this requires a little work but isn't too tedious. Setting $d\epsilon/d\theta = 0$ to find the maximum $\theta$, we get (after some algebra) $$ (1-\alpha)\tan^2(\theta_{\rm max}) = 1 \implies \theta_{\rm max}=\arctan\left(\sqrt{\frac{1}{1-\alpha}}\right). $$ Note that this implies the constraint $\alpha < 1$, or $R\Omega^2 < g$.


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