Q) An insect crawls up a hemispherical surface very slowly.The coeffiecient of friction is $\mu$ between surface and insect.If line joining the centre of hemispherical surface to the insect makes an angle $\alpha $ with the vertical, find the maximum possible value of $\alpha$. enter image description here

With the force method, the solution can be found as at the highest point the frictional force would be equal to gravitational force.Therefore, $$\mu mg\cos\alpha=mg \sin\alpha$$ $$\implies \cot \alpha=1/\mu$$

However, when I tried to do this by energy conservation,equating the total frictional force with potential energy the answer was different.
Let $\theta$ be angle covered by it and $d\theta$ be a small angle covered by it. $$mgr(1-\cos\alpha)=\int_0^\alpha \mu (mg\cos\theta )*rd\theta$$ $$mgr(1-\cos\alpha)=\mu mgr \sin\alpha$$ $$2\sin^2\frac{\alpha}{2}=\mu 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}$$ $$\cot\frac{\alpha}{2}=1/\mu$$
Why is the answer different if I used force or if i use energy conservation?

  • $\begingroup$ @gandalf61 Let us continue this discussion in chat. $\endgroup$ Commented Apr 6, 2020 at 15:35
  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Commented Apr 7, 2020 at 4:23

2 Answers 2


First, I am assuming that there is no kinetic friction acting on the insect as it moves up the bowl. If kinetic friction were involved, you would have energy dissipation, but I will not consider that here.

Your mistake is in assuming that the static friction force is equal to its maximum value during the entire process. $\mu N$ only determines the maximum magnitude the static friction force can have before slipping occurs; it doesn't always hold for the static friction force magnitude. Before slipping, the static friction force is just equal to the force needed to prevent slipping, i.e. $mg\sin\theta$.

Doing this correctly, you will then see that the integral will give you a true expression, but it won't help you find where the ant slips because the integral is true for any angle $\alpha$ before slipping occurs, and the integral doesn't tell you anything about when the static friction force fails. i.e. energy conservation doesn't apply only when slipping occurs, so energy conservation won't help you solve this problem.

Also, technically the static friction force can't do work because the point of contact between the ant and the bowl doesn't move as the force is being applied, but that point isn't important here, as the (correct) integral will still give the work done by the insect's legs on the rest of the insect, even if the physical interpretation isn't correct.

  • $\begingroup$ "you will then see that the integral will give you a true expression" : could you please perhaps give this correct integral to clarify how it should be done? $\endgroup$ Commented Apr 6, 2020 at 15:42
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    $\begingroup$ @user8736288 $$\int_0^\alpha\sin\theta\,\text d\theta=1-\cos\alpha$$ $\endgroup$ Commented Apr 6, 2020 at 15:45
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    $\begingroup$ I was taught $F\leq \mu N$; perhaps it would be useful for other readers to see it explicitly stated like that. $\endgroup$ Commented Apr 7, 2020 at 10:53

The difference in energy between the two static equilibrium positions may only be some potential energy difference. You may assume the friction force is $F=\mu N$ during sliding, where $\mu$ is the kinetic friction coefficient (taken equal to the static friction coefficient) but since this force is non conservative, the work done this force will not account for any potential energy change, instead, it's lost. The balance in energy between the two positions will thus only tell you that the change in potential energy is the work of the weight force, which is not helpful for the determination of $\alpha$.


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