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What assumptions do I need for a simple gravity pendulum?

I know the bob needs to be regarded as a particle. The rod needs to be massless and of constant length. And we ignore air resistance and friction.

But what other assumptions do I need in order to allow the derivation of an equation of motion from, say, Newtonian mechanics?

EDIT

After searching some more online, the only seemingly useful information I could find is this https://en.wikipedia.org/wiki/Pendulum_(mathematics)#Simple_gravity_pendulum

Is it accurate?

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  • $\begingroup$ The rod must not bend... You want no forces other than gravity, but I think you've covered that. $\endgroup$
    – garyp
    Commented Apr 24, 2018 at 22:53
  • $\begingroup$ Rigid systems with non-trivial mass distribution are just as easy as a simple pendulum if treated in terms of moment of inertia and torque, so you don't even need a point-like bob or a massless support. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Apr 24, 2018 at 23:43
  • $\begingroup$ @dmckee I see, yes. But the energy conservation properties of a simple pendulum, when those two assumptions are made, is much simpler to apply. And results obtained with them don't deviate too much from experimental data. $\endgroup$
    – Stephen
    Commented Apr 24, 2018 at 23:58
  • $\begingroup$ Steven I can formulate the so-called "physical pendulum" in exactly the same way. Not that it's obviously better, but it is more general, and you seemed to be interested in the minimal assumptions. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Apr 25, 2018 at 1:20

2 Answers 2

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You need to assume small swing amplitude or face nonlinear behaviour :-) .

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  • $\begingroup$ Really? Even without friction or air resistance? Why? $\endgroup$
    – Stephen
    Commented Apr 24, 2018 at 22:41
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    $\begingroup$ The equation of motion is $ml\ddot{\theta}=mg \sin \theta$. This reduces to a harmonic oscillator only for small $\theta$. $\endgroup$
    – my2cts
    Commented Apr 24, 2018 at 23:01
  • $\begingroup$ I see. I can't use small angles, so I'll face the non-linear behaviour. What are the assumptions of that? $\endgroup$
    – Stephen
    Commented Apr 24, 2018 at 23:49
  • $\begingroup$ After searching some more online, the only seemingly useful information I could find is this en.wikipedia.org/wiki/… Is it accurate? $\endgroup$
    – Stephen
    Commented Apr 25, 2018 at 0:30
  • $\begingroup$ I think you can safely rely on it. $\endgroup$
    – my2cts
    Commented Apr 25, 2018 at 17:07
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Well, most of the times we assume that gravity is purely acting downwards, while due to the earth not being flat it will vary a bit if the pendulum moves. It goes without saying that this effect is negligible and seldom considered.
For a perfect pendulum you also demand that the rod does not stretch during the oscillations.

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  • $\begingroup$ I'm not sure : ( I understand now that gravity needs to be assumed to be constant everywhere, and the ground flat. But how do I know that's all we need to assume in order to allow the derivation of an equation of motion from Newtonian mechanics? $\endgroup$
    – Stephen
    Commented Apr 24, 2018 at 22:45
  • $\begingroup$ After searching some more online, the only seemingly useful information I could find is this en.wikipedia.org/wiki/… Is it accurate? $\endgroup$
    – Stephen
    Commented Apr 25, 2018 at 0:30

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