- #36
Delta2
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Yeap, @haruspex is great and he has very great patience too :D.
Yes, that's ok, but I don't think it is that helpful. A further approximation will be required.kuruman said:We can approximate the bottom of the cup to a circle of radius R,
I don't see where a further approximation is required. The expression $$y=\frac{x^2}{R+\sqrt{R^2-x^2}}$$ is exact for a circular path. The only approximation comes in when one says that for ##x<<R##, one gets ##y\approx\dfrac{x^2}{2R}##, still for a circular path. Now it is always true that, for the actual parabolic path, ##y=Ax^2## for any ##x##. Therefore for small ##x## (in the parabolic path) one can match the parabola to a circle if one sets $$\frac{x^2}{2R}=Ax^2~\Rightarrow~R=\frac{1}{2A}.$$ For a pendulum following a circular path of radius ##R## and in the small angle approximation ##\omega=\sqrt{g/R}##. The answer follows.haruspex said:Yes, that's ok, but I don't think it is that helpful. A further approximation will be required.
That's the one.kuruman said:The only approximation comes in when one says that for x<<R,
Difficulty is in the eye of the solver. The benefit I see is that the method I outlined above, although less elegant than yours, makes the explanation more accessible to a beginner. I'll leave it at that because it's not a point worth belaboring.haruspex said:That's the one.
I am saying I don't see the benefit in approximating it as circular first. There is no great difficulty in writing the exact ODE for the parabola and then doing the small displacement approximation.