I want to find an equation of motion for non-uniform circular motion under the inlfuence of gravity (like, say, a pendulum) of the form r(t)=R cos⁡(θ(t)) i + R sin⁡(θ(t)) j where $\theta (t) $ is only influenced by gravity, and is measured like a trigonometric angle in the unit circle (counter-clockwise from x axis).

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The Lagrangian of the system would be:

$$ L = \frac{1}{2}m{R^2}{\dot \theta ^2} - mgR\sin \theta $$

I'd then apply the E-L equation and get:

$$ \frac{{\partial L}}{{\partial \theta }} = - mgR\cos \theta $$ $$ \frac{{\partial L}}{{\partial \dot \theta }} = m{R^2}\dot \theta $$

$$ \frac{d}{{dt}}\left( {\frac{{\partial L}}{{\partial \dot \theta }}} \right) = m{R^2}\ddot \theta $$

And the differential equation I'd need to solve would be:

$$ \ddot \theta = - \frac{g}{R}\cos \theta $$

The initial conditions I had in mind were that $ \dot \theta (t=0) = \dot \theta _0 $ while $ \theta (t=0) = 0$.

I, however, have no clue how to deal with the cosine term. I know that this non-linear ODE has no analytical solution, but I have no clue how to approximate it either. I cannot use small angles because I want $ \theta $ to be able to do a full circle.

Any help would be appreciated, cheers.

  • $\begingroup$ If you don't want to make an approximation you could just solve it numerically using Eulers method or RK4 $\endgroup$
    – bemjanim
    Mar 29, 2020 at 21:22
  • $\begingroup$ since there is no analytical solution you have to do it numerically. but you can find numerical solutions as graphs in the net $\endgroup$
    – trula
    Mar 29, 2020 at 21:22
  • $\begingroup$ I know that this non-linear ODE has no analytical solution It does have an analytic solution in terms of the Jacobi elliptic sine function sn. See pgccphy.net/ref/nonlin-pendulum.pdf $\endgroup$
    – G. Smith
    Mar 29, 2020 at 21:28
  • $\begingroup$ Your initial conditions are strange. You want to call the initial angular velocity $\theta_0$? $\endgroup$
    – G. Smith
    Mar 29, 2020 at 21:36
  • 1
    $\begingroup$ Exact solution for thé non-linear pendulum. $\endgroup$
    – Farcher
    Mar 29, 2020 at 22:48

1 Answer 1


You can use $E=\frac 12 \dot \theta^2+ V(\theta)$ being constant and separate variables to get something like $$ \int_0^t dt +const. = \int_0^{\theta(t)} \frac{ d\theta}{\sqrt{2(V(\theta)-E)}}, $$ but, except for the case that the pendulum starts at rest at the topmost point, you are looking at an elliptic integral on the RHS. The start-at-top case is easy though. The answer (measuring $\theta$ from the topmost point) is $$ \theta(t)= 4 \tan^{-1} (e^{At}) $$ for some constant $A$ depending on $m$, $g$, $R$ etc. The general solution requires Jacobi functions.

  • $\begingroup$ Pretty sure the constant $A$ won't depend on $m$. Also, the kinetic energy should be $\frac{1}{2} m R^2 \dot{\theta}^2$. $\endgroup$ Mar 29, 2020 at 22:57
  • $\begingroup$ I set all constants to unity. The OP should do the algbera :) $\endgroup$
    – mike stone
    Mar 29, 2020 at 23:00
  • $\begingroup$ In that case, the constant $A$ should be a function of 1, 1, 1, etc. :-) $\endgroup$ Mar 29, 2020 at 23:00
  • $\begingroup$ If it starts at rest at the top, it just stays there, in unstable equilibrium, as shown in the 6th example here. $\endgroup$
    – G. Smith
    Mar 30, 2020 at 0:04
  • $\begingroup$ @G.Smith At top plus $\epsilon$... It takes infinitely long for the $\theta(t)=4\tan^{-1} (e^t)$ to make the round trip. This is the space part of the Sine-Gordon Soliton solution. $\endgroup$
    – mike stone
    Mar 30, 2020 at 11:58

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