Exploiting conservation laws it is possible to show that if we have a bar pendulum in a massive box on plane without friction,

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the equation that rules the function $\theta=\theta(t)$ (pendulum angle with respect to vertical) is \begin{equation} \frac{2}{3} l \ddot{\theta} - \frac{1}{2} \frac{m}{m+M} l \ddot{\theta} \cos^2 \theta +\frac{1}{2} \frac{m}{m+M} l \dot{\theta}^2 \sin \theta \cos \theta + g \sin \theta = 0 \end{equation} But at this stage the book says that for small oscillations the equation became \begin{equation} \frac{4M+m}{6(M+m)} l \ddot{\theta} = -g \theta \end{equation} from which we easily get $T \approx 2\pi \sqrt{ \frac{4M + m}{6(M+m)} \frac{l}{g}}$. But how works the simplification of this differential equation for small angles? What about the missing term $\frac{1}{2}\frac{m}{m+M}l\dot{\theta}^2\theta$, why it is negligible for small oscillations? Making approximations on solutions is easy and safe, but find solutions can be prohibitive and this is the case: we need simplifications in solving equation to solve it. But these are delicate and dangerous steps, I'd like to see clearer how to move in this situation, without finding a solving equation badly simplified, giving wrong solutions.


Sameer suggests this interesting way to solve. The solution of simplified differential equation satisfy both $|\dot{\theta}^2 \theta| \ll \theta$ and $|\dot{\theta}^2 \theta| \ll \ddot{\theta}$ (this is decidely true if amplitude of oscillations are small), that we used to simplify differential equation. But I wonder if this way of solving the problem is legitimate, always safe. I upvoted but to accept I would be sure that this is the answer (the starting of the answer "I think..." doesn't leave me sure). I hope someone will confirm.


I think it's like the chemistry thing one does in solving problems of chemical equilibrium. One assumes that a variable is small, one gets the result, one checks if the variable that was assumed to be small is indeed small. If it's small, then viola!

Here, after getting $\theta=\theta_0\sin(\omega t+\phi)$, we check $\dot \theta=\theta_0\omega\cos(\omega t+\phi)\le \theta_0\omega$ is not so greater than $\theta_0$ for situations in which $\omega=\sqrt{\frac{6(M+m)}{4M+m}\frac gl}$ is not so large.

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The term is negligible because three small values are multiplied - $\dot{\theta}$, $\dot{\theta}$ and $\theta$, whereas in the other terms, only one such term appears - $\theta$ or $\ddot{\theta}$. Because these values are of the same order (related by multiplication by a constant*), in the limit case of $\theta \rightarrow 0$, the ration of this term to either of the other two vanishes.

*Constant here meaning "value independent of $\theta$, but possibly dependent on time".

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  • $\begingroup$ To choose which ignore I should know how much are infinitesimal terms, and this doesn't look self evident to me in the question. $\endgroup$ – Fausto Vezzaro Jul 11 at 12:26
  • $\begingroup$ Do you mean that you want actual values you can use to judge if the term is negligible in a given situation? $\endgroup$ – Kotlopou Jul 11 at 12:37
  • $\begingroup$ I can't undetstand what do you mean by "actual value". I can't understand how we can be sure they are of the same order, this is fundamental and it is not self evident to me. $\endgroup$ – Fausto Vezzaro Jul 11 at 12:56
  • $\begingroup$ I cannot give a rigorous argument. The reason I assumed that was by analogy with other oscillators, such as simple harmonic motion, where this holds. Theta and its derivatives are all cyclical in time with the same phase, since knowing just theta defines the system by conservation of energy, except for sign of velocity. (EDIT: We also need conservation of horizontal momentum) $\endgroup$ – Kotlopou Jul 11 at 13:07
  • $\begingroup$ Then, to make the phasing work, you need all three values to be of the same order to cause changes. Acceleration must be big enough to cycle velocity, which must be big enough to cycle position, and the same argument goes backwards (acceleration cannot be of higher order than velocity and velocity of higher order than position). $\endgroup$ – Kotlopou Jul 11 at 13:15

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