I'm working on a problem where a particle of mass $m$ is confined to the surface of an inverted half cone (and is circling downwards due to gravity), with the cone's half angle $\alpha$. I chose to use cylindrical coordinates $(z,\phi,\rho)$ and I used the Lagrangian to solve this problem.

After going through some math, I find the equation of motion for $z$, from which I can write that

$$\ddot{z}\sec(\alpha) - \frac{p^2_{\phi}}{m^2z^3tan^2(\alpha)}+ g = 0$$

Here, $p_{\phi}$ is the angular momentum, which is conserved. Only $z$ depends on time, the other expressions are all constants.

At this point, I've been told that it 'can be seen' that one solution to this is given by a circular motion at constant height $z_c$. I am then asked to impose a small perturbation $z = z_c + \eta$, and (keeping only first order terms in $\eta$) find the period with which $z$ will oscillate around $z_c$.

Now I am pretty clueless how to do this. First of all, how can you see that there is a circular motion at constant height $z_c$? I mean, I can plug in $z = z_c$ and solve for it, but then I don't see how to find the period of something like that with the small perturbation. All the perturbation does is add some terms, but I don't see how they are time dependent and I certainly don't see how to extract a period from it. Could someone perhaps suggest a 'plan of attack'?

If I do simply plug in $z = z_c$ I find that


which at least has the right units.

Moreover, plugging $z = z_c + \eta$ into the first equation and keeping only first order terms of $\eta$, I find that

$$z = \frac{2z_c}{3} - \frac{p^2_{\phi}}{3z_c^2gm^2\tan^2(\alpha)}$$

But I don't see any period in that.

  • $\begingroup$ This is actually a fine question. When we say homework questions are not allowed, we're talking about things like "how do I do this problem?" or "I'm not sure where to start" or "did I do this correctly?" Here you've shown your work and narrowed down the problem to the specific conceptual step you're stuck on, which is exactly the kind of question we like. It doesn't even matter, really, that it arose in the context of a homework question. $\endgroup$
    – David Z
    Mar 13, 2014 at 21:15
  • $\begingroup$ Thanks for the comment, that's very good to know. I think I might already have an idea for why my answer is wrong: I'm assuming $\eta$ to be constant, while maybe it doesn't have to be? I set its time derivative to zero, but not doing so might get me something more interesting. $\endgroup$
    – user129412
    Mar 13, 2014 at 21:29
  • $\begingroup$ Yes, this does seem to work, I'll write it out in a bit. $\endgroup$
    – user129412
    Mar 13, 2014 at 21:45

2 Answers 2


If you want to see whether a particular function $z(t)$ represents an allowed motion of the particle, all you need to do is check whether it satisfies the equation of motion (the differential equation in your question). If you plug the function in and you get a mathematical contradiction, it is not a solution. Otherwise, it is. (Sometimes you have to be careful about corner cases, but this is not one of those times.)

Perhaps it'll help you to think about it this way: when the problem says

it 'can be seen' that one solution to this is given by a circular motion at constant height $z_c$

that means there is some constant $z_c$ such that $z(t) = z_c$ is a solution to the differential equation. Now, in theory, you could systematically test every possible height until you found one that works - that is, plug $z(t) = 1\text{ m}$, $z(t) = 2\text{ m}$, etc. into the differential equation and see if it works out to be equal to zero, but of course the smarter way is to use algebra to identify the only value that might work, which you did. You found that


If you're not clear about how this shows that circular motion is a possible solution, I'd suggest plugging

$$z(t) = \left(\frac{p^2_{\phi}}{gm^2\tan^2(\alpha)}\right)^{\frac{1}{3}}$$

into the differential equation and checking for yourself that the left side does simplify to zero when you do this.

Now to the part about the perturbation. Forget the cone for a moment and think about a ball rolling along the bottom of a valley of some kind (a ditch or channel or tube). One way this can happen, of course, is that the ball rolls straight down the center. But another allowable motion is that the ball is a little off-center and that it moves slightly side-to-side as it rolls, tracing out some kind of oscillatory pattern centered on the bottom of the valley.

This is a common pattern for any sort of physical system in a stable equilibrium centered on some coordinate $x_c$: while one allowable motion is just being stuck at $x_c$, another allowable motion is some kind of small oscillation around $x_c$. So instead of solving for $x(t)$ directly, you change variables to $\delta(t) = x(t) - x_c$, It's frequently easier to solve for $\delta(t)$ than it is for $x(t)$, because you know that $\delta(t)$ is centered around zero and thus small, and when you write your formulas in terms of $\delta$ instead of $x$ you can expand them in Taylor series and throw away everything but the largest nontrivial terms.

In your case, you're doing this with $z(t) = z_c + \eta(t)$, instead of $x(t) = x_c + \delta(t)$. Different names (and meanings) for the variables, but the procedure is the same. You change variables from $z$ to $\eta$. Then you can expand in a Taylor series in $\eta$ and keep only the lowest-order nontrivial terms in $\eta$. Note that I do say nontrivial because you have to keep some terms which actually do involve $\eta$ in order to solve for it. Usually this means keeping up to order $\eta^1$, but in some cases there is a reason to keep higher-order terms as well - say, if all the $O(\eta^1)$ terms cancel out, or if you want a better approximation.

  • $\begingroup$ Thank you for the extensive answer. I overcomplicated the situation by assuming that there was something to show about circular motion (that x^2 + y^2 would be equal to a constant for this specific height and such), rather than just showing that there is a constant z that solves the equation. For the perturbation, that makes a lot of sense. The example clarifies it quite a bit! Making it time dependent solves the problem. Rather than just writing out the solutions, which is just math, I'll leave it at this so that others are not tempted to simply copy it. Your answer should suffice! $\endgroup$
    – user129412
    Mar 13, 2014 at 22:18
  • $\begingroup$ @user129412 Also note that the idea of keeping first order terms only is that the perturbation is small, so second order and higher terms are very small. For some systems this is okay to ignore and doesn't cause much problem. For other systems, including that very small higher order term may give more insight into the system. It all depends on the context and what you are looking for from the solution. $\endgroup$
    – tpg2114
    Mar 13, 2014 at 22:35
  • $\begingroup$ Hm yes, it's all about the ratio $\frac{\delta z}{z_c}$ I suppose. $\endgroup$
    – user129412
    Mar 13, 2014 at 22:44
  • $\begingroup$ @user129412 (3 comments up) ah, well remember the motion is restricted to the surface of a cone so it's automatically true that $x^2 + y^2 = \text{const}$ at a particular height. You don't need to show that. $\endgroup$
    – David Z
    Mar 13, 2014 at 23:43

Any time you linearize something (which is what you're doing with $z = z_c + \eta$), you are substituting in a constant value $z_c$ and a perturbation $\eta$. The constant $z_c$ is independent of time while $\eta$ is a function of time. Additionally, the time average of $\eta$ is 0 because $z_c$ is defined to be the mean of $z$ in time.

So with that out of the way, when you plug into the governing equation, you will get a function that has $\ddot{\eta}$ in it. This is the governing equation for the perturbation $\eta$ about $z_c$.

This is the governing equation you need to use to find the period of motion.


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