How can I tell that circular motion is a solution for a particle confined to the surface of a cone?

Question

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

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

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.

Answer 1

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.

Answer 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

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

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.