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This is the question from Goldstein's Classical Mechanics, 2nd edition. Chapter 3 problem 1.

A particle of mass $m$ is constrained to move under gravity without friction on the inside of a paraboloid of revolution whose axis is vertical. Find the one-dimensional problem equivalent to its motion. What is the condition on the particle's initial velocity to produce circular motion? Find the period of small oscillations about this circular motion.

The equation of motion is found to be, $$m\ddot{r}+4mc^2r^2\ddot{r}+4mc^2r{\dot{r}}^2-mr{\dot{\theta}}^2+2mgcr=0$$

The condition for circular orbit is found to be, $$\dot{\theta}=\sqrt{2gc}$$ on substituting $\dot{\theta}$ in EOM we get, $$(1+4mc^2r^2)\ddot{r}+4mc^2r\dot{r}^2=0$$

I dont think the particle execute harmonic motion in radial co-ordinate $r$, because the first term consists of $\ddot{r}$ and second term consists of $r\dot{r}^2$ (not simply $r$), so I dont think that I can be able to reduce this expression into a simple harmonic oscillation form and the particle cannot execute harmonic oscillation in radial direction? Am I right?

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  • $\begingroup$ Write the Lagrangian and isolate the term that is not the kinetic energy, what do you get? $\endgroup$
    – caverac
    Commented May 17, 2017 at 16:15

2 Answers 2

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This is an overview of what I would do.

  1. Write down the Lagrangian of a particle under the gravitational potential in cilindrical coordinates. Use the paraboloid equation $z = r^2 c$ to constraint it on the surface of the paraboloid.

  2. Find the EOM for $r$ and $\theta$. If you set in them $r = R_0$, where $R_0$ is a constant, you find the condition for a circular movement. It will be $\sqrt{2gc}$ and the velocity is thus $\vec{v}=R_0 \hat{\theta} \sqrt{2gc}$.

  3. Notice the system has rotational symmetry and thus the angular momentum will be conserved. Use this to eliminate $\dot{\theta}$ in the EOM for $r$ and you will get the EOM for the 1D equivalent problem.

  4. Perform a Legendre transformation to find the Hamiltonian. All terms not invoving the radial momentum $p_r = m \dot{r}$ will be your effective potential.

  5. Find the mimimun of the effective potential (derive it and equate to 0). Then Taylor expand it around this minimun. The first term will be am irrelevant constant, the second is zero (since you are expanding around a minimun) and the third one has the form $8mgc(r-R_0)^2$. From the coeficient $8mgc$ you can get the period.

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  • $\begingroup$ If $c$ has units of $L^{-1}$, then aren't the units of $\sqrt{2gc}$ equal to $\sqrt{ [L/T^2] [L]^{-1}} = [T]^{-1}$ as desired? $\endgroup$ Commented May 17, 2017 at 19:48
  • $\begingroup$ Of course! My mistake. I am editing the answer. $\endgroup$
    – WilhelmM
    Commented May 17, 2017 at 21:10
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You forgot the fact that the $\theta$ motion will no longer be uniform once the particle is no longer in circular motion: angular momentum is conserved in this situation, which implies that if $r$ changes then $\dot{\theta}$ must change as well. So you have to look for a solution of the equations that takes both of these perturbations into account.

If we write out the full Euler-Lagrange equations of motion for this situation, the $r$ equation is as you've found above, while the $\theta$ equation is $$ \frac{d}{dt} \left( m r^2 \dot{\theta} \right) = 0 \qquad \Rightarrow \qquad \dot{\theta} = \frac{L_z}{m r^2}, $$ where $L_z$ is a constant of the motion. This gives us a relationship between $r$ and $\dot{\theta}$ for any type of motion, circular or not. We can use this relationship to eliminate $\dot{\theta}$ from the $r$ equation, obtaining $$ m\ddot{r}+4mc^2r^2\ddot{r}+4mc^2r{\dot{r}}^2-\frac{L_z^2}{m r^3}+2mgcr=0 $$ This is now an effective one-dimensional problem. To find circular motion, you can try an ansatz of the form $r(t) = r_0$ and figure out what $L_z$ (and thereby $v$) must be for this to occur. You can then look for solutions of the form $r(t) = r_0 + \epsilon(t)$, discarding all terms that are $\mathcal{O}(\epsilon^2)$ and higher, to see if harmonic oscillations are possible. (Spoiler alert: they are.)

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