# Constant solutions to the Lagrange equations of motion

1. Apr 24, 2015

### bobred

1. The problem statement, all variables and given/known data
A particle moves on the surface of an inverted cone. The Lagrangian is given by

Show that there is a solution of the equations of motion where and take constant values and respectively

2. Relevant equations
The equations of motion and are

(1)

(2)

So can be eliminated from equation (1)

3. The attempt at a solution

Last edited by a moderator: Apr 17, 2017
2. Apr 24, 2015

### MarcusAgrippa

It is not $\ddot{r}, \dot{\phi}$ that you were asked to show provide a solution, but ${r}, \dot{\phi}$. These requirements define uniform circular motion. The question is thus asking you to use the equations to check that uniform circular motion is a solution.

3. Apr 24, 2015

### Ray Vickson

If $r(t) = r_0$ (a constant), what is $\ddot{r}(t)$?

4. Apr 24, 2015

### MarcusAgrippa

What do you get when you differentiate a constant?

5. Apr 24, 2015

### bobred

Differentiating a constant is zero.
I was thinking about the equations in isolation to each other and not as part of a system. So

$$v=r \dot{\phi}=r_0 \Omega$$

I just need to show that the equations of motion lead to this.

Last edited: Apr 24, 2015
6. Apr 24, 2015

### Ray Vickson

No: they do not "lead to" this; they merely allow this.

There is a big difference, since (due to the lack of friction) a solution with non-constant $r$ and $\dot{\phi}$ never settles down to a limiting orbit of constant $r$ and $\dot{\phi}$. However, if you start off with the correct initial conditions, your orbit maintains constant $r$ and $\dot{\phi}$ forever. Essentially, that is what you are being asked to verify.

7. Apr 24, 2015

### bobred

So setting the acceleration $\ddot{r}=0$ leaves a constant $r$ and $\dot{\phi}$ which allows the uniform motion.