Constant solutions to the Lagrange equations of motion

[bobred]

Homework Statement

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

Homework Equations

The equations of motion

and

are

(1)

(2)

So

can be eliminated from equation (1)

The Attempt at a Solution

Would I go about this by setting

equal to a constant,

, and similarly for

?

 

[MarcusAgrippa]

Would I go about this by setting

equal to a constant,

, and similarly for

?

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.

 

[Ray Vickson]

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.

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

 

[MarcusAgrippa]

What do you get when you differentiate a constant?

 

[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.

 

[Ray Vickson]

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.

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.

 

[bobred]

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