Equations of motion of a particle over a cone+conserved quantities

[fluidistic]

Homework Statement

Exactly the same problem as https://www.physicsforums.com/showthread.php?p=3335113#post3335113 but instead of a cylinder, the surface is a cone.

Homework Equations

Same as previous thread.

The Attempt at a Solution

I used cylindrical coordinates $(r, \phi , z)$.
By intuition I know the angular momentum with respect to the z-axis (axis of symmetry of the cone) must be conserved, thus the Lagrangian must not contain $\phi$.
I reached that the Lagrangian $L=\frac{m}{2}(\dot r ^2 + r^2 \dot \phi ^2 + \dot z ^2)$.
Now, I believe I must express r in function of z. I notice that if $\theta$ is the angle worth half the angle of the vertex of the cone, then $\tan \theta = constant = \frac{r}{z} \Rightarrow r=k_1z$.
So now I have 2 cyclic coordinates and it means that the momentum $P_r$ is a constant. Since r is directly related to z, this also means that the z-component of the linear momentum is constant... Well I think so.

Calculating Lagrange's equations, I reach as equation of motion: $\ddot z (K+1) - k_2 z \dot \phi ^2=0$. This can be written as $K_3 \ddot z + K_2 z \dot \phi =0$. Where $K_3 >1$.
Am I right?
If so, how can I solve it?! What method would do the job?

 

[ideasrule]

So now I have 2 cyclic coordinates and it means that the momentum $P_r$ is a constant. Since r is directly related to z, this also means that the z-component of the linear momentum is constant... Well I think so.

Your Lagrangian and your equation for r are both right, but this conclusion isn't. The Lagrangian's variables are not independent, so you need to somehow include the restraint r=k_1*z. There are two ways to do this: by directly substituting r=k_1*z into the Lagrangian, or by using Lagrange multipliers. If you use the first method, you get rid of r, so it becomes meaningless to talk about P_r. You also introduce an explicit dependence on z, so P_z isn't constant. If you use the second method, the Lagrangian would have an explicit dependence on both r and z, so neither P_r nor P_z are constant. Either way, P_r and P_z are not constants of motion.

Calculating Lagrange's equations, I reach as equation of motion: $\ddot z (K+1) - k_2 z \dot \phi ^2=0$. This can be written as $K_3 \ddot z + K_2 z \dot \phi =0$. Where $K_3 >1$.
Am I right?
If so, how can I solve it?! What method would do the job?

That's right, but you need another equation to deal with the phi dependence--the equation you get by differentiating the Lagrangian with respect to phi and its time derivative.

 

[fluidistic]

Your Lagrangian and your equation for r are both right, but this conclusion isn't. The Lagrangian's variables are not independent, so you need to somehow include the restraint r=k_1*z. There are two ways to do this: by directly substituting r=k_1*z into the Lagrangian, or by using Lagrange multipliers. If you use the first method, you get rid of r, so it becomes meaningless to talk about P_r. You also introduce an explicit dependence on z, so P_z isn't constant. If you use the second method, the Lagrangian would have an explicit dependence on both r and z, so neither P_r nor P_z are constant. Either way, P_r and P_z are not constants of motion.

Thank you very much. I hadn't been clear enough, I had replaced $r$ by $k_1z$ into my Lagrangian and I used it in order to calculate the Lagrange's equations.

That's right, but you need another equation to deal with the phi dependence--the equation you get by differentiating the Lagrangian with respect to phi and its time derivative.

Ah I see. This gives me $2 \dot r \dot \phi + r \ddot \phi =0$.
If I do the same for the $r$ coordinate, I get $\ddot z =0$. I am not sure I can use this coordinate for the Lagrange's equations however. (In the link I gave in the first post of this thread, I've learned that if the coordinate is constant then I must NOT use it for the Lagrange's equations. Now, if a coordinate is dependent from another, I do not know, hence my doubt).

 

[ideasrule]

Ah I see. This gives me $2 \dot r \dot \phi + r \ddot \phi =0$.
If I do the same for the $r$ coordinate, I get $\ddot z =0$. I am not sure I can use this coordinate for the Lagrange's equations however. (In the link I gave in the first post of this thread, I've learned that if the coordinate is constant then I must NOT use it for the Lagrange's equations. Now, if a coordinate is dependent from another, I do not know, hence my doubt).

Why is r still in the Lagrangian? I thought you replaced r by k_1*z.

 

[fluidistic]

Why is r still in the Lagrangian? I thought you replaced r by k_1*z.

My bad... sorry about this error. I did replace it but when I derived the Lagrangian with respect to phi I took my old Lagrangian by error.
I'll fix this as soon as possible. Thanks for pointing this out.

 

[fluidistic]

So I get $2 \dot z \dot \phi +z \ddot \phi =0$.
For z, I get $\ddot z K_3+K_2 z \dot \phi ^2 =0$.

 

[ideasrule]

So I get $2 \dot z \dot \phi +z \ddot \phi =0$.

I think you made an algebra error somewhere. The Lagrangian has no explicit dependence on phi, so the partial of L wrt the time derivative of phi should be constant.

 

[fluidistic]

I think you made an algebra error somewhere. The Lagrangian has no explicit dependence on phi, so the partial of L wrt the time derivative of phi should be constant.

Here is my Lagrangian: $\frac{m}{2}(K \dot z ^2 + k_2 z^2 \dot \phi ^2 + \dot z ^2)$.
$\frac{\partial L}{\partial \dot \phi}= mk_2 z^2 \dot \phi$.
Thus $\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot \phi} \right )=mk_2(2\dot z z \dot \phi+ z^2 \ddot \phi )$. Since $\frac{\partial L}{\partial \phi}=0$, I have that $mk_2(2\dot z z \dot \phi+ z^2 \ddot \phi )=0$ and since $m$, $z$ and $k_2 \neq 0$, I reach the expression I wrote in my previous post.

 

[ideasrule]

Here is my Lagrangian: $\frac{m}{2}(K \dot z ^2 + k_2 z^2 \dot \phi ^2 + \dot z ^2)$.
$\frac{\partial L}{\partial \dot \phi}= mk_2 z^2 \dot \phi$.
Thus $\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot \phi} \right )=mk_2(2\dot z z \dot \phi+ z^2 \ddot \phi )$. Since $\frac{\partial L}{\partial \phi}=0$, I have that $mk_2(2\dot z z \dot \phi+ z^2 \ddot \phi )=0$ and since $m$, $z$ and $k_2 \neq 0$, I reach the expression I wrote in my previous post.

My bad, that equation is actually correct. However, I don't think it's as useful as writing $mk_2 z^2 \dot \phi=C$, since the time derivative of $\frac{\partial L}{\partial \dot \phi}$ is 0. That way, you'd be able to express the time derivative of phi in terms of z, and substitute the result into the equation of motion for z.

 

[fluidistic]

My bad, that equation is actually correct. However, I don't think it's as useful as writing $mk_2 z^2 \dot \phi=C$, since the time derivative of $\frac{\partial L}{\partial \dot \phi}$ is 0. That way, you'd be able to express the time derivative of phi in terms of z, and substitute the result into the equation of motion for z.

Oh this is bright.
I reach $\ddot z z K_3+C_2=0$. Is this a Cauchy-Euler equation?
So I have only 1 equation of motion and I have to solve it in order to get the trajectory (since they ask for it)?