Mass point sliding on a rotating line (Lagrangian)

[JulienB]

Homework Statement

Hi everybody! I have a crazy problem about Lagrangian on which I've been working on for two days without being able to figure it out. I have a solution, but I think there is a flaw in it. First here is the problem:

A line rotates with constant angle velocity $\omega$ about a vertical axis, which it intersects at a fixed angle $\upsilon$. Find the equation of motion of a mass point $m$ sliding without friction on the line under the influence of gravity.

The Attempt at a Solution

I think the whole twist of the problem is about using the right coordinates. I chose cylindrical coordinates as they seemed to help me describe best the system, and here is what I've done:

$q = \rho$
$\varphi = \omega \cdot t$
$z = r \cdot \cos \upsilon$

Now since I use $\rho$ (distance between $m$ and the axis $z$) as my $q$, I want to get rid of that $r$ in $z$:

$\rho = r \cos (\frac{\pi}{2} - \upsilon) = r \sin \upsilon$
$z = r \sin (\frac{\pi}{2} - \upsilon) = r \cos \upsilon$
$\implies \frac{\rho}{z} = \frac{\sin \upsilon}{\cos \upsilon}$
$\implies z = \rho \cot \upsilon$

Here is my main question... To obtain that $z$, I had to take the moment where the line is perpendicular to the y-axis, but am I allowed to do that? I mean, when the line has rotated out of this position, the $r$ I project on my x-axis (see second graph in attached picture) is not anymore the "real" $r$, if you see what I mean.

Then to find the Lagrange function and the equation of motion was not too painful:

$L = \frac{1}{2} m \cdot \dot{\rho}^2 + \frac{1}{2} I \cdot \omega^2 + \frac{1}{2} \dot{z}^2 + m \cdot g \cdot z$
$= \frac{1}{2} m (\dot{\rho}^2 + \rho^2 \dot{\varphi}^2 + \dot{\rho}^2 \cot^2 \upsilon) + m \cdot g \cdot \rho \cdot \cot \upsilon$

$\implies \frac{\partial L}{\partial \rho} = m (\rho \dot{\varphi}^2 + g \cot \upsilon)$
$\mbox{and } \frac{d}{dt} \frac{\partial L}{\partial \dot{\rho}} = m\ddot{\rho} (1 + \cot^2 \upsilon)$

$\implies \ddot{\rho} - \frac{\dot{\varphi}^2 + g \cot \varphi}{1 + \cot^2 \varphi} \rho = 0$

Is that correct? I've done two things to check the result:

(1) See what happens when $\upsilon = \frac{\pi}{2}$, that is when the line is perfectly horizontal. The equation of motion becomes $\ddot{\rho} = \dot{\varphi}^2 \rho = \frac{v^2}{\rho} =$ centripetal acceleration which I guess makes sense since the line is still rotating.

(2) I found the equation of motion with respect to $\varphi$ and found $\ddot{\varphi} = 0$ which again makes sense since $\varphi = \mbox{ const}$. But I can't explain myself how the equation "knows" that, I don't have the feeling I have really inputed the fact that the angular velocity is constant anywhere! Or did I?

Another student found an equation of motion with respect to $r$, which I guess could also work though it looks quite a bit more difficult. That's what he got:

$\ddot{r} - \frac{\omega^2}{1 + \cos^2 \upsilon} r + \frac{g \cdot \cos \upsilon}{1 + \cos^2 \upsilon} = 0$

I don't know what to think about it. The verification (1) I did also works for his equation, it might even be the same in another coordinates system! But as far as I know, he also considered $r$ when the line is aligned with the x-axis, which I am still not sure is correct...

You guys always have good advices, hopefully you have an idea. :)Thanks a lot in advance for your answers.Julien.

 

[JulienB]

Actually in the case of the equation of motion of my friend, the verification (1) provides a slightly different result as mine:

$\ddot{r} = \omega^2 \cdot r = \frac{v^2}{r}$

That also looks like the centripetal acceleration, but with $r$ instead of $\rho$ (which is the distance between $m$ and the z-axis). I think mine is correct, because the distance between $m$ and the rotation axis is $\rho$, but I would not put my hand in the fire for that... What do you guys think?Julien.

EDIT: I just realized that none of our equations produce $\ddot{\rho}/\ddot{r} = -g$ for $\upsilon = 0$... Are we both wrong, or is there something I am missing here?

 

[TSny]

$\varphi$ is not a degree of freedom in this problem since $\dot{\varphi} = \omega =$ constant. There is only one degree of freedom. For me, it seemed more natural to let $r$ be the variable. It works out nicely with $r$ , but you can just as well use $\rho$. $r$ and $\rho$ are proportional to each other. A nice thing about the Lagrangian approach is that you do not need to adopt any particular coordinate system, such as Cartesian, cylindrical, spherical, etc. You can use any convenient "generalized coordinates."

Looks like you forgot the negative sign for the potential energy when setting up $L = T - V$.

You must have made an algebra error in solving for the equation of motion in terms of $\ddot{\rho}$. Notice that your equation contains $\dot{\varphi}^2+g \cos \varphi$ which is dimensionally inconsistent.

Also, you can let $\frac{1}{1+\cot \varphi ^2} = \sin ^2 \varphi$.

 

[TSny]

Another student found an equation of motion with respect to $r$, which I guess could also work though it looks quite a bit more difficult. That's what he got:

$\ddot{r} - \frac{\omega^2}{1 + \cos^2 \upsilon} r + \frac{g \cdot \cos \upsilon}{1 + \cos^2 \upsilon} = 0$
.

The last term on the left is incorrect.

 

[TSny]

EDIT: I just realized that none of our equations produce $\ddot{\rho}/\ddot{r} = -g$ for $\upsilon = 0$... Are we both wrong, or is there something I am missing here?

For $\upsilon = 0$, $\rho = 0 =$ a constant. So, $\dot{\rho} = \ddot{\rho} = 0$. The equation of motion for $\rho$ will just reduce to $0=0$. However, the equation of motion for $r$ should reduce to $\ddot{r} = -g$.

 

[JulienB]

@TSny Hi and thank you for your answer!

Taking in consideration all your remarks, I come up now with:

$\ddot{\rho} - \frac{\rho \dot{\varphi}^2 - g \cot \upsilon}{\sin^2 \upsilon} = 0$

I'm also going to try with r, though I am unsure how to treat the rotational energy then. I'm at a party, I sat in a corner to do the equation again so I'm only writing a short message now. Thanks for your answer! :P

Julien.

 

[TSny]

@TSny
$\ddot{\rho} - \frac{\rho \dot{\varphi}^2 - g \cot \upsilon}{\sin^2 \upsilon} = 0$

The $\sin^2 \upsilon$ factor is in the wrong place.

I'm also going to try with r, though I am unsure how to treat the rotational energy then.

Your expression for $T$ that you had in terms of $\rho$ is correct. You can easily rewrite it in terms of $r$ since $\rho$ is proportional to $r$.

I'm at a party,

Enjoy!

 

[JulienB]

The $\sin^2 \upsilon$ factor is in the wrong place.

Really? It was a factor of $\ddot{\rho}$ but then I divided the whole expression by $\sin^2 \upsilon$.

 

[TSny]

Really? It was a factor of $\ddot{\rho}$ but then I divided the whole expression by $\sin^2 \upsilon$.

The factor of $\ddot{\rho}$ was $1+\cot^2 \upsilon$, which does not equal $\sin^2 \upsilon$.

 

[JulienB]

Yes my mistake, I've realized that: $\ddot{\rho} - \sin^2 \upsilon (\rho \dot{\varphi}^2 - g \cot\upsilon) = 0$

 

[TSny]

$\ddot{\rho} - \sin^2 \upsilon (\rho \dot{\varphi}^2 - g \cot\upsilon) = 0$

Looks good.

 

[JulienB]

@TSny Thanks a lot, that's great. ^^ And with $r$ I get:

$\ddot{r} - \frac{\sin^2 \upsilon \dot{\varphi}^2}{1 + cos^2 \varphi} r + \frac{g \cos \upsilon}{1 + \cos^2 \upsilon} = 0$

To get that I wrote $\rho = r \cos (\frac{\pi}{2} - \upsilon) = r \sin \upsilon$. There might be a way to simplify that, I haven't looked into it yet.Julien.

 

[TSny]

And with $r$ I get:

$\ddot{r} - \frac{\sin^2 \upsilon \dot{\varphi}^2}{1 + cos^2 \varphi} r + \frac{g \cos \upsilon}{1 + \cos^2 \upsilon} = 0$

I don't think those denominators should be there.

To get that I wrote $\rho = r \cos (\frac{\pi}{2} - \upsilon) = r \sin \upsilon$.

Yes.

 

[JulienB]

Yeah I just did it again starting from x, y and z and I don't get those denominators anymore. Instead just:

$\ddot{r} - r \dot{\varphi}^2 \sin^2 \upsilon + g cos \upsilon = 0$

 

[TSny]

OK. Good work.

 

[JulienB]

@TSny Thanks a lot. At the end it was a good exercise since I solved the problem using cartesian, cylindrical and spherical coordinates. Using spherical coordinates was by far the easiest (the Lagrangian comes after two lines of calculation).Julien.