Barbell in gravitational field (Lagrange)

[JulienB]

@TSny Thanks again for your answer. I tried integrating without making any decision for the values to see where it goes:

$\dot{\theta} = k^2 \int \frac{k^2}{\sin^2 \theta \tan \theta} dt$
$\dot{\theta} = k^2 \big(\frac{-1}{2 \sin^2 \theta} + \dot{\theta}_0\big)$

$\implies \theta = k^2 \int \big(\frac{-1}{2 \sin^2 \theta} + \dot{\theta}_0\big) dt$
$\theta = \frac{k^2}{2 \tan \theta} + \dot{\theta}_0 t + \theta_0$

which by substituting back $k = \sin^2 \theta \dot{\varphi}$ gives

$\theta = \frac{1}{2} \frac{\sin^4 \theta}{\tan \theta} \dot{\varphi}^2 + \sin^4 \theta \dot{\varphi}^2 \dot{\theta}_0 + \sin^4 \theta \dot{\theta}^2 \theta_0$

Is that correct? Seems a bit crazy. If I define the initial conditions as $\theta_0 = 0$ and $\dot{\theta} = 0$, then I would get the special solution

$\theta = \frac{1}{2} \frac{\sin^4 \theta}{\tan \theta} \dot{\varphi}^2$.

Is that going the right direction? I'm a little skeptical :)Julien.

 

[TSny]

@TSny Thanks again for your answer. I tried integrating without making any decision for the values to see where it goes:

$\dot{\theta} = k^2 \int \frac{k^2}{\sin^2 \theta \tan \theta} dt$
$\dot{\theta} = k^2 \big(\frac{-1}{2 \sin^2 \theta} + \dot{\theta}_0\big)$

You integrated the integrand with respect to $\theta$ instead of $t$. So, this isn't right.

But you are not asked to solve the equations of motion in general. Just find one or more specific solutions corresponding to specific initial conditions.

Here's a rather trivial example just to illustrate. Suppose you held the barbell at rest such that the center of mass is located at $x = 0$, $y = 0$, and $z = 1$ m, and such that the barbell is initially oriented with $\theta = \pi/4$ and $\varphi = \pi/6$. You release it from rest.

So the initial conditions are

$x_0 = 0, \dot{x}_0 = 0$
$y_0 = 0, \dot{y}_0 = 0$
$z_0 = 1, \dot{z}_0 = 0$
$\theta_0 = \pi/4, \dot{\theta}_0 = 0$
$\varphi_0 = \pi/6, \dot{\varphi}_0 = 0$.

Solve the equations of motion for these initial conditions.

 

[JulienB]

$x_0 = 0, \dot{x}_0 = 0$
$y_0 = 0, \dot{y}_0 = 0$
$z_0 = 1, \dot{z}_0 = 0$
$\theta_0 = \pi/4, \dot{\theta}_0 = 0$
$\varphi_0 = \pi/6, \dot{\varphi}_0 = 0$.

Solve the equations of motion for these initial conditions.

Okay. Sorry for making so many mistakes, those concepts are still very abstract to me. I should be able to determine $k$ with the initial conditions, shouldn't I? But then I have $k = \sin^2 \theta \dot{\varphi} = \sin^2 \theta_0 \dot{\varphi_0} = \sin^2 \frac{\pi}{4} \cdot 0 = 0$ which would mean that the angle between the z-axis and the barbell is not accelerating under those initial conditions, or $\ddot{\theta} = 0$. But then, it means in other words that the velocity of $\theta$ is constant and since $\theta_0 = 0$ the angle would stay constant during the motion.

Are those correct assumptions or am I off again?

Thanks a lot.Julien.

 

[TSny]

Okay. Sorry for making so many mistakes, those concepts are still very abstract to me. I should be able to determine $k$ with the initial conditions, shouldn't I? But then I have $k = \sin^2 \theta \dot{\varphi} = \sin^2 \theta_0 \dot{\varphi_0} = \sin^2 \frac{\pi}{4} \cdot 0 = 0$ which would mean that the angle between the z-axis and the barbell is not accelerating under those initial conditions, or $\ddot{\theta} = 0$. But then, it means in other words that the velocity of $\theta$ is constant and since $\theta_0 = 0$ the angle would stay constant during the motion.

Yes. Except I chose $\theta_0 = \pi/4$.

 

[JulienB]

@TSny Oops yes. I meant $\dot{\theta}$. I think I understood a little now, I just need to practice. So then using your initial conditions again, I got for $\varphi$:

$\ddot{\varphi} = mr^2 \sin^2 \theta \dot{\varphi} = mr^2 k = 0$
$\implies \dot{\varphi} = const.$
$\dot{\varphi}_0 = 0 \implies \dot{\varphi} = 0 \implies \varphi = \varphi_0 = \frac{\pi}{6}$

Is that correct? I'm going to try to solve c and come back with a suggestion for the answer. Thanks a lot for your valuable help. If you ever come to Berlin let me know and you'll get a piece of cake and a coffee for free in our café :)Julien.

 

[JulienB]

I gave it a go and that's what I came up with so far. Since I am in the middle of that lesson, I attempted solving it with informations found here and there on the internet and in our script, so I hope it makes sense.

$\frac{\partial V}{\partial t} = 0 \implies E = T+V = const. \implies \mbox{ Energy is conserved.}$

$\mbox{Total (linear) momentum: } P = \sum p_i = (m_1 + m_2)(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)$
$\frac{d}{dt} P = (m_1 + m_2)(\ddot{x}^2 + \ddot{y}^2 + \ddot{z}^2) = (m_1 + m_2)(0 + 0 + g^2) \neq 0$
$\implies \mbox{Linear momentum is not conserved.}$

$\mbox{Angular momentum: } \frac{d}{dt} \frac{\partial L}{\partial \dot{\varphi}} = 0$
$\implies \frac{\partial L}{\partial \dot{\varphi}} = const. \implies \mbox{ Angular momentum is conserved.}$

Now I am not sure what represents $\theta$. Could it be that $\frac{\partial L}{\partial \dot{\varphi}} = mr^2 \dot{\varphi} \sin^2 \theta$ refers to the angular momentum with respect to the z-axis while $\frac{\partial L}{\partial \dot{\theta}} = mr^2 \dot{\theta}$ would refer to the angular momentum with respect to the x,y-plane? I would add that whatever it is I think it is not conserved, because $\frac{\partial L}{\partial \dot{\theta}} \neq 0$.

I have some doubts about the linear momentum. I thought that $x$, $y$, $z$ were the only translations possible for $m_1$ and $m_2$. Could that be right?Julien.

 

[TSny]

I got for $\varphi$:

$\ddot{\varphi} = mr^2 \sin^2 \theta \dot{\varphi} = mr^2 k = 0$

The first equality is not true (I don't think you meant to type that). However, $mr^2 \sin^2 \theta \dot{\varphi} = mr^2 k$ is correct.

$\implies \dot{\varphi} = const.$

This is true, but I'm not sure how you deduce that. Note that $mr^2 \sin^2 \theta \dot{\varphi} = mr^2 k$ implies that $\sin^2 \theta \; \dot{\varphi}$ is constant. But this by itself does not imply that $\dot{\varphi}$ is constant.

$\dot{\varphi}_0 = 0 \implies \dot{\varphi} = 0 \implies \varphi = \varphi_0 = \frac{\pi}{6}$

OK. Here's how I would get those results. The initial condition $\dot{\varphi}_0 = 0$ implies that the constant $k = 0$ (using $mr^2 \sin^2 \theta \dot{\varphi} = mr^2 k$ ). Therefore, $mr^2 \sin^2 \theta \dot{\varphi} = 0$ at all times.

You also have $\ddot{\theta} = \frac{k^2}{\sin^2 \theta \tan \theta}$. Since $k = 0$ in this example, $\ddot{\theta} = 0$ at all times. Since you also have the initial condition $\dot{\theta} = 0$, $\theta$ remains constant at the initial value $\pi/4$.

Then, going back to $mr^2 \sin^2 \theta \dot{\varphi} = 0$ at all times, you can conclude that $\dot{\varphi} = 0$ at all times. So, $\dot{\varphi}$ remains constant at $\pi/6$.

Thanks a lot for your valuable help. If you ever come to Berlin let me know and you'll get a piece of cake and a coffee for free in our café :)

Thanks! Maybe I will.

 

[TSny]

I gave it a go and that's what I came up with so far. Since I am in the middle of that lesson, I attempted solving it with informations found here and there on the internet and in our script, so I hope it makes sense.

$\frac{\partial V}{\partial t} = 0 \implies E = T+V = const. \implies \mbox{ Energy is conserved.}$

OK, energy is conserved.

$\mbox{Total (linear) momentum: } P = \sum p_i = (m_1 + m_2)(\dot{x}^2 + \dot{y}^2 + \dot{z}^2)$
$\frac{d}{dt} P = (m_1 + m_2)(\ddot{x}^2 + \ddot{y}^2 + \ddot{z}^2) = (m_1 + m_2)(0 + 0 + g^2) \neq 0$
$\implies \mbox{Linear momentum is not conserved.}$

Linear momentum is a vector. $\vec{P} = (m_1+m_2) \left( \dot{x} \hat{i} + \dot{y} \hat{j} +\dot{z} \hat{k} \right)$
Are there any components of the total linear momentum that are conserved?

$\mbox{Angular momentum: } \frac{d}{dt} \frac{\partial L}{\partial \dot{\varphi}} = 0$
$\implies \frac{\partial L}{\partial \dot{\varphi}} = const. \implies \mbox{ Angular momentum is conserved.}$

As you note below, this is showing that the angular momentum about the z-axis is conserved. More specifically, it is the z-component of the angular momentum of the two masses as measured relative to the center of mass of the barbell. That is, it's the z-component of angular momentum due to the rotation of the barbell about its center of mass. It does not include any angular momentum due to translation of the center of mass of the barbell.

Now I am not sure what represents $\theta$. Could it be that $\frac{\partial L}{\partial \dot{\varphi}} = mr^2 \dot{\varphi} \sin^2 \theta$ refers to the angular momentum with respect to the z-axis while $\frac{\partial L}{\partial \dot{\theta}} = mr^2 \dot{\theta}$ would refer to the angular momentum with respect to the x,y-plane? I would add that whatever it is I think it is not conserved, because $\frac{\partial L}{\partial \dot{\theta}} \neq 0$.

Yes this a little tricky. The quantity $\frac{\partial L}{\partial \dot{\theta}} = mr^2 \dot{\theta}$ can be interpreted as the angular momentum relative to the center of mass about an axis parallel to the xy plane. This axis is perpendicular to the vertical plane that contains the z-axis and instantaneous position of the particle. But this axis is not fixed. This axis changes direction as $\varphi$ changes. So, $mr^2 \dot{\theta}$ does not represent a component of angular momentum about a fixed axis in the inertial (earth-based) frame. As you note, this quantity is not conserved.

 

[JulienB]

The first equality is not true (I don't think you meant to type that).

The initial condition $\dot{\varphi}_0 = 0$ implies that the constant $k = 0$ (using $mr^2 \sin^2 \theta \dot{\varphi} = mr^2 k$ ). Therefore, $mr^2 \sin^2 \theta \dot{\varphi} = 0$ at all times.

I've pretty much done the same, I think you misinterpreted my first equality, or maybe I should have it described it more precisely. Anyway that was very instructive for me.

Linear momentum is a vector. $\vec{P} = (m_1+m_2) \left( \dot{x} \hat{i} + \dot{y} \hat{j} +\dot{z} \hat{k} \right)$
Are there any components of the total linear momentum that are conserved?

I think all linear momentum taking place in the x and y directions are conserved, but not in the z direction. That makes sense, since an external force (gravity) is applying in that direction.

Yes this a little tricky. The quantity $\frac{\partial L}{\partial \dot{\theta}} = mr^2 \dot{\theta}$ can be interpreted as the angular momentum relative to the center of mass about an axis parallel to the xy plane. This axis is perpendicular to the vertical plane that contains the z-axis and instantaneous position of the particle. But this axis is not fixed. This axis changes direction as $\varphi$ changes. So, $mr^2 \dot{\theta}$ does not represent a component of angular momentum about a fixed axis in the inertial (earth-based) frame. As you note, this quantity is not conserved.

If I understand correctly, for this angular momentum to not be conserved, there must be an external force applying and the only one I see is gravity. It is indeed a little difficult to grasp, but it would make sense since you describe the axis parallel to the xy plane.

Thanks a lot, it's always nice to get some physical intuition! See you in another post. ^^

Thanks! Maybe I will.

Anytime!Julien.