Equivalence of Euler-Lagrange equations and Cardinal Equations for a rigid planar system

[l4teLearner]

Homework Statement

Check the equivalence of cardinal equations w.r.t Euler-Lagrange equations for a simple rigid planar system of a rod freely moving in space.

Consider the system depicted in the figure (see solution attempt) where a (thin) uniform rod of mass $m$ and length $l$ (section of the rod is negligible) is freely moving in two dimensional space. Given the lagrangian coordinates $r$, $\varphi$ and $\psi$ shown in the picture, calculate the equations of motion for the body using Lagrange-Euler equations, then using the cardinal equations and finally compare the result. The two descriptions must be equivalent.

Relevant Equations

Koenig theorem for the Kinetic energy of a rigid body
$$T=\frac{1}{2}mv_p^2+\frac{1}{2}mI{\omega}^2$$

Euler Lagrange equations
$$L_r=\frac{d}{dt}\left(\frac{\partial}{\partial \dot r}(L) \right)-\frac{\partial}{\partial r}(L)=0$$
$$L_\varphi=\frac{d}{dt}\left(\frac{\partial}{\partial \dot \varphi}(L) \right)-\frac{\partial}{\partial \varphi}(L)=0$$
$$L_\psi=\frac{d}{dt}\left(\frac{\partial}{\partial \dot \psi}(L) \right)-\frac{\partial}{\partial \psi}(L)=0$$

Cardinal equations for a rigid body.
$$C_x=m\frac{dv_x}{dt}=0$$
$$C_y=m\frac{dv_y}{dt}=0$$
$$C_\omega=I\frac{d}{dt}(\omega)=0$$

I express the total kinetic energy of the body, via König theorem, as

$$T=\frac{1}{2}mv_p^2+\frac{1}{2}mI{\omega}^2$$

where $$v_p=(v_x,v_y)=(\dot{r}\cos\varphi-r\dot{\varphi}\sin\varphi-\frac{l}{2}(\dot\varphi-\dot\psi)\sin(\varphi-\psi),\dot r \sin\varphi+r\dot\varphi \cos\varphi+\frac{l}{2}(\dot\varphi-\dot\psi)\cos(\varphi-\psi))$$ is the velocity of the center of mass, $\omega=\dot\varphi-\dot\psi$ is the angular velocity and $I=\frac{1}{12}ml^2$ is the moment of inertia of the rod with respect to its center of mass.

Doing the calculations (which I double checked with various platforms whose name starts with "M", so I am not sure the problem lies in the calculations) I get

$$T=\frac{1}{2}(\dot r^2+r^2\dot\varphi^2+\frac{l^2}{3}(\dot\varphi-\dot\psi)^2+l(\dot\varphi-\dot\psi)(r\dot\varphi \cos\psi+\dot r \sin\psi))$$

and the following Euler-Lagrange equations

$$L_r=\frac{d}{dt}\left(\frac{\partial}{\partial \dot r}(L) \right)-\frac{\partial}{\partial r}(L)=m (-2 \dot{\varphi}^2 r + 2 \ddot{r} +l \sin(\psi) (\ddot{\varphi} - \ddot{\psi} ) +l \cos(\psi)(\dot{\varphi}-\dot{\psi})^2 )/2=0$$

$$L_\varphi=\frac{d}{dt} \left(\frac{\partial}{\partial \dot \varphi}(L)\right)-\frac{\partial}{\partial \varphi}(L)=
\frac{m}{3} l^2(\ddot{\varphi} - \ddot{\psi}) +
m l \cos(\psi)\left(\dot{\varphi} \dot{r} + \ddot{\varphi} r - \frac{\ddot{\psi} r}{2}\right)
+ m l \sin(\psi) \left( \frac{\dot{\psi}^2 r}{2} - \dot{\varphi} \dot{\psi} r+ \frac{\ddot{r}}{2} \right) + m\ddot{\varphi} r^2+ 2m\dot{\varphi} \dot{r} r
=0$$

$$L_\psi=\frac{d}{dt}\left(\frac{\partial}{\partial \dot \psi}(L)\right)-\frac{\partial}{\partial \psi}(L)=
-\frac{m}{3} l^2 (\ddot{\varphi} - \ddot{\psi})- m l \cos(\psi) \left(\dot{r}\dot{\varphi} + \frac{\ddot{\varphi} r}{2}\right) -\frac{m}{2} l \sin(\psi)(\ddot{r}- r \dot{\varphi}^2 )
=0$$

If instead I calculate the cardinal equations, where I indicate quantities as:

- $C_x=0$ for the conservation of linear momentum along the x-axis
- $C_y=0$ for the conservation of linear momentum along the y-axix
- $C_\omega=0$, for the conservation of angular momentum

I get the following equations:

$$C_x=m\frac{dv_x}{dt}=m(\ddot r \cos \varphi -2 \dot r \dot \varphi \sin \varphi -r \ddot \varphi \sin \varphi -r \dot \varphi^2 \cos \varphi-\frac{l}{2}(\ddot \varphi - \ddot \psi)\sin(\varphi-\psi)-\frac{l}{2}(\dot\varphi-\dot\psi)^2\cos(\varphi-\psi)=0$$

$$C_y=m\frac{dv_y}{dt}=m(\ddot r \sin \varphi + 2 \dot r \dot \varphi \cos \varphi + r \ddot \varphi \cos \varphi -r \dot \varphi^2 \sin \varphi+\frac{l}{2}(\ddot \varphi - \ddot \psi)\cos(\varphi-\psi)-\frac{l}{2}(\dot\varphi-\dot\psi)^2\sin(\varphi-\psi)=0$$

$$C_\omega=I\frac{d}{dt}(\omega)=\frac{1}{12}ml^2\dot\omega=\frac{1}{12}ml^2(\ddot\varphi-\ddot\psi)=0$$

In particular, I do get $$C_x \cos\varphi+C_y \sin\varphi=L_r$$ which seems promising to me, but I cannot find any way to retrieve $C_\omega$ from $L_r$, $L_\varphi$, $L_\psi$.

I need to understand, am I just having an algebraic computation problem here, or is there a flaw in my reasoning? I am not asking to check my calculations, (you are welcome of course if you want) but to check the methodology. Maybe I am making some wrong assumptions on the $T$, $C$ and $\omega$ quantities?

thanks

 

[anuttarasammyak]

I observe that x,y coordinates of center of inertia is
(r \cos\phi+\frac{l}{2} \cos \theta, r \sin\phi+\frac{l}{2} \sin \theta)
\omega = \dot{\theta}
where
\theta = \phi - \Psi
I hope it could be your help to check your result.
L(r,\dot{r},\phi,\dot{\phi},\theta,\dot{\theta})=\frac{m}{2}[ \ \dot{r}^2+r^2\dot{\phi}^2+\frac{l^2}{3}\dot{\theta}^2+l\dot{\theta}\cos(\phi-\theta)r\dot{\phi}+l\dot{\theta}\sin(\phi-\theta)\dot{r} \ ]

 

[l4teLearner]

I observe that x,y coordinates of center of inertia is
(r \cos\phi+\frac{l}{2} \cos \theta, r \sin\phi+\frac{l}{2} \sin \theta)
\omega = \dot{\theta}
where
\theta = \phi - \psi
I hope it could be your help to check your result.

hi anuttarasammyak, thanks for your help. yes, I evaluated the angular velocity as $\dot \phi - \dot \psi$, it's one of the formulas I wrote just under the picture, in the first paragraph.

 

[anuttarasammyak]

I think the variables are complex and indirect. By changing the variables, center of inertia is
(r \cos\phi+\frac{l}{2} \cos \theta, r \sin\phi+\frac{l}{2} \sin \theta)=(R\cos \xi, R\sin\xi)
Lagrangian should be
L(R,\dot{R},\xi,\dot{\xi},\theta,\dot{\theta})=\frac{m}{2}[ \ \dot{R}^2+R^2\dot{\xi}^2+\frac{l^2}{12}\dot{\theta}^2\ ]
where
R=\sqrt{r^2+\frac{l^2}{4}+rl\cos\Psi}
\xi=\tan^{-1}\frac{r \sin\phi+\frac{l}{2} \sin (\phi-\Psi)}{r \cos\phi+\frac{l}{2} \cos(\phi-\Psi)}
\theta=\phi-\Psi

 

[l4teLearner]

I think the variables are complex and indirect. By changing the variables, center of inertia is
(r \cos\phi+\frac{l}{2} \cos \theta, r \sin\phi+\frac{l}{2} \sin \theta)=(R\cos \xi, R\sin\xi)
Lagrangian should be
L(R,\dot{R},\xi,\dot{\xi},\theta,\dot{\theta})=\frac{m}{2}[ \ \dot{R}^2+R^2\dot{\xi}^2+\frac{l^2}{12}\dot{\theta}^2\ ]
where
R=\sqrt{r^2+\frac{l^2}{4}+rl\cos\Psi}
\xi=\tan^{-1}\frac{r \sin\phi+\frac{l}{2} \sin (\phi-\Psi)}{r \cos\phi+\frac{l}{2} \cos(\phi-\Psi)}
\theta=\phi-\Psi

thank you! I will attempt the calculations with the lagrangian that you provided and will let you know.