Watt Rotational Speed Regulator's Lagrangian

[Hari Seldon]

Homework Statement

I am trying to find the Lagrangian of a system called "Watt Regulator". My kinetic energy doesn't match the kinetic energy on the book. I would like to know if I am calculating the coordinates of the system in the right way.

Relevant Equations

Kinetic energy of the whole system: $T = m v^2 = m (\dot x^2+\dot y^2)$

I understand that it is a system with two degrees of freedom. And I chose as generalized coordinates the two angles shown in the pic I posted. I am having troubles in finding the kinetic energy of this system, cause the book tells me that the kinetic energy is something different then what I calculated.
So very likely I am wrongly calculating the $x$ and the $y$.
Indeed I calculated the $x$ as
$$x = l \sin{\varphi}+l \sin{\theta}$$
and the $y$ as
$$y = l \cos{\varphi}+l cos{\theta}$$
Then I calculated the derivative of $x$ and $y$:
$$\dot x=l \dot \varphi \cos{\varphi}+l \dot \theta \cos{\theta}$$
$$\dot y=-l \dot \varphi \sin{\varphi}-l \dot \theta \sin{\theta}$$
So my kinetic energy would be
$$T = m (\dot x^2+\dot y^2)=m l^2 [\dot \theta^2 +\dot \varphi^2+2\dot \theta \dot \varphi (\cos{\varphi} \cos{\theta}-\sin{\varphi} \sin{\theta})]$$
While the kinetic energy that the book calculated is
$$T=m l^2 (\dot \theta^2+\dot \varphi^2 \sin^2{\theta})$$
What am I doing wrong?

 

[PhDeezNutz]

Homework Statement:: I am trying to find the Lagrangian of a system called "Watt Regulator". My kinetic energy doesn't match the kinetic energy on the book. I would like to know if I am calculating the coordinates of the system in the right way.
Relevant Equations:: Kinetic energy of the whole system: $T = m v^2 = m (\dot x^2+\dot y^2)$

View attachment 295144
I understand that it is a system with two degrees of freedom. And I chose as generalized coordinates the two angles shown in the pic I posted. I am having troubles in finding the kinetic energy of this system, cause the book tells me that the kinetic energy is something different then what I calculated.
So very likely I am wrongly calculating the $x$ and the $y$.
Indeed I calculated the $x$ as
$$x = l \sin{\varphi}+l \sin{\theta}$$
and the $y$ as
$$y = l \cos{\varphi}+l cos{\theta}$$
Then I calculated the derivative of $x$ and $y$:
$$\dot x=l \dot \varphi \cos{\varphi}+l \dot \theta \cos{\theta}$$
$$\dot y=-l \dot \varphi \sin{\varphi}-l \dot \theta \sin{\theta}$$
So my kinetic energy would be
$$T = m (\dot x^2+\dot y^2)=m l^2 [\dot \theta^2 +\dot \varphi^2+2\dot \theta \dot \varphi (\cos{\varphi} \cos{\theta}-\sin{\varphi} \sin{\theta})]$$
While the kinetic energy that the book calculated is
$$T=m l^2 (\dot \theta^2+\dot \varphi^2 \sin^2{\theta})$$
What am I doing wrong?

I think you are adding each of the individual projections instead of projecting twice (i.e. multiplying)

(Interesting that they would define $\phi$ from $y$-axis but it should work out)

Also you have two particles not just one. (merely multiply $\frac{1}{2}$ by $2$ won't work because their coordinates are not the same)

$x_1$ (The closer one in the pic) should be calculated as follows

Project into the $xy$-plane

$\ell \sin \theta$

now project again onto the $x-axis$

$x_1 = \ell \sin \theta \sin \phi$

Do this for $y_1, z_1, x_2, y_2, z_2$

 

[Hari Seldon]

Thank you very much for your help!
Yes, I was adding instead then projecting. So the coordinates should be the following?
$$ x_{1}=l\sin{\theta}\sin{\varphi}$$
$$ y_{1}=l\cos{\theta}\cos{\varphi}$$
$$ x_{2}=-l\sin{\theta}\sin{\varphi}$$
$$ y_{2}=l\cos{\theta}\cos{\varphi}$$
Do I need also $z_{1}$ and $z_{2}$?

 

[PhDeezNutz]

Thank you very much for your help!
Yes, I was adding instead then projecting. So the coordinates should be the following?
$$ x_{1}=l\sin{\theta}\sin{\varphi}$$
$$ y_{1}=l\cos{\theta}\cos{\varphi}$$
$$ x_{2}=-l\sin{\theta}\sin{\varphi}$$
$$ y_{2}=l\cos{\theta}\cos{\varphi}$$
Do I need also $z_{1}$ and $z_{2}$?

I would think so because as $\theta$ changes so would $z_1$ and $z_2$ (which are the same)

 

[PhDeezNutz]

@Hari Seldon I just realized I didn't full respond to your previous post

I agree with your $x_1$ and $x_2$ but not your $y_1$ and $y_2$.

I'm getting

$x_1 = \ell \sin \theta \sin \phi$

$y_1 = \ell \sin \theta \cos \phi$

$z_1 = \ell \cos \theta$

$x_2 = - \ell \sin \theta \sin \phi$

$y_2 = - \ell \sin \theta \cos \phi$

$z_1 = \ell \cos \theta$

Remember finding the $x$-projection involves two steps 1) projecting onto the $xy$-plane and 2) projecting onto the $x-axis$

Likewise

Remember finding the $y$-projection involves two steps 1) projecting onto the $xy$-plane and 2) projecting onto the $y-axis$

The first two steps are the same so they should both have the same first factor.

 

[Hari Seldon]

Perfect!
So given the following coordinates of the two masses:
$$x_{1}=l\sin{\theta}\sin{\varphi}~~~x_{2}=-l\sin{\theta}\sin{\varphi}$$
$$y_{1}=l\sin{\theta}\cos{\varphi}~~~y_{2}=-l\sin{\theta}\cos{\varphi}$$
$$z_{1}=l\cos{\theta}~~~~~z_{2}=l\cos{\theta}$$
We can calculate the derivative of them
$$\dot x_{1}=l\dot \theta \cos{\theta}\sin{\varphi}+l\dot \varphi \cos{\varphi} \sin{\theta}$$
$$\dot y_{1}=l\dot \theta \cos{\theta}\cos{\varphi}-l\dot \varphi \sin{\varphi} \sin{\theta}$$
$$\dot z_{1}=-l\dot \theta \sin{\theta}$$

$$\dot x_{2}=-l\dot \theta \cos{\theta}\sin{\varphi}-l\dot \varphi \cos{\varphi} \sin{\theta}$$
$$\dot y_{2}=-l\dot \theta \cos{\theta}\cos{\varphi}+l\dot \varphi \sin{\varphi} \sin{\theta}$$
$$\dot z_{2}=-l\dot \theta \sin{\theta}$$

Then we can calculate the kinetic energy of the system
$$T=T_{1}+T_{2}$$
$$T_{1}=\frac {1} {2} m v_{1}^{2}=\frac {1} {2} m (\dot x_{1}^{2}+\dot y_{1}^{2}+\dot z_{1}^{2})=\frac {1} {2} m l^2 (\dot \theta^2 \cos^{2}{\theta} \sin^{2}{\varphi}+\dot \varphi^2 \cos^{2}{\varphi} \sin^{2}{\theta}+2 \dot \theta \dot \varphi \cos{\theta} \cos{\varphi} \sin{\theta} \sin{\varphi}+\dot \theta^2 \cos^{2}{\theta} \cos^{2}{\varphi}+\dot \varphi^2 \sin^{2}{\varphi} \sin^{2}{\theta}-2 \dot \theta \dot \varphi \cos{\theta} \cos{\varphi} \sin{\theta} \sin{\varphi}+\dot \theta^2 \sin^{2}{\theta})$$
$$T_{1}=\frac {1} {2} m l^2(\dot \theta^2 \cos^{2}{\theta}+\dot \varphi^2 \sin^{2}{\theta}+\dot \theta^2 \sin^{2}{\theta})=$$
$$=\frac {1} {2} m l^2(\dot \theta^2+\dot \varphi^2 \sin^{2}{\theta})$$Similarly for $T_{2}$ we find
$$T_{2}=\frac {1} {2} m l^2(\dot \theta^2+\dot \varphi^2 \sin^{2}{\theta})$$
And then
$$T=m l^2 (\dot \theta^2+\dot \varphi^2 \sin^{2}{\theta})$$

Now we calculate the potential energy
$$V=-mgz_{1}-mgz_{2}=-2mgl \cos{\theta}$$
Finally we can write the Lagangian
$$L=T-V=ml^2(\dot \theta^2+\dot \varphi^2 \sin^{2}{\theta})+2mgl \cos{\theta}$$

Thank you very much for your help @PhDeezNutz !

 

[PhDeezNutz]

Looks great!