# Watt Rotational Speed Regulator's Lagrangian

• Hari Seldon
In summary, the conversation was about finding the Lagrangian of a system called "Watt Regulator". The individual was having trouble finding the kinetic energy of the system, as the book's calculation did not match their own. They realized they had been adding instead of projecting for the coordinates, and were able to correctly calculate the kinetic energy and 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?

Hari Seldon said:
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##

vanhees71 and 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}##?

Hari Seldon said:
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)

@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!

Hari Seldon

## 1. What is a Watt Rotational Speed Regulator's Lagrangian?

The Watt Rotational Speed Regulator's Lagrangian is a mathematical expression used to describe the motion of a Watt governor, which is a type of centrifugal governor used to regulate the speed of a steam engine. It is derived from the Lagrangian mechanics, a branch of classical mechanics.

## 2. How does the Watt Rotational Speed Regulator's Lagrangian work?

The Lagrangian takes into account the kinetic and potential energies of the governor system, as well as any external forces acting on it. By solving the equations of motion derived from the Lagrangian, the rotational speed of the governor can be determined.

## 3. What is the significance of the Watt Rotational Speed Regulator's Lagrangian?

The Lagrangian provides a more comprehensive and accurate description of the governor's motion compared to traditional methods. It also allows for the analysis of the system's stability and the effects of different parameters on its behavior.

## 4. How is the Lagrangian different from other mathematical models?

The Lagrangian is a generalized approach that can be applied to a wide range of mechanical systems, including the Watt governor. It takes into account all forces and energies involved in the system, making it more accurate and versatile compared to other models.

## 5. Can the Watt Rotational Speed Regulator's Lagrangian be applied to other types of governors?

Yes, the Lagrangian can be used to analyze the motion of other types of governors, such as the Porter governor and the Hartnell governor. However, the specific form of the Lagrangian may vary depending on the design and parameters of the governor system.

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