# Lagrangian of 2 rotating masses on a spring, sliding down plane

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1. Dec 9, 2016

### 1v1Dota2RightMeow

1. The problem statement, all variables and given/known data
2 masses are connected by a spring. They are on a frictionless plane inclined relative to the horizontal by $\alpha$. The masses are free to slide, rotate about their center of mass, and oscillate.

1. Find the Lagrangian as a sum of the Lagrangian for the COM motion and a Lagrangian for the relative motion.

2. From this equation, determine and solve the equations of motion for the COM and the relative motion.

3. Determine the constants of the motion

4. Determine the motions of the COM and the relative coordinate

2. Relevant equations
$L_{total}=L_{COM}+L_{rel}$

$L=KE-PE$

$KE=(1/2)mv^2$

$PE=mgh$

$KE_{rot}=(1/2)I\omega ^2$

$PE_{spring}=(1/2)kx^2$

$I=mr^2$ because we are dealing with point masses

$\omega = \frac{v}{r}$

3. The attempt at a solution

1.

$L_{COM} = KE_{COM}-PE_{COM}$

$KE_{COM} = (1/2)Mv^2=(1/2)(\frac{m_1m_2}{m_1+m_2})v^2$ , but what would $v$ be here? And am I correct in using the reduced mass?

$PE_{COM} = Mgh(t) = (\frac{m_1m_2}{m_1+m_2})gh(t)$ , but how do I determine $h(t)$?

$L_{rel}=KE_{rel}-PE_{rel}$

$KE_{rel}$ is comprised of both rotational motion and oscillatory motion:

$KE_{rel} = (1/2)I_1\omega_1^2 + (1/2)I_2\omega_2^2 + (1/2)m_1v_1^2+(1/2)m_2v_2^2$

$PE_{rel} = (1/2)kd_1^2+(1/2)kd_2^2$, where $d$ is the distance past the equilibrium point.

I can't move on to the other parts without understanding this first part. Could someone please help me understand?

2. Dec 10, 2016

### Orodruin

Staff Emeritus
Let us stop and think for a while. How does the CoM move? What would you think would be the corresponding velocity?

Again, let us stop and think. What would happen to the kinetic energy if the masses were not moving relative to each other (i.e., if there is only the CoM motion) and one of the masses was much larger than the other one, say $m_1 \gg m_2$?

If you have problems with this, I suggest that you first write the Lagrangian down by using the coordinates of the masses instead of the CoM and the relative motion. Once you have done that, you can make a coordinate transformation to find the Lagrangian in terms of the CoM coordinates and the separation of the masses.

3. Dec 10, 2016

### 1v1Dota2RightMeow

I guess the CoM would slide straight down the plane and accelerate with an acceleration magnitude proportional to gravity. Although the problem doesn't explicitly state it, I assume that it has a nonzero initial velocity.

Wait, if they're not moving relative to each other, i.e. not rotating or oscillating, then it would just move down the plane in whatever orientation it started with. In that case, a more massive system just means more kinetic energy.

4. Dec 10, 2016

### Orodruin

Staff Emeritus
The initial velocity does not really matter for the formulation of the Lagrangian. What I am asking is that you write down an expression for the position of the centre of mass and then think about what velocity should go into the kinetic energy. Hint: You are allowed to use the centre of mass position here!

I suggest you do what I said and consider what the kinetic energy would be in that case. Does it match your expression? It has nothing to do with the potential or the actual motion, just care about the kinetic terms.

Edit: Also, pay attention to my last comment. Do it using the mass positions and then make a change of variables if you are having trouble.

5. Dec 10, 2016

### 1v1Dota2RightMeow

Ok so the position of the center of mass is just $x_{COM} = \frac{m_1x_1+m_2x_2}{m_1+m_2}$. Looking at this equation, I guess the velocity would just be the derivative of this, so $v_{COM} = \dot x_{COM}$, right?

So then using the above, the KE is just $(1/2)Mv_{COM} ^2 =(1/2)(\frac{m_1m_2}{m_1+m_2})(\dot x_{COM})^2$ where $m_1 >> m_2$. I'm sorry, I don't understand the point of making one mass extremely large. To me, it just means a larger KE at the bottom.
*Edited to use reduced mass. I read on (Wikipedia)[https://en.wikipedia.org/wiki/Reduced_mass] that the reduced mass is used to convert a two-body problem into a one-body problem, which is what we're doing here by interpreting this problem as a COM problem, so it seems like the correct mass to use.

6. Dec 10, 2016

### Orodruin

Staff Emeritus
Right.

No, it is not the correct mass to use. Consider the case when $m_1 \gg m_2$. You then find
$$\mu = \frac{m_1m_2}{m_1+m_2} \simeq m_2 \ll m_1+m_2$$
so the reduced mass is much less than the total mass of the system. Now consider the case when there is no relative motion, which means that both particles are moving with the centre of mass velocity. What is the kinetic energy in this case? What does your kinetic energy evaluate to?

Again, we do not care about the actual motion at this stage. We just want to find an expression for the kinetic energy.

7. Dec 10, 2016

### 1v1Dota2RightMeow

Ok so then using the correct mass of the COM, then $KE = (1/2)(m_1+m_2)(\dot x_{COM})^2=(1/2)(m_1+m_2)(\frac{m_1\dot x_1 + m_2 \dot x_2}{m_1 + m_2})^2=(1/2)\frac{(m_1\dot x_1 + m_2 \dot x_2)^2}{m_1+m_2}$

8. Dec 10, 2016

### Orodruin

Staff Emeritus
Yes, but keep it on the first form. You are going to want to use the CoM coordinates to solve the problem.

Also, it is not clear from your post that you have understood why the reduced mass is not the correct one. Again the best way of doing this is to write everything down in terms of $x_1$ and $x_2$ first and then make the coordinate change. It will help you in understanding where the reduced mass will appear (because it will!) and why.

9. Dec 10, 2016

### 1v1Dota2RightMeow

Oh yes, sorry! I see what you meant by it - I plugged it into my calculator and saw that reduced mass gives me a really small number, close to what I used for $m_2$. So using the reduced mass kinda ignores the other gigantic mass and is therefore not the correct mass to use.

10. Dec 10, 2016

### 1v1Dota2RightMeow

Wait I feel as though what you say is a bit contradictory. You say to keep it in the first form with $x_{COM}$, but then say to write it instead with the $x_1$ and $x_2$. Am I misinterpreting or...?

11. Dec 10, 2016

### Orodruin

Staff Emeritus
I am saying that you will need it in terms of the centre of mass to solve more easily it later on. However, when you want to find the expression for the Lagrangian, it might be easier for you to first write it down in terms of $x_1$ and $x_2$ and then make the transformation to the CoM position and relative displacement. It will likely be more intuitive to you than trying to figure out how to interpret the mass in the CoM, the moments of inertia, etc.

12. Dec 10, 2016

### 1v1Dota2RightMeow

Ok so then how would I find $h(t)$ for the potential energy of the center of mass? I think I might be over complicating this part - most of the time the potential energy is just given by a fixed height ($mgh$) and we just say that it decreases from then on, right? So am I unnecessarily trying to find the height as a function of time?

13. Dec 10, 2016

### 1v1Dota2RightMeow

1. The problem statement, all variables and given/known data
I am given the following scenario: 2 masses are connected by a spring and are on a frictionless inclined plane. They are free to rotate, oscillate, and slide down the plane.

For the potential energy of the center of mass of the system, it is $Mgh=(m_1+m_2)gh$. But isn't height also a function of time? As it slides down, it will need to change. How do I represent this? Is it even necessary?

2. Relevant equations
$PE=mgh$

3. The attempt at a solution
I would assume that it is not needed to do this.

14. Dec 10, 2016

### PeroK

The PE of a particle or system free to move in a gravitational field will always be a function of time.

15. Dec 10, 2016

### Orodruin

Staff Emeritus
Again, if you have problems, I suggest that you write down the potential energy in the individual coordinates of the masses and then transform it to the CoM-separation coordinates.

16. Dec 10, 2016

### 1v1Dota2RightMeow

Ok so then $PE=m_1gr_{1,z}+m_2g r_{2,z}$ ? I changed to $r$ because I just realized that the masses have 3 coordinates.

But I don't really understand how to transform to CoM separation coordinates.

17. Dec 10, 2016

### Orodruin

Staff Emeritus
How do you express the CoM position in the coordinates of the masses?

Well, they do not, they only have two coordinates each since they are restricted to move in a plane.

18. Dec 10, 2016

### 1v1Dota2RightMeow

What, this confuses me greatly. They slide down the plane, of course, but their potential energy comes from how far up the plane they are, no? Which would correspond to the height from the bottom (assuming there is a bottom).

19. Dec 10, 2016

### Orodruin

Staff Emeritus
I am sorry, I do not see why this is a problem. You assign your coordinates in the plane such that one coordinate corresponds to movement in the sloped direction. You would do something similar on an inclined plane without a second direction to move in.

20. Dec 10, 2016

### vela

Staff Emeritus
Because the masses are moving on an inclined plane, which is a two-dimensional surface, you only need two coordinates to specify the position of each mass. The height of a mass can be expressed as some function of its two coordinates. There's no need to introduce a third, independent coordinate.