# Double pulley problem using Lagrangian

• mcconnellmelany
In summary, the conversation revolved around setting up coordinates for a problem involving a system of two blocks connected by two ropes and a pulley. The participants discussed different methods of solving the problem, including using Euler-Lagrange equations and Newtonian methods. They also discussed the use of constraints in the problem and potential errors in the kinetic energy term. Ultimately, the conversation ended with the request for the individual to share their working and calculations for further analysis.
mcconnellmelany
Homework Statement
The diagram of the problem is attached on the body and strings are massless. Find the acceleration of M_1
Relevant Equations
L=T-U
Euler-Lagrange equation

Setting up coordinates for the problem

##L=\frac{1}{2}M_1 \dot{x}^2+\frac{1}{2}M_2(\dot y-\dot x)^2+M_1gx+M_2g(l_a-x+y)##

After using Euler Lagrange for x component and y component separate and substitute one to another then I get that ##\ddot{x}=\frac{M_1-2M_2}{M_1}g##

whereas on the solution they use Newtonian method and get that ##\ddot{x}=\frac{2M_2-M_1}{4M_2+M_1}##

And I thought my method of solving the question was wrong so I tried applying M_3=0 kg on [the solution](https://youtu.be/HbmN5Xb6dYg?t=566). I still can't get the answer the book wrote.

I was trying to write constraint another way...
##2x=l_A##
##l_A-x+2y=0##
for first and second rope respectively. substituted l_A and got x=-2y.. I end up with ##\ddot{x}=\frac{2M_1-M_2}{8M_1+2M_2}##. too close. Am I applying constraint correctly?

##y=2x##

Welcome!

Delta2
Lnewqban said:
##y=2x##

Welcome!

Delta2
What constraint did you use for the Newtonian approach? You should use the same for the Lagrangian approach too, you just can't pick different constraint in order to make the answers match.

I think the second kinetic energy term in your lagrangian is wrong, according to my opinion it should be ##\frac{1}{2}M_2(\dot x+\dot y)^2##

But you haven't tell us exactly how you measure ##y##. Do you measure y relative to the position of the second pulley (the one that the body M_2 hangs from)?

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Lnewqban
@haruspex maybe you want to take a look at this and offer your lights.

mcconnellmelany said:
I was trying to write constraint another way...
##2x=l_A##
##l_A-x+2y=0##
for first and second rope respectively. substituted l_A and got x=-2y.. I end up with ##\ddot{x}=\frac{2M_1-M_2}{8M_1+2M_2}##. too close. Am I applying constraint correctly?

Delta2
I would place the datum at the center of the pulley that holds ##M_1##.

Call the length of rope on the left ##x## as you have done, and on the right ##s##.

Then ##\dot{s} = -\dot{x}##

The kinetic energy for block ##M_2## I think is given by ## \frac{1}{2}M_2 \left( \dot s + \dot y \right)^2 = \frac{1}{2}M_2 \left( \dot y - \dot x \right)^2 ##

Then redefine all of your potential energies in from that datum

Edit:
Wait, I see that is what you have done! So I’m not sure. Should your potentials be negative?

I feel like I’m probably wrong and what @Delta2 said is correct for the kinetic energy of block 2. One seems to have ## 4 \dot x \dot y ## more kinetic energy than the other, and without using Newton’s second to confirm I’m not sure. I’m tapping out!

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Delta2
I took constraint differently but time derivative of position is same as the solution manual (they took second derivative for acceleration but the result is same for first order also)

Delta2
Delta2 said:
I think the second kinetic energy term in your lagrangian is wrong, according to my opinion it should be
took first time derivative of x_2 (the equation is on first picture I attached)

mcconnellmelany said:
took first time derivative of x_2 (the equation is on first picture I attached)
Yes ok I see now.

erobz said:
I feel like I’m probably wrong and what @Delta2 said is correct for the kinetic energy of block 2. One seems to have 4x˙y˙ more kinetic energy than the other, and without using Newton’s second to confirm I’m not sure. I’m tapping out!
Does #10 answer your question? If it doesn't than I will say that motion of first body doesn't depend on y and potential for the body is -m1gx (but for the Lagrangian the negative becomes positive) but second body depends on both x and y.

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haruspex said:
##L=1/2 M1\dot x^2+1/2M2(\dot x+\dot y)^2+M1 gx-M2g\frac{x}{2}##
To shorten the Lagrangian I ignored other constant terms for second body (since they will vanish for derivative with respect to x)

Delta2 said:
Do you measure y relative to the position of the second pulley (the one that the body M_2 hangs from)?
I have written all of them in my first diagram. If I understood your question correctly then the answer is yes.

I think @haruspex wants to see your work on how you worked with the Euler-Lagrange equations for this system and how you used with them the constraint ##\dot x=-2\dot y##

I know its going to be a bit tiresome to type in Latex your exact workings, but you got to do it if you want us to pinpoint your mistake.

Delta2 said:
I think @haruspex wants to see your work on how you worked with the Euler-Lagrange equations for this system and how you used with them the constraint ##\dot x=-2\dot y##
Since I have written down the first one so I am going to write second one.
##L=\frac{1}{2}M_1\dot{x}^2+\frac{1}{2} M_2(\dot{x}-\frac{\dot x}{2})^2+M_1gx-M_2g\frac{x}{2}##
I have written first line on #14.
Second term on the above equation becomes 1/8 and last term becomes 1/2.
From Euler-Lagrange ##M_1\ddot x+\frac{1}{4}M_2\ddot x=M_1g-\frac{M_2g}{2}##

If you solve for ddot x then you will get what I just said earlier.

Btw I can confirm the Newtonian result is correct I also got it working on my own.

Delta2 said:
Btw I can confirm the Newtonian result is correct I also got it working on my own.
What's your opinion on the Lagrangian? I think I should have learned Newtonian method first.

mcconnellmelany said:
What's your opinion on the Lagrangian? I think I should have learned Newtonian method first.
I think the Lagrangian is fine.

mcconnellmelany
The constraint is actually ##\dot y=-2\dot x## NOT ##\dot x=-2\dot y##.

Lnewqban and mcconnellmelany
Hmm there must be something wrong with lagrangian after all , I can't reach to the Newtonian result using this Lagrangian and the constraint ##y=c-2x\Rightarrow \dot y=-2\dot x##...

Delta2 said:
I can't reach to the Newtonian result using this Lagrangian and the constraint y=c−2x⇒y˙=−2x˙...
Sorry, are you saying that the constraint isn't working for you? But I have got the correct answer using the constraint.. Our x is the "solution manual's" x_1.

And after this I have solved some others problems on my own, thanks for the guide. (I was misreading solution manual from the beginning)

mcconnellmelany said:
Sorry, are you saying that the constraint isn't working for you? But I have got the correct answer using the constraint.. Our x is the "solution manual's" x_1.

And after this I have solved some others problems on my own, thanks for the guide. (I was misreading solution manual from the beginning)
I know the constraint is correct so it must be the lagrangian. Btw what you doing at post #17 you seem to be using the constraint ##y=\frac{x}{2}## which is wrong.

Lnewqban

How do you get the constraints from this datum setup? I get the following equations:

$$x + s = c \implies \dot x = - \dot s$$

$$w + y = k \implies \dot w = - \dot y$$

And

$$s + w = L \implies \dot s = -\dot w$$

So what am I missing or messing up?

Delta2 and haruspex
mcconnellmelany said:
##L=1/2 M1\dot x^2+1/2M2(\dot x+\dot y)^2+M1 gx-M2g\frac{x}{2}##
To shorten the Lagrangian I ignored other constant terms for second body (since they will vanish for derivative with respect to x)
That's just one equation. I asked to see your working; that means all of it.
i suspect that at some point you have become confused between the ##y## in there and in your post #1 diagrams and the ##x_2## in post #9. In their relationships to ##x## and ##x_1,## one has a factor 2, the other does not.

I managed to solve it via Lagrangian method , using as Lagrangian $$L=\frac{1}{2}M_1\dot x^2+\frac{1}{2}M_2\dot y^2-M_1gx-M_2gy$$ and the constraint $$y=c-2x\Rightarrow \dot y=-2\dot x$$ where ##c## "proper" constant...

PS. I took ##x## and ##y## as ##x_1## and ##x_2## respectively in figure of post #9, so ##y## is defined differently which allows our Lagrangian terms to have a simple form.

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erobz

## 1. What is the double pulley problem using Lagrangian?

The double pulley problem using Lagrangian is a physics problem that involves two masses connected by a string that passes over two pulleys. The goal is to find the equations of motion for the two masses using the Lagrangian approach, which is a mathematical method for solving systems of particles.

## 2. Why is Lagrangian used to solve this problem?

Lagrangian is used to solve the double pulley problem because it is a more efficient and elegant way to solve complex systems of particles. It allows for the use of generalized coordinates, which simplifies the equations of motion and makes the problem easier to solve.

## 3. How do you set up the Lagrangian for the double pulley problem?

The Lagrangian for the double pulley problem is set up by first defining the kinetic and potential energies for each mass using their respective generalized coordinates. Then, the Lagrangian is calculated as the difference between the kinetic and potential energies.

## 4. What are the equations of motion for this problem?

The equations of motion for the double pulley problem using Lagrangian can be found by taking the partial derivatives of the Lagrangian with respect to each generalized coordinate. This will result in a set of differential equations that describe the motion of the two masses.

## 5. What are some real-world applications of the double pulley problem using Lagrangian?

The double pulley problem using Lagrangian has many real-world applications, such as in the design of mechanical systems, robotics, and even in analyzing the motion of celestial bodies. It can also be used to study the behavior of systems under tension or to optimize the efficiency of pulley systems.

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