Is it possible to find Tensional force from Lagrange?

In summary, the Lagrangian principle provides an easier method for solving problems, but it is important to take the tension in a system into account. This can be done by obtaining the equation of motion from Newton's 2nd law, but it can also be done through the use of Lagrange multipliers. The Lagrange multiplier method is associated with constraints in the system, and in some cases, the potential energy in the connectors may also affect the Euler-Lagrange equations. The presence of terms such as ##T_1x_1## and ##T_2x_2## in the Lagrangian can be justified through the use of Lagrange multipliers, which take the role of constraint forces.
  • #1
mcconnellmelany
21
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Homework Statement
Is it possible to find Tensional force from Lagrange?
Relevant Equations
L=T-U
Lagrangian principle is easier to solve any kind of problem. But we always "forget" (not really. But we don't take it into account directly.) of Tension in a system when looking at Lagrangian. But some questions say to find Tension. Since we can get the equation of motion from Newton's 2nd law (where lots of force are taken into account, with too messy equation). Then can we get only tension force from the equation of motion without looking at Newton's 2nd law.
 
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  • #2
Consider: [tex]
L = \tfrac12(m_1\dot x_1^2 + m_2 \dot x_2^2) + T_1x_1 + T_2x_2 - U(x_1,x_2).[/tex] This yields the Euler-Lagrange equations [tex]
\begin{split}
m_1\ddot x_1 &= T_1 - \frac{\partial U}{\partial x_1}\\
m_2 \ddot x_2 &= T_2 - \frac{\partial U}{\partial x_2}
\end{split}[/tex] and with suitable choice of [itex]T_1[/itex] and [itex]T_2[/itex] you can get the EOM for two massive particles connected by a light rigid rod and subject to an external conservative force. But in that case you have the constraint [itex]x_2 = x_1 + d[/itex] where [itex]x_2 > x_1[/itex] and [itex]d[/itex] is the length of the rod, so [tex]
L = \tfrac12(m_1+m_2)\dot x_1^2 + (T_1 + T_2)x_1 + T_2d - U(x_1,x_1 + d).[/tex] The [itex]T_2d[/itex] term is constant, so neglecting it won't change the Euler-Lagrange equations.

If you have something other than a rigid rod then the potential energy in the connectors depends on the state of the system, so it will turn up in the EL equations.
 
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  • #3
pasmith said:
Consider: [tex]
L = \tfrac12(m_1\dot x_1^2 + m_2 \dot x_2^2) + T_1x_1 + T_2x_2 - U(x_1,x_2).[/tex] This yields the Euler-Lagrange equations [tex]
\begin{split}
m_1\ddot x_1 &= T_1 - \frac{\partial U}{\partial x_1}\\
m_2 \ddot x_2 &= T_2 - \frac{\partial U}{\partial x_2}
\end{split}[/tex] and with suitable choice of [itex]T_1[/itex] and [itex]T_2[/itex] you can get the EOM for two massive particles connected by a light rigid rod and subject to an external conservative force. But in that case you have the constraint [itex]x_2 = x_1 + d[/itex] where [itex]x_2 > x_1[/itex] and [itex]d[/itex] is the length of the rod, so [tex]
L = \tfrac12(m_1+m_2)\dot x_1^2 + (T_1 + T_2)x_1 + T_2d - U(x_1,x_1 + d).[/tex] The [itex]T_2d[/itex] term is constant, so neglecting it won't change the Euler-Lagrange equations.

If you have something other than a rigid rod then the potential energy in the connectors depends on the state of the system, so it will turn up in the EL equations.
How can you justify the presence of terms ##T_1x_1## and ##T_2x_2## in the Lagrangian? Are they kinetic or potential energy terms? I am afraid neither of two...
 
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  • #4
Delta2 said:
How can you justify the presence of terms ##T_1x_1## and ##T_2x_2## in the Lagrangian? Are they kinetic or potential energy terms? I am afraid neither of two...
They follow from a Lagrange multiplier approach to finding the extremum of the action under the condition that the rod’s length is fixed.
 
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  • #5
Delta2 said:
I am afraid neither of two...
I think he just used T for tension and W=-Fx. Here, W is potential energy. And F=T.
 
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  • #6
Orodruin said:
They follow from a Lagrange multiplier approach to finding the extremum of the action under the condition that the rod’s length is fixed.
Well ok, if this is the case then it isn't something too obvious from first glance so I feel I can ask.
 
  • #7
Cross posted in PSE
 
  • #8
As @Orodruin suggests, the Lagrange multiplier method is associated with "constraints" (constraint forces).

Possibly useful:




 
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  • #9
mcconnellmelany said:
Cross posted in PSE

For the example you posted there (two masses connected by a light inextensible rope over a frictionless pulley):

Consider the Lagrangian [tex]
L_1(x_1,x_2,\dot x_1, \dot x_2) = \tfrac12(m_1 \dot x_1^2 + m_2 \dot x_2^2) + m_1gx_1 + m_2gx_2 - T(x_1 + x_2 - l),[/tex] where [itex]x_1[/itex] and [itex]x_2[/itex] are measured vertically downwards from the center of the pulley, [itex]l[/itex] is the length of the rope less half the circumference of the pulley, and [itex]T[/itex] is a Lagrange multiplier. Obtaining the Euler-Lagrange equations shows that [itex]T[/itex] takes the role of a force directed vertically upwards and acting on both masses, which in this physical context must be the tension in the rope.

Alternatively, consider [tex]
\begin{split}
L_2(x_1, \dot x_1) &= L_1(x_1, l - x_1, \dot x_1, -\dot x_1) \\
&=\tfrac12 (m_1 + m_2) \dot x_1^2 + m_1gx_1 + m_2g(l - x_1),\end{split}[/tex] which is the result of eliminating [itex]x_2[/itex] and [itex]\dot x_2[/itex] from [itex]L_1[/itex] using the constraint. Note that [itex]T[/itex] does not appear in [itex]L_2[/itex].

If you want to find the tension, then you have two unknowns: the tension and the acceleration of one of the masses. Starting from [itex]L_1[/itex], obtaining the Euler-Lagrange equations for [itex]x_1[/itex] and [itex]x_2[/itex], and only then imposing the constraint [itex]x_1 + x_2 = l[/itex] gives you the two equations necessary to solve for [itex]T[/itex] and [itex]\ddot x_1 = -\ddot x_2[/itex]. Starting from [itex]L_2[/itex] only gives you a single Euler-Lagrange equation from which you can find [itex]\ddot x_1[/itex].
 
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  • #10
pasmith said:
For the example you posted there (two masses connected by a light inextensible rope over a frictionless pulley):

Consider the Lagrangian [tex]
L_1(x_1,x_2,\dot x_1, \dot x_2) = \tfrac12(m_1 \dot x_1^2 + m_2 \dot x_2^2) + m_1gx_1 + m_2gx_2 - T(x_1 + x_2 - l),[/tex] where [itex]x_1[/itex] and [itex]x_2[/itex] are measured vertically downwards from the center of the pulley, [itex]l[/itex] is the length of the rope less half the circumference of the pulley, and [itex]T[/itex] is a Lagrange multiplier. Obtaining the Euler-Lagrange equations shows that [itex]T[/itex] takes the role of a force directed vertically upwards and acting on both masses, which in this physical context must be the tension in the rope.

Alternatively, consider [tex]
\begin{split}
L_2(x_1, \dot x_1) &= L_1(x_1, l - x_1, \dot x_1, -\dot x_1) \\
&=\tfrac12 (m_1 + m_2) \dot x_1^2 + m_1gx_1 + m_2g(l - x_1),\end{split}[/tex] which is the result of eliminating [itex]x_2[/itex] and [itex]\dot x_2[/itex] from [itex]L_1[/itex] using the constraint. Note that [itex]T[/itex] does not appear in [itex]L_2[/itex].

If you want to find the tension, then you have two unknowns: the tension and the acceleration of one of the masses. Starting from [itex]L_1[/itex], obtaining the Euler-Lagrange equations for [itex]x_1[/itex] and [itex]x_2[/itex], and only then imposing the constraint [itex]x_1 + x_2 = l[/itex] gives you the two equations necessary to solve for [itex]T[/itex] and [itex]\ddot x_1 = -\ddot x_2[/itex]. Starting from [itex]L_2[/itex] only gives you a single Euler-Lagrange equation from which you can find [itex]\ddot x_1[/itex].
Applying Euler-Lagrange of ##L_1##in terms of ##x_1##, assuming ##x_2## has nothing to do with ##x_1## (If I use constraint on first Lagrangian then first and second Lagrangian becomes same).

##m_1 \ddot{x_1}=m_1 g-T##

Applying Euler-Lagrange of ##L_2##.

##(m_1+m_2)\ddot{x_1}=m_1 g-m_2 g##

If I solve for ##T## I get
##T=((\frac{m_2 -m_1}{m_1+m_2})+1)m_1 g##

But that's totally wrong.

Solving using Newton's second law
##m_1\ddot x=m_1g-T \tag{1}##
##m_2\ddot x=T-m_2 g \tag{2}##
Solving for ##T## :

##T=\frac{2m_1m_2g}{m_1+m_2}##

<hr/>

Using the first Lagrangian I can get equation (1). Using the same Lagrangian (if I apply Euler-Lagrange for x_2) I can get similar equation as (2) but there should a negative sign on the RHS.
 
  • #11
Ehm are you using a system of equations for ##\ddot x_1,T## which one equation is from one lagrangian, and the other from the other lagrangian? That's not valid you are supposed to use only one lagrangian and Euler-Lagrange on one Lagrangian to deduce a system of equations.
 
  • #12
Delta2 said:
That's not valid you are supposed to use only one lagrangian and Euler-Lagrange on one Lagrangian to deduce a system of equations.
Than I don't think I understood pasmith. I would have watched those videos but lost my sound system hence can't do anything. Will you please explain the Lagrange Multipliers?
 
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  • #13
mcconnellmelany said:
Than I don't think I understood pasmith. I would have watched those videos but lost my sound system hence can't do anything. Will you please explain the Lagrange Multipliers?
On second thought I am not so sure about if you can mix equations from different Lagrangians, let's hear what @Orodruin, @pasmith and others have to say about this.
 
  • #14
mcconnellmelany said:
Applying Euler-Lagrange of ##L_1##in terms of ##x_1##, assuming ##x_2## has nothing to do with ##x_1## (If I use constraint on first Lagrangian then first and second Lagrangian becomes same).

##m_1 \ddot{x_1}=m_1 g-T##

Applying Euler-Lagrange of ##L_2##.

##(m_1+m_2)\ddot{x_1}=m_1 g-m_2 g##

If I solve for ##T## I get
##T=((\frac{m_2 -m_1}{m_1+m_2})+1)m_1 g##

But that's totally wrong.

You can simplify this: [tex]
\begin{split}
\left(\frac{m_2 - m_1}{m_1 + m_2} + 1\right)m_1g &=
\left(\frac{m_2-m_1}{m_1 + m_2} + \frac{m_1 + m_2}{m_1 + m_2}\right)m_1g \\
&= \left(\frac{m_2 - m_1 + m_1 + m_2}{m_1 + m_2}\right)m_1g \\
&= \frac{2m_2}{m_1 + m_2}m_1g \\ &= \frac{2m_1m_2g}{m_1 + m_2}.\end{split}[/tex]

Solving using Newton's second law
##m_1\ddot x=m_1g-T \tag{1}##
##m_2\ddot x=T-m_2 g \tag{2}##
Solving for ##T## :

##T=\frac{2m_1m_2g}{m_1+m_2}##

... agreeing with the first method, if you fully simplify the expression!
 
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Related to Is it possible to find Tensional force from Lagrange?

1. What is Tensional force?

Tensional force is a type of force that occurs when an object is being pulled or stretched. It is a force that is transmitted through a medium, such as a rope or cable, and is directed along the length of the medium.

2. What is Lagrange?

Lagrange, also known as Lagrange's equations, is a mathematical tool used to analyze the motion of a system. It is based on the principle of least action and can be used to find the equations of motion for a system.

3. Can Tensional force be found from Lagrange?

Yes, it is possible to find Tensional force from Lagrange's equations. However, it depends on the specific system and its constraints. In some cases, Lagrange's equations may not be able to fully describe the Tensional force present in a system.

4. What is the process for finding Tensional force from Lagrange?

The process for finding Tensional force from Lagrange's equations involves setting up the equations of motion for the system using Lagrange's equations. This involves identifying the generalized coordinates and applying the principle of least action to find the equations of motion. From there, Tensional force can be found by solving for the appropriate variables.

5. Are there any limitations to using Lagrange's equations to find Tensional force?

Yes, there are some limitations to using Lagrange's equations to find Tensional force. These equations may not be applicable to all types of systems, and they may not accurately describe the Tensional force in highly complex systems. Additionally, the process of setting up and solving Lagrange's equations can be time-consuming and may require advanced mathematical skills.

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