Is it possible to find Tensional force from Lagrange?

[mcconnellmelany]

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.

 

[pasmith]

Consider: $$L = \tfrac12(m_1\dot x_1^2 + m_2 \dot x_2^2) + T_1x_1 + T_2x_2 - U(x_1,x_2).$$ This yields the Euler-Lagrange equations $$\begin{split} m_1\ddot x_1 &= T_1 - \frac{\partial U}{\partial x_1}\\ m_2 \ddot x_2 &= T_2 - \frac{\partial U}{\partial x_2} \end{split}$$ and with suitable choice of $T_1$ and $T_2$ 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 $x_2 = x_1 + d$ where $x_2 > x_1$ and $d$ is the length of the rod, so $$L = \tfrac12(m_1+m_2)\dot x_1^2 + (T_1 + T_2)x_1 + T_2d - U(x_1,x_1 + d).$$ The $T_2d$ 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.

 

[Delta2]

Consider: $$L = \tfrac12(m_1\dot x_1^2 + m_2 \dot x_2^2) + T_1x_1 + T_2x_2 - U(x_1,x_2).$$ This yields the Euler-Lagrange equations $$\begin{split} m_1\ddot x_1 &= T_1 - \frac{\partial U}{\partial x_1}\\ m_2 \ddot x_2 &= T_2 - \frac{\partial U}{\partial x_2} \end{split}$$ and with suitable choice of $T_1$ and $T_2$ 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 $x_2 = x_1 + d$ where $x_2 > x_1$ and $d$ is the length of the rod, so $$L = \tfrac12(m_1+m_2)\dot x_1^2 + (T_1 + T_2)x_1 + T_2d - U(x_1,x_1 + d).$$ The $T_2d$ 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...

 

[Orodruin]

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.

 

[mcconnellmelany]

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.

 

[Delta2]

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.

 

[mcconnellmelany]

Cross posted in PSE

 

[robphy]

As @Orodruin suggests, the Lagrange multiplier method is associated with "constraints" (constraint forces).

Possibly useful:

 

[pasmith]

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 $$L_1(x_1,x_2,\dot x_1, \dot x_2) = \tfrac12(m_1 \dot x_1^2 + m_2 \dot x_2^2) + m_1gx_1 + m_2gx_2 - T(x_1 + x_2 - l),$$ where $x_1$ and $x_2$ are measured vertically downwards from the center of the pulley, $l$ is the length of the rope less half the circumference of the pulley, and $T$ is a Lagrange multiplier. Obtaining the Euler-Lagrange equations shows that $T$ 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 $$\begin{split} L_2(x_1, \dot x_1) &= L_1(x_1, l - x_1, \dot x_1, -\dot x_1) \\ &=\tfrac12 (m_1 + m_2) \dot x_1^2 + m_1gx_1 + m_2g(l - x_1),\end{split}$$ which is the result of eliminating $x_2$ and $\dot x_2$ from $L_1$ using the constraint. Note that $T$ does not appear in $L_2$.

If you want to find the tension, then you have two unknowns: the tension and the acceleration of one of the masses. Starting from $L_1$, obtaining the Euler-Lagrange equations for $x_1$ and $x_2$, and only then imposing the constraint $x_1 + x_2 = l$ gives you the two equations necessary to solve for $T$ and $\ddot x_1 = -\ddot x_2$. Starting from $L_2$ only gives you a single Euler-Lagrange equation from which you can find $\ddot x_1$.

 

[mcconnellmelany]

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

Consider the Lagrangian $$L_1(x_1,x_2,\dot x_1, \dot x_2) = \tfrac12(m_1 \dot x_1^2 + m_2 \dot x_2^2) + m_1gx_1 + m_2gx_2 - T(x_1 + x_2 - l),$$ where $x_1$ and $x_2$ are measured vertically downwards from the center of the pulley, $l$ is the length of the rope less half the circumference of the pulley, and $T$ is a Lagrange multiplier. Obtaining the Euler-Lagrange equations shows that $T$ 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 $$\begin{split} L_2(x_1, \dot x_1) &= L_1(x_1, l - x_1, \dot x_1, -\dot x_1) \\ &=\tfrac12 (m_1 + m_2) \dot x_1^2 + m_1gx_1 + m_2g(l - x_1),\end{split}$$ which is the result of eliminating $x_2$ and $\dot x_2$ from $L_1$ using the constraint. Note that $T$ does not appear in $L_2$.

If you want to find the tension, then you have two unknowns: the tension and the acceleration of one of the masses. Starting from $L_1$, obtaining the Euler-Lagrange equations for $x_1$ and $x_2$, and only then imposing the constraint $x_1 + x_2 = l$ gives you the two equations necessary to solve for $T$ and $\ddot x_1 = -\ddot x_2$. Starting from $L_2$ only gives you a single Euler-Lagrange equation from which you can find $\ddot x_1$.

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.

 

[Delta2]

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.

 

[mcconnellmelany]

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?

 

[Delta2]

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.

 

[pasmith]

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: $$\begin{split} \left(\frac{m_2 - m_1}{m_1 + m_2} + 1\right)m_1g &= \left(\frac{m_2-m_1}{m_1 + m_2} + \frac{m_1 + m_2}{m_1 + m_2}\right)m_1g \\ &= \left(\frac{m_2 - m_1 + m_1 + m_2}{m_1 + m_2}\right)m_1g \\ &= \frac{2m_2}{m_1 + m_2}m_1g \\ &= \frac{2m_1m_2g}{m_1 + m_2}.\end{split}$$

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!