Day 33 - Lagrangian Examples

Day 33 - Lagrangian Examples#

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Interview Study - GenAI#

  • Basma AlMahmood is a PER PhD student at MSU.

  • She is doing research on how students use generative AI in their (astro)physics classes.

  • She is piloting an interview protocol to understand how students are using generative AI in their (astro)physics classes. bg right:45% width:500px

Announcements#

  • Homework 8 has been posted (Last HW; Due 17 Apr, “Late” 24 Apr)

    • Last Exercise 0: Reflect Learning Outcomes

  • Final Project is posted

    • Video Presentations due 27 Apr

    • Computational Essay due 1 May

    • Rubric for both are posted

  • No class (20 Apr - 24 Apr) - DC out of country

    • Make appointment for project help (clicker extra credit)

Reminder: The Lagrangian#

The Lagrangian \(\mathcal{L}\) is a function that summarizes the dynamics of the system:

\[ \mathcal{L}(q, \dot{q}, t) = T - V \]

where:

  • \(T\) is the kinetic energy of the system (depends on gen. vel., \(\dot{q}\)),

  • \(V\) is the potential energy of the system (depends on gen. pos., \(q\)).

The equation of motion is recovered by applying the Euler-Lagrange equation to the Lagrangian (minimizing the action integral).

\[ \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) - \frac{\partial \mathcal{L}}{\partial q} = 0 \]

Clicker Question 33-1#

For a 1D SHO, the kinetic and potential energy are given by:

\[ T = \frac{1}{2} m \dot{x}^2 \quad \text{and} \quad V = \frac{1}{2} k x^2 \]

What are the derivatives of the Lagrangian \(\mathcal{L} = T - V\) with respect to \(x\) and \(\dot{x}\)?

  1. \(\frac{\partial \mathcal{L}}{\partial x} = kx\) and \(\frac{\partial \mathcal{L}}{\partial \dot{x}} = m\dot{x}\)

  2. \(\frac{\partial \mathcal{L}}{\partial x} = -kx\) and \(\frac{\partial \mathcal{L}}{\partial \dot{x}} = m\dot{x}\)

  3. \(\frac{\partial \mathcal{L}}{\partial x} = kx\) and \(\frac{\partial \mathcal{L}}{\partial \dot{x}} = -m\dot{x}\)

  4. \(\frac{\partial \mathcal{L}}{\partial x} = -kx\) and \(\frac{\partial \mathcal{L}}{\partial \dot{x}} = -m\dot{x}\)

  5. None of the above.

Clicker Question 33-2#

For the plane pendulum, with \(\mathcal{L}(x, \dot{x}, y, \dot{y}, t) = \frac{1}{2} m \left( \dot{x}^2 + \dot{y}^2 \right) - mgy\)

We found:

\[\frac{d}{dt}\left(m\dot{x}\right) = 0 \qquad \text{and} \quad \ddot{y} = - g\]

Does that seem right?

  1. Yes, it’s fine.

  2. Maybe, but I’m not sure I can tell you why.

  3. No, I know this is wrong, but I’m not sure why.

  4. No, this is definitely wrong and I can prove it!

Clicker Question 33-3#

For the plane pendulum, we changed the Lagrangian from Cartesian coordinates to plane polar coordinates. In Cartesian, we found the Lagrangian depended on \(y,\dot{x},\dot{y}\). In polar, it only depended on \(\phi\) and \(\dot{\phi}\).

\[\mathcal{L}(x,y,\dot{y}) \longrightarrow \mathcal{L}(\phi, \dot{\phi})\]

What does that tell you about the dimensions of the system? The system is:

  1. in 3D space, so it’s 3D.

  2. described by two spatial dimensions (\(x,y\)), so it’s 2D.

  3. described by one spatial dimension (\(\phi\)), so it’s 1D.

We chose our generalized coordinates poorly#

We used the Lagrangian formalism to derive the equations of motion for a plane pendulum. We chose the \(x\) and \(y\) coordinates.

\[T(\dot{x}, \dot{y}) = \dfrac{1}{2} m (\dot{x}^2 + \dot{y}^2), \quad V(y) = mgy \longrightarrow \mathcal{L} = T - V = \dfrac{1}{2} m (\dot{x}^2 + \dot{y}^2) - mgy\]

This gave us the following derivatives for the Lagrangian:

\[\frac{\partial \mathcal{L}}{\partial x} = 0 \quad \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) = \frac{d}{dt} \left( m\dot{x} \right) = 0\]
\[\frac{\partial \mathcal{L}}{\partial y} = -mg \quad \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{y}} \right) = -m\ddot{y}\]

We made a mistake by not including the constraint#

We made a mistake by not including the constraint \(x^2 + y^2 = L^2\) in our Lagrangian.

We can change variables to \(r\) and \(\phi\). $\(x = r \cos(\phi) \quad y = r \sin(\phi)\)$

\[T(\dot{x}, \dot{y}) = \dfrac{1}{2} m (\dot{x}^2 + \dot{y}^2) = \dfrac{1}{2} m \left( r^2 \dot{\phi}^2 + 2r\dot{r}\dot{\phi} + \dot{r}^2 \right) = T(r, \dot{r}, \phi, \dot{\phi})\]
\[V(y) = mgy = mg r \sin(\phi) = V(r, \phi)\]

We made a mistake by not including the constraint#

Now we include the constraint \(r = L\), so that \(\dot{r} = 0\).

\[T(\phi, \dot{\phi}) = \dfrac{1}{2} m L^2 \dot{\phi}^2 \quad V(\phi) = mgL \cos(\phi)\]
\[\mathcal{L} = \dfrac{1}{2} m L^2 \dot{\phi}^2 - mgL \cos(\phi)\]

Clicker Question 33-4#

For the plane pendulum, we changed the Lagrangian from Cartesian coordinates to plane polar coordinates. In Cartesian, we found the Lagrangian depended on \(y,\dot{x},\dot{y}\). In polar, it only depended on \(\phi\) and \(\dot{\phi}\).

\[\mathcal{L}(x,y,\dot{y}) \longrightarrow \mathcal{L}(\phi, \dot{\phi})\]

What does that tell you about the dimensions of the system? The system is:

  1. in 3D space, so it’s 3D.

  2. described by two spatial dimensions (\(x,y\)), so it’s 2D.

  3. described by one spatial dimension (\(\phi\)), so it’s 1D.

Clicker Question 33-5#

With \(\mathcal{L} = \dfrac{1}{2} m L^2 \dot{\phi}^2 - mgL \cos(\phi)\), we can find the equations of motion.

\[\dfrac{\partial \mathcal{L}}{\partial \phi} - \dfrac{d}{dt} \left( \dfrac{\partial \mathcal{L}}{\partial \dot{\phi}} \right) = 0\]

Which of the following equations of motion is correct?

\(1.\quad \ddot{\phi} = -\frac{g}{L} \sin(\phi) \qquad 2. \quad \ddot{\phi} = -\frac{g}{L} \cos(\phi)\) \(3.\quad \ddot{\phi} = -\sqrt{\frac{g}{L} \sin(\phi)} \qquad 4.\quad \ddot{\phi} = -\sqrt{\frac{g}{L} \cos(\phi)}\) \(5.\quad \textrm{None of these}\)

Clicker Question 33-6#

For the Atwood’s machine, \(M\) is connected to \(m\) by a string of length \(l\). Each mass has a length of string extended as measured from the center of the pulley (\(R\)) of \(y_1\) and \(y_2\), respectively. The string wraps around half the pulley.

Which of the following represents the equation of constraint for the system?

\(1. \quad y_1 + y_2 = l - R \phi \qquad 2. \quad y_1 - y_2 = l + R \phi\) \(3. \quad y_1 + y_2 = l - \pi R \qquad 4. \quad y_1 - y_2 = l + \pi R\) \(5. \quad \textrm{None of these}\)

Take the time derivative of the constraint equation. What do you notice?

Clicker Question 33-7#

With a Lagrangian of the form \(\mathcal{L} = \frac{1}{2}(M+m)\dot{y}^2_1 - (M-m)gy_1\), we can find the generalized forces and generalized momenta.

\[F_{y_1} = \frac{\partial \mathcal{L}}{\partial y_1} = -\frac{\partial V}{\partial y_1} \quad p_{y_1} = \frac{\partial \mathcal{L}}{\partial \dot{y}_1} = \frac{\partial T}{\partial \dot{y}_1}\]

What are \(F_{y_1}\) and \(p_{y_1}\) for the Atwood’s machine?

\(1. \quad F_{y_1} = -mg \textrm{ and } p_{y_1} = m\dot{y}_1 \qquad 2. \quad F_{y_1} = -Mgy_1 \textrm{ and } p_{y_1} = M\dot{y}_1\) \(3. \quad F_{y_1} = -(M-m)g \textrm{ and } p_{y_1} = (M+m)\dot{y}_1\) \(4. \quad F_{y_1} = -(M+m)g \textrm{ and } p_{y_1} = (M-m)\dot{y}_1\) \(5. \quad \textrm{None of these}\)

Clicker Question 33-8#

Now, we allow the pulley (mass, \(M_p\)) to rotate. The Lagrangian is given by: $\(\mathcal{L} = \frac{1}{2}(M+m)\dot{y}_1^2 + \frac{1}{2}I\dot{\phi}^2 - (M-m)gy_1\)$

Where \(I\) is the moment of inertia of the pulley. What is the moment of inertia of the pulley?

\(1. \quad I = \frac{1}{2}M_pR^2 \qquad 2. \quad I = \frac{1}{3}M_pR^2\) \(3. \quad I = M_pR^2 \qquad 4. \quad I = \frac{1}{4}M_pR^2\) \(5. \quad \textrm{None of these}\)

Clicker Question 33-9#

The rope moves without slipping on the pulley. A rotation of \(R d\phi\) corresponds to a displacement of \(dy_1\) for the first mass, \(M\). What is the new equation of constraint for the system?

  1. \(y_1 + y_2 = l - R \phi\)

  2. \(dy_1 = R d\phi\)

  3. \(y_1 = R\phi\)

  4. \(\dot{y}_1 = R \dot{\phi}\)

  5. More than one of these