Day 35 - Lagrangian Examples II

Day 35 - Lagrangian Examples II#

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Announcements#

  • Last “Class” Week

  • Homework 8 due Friday, Nov 21st (late after Nov 26th)

  • Next Week: Project Work and Discussion

  • Last Week: Presentations

  • Final Project Due Dec 8th (no later than 11:59 pm)

  • No Final Exam

Complete Google Form#

By November 21st#

Reporting your group members for the final project and a short summary of your project idea for sharing with the class.

https://forms.gle/iPKR9EDAaHW3GirN7

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Reminders#

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

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

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 \(\phi\). $\(x = r \cos(\phi) \quad y = r \sin(\phi)\)$

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

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 35-1#

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. \(\ddot{\phi} = -\frac{g}{L} \sin(\phi)\)

  2. \(\ddot{\phi} = -\frac{g}{L} \cos(\phi)\)

  3. \(\ddot{\phi} = -\sqrt{\frac{g}{L} \sin(\phi)}\)

  4. \(\ddot{\phi} = -\sqrt{\frac{g}{L} \cos(\phi)}\)

  5. None of these

Clicker Question 35-2#

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

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

  3. \(y_1 + y_2 = l - \pi R\)

  4. \(y_1 - y_2 = l + \pi R\)

  5. None of these

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

Clicker Question 35-3#

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. \(F_{y_1} = -mg\) and \(p_{y_1} = m\dot{y}_1\)

  2. \(F_{y_1} = -Mgy_1\) and \(p_{y_1} = M\dot{y}_1\)

  3. \(F_{y_1} = -(M-m)g\) and \(p_{y_1} = (M+m)\dot{y}_1\)

  4. \(F_{y_1} = -(M+m)g\) and \(p_{y_1} = (M-m)\dot{y}_1\)

  5. None of these

Clicker Question 35-4#

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. \(I = \frac{1}{2}M_pR^2\)

  2. \(I = \frac{1}{3}M_pR^2\)

  3. \(I = M_pR^2\)

  4. \(I = \frac{1}{4}M_pR^2\)

  5. None of these

Clicker Question 35-5#

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

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