Day 14 - Potential Energy and Stability

Day 14 - Potential Energy and Stability#

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Mexican Hat/Sombrero Potential \(\longrightarrow\)

Mexican Hat Potential#

\(V(\phi) = -5|\phi|^2 + |\phi|^4\)

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Welcome Prof. Rachel Henderson!#

Announcements#

  • Midterm 1 is available today (Due 25 Feb; late 27 Feb)

  • DC will say more about this on Wednesday, but:

    • You may work in larger groups, but solutions are submitted like homework (max 3 group members) on Gradescope

    • Exercise 0 is for project planning; and can be submitted individually or as a different group on D2l

This Week’s Goals#

  • Understand the concept of potential energy

  • Determine the equilibrium points of a system using potential energy

  • Analyze the stability of equilibrium points

  • Define and begin to apply conservation of linear and angular momentum

Reminder: The Gradient Operator \(\nabla\)#

\(\nabla\) is a vector operator. In Cartesian coordinates: $\(\nabla = \hat{x}\dfrac{\partial}{\partial x}+\hat{y}\dfrac{\partial}{\partial y}+\hat{z}\dfrac{\partial}{\partial z} = \left\langle \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z} \right\rangle\)$

Acting on a scalar function \(f(x,y,z)\) produces a vector:

\[\nabla f(x,y,z) = \hat{x}\dfrac{\partial f}{\partial x}+\hat{y}\dfrac{\partial f}{\partial y}+\hat{z}\dfrac{\partial f}{\partial z} = \left\langle \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z} \right\rangle\]

Reminder: The Gradient Operator \(\nabla\)#

\(\nabla\) can act on vector field (function), \(\mathbf{F}(x,y,z)\) with both dot and cross products.

Divergence (Scalar Product) - How does the vector field change in the direction of the vector?#

\[\nabla \cdot \mathbf{F}(x,y,z) = \left\langle \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z} \right\rangle \cdot \langle F_x, F_y, F_z \rangle\]
\[\nabla \cdot \mathbf{F}(x,y,z) = \dfrac{\partial F_x}{\partial x} + \dfrac{\partial F_y}{\partial y} + \dfrac{\partial F_z}{\partial z}\]

Clicker Question 14-1a#

Which of the following fields have no divergence?

A. A B. B

  1. A

  2. B

  3. Both A and B

  4. Neither A nor B

Reminder: The Gradient Operator \(\nabla\)#

Curl (Vector Product) - How does the vector field change in the direction perpendicular to the vector?#

\[\begin{split} \nabla \times \mathbf{F}(x,y,z) = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ F_x & F_y & F_z \end{vmatrix} \end{split}\]
\[\nabla \times \mathbf{F}(x,y,z) = \left\langle \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z},\ \dfrac{\partial F_x}{\partial z} - \dfrac{\partial F_z}{\partial x},\ \dfrac{\partial F_y}{\partial x} - \dfrac{\partial F_x}{\partial y} \right\rangle\]

Clicker Question 14-1b#

Which of the following fields have no curl?

A. A B. B

  1. A

  2. B

  3. Both A and B

  4. Neither A nor B

Clicker Question 14-1c#

Consider a vector field with zero curl: \(\nabla \times \vec{F} = 0\). Which of the following statements is true?

  1. The field is conservative

  2. \(\int \nabla \times \vec{F} \cdot d\vec{A} = 0\)

  3. \(\oint \vec{F} \cdot d\vec{r} \neq 0\)

  4. \(\vec{F}\) is the gradient of some scalar function, e.g., \(\vec{F} = - \nabla U\)

  5. Some combination of the above

Reminders: Conservative Forces#

  • Conservative forces are those with zero curl

\[\nabla \times \vec{F} = 0\]

  • The work done by a conservative force is path-independent; on a closed path, the work done is zero

\[\oint \vec{F} \cdot d\vec{r} = 0\]

  • The force can be written as the gradient of a scalar potential energy function

\[\vec{F} = - \nabla U\]

Clicker Question 14-2#

Here’s the graph of the potential energy function \(U(x)\) for a pendulum.

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What can you say about the equilibrium points? There is/are:

  1. One stable point

  2. Two stable points

  3. One stable and one unstable point

  4. Two unstable and one stable point

Clicker Question 14-3#

Here’s a potential energy function \(U(x)\) for a pendulum:

\[U(\phi) = -mgL\cos(\phi) + U_0\]

  1. Find the equilibrium points (\(\phi^*\)) of the pendulum by setting:

\[\frac{dU(\phi^*)}{d\phi} = 0.\]

  1. Characterize the stability of the equilibrium points (\(\phi^*\)) by examining the second derivative:

\[\frac{d^2U(\phi^*)}{d\phi^2}>0? \qquad \frac{d^2U(\phi^*)}{d\phi^2}<0?\]

Click when done.

Clicker Question 14-4#

A double-well potential energy function \(U(x)\) is given by

\[U(x) = -\frac{1}{2}kx^2 + \frac{1}{4}kx^4.\]

We assume we have scaled the potential energy so that all the units are consistent.

How many equilibrium points does this system have?

  1. 1

  2. 2

  3. 3

  4. 4

Clicker Question 14-5#

A double-well potential energy function \(U(x)\) is given by

\[U(x) = -\frac{1}{2}kx^2 + \frac{1}{4}kx^4.\]

  1. Find the equilibrium points (\(x^*\)) of the pendulum by setting:

\[\frac{dU(x^*)}{dx} = 0.\]

  1. Characterize the stability of the equilibrium points (\(x^*\)):

\[\frac{d^2U(x^*)}{dx^2}>0? \qquad \frac{d^2U(x^*)}{dx^2}<0?\]

Click when done.

Clicker Question 14-6#

Here’s a graph of the potential energy function \(U(x)\) for a double-well potential.

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Describe the motion of a particle with the total energy, \(E=\)

  1. \(0.4\,\mathrm{J}\), \(<\) barrier height

  2. \(1.2\,\mathrm{J}\), \(>\) barrier height

  3. \(1.0\,\mathrm{J}\), \(=\) barrier height

Click when done.

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