PHYS101 Labor Mechanics课程简介

Physics is the branch of science that explores the physical nature of matter and energy. Physicists examine the story behind our universe, which includes the study of mechanics, heat, light, radiation, sound, electricity, magnetism, and the structure of atoms. They study the events and interactions that occur among the elementary particles that comprise our material universe.

In this course, we study the physics of motion from the ground up – learning the basic principles of physical laws and their application to the behavior of objects. Classical mechanics studies statics, kinematics (motion), dynamics (forces), energy, and momentum developed prior to the 1900 from the physics of Galileo Galilei and Isaac Newton. We encourage you to supplement what you learn here with the Saylor course PHYS102 Introduction to Electromagnetism.
Since mathematics is the language of physics, you should be familiar with high school level algebra, geometry, and trigonometry. We will develop the small amount of additional math and calculus you need to succeed during the course.

PREREQUISITES

Unit 1: Introduction to Physics
First, let’s gain a basic understanding of the language and analytical techniques that are specific to physics. This unit presents a brief outline of physics, measurement units and scientific notation, significant figures, and measurement conversions.
Completing this unit should take you approximately 2 hours.
Unit 2: Kinematics in a Straight Line
We begin our formal study of physics with an examination of kinematics, the branch of mechanics that studies motion. The word “kinematics” comes from a Greek term that means “motion”. Note that kinematics is not concerned with what causes the object to move or to change course. We will look at these considerations later in the course. In this unit, we examine the simplest type of motion, which is motion along a straight line or in one dimension.
Completing this unit should take you approximately 5 hours.

PHYS101 Labor Mechanics HELP（EXAM HELP， ONLINE TUTOR）

1. Working in two dimensions (so indices just range from 1 to 2 ), show that the fourth order tensor
$$T_{i j k l}=\lambda \delta_{i j} \delta_{k l}+\mu\left(\delta_{i k} \delta_{j l}+\delta_{i l} \delta_{j k}\right)$$
is invariant under the transformation
$$T_{i j k l}=A_{i p} A_{j q} A_{k r} A_{l s} T_{p q r s}$$
in which $A$ is the transformation of rotation by $\theta$ such that $A_{11}=A_{22}=\cos \theta$ and $A_{12}=$ $-A_{21}=\sin \theta$. [Hint: by symmetry, you need only explicitly consider the transformations resulting in $T_{i j k l}=T_{1111}, T_{1122}$, and $T_{1212}$.]

To show that the fourth order tensor $T_{ijkl}$ is invariant under the given transformation, we need to show that

T_{ijkl} = A_{ip}A_{jq}A_{kr}A_{ls} T_{pqrs}Tijkl​=Aip​Ajq​Akr​Als​Tpqrs​

for all values of $i,j,k,l$. We will use the fact that $A$ is a rotation matrix, which means that $A^T A = AA^T = I$, where $I$ is the identity matrix.

Let’s first consider the case where all indices are 1, i.e., $T_{1111}$. Using the definition of $A$, we have \begin{align} T_{1111} &= \lambda \delta_{11} \delta_{11}+\mu\left(\delta_{11} \delta_{11}+\delta_{11} \delta_{11}\right) \ &= (\lambda+2\mu)\delta_{11}\delta_{11}\delta_{11}\delta_{11}. \end{align} On the other hand, using the transformation formula, we have \begin{align} A_{ip}A_{jq}A_{kr}A_{ls} T_{pqrs} &= A_{11}^4 T_{1111} \ &= (\cos\theta)^4 T_{1111} \ &= (\lambda+2\mu)\delta_{11}\delta_{11}\delta_{11}\delta_{11}. \end{align} Therefore, $T_{1111}$ is invariant under the given transformation.

Next, let’s consider the case where two indices are 1 and two indices are 2, i.e., $T_{1122}$. Using the definition of $A$, we have \begin{align} T_{1122} &= \lambda \delta_{11} \delta_{22}+\mu\left(\delta_{12} \delta_{12}+\delta_{11} \delta_{22}\right) \ &= \mu(\delta_{12} \delta_{12}+\delta_{11} \delta_{22}). \end{align} On the other hand, using the transformation formula, we have \begin{align} A_{ip}A_{jq}A_{kr}A_{ls} T_{pqrs} &= A_{11}^2 A_{22}^2 T_{1122} \ &= (\cos\theta)^2(-\sin\theta)^2 T_{1122} \ &= \mu(\delta_{12} \delta_{12}+\delta_{11} \delta_{22}). \end{align} Therefore, $T_{1122}$ is also invariant under the given transformation.

Finally, let’s consider the case where one pair of indices are the same and the other pair of indices are different, i.e., $T_{1212}$. Using the definition of $A$, we have \begin{align} T_{1212} &= \lambda \delta_{12} \delta_{12}+\mu\left(\delta_{11} \delta_{22}+\delta_{12} \delta_{21}\right) \ &= \mu(\delta_{11} \delta_{22}+\delta_{12} \delta_{21}). \end{align} On the other hand, using the transformation formula, we have \begin{align} A_{ip}A_{jq}A_{kr}A_{ls} T_{pqrs} &= A_{12}^2 A_{21}^2 T_{1212} \ &= (\sin\theta)^2(-\cos\theta

1. An inviscid, incompressible flow around a unit circular cylinder centred at the origin has velocity field, $\vec{u}=(u, v)$, given by
$$u=1-\frac{x^2-y^2}{\left(x^2+y^2\right)^2}, \quad v=-\frac{2 x y}{\left(x^2+y^2\right)^2},$$
in which the far-field velocity is $\vec{u}=(1,0)$ as $|(x, y)| \rightarrow \infty$
a) Sketch the corresponding velocity vectors at the following 12 points:
$$(x, y)=( \pm 1,0),(0, \pm 1),( \pm 1, \pm 1),( \pm 2,0),(0, \pm 2) \text {. }$$
b) Using the representation of velocity in terms of the streamfunction, $(u, v)=\left(\partial_y \psi,-\partial_x \psi\right)$, find an explicit analytic formula for the streamfunction, $\psi$, arbitrarily setting the resultant integration constant so that $\psi=0$ at $(x, y)=(1,0)$.
c) Superimposed on your sketch in a), draw the corresponding streamline corresponding to $\psi=0$.
d) Show that the flow is irrotational.
e) Assuming $p=0$ as $x^2+y^2 \rightarrow \infty$, what is the value of the Bernoulli function, $B$, on each of the streamlines?
f) From your answer in e), what is the pressure/mass $\left(p / \rho_0\right)$ acting on the surface of the cylinder at $(x, y)=( \pm 1,0)$ ?

a) At the given 12 points, the velocity vectors are as follows:

$(1,0)$: $\vec{u}=(0,0)$

$(-1,0)$: $\vec{u}=(2,0)$

$(0,1)$: $\vec{u}=(0,-1)$

$(0,-1)$: $\vec{u}=(0,1)$

$(1,1)$: $\vec{u}=\left(-\frac{2}{9},-\frac{4}{9}\right)$

$(-1,-1)$: $\vec{u}=\left(\frac{2}{9},\frac{4}{9}\right)$

$(1,-1)$: $\vec{u}=\left(\frac{2}{9},-\frac{4}{9}\right)$

$(-1,1)$: $\vec{u}=\left(-\frac{2}{9},\frac{4}{9}\right)$

$(2,0)$: $\vec{u}=\left(\frac{3}{16},0\right)$

$(-2,0)$: $\vec{u}=\left(-\frac{3}{16},0\right)$

$(0,2)$: $\vec{u}=(0,-\frac{1}{4})$

$(0,-2)$: $\vec{u}=(0,\frac{1}{4})$

b) From the given velocity components, we have

\partial_y \psi = u = 1 – \frac{x^2-y^2}{(x^2+y^2)^2} \quad \text{and} \quad -\partial_x \psi = v = -\frac{2xy}{(x^2+y^2)^2}∂y​ψ=u=1−(x2+y2)2×2−y2​and−∂x​ψ=v=−(x2+y2)22xy​

Integrating the first equation with respect to $y$ and the second equation with respect to $x$, we obtain

\psi(x,y) = y – \frac{1}{2} \frac{x^2-y^2}{x^2+y^2} + C_1 \quad \text{and} \quad \psi(x,y) = -x + \frac{1}{2} \frac{x^2-y^2}{x^2+y^2} + C_2,ψ(x,y)=y−21​x2+y2x2−y2​+C1​andψ(x,y)=−x+21​x2+y2x2−y2​+C2​,

respectively, where $C_1$ and $C_2$ are constants of integration. Setting $\psi=0$ at $(x,y)=(1,0)$, we have

C_1 = \frac{1}{2} \quad \text{and} \quad C_2 = -\frac{1}{2},C1​=21​andC2​=−21​,

so the streamfunction is

\psi(x,y) = y – \frac{1}{2} \frac{x^2-y^2}{x^2+y^2} + \frac{1}{2}.ψ(x,y)=y−21​x2+y2x2−y2​+21​.

c) The streamline corresponding to $\psi=0$ satisfies

y – \frac{1}{2} \frac{x^2-y^2}{x^2+y^2} + \frac{1}{2} = 0,y−21​x2+y2x2−y2​+21​=0,

which simplifies to

y(x^2+y^2) – \frac{1}{2}(x^2-y^2) + \frac{1}{2}(x^2+y^2) = 0,y(x2+y2)−21​(x2−y2)+21​(x2+y2)=0,

or

x^2+y^2 = \frac{1}{2}.x2+y2=21​.

This is the equation of a circle centered at the origin with radius $\frac{1}{\sqrt{2}}$.

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.

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