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\begin{center}
{\bf Notes I on Introductory Mechanics \\
Statics and Describing Motion}
\end{center}
These notes are meant to help students understand the basic "physics" behind
introductory "Newtonian" mechanics. The ideas are presented in the order in
which they are taught in my first year class, and are designed to supplement
the text.
The approach we take is somewhat historical. In this section we first study the
conditions for static equilibrium for a rigid object: how forces combine such that
a rigid object does not move or rotate. The branch of mathematics
that is relevant for these topics is Vectors. When we have finished statics, we
will learn the language for describing motion.
\bigskip
\centerline{\bf Mechanics before 1500}
\bigskip
In early times the balancing of forces was important for building, and was
the first "physics" to be investigated.
Later on in the quarter we will analyze mechanical systems
precisely. We will discuss the concepts of inertial mass, force and
motion, ideas that were discovered in the 1600's. For now,
we will consider a simple set of experiments with forces.
Simply speaking, a force is a push (or pull). Let's consider forces due
to an objects weight, i.e. the gravitational force due to the attraction of
the earth. For this discussion, we limit ourselves to objects that are
made of pure elements, i.e. copper, iron, etc. We will make the assumption
that for a pure homogenious substance the weight is proportional to the
volume of the object. That is, if object A is twice as large object B, and
both are made of the same substance, then object A has twice the weight
as object B. If one object has a weight of 10 units, then an object of
twice (or x times) the volume will have a weight of 20 units (or 10x units).
The weights in our weight set are so labeled.
\bigskip
\noindent {\it Combining forces in the same direction}
In our first experiments we will combine forces on a ring by using pulleys and
strings. A string attached to the ring will pass over a pulley. The
other end of the string will be attached to a weight. Suppose two weights,
each of magnitude 10 units, pulls on the ring to the right (the + direction).
How much weight and where should it be placed to balance the two weights
of 10 units. Your guess is probably one weight of magnitude 20 units
pulling to the left. We will check this result in class. (Your guess
was correct). This means that two forces pulling in the same direction
add like real numbers. Similarly, we will show that two forces pulling
in the opposite direction subtract. Thus for the case of one dimension, we
can assign a (+ or -) to a force to signify the direction (right or left).
\bigskip
\noindent {\it Combining forces in different directions}
A forces can act in any direction. What role does the direction of a force
play when combining forces? Consider the following experiment:\\
\bigskip
\noindent {\it A force of 40 units pulls on the ring to the east and a force
of 30 units pulls on the ring to the north. What force on the ring will
balance these two forces?}\\
\bigskip
\noindent After doing the experiment in class, we will find that the
balancing force is a force that pulls on the ring with an amount
of 50 units at a direction of $36.9^\circ$ S of W. Thus the force
of 40 units to the east plus the force of 30 units to the north
is equivalent to a single force of magnitude 50 units at an angle
of $36.869^\circ$ N of E. In this case 30 plus 40 equals 50.
You probably remember from trig that lengths of 30, 40, and 50 form
a right triangle (3-4-5 triangle). Thus, if we represent each
force by an arrow, whose length is proportional to its magnitude and
whose direction is in the direction of the force, the two forces
combine by placing the tail of one at the tip of the other. The
resultant force is represented by the arrow along the hypotenuse.
In lab, you will carry out a number of experiments that demonstrate
that any two forces combine using the "arrow" tail-to-tip method described
above. The mathematics that best describes how forces combine is
the mathematics of {\it vector addition}. Since a number of physical
quantities have the properties of mathematical vectors, we will spend
some time in lecture discussing vectors.
Vectors are discussed in many texts and usually are defined (initially)
as something with direction and magnitude. Quantities with direction and magnitude are not
necessarily vectors. They must also have certain mathematical
properties. An addition operation must be defined, and the sum of two vectors must
also be a vector. If $\vec{A}$ and $\vec{B}$ represent any two
vectors, then it must be true that $\vec{A} + \vec{B} = \vec{B}
+ \vec{A}$. Vectors are also defined over a field, which in our
class will be the real numbers. To be an "inner-product" vector
space, a scalar product between two vectors must be defined with
certain properties. We will talk about this later when we discuss
energy.
The experimental result pertaining to adding "static" forces is
summarized by the following experiment:\\
\bigskip
\noindent {\bf Experiment:} If two forces, $\vec{F}_1$ and $\vec{F}_2$,
act on an object that doesn't move, the resulting force is the same as if the
object were subject to the force $\vec{F}$ where $\vec{F}$ is the
{\bf vector sum} of the two forces: $\vec{F} = \vec{F}_1 + \vec{F}_2$. The
same vector addition applies if more than two forces are acting on the object.\\
\bigskip
\noindent This is a wonderful experimental result! Forces didn't have to
"add" in such a simple way, but they do. It demonstrates that certain aspects of nature
can be understood using geometry (trig). These ideas were developed
over 3000 years ago, and helped in the building of magnificant structures.
We will do many examples in class demonstrating this remarkable property of forces, but
first we review the mathematical properties of vector addition.
\bigskip
\noindent {\it Multiplication of a vector by a scalar}
\bigskip
Multiplication of a vector by a scalar $c$ changes the magnitude of the vector by the factor
$|c|$. If $c>0$, then the direction remains the same, but if $c<0$ the
direction is reversed.
\bigskip
\noindent {\it Unit Vectors}
\bigskip
Unit vectors are vectors that have a magnitude of one.
In three dimensional cartesian space, one usually defines the unit vector
$\hat{i}$ to point in the +x direction, the unit vector $\hat{j}$ to point
in the +y direction, and the unit vector $\hat{k}$ to point in the +z direction.
Any vector in cartesian 3D space can be expressed as linear combinations
of these unit vectors:
\begin{equation}
\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}
\end{equation}
\noindent where $A_x$, $A_y$, and $A_z$ are the coordinates of the tip
of the vector. These unit vectors form an orthonormal basis,
whose linear combinations span the vector space. The unit vectors are
like "building blocks": any vector can be "built" from these three
elements.
\bigskip
\noindent{\it Vector Addition}
\bigskip
We explain vector addition with an example. Let vector $\vec{A}$ have a
magnitude of 150 units at an angle of 30 degrees N of E, vector $\vec{B}$
have a magnitude of 100 units at an angle of 45 degrees N of W, and vector
$\vec{C}$ have a magnitude of 200 units at an angle of 80 degrees S of E.
Find the vector $\vec{D} = \vec{A} + \vec{B} + \vec{C}$, which is the sum of
the three vectors. One could use the tail-to-tip method to add the three
vectors. We will show in class that the tail-to-tip method is equivalent
to adding the cartesian components after expressing the vectors in terms
of the unit vectors. So, let's add the three vectors using the
unit vector basis. First, express $\vec{A}$ as a
linear combination of the unit vectors $\hat{i}$ and $\hat{j}$:
\begin{eqnarray*}
\vec{A} & = & 150 cos(30) \hat{i} + 150 sin(30) \hat{j}\\
& = & 130 \hat{i} + 75 \hat{j}
\end{eqnarray*}
\noindent from the definition of the unit vectors and scalar multiplication.
Then, express $\vec{B}$ in terms of the unit vectors:
\begin{eqnarray*}
\vec{B} & = & - 100 cos(45) \hat{i} + 100 sin(45) \hat{j}\\
& = & - 70.7 \hat{i} + 70.7 \hat{j}
\end{eqnarray*}
\noindent Finally, express $\vec{C}$ in terms of the unit vectors:
\begin{eqnarray*}
\vec{C} & = & 200 cos(80) \hat{i} - 200 sin(80) \hat{j}\\
& = & 34.7 \hat{i} - 197 \hat{j}
\end{eqnarray*}
\noindent To add the three vectors, one simple adds the components of the
unit vectors:
\begin{eqnarray*}
\vec{A} + \vec{B} + \vec{C} & = & (130-70.7+34.7) \hat{i} +
(75+70.7-197) \hat{j}\\
\vec{D} & = & 94 \hat{i} - 51.3 \hat{j}
\end{eqnarray*}
\noindent The sum vector $\vec{D}$ can be expressed in terms of the unit vectors, or
as a magnitude and direction: $|\vec{D}| = \sqrt{94^2 + 51.3^2} =
107$ units. The vector $\vec{D}$ points in the fourth quadrant, with the
angle $tan^{-1} \theta = 51.3/94$, or $\theta = 28.6^\circ$. Since
$\vec{D}$ is in the fourth quadrant, $\theta = 28.6^\circ$ S of E.
The calculation is summarized in the figure.
\begin{figure}
\includegraphics[width=14cm]{fig1311a.png}
\end{figure}
{\bf The connection that the mathematics of this example has to physics
is the following. If an
object is subject to three forces: a force of
magnitude 150 units at an angle of 30 degrees N of E, a force of
magnitude 100 units at an angle of 45 degrees N of W, and a force
of magnitude 200 units at an angle of 80 degrees S of E, then the
net force on the object is $107$ units at an angle of $28.6^\circ$
S of E.} If one wants to balance these three forces, then one needs
a force of $107$ units in the opposite direction at $28.6^\circ$ N of
W.
Expressing the vectors using unit (or basis) vectors is probably the
most convenient way to perform operations with them. In general,
if $\vec{A} = A_x \hat{i} + A_y \hat{j}$, and
$\vec{B} = B_x \hat{i} + B_y \hat{j}$, then the sum
$\vec{A} + \vec{B}$ equals
\begin{equation}
\vec{A} + \vec{B} = (A_x + B_x) \hat{i} +
(A_y + B_y) \hat{j}
\end{equation}
\noindent for subtraction, replace the "+" with a "-". Using unit
(or basis) vectors is particulary useful when adding (or subtraction)
more than two vectors.
Some final points on vectors and vector addition:
\begin{enumerate}
\item It is important to remember that {\bf for a complete description of a
vector in two dimensions, two numbers are needed}: a magnitude
plus direction, two components, etc. For example, the vector
$\vec{D}$ above can be expressed as: $94 \hat{i} - 51.3 \hat{j}$,
or 107 units at a direction $28.6^\circ$ S of E, or 107 units
at an angle of $331.4^\circ$ clockwise from the x-axis.
\item Here we have limited our applications to
two and three dimensional vectors. These ideas can be generalized to any number of dimensions
(including infinity). For vector spaces with dimension greater than 3, it is not always
useful to think of the direction of a vector. The generalization of the ideas
presented here is the branch of mathematics called Abstract Linear Algebra.
\item In order for a quantity to be a vector,
the quantity must add (or combine) according to the rules of vector addition.
Not all quantities with magnitude and direction add accoring to the
rules of vector addition. An example of such a quantity is
rotations. Any rotation in three dimensions can be represented
by an arrow with magnitude and direction: the direction is the
axis of rotation, and the magnitude is the amount of rotation
(e.g. in radians). Let $A$ be a rotation about the
x-axis of $\pi/2$ radians, and $B$ be a rotation about the
y-axis of $\pi/2$ radians. You can show by rotating
your physics book that $\vec{A} + \vec{B} \neq \vec{B} +
\vec{A}$. That forces add according to the laws of simple
vector addition is remarkable.
\end{enumerate}
There are a number of quantities in classical physics that behave as vectors: displacement,
velocity, acceleration, force, the electric field and the magnetic field to name a few. You will encounter
these during your first year of physics. Although they represent different physical quantites,
they all add like the method described here. Vector addition pertains to many different
physical quantites, it is a "universal mathematics", and its importance cannot be understated.
\bigskip
\centerline{\bf Torque: twisting force}
\vspace{4mm}
We all have experience tightening a bolt. If we want it tight, we use
a wrench. To apply a large amount of "twisting force", we hold the
wrench as far as we can from the bolt and push with a force perpendicular
to the wrench arm. The name we call "twisting force" is {\bf torque}.
In order to determine the amount of torque, we need to specify three
things:
\vspace{2mm}
\noindent 1) The axis of rotation. In the case of the bolt, this is where
the bolt is.\\
\noindent 2) The force that is doing the twisting. (The force, direction
and magnitude, that you apply to the wrench).\\
\noindent 3) The location of the force. (How far from the bolt that you
apply the force.) \\
\vspace{2mm}
\noindent When we speak of torque, we usually say "the torque {\it about}
the axis (or point) $O$ {\it due} to the force $F$ {\it applied} at the location $P$."
All three things matter in determining the torque. We know from experience that
a larger force produces a larger torque, and the further away from the axis
that the force is applied, the larger the torque, but what is the correct
mathematical formula for torque? We will show in class that:
\begin{center}
torque = $|\vec{r}|$ times the component of the force perpendicular to $\vec{r}$
\end{center}
\noindent where $\vec{r}$ is the vector from the axis to the location where the force
is applied. The component of the force perpendicular to $\vec{r}$ is the component
of the force that does the twisting. The component of the force, $\vec{F}$, that
is parallel to $\vec{r}$ just pushes on the bolt. This component doesn't cause
any twisting about the axis. The symbol for torque used in most physics texts is
$\tau$:
\begin{equation}
|\vec{\tau}| = |\vec{r}| F_\bot
\end{equation}
\noindent If we let $\theta$ be the angle between the vector $\vec{r}$ and the
vector $\vec{F}$, the torque equation can be written as:
\begin{equation}
|\vec{\tau}| = |\vec{r}| |\vec{F}| sin(\theta)
\end{equation}
\noindent Remember, the vector $\vec{r}$ goes from the axis of rotation
to the location at which the force is applied.
Is torque a vector, a scalar, or something else? Torque certainly has
a magnitude, which is the amount of twisting force. We can also
associate a direction with torque: The axis of rotation of the twist.
The axis of rotation is in the direction perpendicular to both the
force $\vec{F}$ and the position vector $\vec{r}$. Thus, we can
assign a vector to a particular torque: The magnitude of the vector is the
amount of twisting force, $|\vec{r}| |\vec{F}| sin(\theta)$; and
the direction is the direction perpendicular to the plane containing
$\vec{r}$ and $\vec{F}$. There is one more decision to make: which
direction along the axis of the torque corresponds to a clockwise twist,
and which direction to a counter-clockwise twist? Since $sin(-\theta) =
-sin(\theta)$, the vector for clockwise twist will be in the opposite
direction to the one for counter-clockwise twist. The convention that
is agreed upon around the world is the following: the direction of
the torque vector is the direction that the bolt would move (for normal
"right-handed" bolts). Using your right hand, if your fingers curl
in the direction of the twist, your thumb points in the direction
of the torque.
We will show that torques add according to the rules of vector
addition, so it is a true vector. In mathematics, there is an
operation between vectors that yields another vector. It is the
cross product and we can write the torque vector as $\vec{r}$
cross $\vec{F}$:
\begin{equation}
\vec{\tau} = \vec{r} \times \vec{F}
\end{equation}
\noindent In this course, we will only solve problems in which all the location $\vec{r}$ vectors
and all the force $\vec{F}$ vectors lie in the same plane. Thus, we will not need to carry
out the cross product explicitly using vectors. We will only do problems in which the torques
produce a clockwise or counter-clockwise twist, and consquently the torque vectors are either
in or out of the page. Nonetheless, to prepare you for more advanced courses, we review some
of the mathematical properties of the vector cross product.
\bigskip
\noindent {\it Vector Cross Product}
\bigskip
The cross product of vector $\vec{A}$ and $\vec{B}$ is a vector. The magnitude
of $\vec{A} \times \vec{B}$ equals the magnitude of $\vec{A}$, $|\vec{A}|$ times
the magnitude of $\vec{B}$, $|\vec{B}|$, times the sin of the angle between
them:
\begin{equation}
|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| sin(\theta)
\end{equation}
\noindent where the angle $\theta$ is the angle between vector $\vec{A}$ and
vector $\vec{B}$. The angle goes from $\vec{A}$ to the vector $\vec{B}$. The
direction of $\vec{A} \times \vec{B}$ is determined by the "right hand" rule
described above. Note that since $sin(-\theta) = - sin(\theta)$,
\begin{equation}
\vec{B} \times \vec{A} = - \vec{A} \times \vec{B}
\end{equation}
Note that the cross product is maximized when $\vec{A}$ is perpendicular to
$\vec{B}$, and zero when $\theta$ is zero or $180^\circ$. The unit vectors
have simple cross product properties:
\begin{eqnarray*}
\hat{i} \times \hat{j} & = & \hat{k}\\
\hat{j} \times \hat{k} & = & \hat{i}\\
\hat{k} \times \hat{i} & = & \hat{j}\\
\hat{i} \times \hat{i} & = & 0\\
\hat{j} \times \hat{j} & = & 0\\
\hat{k} \times \hat{k} & = & 0
\end{eqnarray*}
\noindent {\it for a right handed coordinate system}.
We can express $\vec{A}$ and $\vec{B}$ in terms of the unit vectors:
$\vec{A} = A_x \hat{i} + A_y \hat{j} +
A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$. If we
use the above forumlas for the cross products of the unit vectors, after
"foil"ing we have:
\begin{eqnarray*}
\vec{A} \times \vec{B} & = & (A_y B_z - A_z B_y) \hat{i} +
(A_z B_x - A_x B_z) \hat{j}\\
& & + (A_x B_y - A_y B_x) \hat{k}
\end{eqnarray*}
\noindent This formula can be succinctly written in terms of the
determinant of a matrix:
\begin{equation}
\vec{A} \times \vec{B} =
\left| \begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k}\\
A_x & A_y & A_z \\
B_x & B_y & B_z
\end{array} \right|
\end{equation}
\bigskip
\centerline{\bf Condition for Static equilibrium for a rigid object}
\vspace{3mm}
In summary, there are two conditions for static equilibrium. In order for an
object to stay at rest, the vector sum of the forces on the object must be zero. This is
all one needs if the object is a point particle. However, if the object has
size, the location of the forces on the object is important. Eventhough
the sum of the forces is zero, the object might rotate.
In order for the object to stay stationary and not rotate, the sum of the torques
must be zero about any axis. These are the two conditions for an object to
remain "static", i.e. not translate and not rotate:
\begin{eqnarray*}
\sum_i \vec{F}_i & = &0\\
\sum_i \vec{\tau}_i & = & 0
\end{eqnarray*}
\noindent where the torque sum can be taken about any axis. Note that for the first
requirement, $\sum_i \vec{F}_i =0$, the location of the forces do not enter in
the calculation. However, for the second requirement, the location of the forces are
very important. Just like in Real Estate, the three most important things are Location, Location,
and Location. With torque, it is just Location, Location: the Location of the axis and the
Location of the forces.
In class we will do many examples of objects in static equilibrium. In evaluating
the torque sum, a proper choice of axis can reduce the mathematical complexity.
\bigskip
\centerline{\bf One Dimensional (straight-line) Frictionless Environment}
\bigskip
Objects are not always stationary. They move. In order to formulate the laws of
motion, we need to come up with a useful language. It turns out that the laws of
motion are most easily described in terms of instantaneous rates of change.
The mathematics of calculus was key to understanding the laws motion, and we start
with the language for describing motion.
\bigskip
\noindent {\bf Describing Motion: Position, Time, Velocity}
\vspace{3mm}
To describe motion of a point particle in one "straight-line" dimension one
needs to come up with a measure of
two quantities: position (or distance) and time.
To measure {\bf position}, we set up
a coordinate system, which in one dimension is just a straight line with a reference point
chosen as the origin. Once a direction for positive
distance and a unit length have been decided upon, equal lengths can be marked on the line.
A particles position is specified by a real number (+ or -) indicating its location on the
line. Two systems of units are used in the class, MKS and British. Length is measured
in meters in the MKS, and in units of feet in the British system.
To measure {\bf time}, we need an instrument (a clock) that can produce equal time intervals.
Whereas we have a physical feeling for equal distances from our hands, arms and eyes, equal time
intervals are difficult to get a feeling for. Physiologically a good movie lasting as long
as a physics lecture might seem much shorter in time. Therefore we can't rely on our senses
to judge equal time intervals. One could use a pendulum or other oscillating device.
However, using the motion of a physical system to determine equal time intervals and then using this clock
to understand motion itself might bias our description and laws of physics. We might wonder if
our method of measuring time is making our equations of physics to complicated. It would be nice to
have a clock that does not rely on a physical system. We will discuss such a possibility
with Newton's first law of motion. In order not to get caught up in circular
arguments and deep philosophy, in this
class we will accept that clocks can be made that give equal time intervals. As discussed in
the text, the accepted definition of a second is 9192631770 oscillations of a particular
transition in the Cesium atom. Although the atom is a physical system, we obtain a workable
definition of time for developing physics theories. The interested student should take
our modern physics course, where Einstein's theory of special relativity addresses these
problems.
Once we have established distance and time units, we can define a {\bf position
function}, $x(t)$. The position function is just the position $x$ for the particle
as a function of time, $t$.
A particles {\bf displacement} between the times $t_1$ and $t_2$ is just equal to
$x(t_2) - x(t_1)$, where $t_2 > t_1$. If the displacement is positive, the
particle has moved to the right (+ direction) from the time $t_1$ to the time
$t_2$. If the displacement is negative, the particle has moved to the left
(- direction). To talk about displacement, you need to refer to {\it two times}.
Velocity is a measure of how fast an object is moving. {\bf Average velocity},
$\bar{v}_{t_1 \rightarrow t_2}$, is defined to be the displacement/(time interval):
\begin{equation}
\bar{v}_{t_1 \rightarrow t_2} \equiv {{x(t_2) - x(t_1)} \over {t_2 - t_1}}
\end{equation}
\noindent As with displacement, to calculate the average velocity of an object
one needs to specify the {\it two} times, $t_1$ AND $t_2$. Average time is not
a particularly good quantity to express the laws of mechanics in a simple way.
For understanding the laws of physics in a simple way, the velocity at an instant,
instantaneous velocity, is a much better quantity. {\bf Instantaneous velocity}
at the time $t_1$, $v(t_1)$, is defined to be:
\begin{equation}
v(t_1) \equiv \lim_{t_2 \rightarrow t_1} {{x(t_2) - x(t_1)} \over {t_2 - t_1}}
\end{equation}
\noindent This can also be written as:
\begin{equation}
v(t_1) \equiv \lim_{\Delta t \rightarrow 0} {{x(t_1 + \Delta t) - x(t_1)} \over {\Delta t}}
\end{equation}
\noindent which you will recognize as the first derivative of $x(t)$ evaluated at the time
$t_1$: $v(t_1) \equiv (dx/dt)|_{t = t_1}$. A nice thing about instaneous velocity is that
it is defined at a single time, $t_1$. For this reason it is a better quantity to use
than average velocity to try and describe how nature behaves. If the instantaneous velocity
is positive (negative) then the particle is moving to the right (left) at that instant.
A particle's {\bf speed} is defined to be the magnitude (or absolute value) of the instantaneous
velocity.
\bigskip
\noindent {\bf Special Case of Constant Velocity}
\vspace{3mm}
In some special cases, a particles velocity may be constant for a long period of time.
If this is so, then the position function takes on a simple form. Let $v_0$ be the
velocity of the particle at the time when we start our clocks, $t=0$. We will often
use the subscript 0 to label the value of a variable at time $t=0$. For a particle moving
with a constant velocity, $v_0$ will be the velocity for all times $t$. From the
definition of velocity: $(x(t) - x(0))/(t-0) = v_0$. If we define $x(0) \equiv x_0$,
we have the simple relationship for $x(t)$:
\begin{equation}
x(t) = x_0 + v_0 t
\end{equation}
\noindent Often one writes $x(t)$ as just $x$, so the equation becomes $x=x_0 + v_0 t$.
Under what situations will an object move with a constant velocity in a straight line?
A person walking in a straight line down the street with a constant speed, or a car
driving down a straight length of road with a constant speed are some examples that
come to mind. However, there is a very important case related to a fundamental Law of Motion.
Consider a "reference frame" which is a box floating in space far from any other objects.
If you were to go inside this box, floating in space, you would feel "at rest". You
could not sense that you were moving. You set up a coordinate system to measure
position (one dimensional) and time. Suppose a particle had an initial velocity $v_0$ with
respect to your coordinate system. There are no forces on the particle. What would happen
to the particle in time? Would it come to rest, or continue to move with the
velocity $v_0$ in a straight line? Newton and Galileo realized that the particle would
continue to move with the velocity $v_0$. If the particle was at rest ($v_0 = 0$) it
would remain at rest. This property of motion is refered to as Newton's first Law
of Motion:
\bigskip
\begin{center}
{\bf If there are \underline {no forces} acting on an object, an object at rest\\
remains at rest and an object in motion continues in a state of\\
uniform motion}
\end{center}
\bigskip
This idea might seem simple to us, but at the time it was proposed it
was profound. It was believed that the natural state of an object was
at rest, and that objects that were moving came to rest on their own.
Newton's first law of motion applies to reference frames that are floating
in space. Only in these frames will an object that is released at rest
in "mid air" stay at rest. The name we give to reference frames for which
this law (Newton's first law) of motion holds is an {\bf inertial reference
frame}.
Imagine two reference frames floating in space. Suppose someone named Bill was in one,
and George in the other. Suppose that George observed that Bill was moving in the
+ direction with a {\it constant velocity} $+v$. George would feel at rest and
say that Bill was moving to the right with a constant velocity. Bill, however, would also feel
at rest and say that George is moving to the left with a constant velocity $-v$.
Who is correct? Both are. Each of these frames is an inertial reference
frame. Bill feels at rest, and so does George. There is something very special
about reference frames floating in space with a constant velocity with respect
to each other. They are all inertial reference frames and have the following
properties:\\
\noindent 1. A reference frame moving with a constant velocity with respect to an inertial
frame is also an inertial reference frame.\\
\noindent 2. In an inertial reference frame, one "feels" at rest.\\
\noindent 3. There is no experiment that one can do in an inertial reference
frame to determine the velocity of the reference frame.\\
\noindent 4. The laws of physics take on the same form in all inertial
reference frames.\\
\noindent 5. There is no absolute reference frame.\\
The equivalance of inertial reference frames is a fundamental property of physics, and
is the basis of Einstein's theory of special relativity. It is a wonderful property
of nature, and one can marvel at its simplicity.
A final note on Newton's first law is that it allows one to define equal time
intervals independent of a physical system. Here is how to do it: Set up your
"x" axis, pick an origin, and a unit length. Use your unit length to mark on your "x" axis
equal distances. Then set an object (with no forces) in motion. It will float along your
"x" axis. A time interval occurs each time that it passes a mark. According to Newton's first
law the time intervals will be equal.
\bigskip
\noindent {\bf Describing Motion: Acceleration}
\vspace{3mm}
Objects don't always move with constant velocity, velocities change. The
change in velocity per unit time is called acceleration. The
average acceleration, $\bar{a}_{t_1 \rightarrow t_2}$, between time $t_1$ and $t_2$
is defined to be
\begin{equation}
\bar{a}_{t_1 \rightarrow t_2} \equiv {{v(t_2) - v(t_1)} \over {t_2 - t_1}}
\end{equation}
\noindent As with average displacement and average velocity, one needs {\it two
times} to calculate the average acceleration. Since two times are needed, the
average acceleration is not a good quantity to describe the laws of physics.
A more useful quantity is the instantaneous acceleration, which is defined
in the same way as instananeous velocity. One takes the limit of the average
acceleration as $t_2$ approaches $t_1$:
\begin{equation}
a(t_1) \equiv \lim_{t_2 \rightarrow t_1} {{v(t_2) - v(t_1)} \over {t_2 - t_1}}
\end{equation}
\noindent Instantaneous acceleration is also written as:
\begin{equation}
a(t_1) \equiv \lim_{\delta t \rightarrow 0} {{v(t_1 + \delta t) - v(t_1)} \over {\delta t}}
\end{equation}
\noindent The limit on the right side is just the first derivative of the velocity evaluated
at the time $t_1$. So $a(t_1) \equiv dv/dt|_{t = t_1}$. Often the $t_1$ is replaced by $t$,
and one writes $a \equiv dv/dt$. A nice property about instananeous acceleration is that
it is determined at {\it one time}. The acceleration is just the second derivative of
position with respect to time: $a = d^2x/dt^2$.
One could also consider more derivatives such as $da/dt$, $d^2 a/ dt^2$, etc... to describe
motion. We need to rely on experiments to determine the simplest way to relate interactions
(forces) and motion.
If the position function, $x(t)$, is known, it is easy to find $v(t)$ and $a(t)$ by differentiation.
If one knows $a(t)$, one needs to integrate with respect to $t$ to find $v(t)$. To find
$x(t)$, one needs to integrate $v(t)$ with respect to $t$. A simple case that is discussed in
many texts is the special case of motion with constant acceleration. Suppose the
acceleration of a particle is constant, label it $a_0$. Then, $dv/dt = a_0$.
Multiplying both sides by $dt$ and integrating we have $ \int_0^t dv = \int_0^t a_0 dt$,
which gives $v(t) - v_0 = a_0 t$ where $v_0$ is the velocity at $t=0$, $v(0)$. This
is often written as $v = v_0 + a_0 t$, where $v$ means $v(t)$.
To solve for $x(t)$ for the case of constant acceleration requires one more integration.
From $dx/dt = v(t) = v_0 + a_0 t$, one can multiply both sides by $dt$ and integrate:
$\int_0^t dx = \int_0^t v_0 dt + \int_0^t a_0 t dt$. After integrating, one
obtains: $x(t) - x(0) = v_0 t + a_0 t^2/2$. This equation is often written as
$x = x_0 + v_0 t + a_0 t^2/2$, where $x$ means $x(t)$.
Summarizing for the {\it special case of constant acceleration}:
\begin{equation}
v = v_0 + a_0 t
\end{equation}
\begin{equation}
x = x_0 + v_0 t + {a_0 \over 2} t^2
\end{equation}
\noindent Eliminating $t$ from these two equations gives:
\begin{equation}
v^2 = v_0^2 + 2 a_0 (x-x_0)
\end{equation}
\noindent In general, motion that has constant (non-zero) acceleration for extended
periods of time is fairly rare. Under certain
circumstances however, the motion of an object can
be {\it approximately} one of constant acceleration for a period of time.
These special cases include situations where the net force is
{\it approximately} constant
(e.g. free fall near the earth's surface in the absence of air friction,
rolling or sliding down a ramp with no friction, ...). Since the
equations for position and velocity are simple for motion of constant
acceleration, many texts use examples and problems of this special case
to test the student's understanding of these concepts.
\bigskip
One final note: the sign's of $x(t)$, $v(t)$, and $a(t)$ are not related to
each other. The acceleration can be in the negative (positive) direction eventhough
the velocity is positive (negative), etc. The sign of $a$ is in the direction of
the change of $v$.
\bigskip
\noindent {\bf Describing Motion in 2 and 3 Dimensions}
\bigskip
The extension of velocity and acceleration to 2 and 3 dimensions is straighforward,
since displacements add like vectors. We can define a position vector,
$\vec{r}(t)$, a velocity vector, $\vec{v}(t)$, and an acceleration vector,
$\vec{a}(t)$ as follows
\begin{eqnarray*}
\vec{r}(t) & = & x(t) \hat{i} + y(t) \hat{j} + z(t) \hat{k} \\
\vec{v}(t) & = & {{d\vec{r}(t)} \over {dt}} \\
& = & v_x(t) \hat{i} + v_y(t) \hat{j} + v_z(t) \hat{k} \\
\vec{a}(t) & = & {{d\vec{v}(t)} \over {dt}} \\
& = & a_x(t) \hat{i} + a_y(t) \hat{j} + a_z(t) \hat{k}
\end{eqnarray*}
\noindent If we pick our coordinate system properly, sometimes we can change
a three dimensional problem into three one dimensional problems. An important point
to mention is that in general {\bf the three vectors $\vec{r}(t)$, $\vec{v}(t)$, and
$\vec{a}(t)$ do not necessarily point in the same direction}.
\end{document}