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\centerline{\bf Mechanics Notes II}
\centerline {\bf Forces, Inertia and Motion}
\vspace{3mm}
The mathematics of calculus, which enables us to work with instantaneous
rates of change, provides a language to describe motion. Our
perception of {\bf force} is that it is a push or a pull. We also have
experienced that it is easier to change the motion of a "smaller" object
than a "larger" one. We call {\bf inertia} the
resistance to a change in motion, and as you might imagine it is related
to how much "matter" (or mass) an object has.
\bigskip
\noindent {\bf Mass} \\
There are three different properties associated with the mass of an object:
\begin{enumerate}
\item The amount of matter an object has. For a pure substance, say iron,
we assume that mass is proportional to the volume of the substance. There
is no law of nature associated with this definition of mass, only that
mass $m=\rho V$, where $\rho$ is the density of the substance and $V$ it's
volume. We will refer to this property of mass as just {\bf mass}, or
"$\rho V$ mass".
\item The resistance to a change of motion. We refer to this property of
mass as {\bf inertial mass}.
\item The property of an object that causes the attractive force of gravity.
We refer to this property of mass as {\bf gravitational mass}.
\end{enumerate}
We will carry out and discuss experiments that relates these three properties
associated with mass to each other. As we will see, these three different
aspects of mass are proportional to each other.
\bigskip
\noindent {\bf Force}\\
We demonstrated that if an object has {\bf no} forces acting on it, it will
stay at rest or move with a constant velocity. Is there any other situation
that an object will move with constant velocity. Yes, there is. If we
push our physics book across the table, we can push it such that it will move
with a constant velocity, say $\vec{v}=v_0 \hat{i}$. Does this mean that there
are no forces on it? NO.
We are pushing with a certain force, the table is exerting a force, and
the earth is attracting the book (gravitational force). These three forces
are acting on the book. The table is actually "pushing up" on the book and
exerting a frictional force as well. Newton realized that if an object
is moving with a {\bf constant velocity} then all the forces that it experiences must
add up to zero (via vector addition). This is consistent with our experiments
on statics, since if we are in a reference frame that is moving with a velocity
of $v_0 \hat{i}$ then the book is at rest. In this case, we know that the
vector sum of the forces must add to zero.
Therefore, {\bf if an object is moving with a constant velocity, the vector sum of the forces
acting on it equals zero}. If the net force on an object is not equal to zero, then
the object will not move with a constant velocity.
At present we have discovered only three basic interactions in our universe:
\begin{enumerate}
\item Gravity
\item Electro-magnetic-weak interaction
\item Strong interaction
\end{enumerate}
As a physicist we are interested in
discovering the Laws of Nature in their most simple form. Since particle interactions result in
changes in the velocity of a particle, we are motivated to try and express
the laws of motion in terms of
the change in velocity of a particle: $d \vec{v} /dt$. Therefore, we believe
that the highest derivative of position with time in the basic equations of motion should be the
second derivative $d^2 \vec{r}(t) / dt^2$. There is another reason to expect that the laws
of classical physics will not need higher derivatives. If the equations
were to contain terms involving $d^3x/dt^3$, one would need
three initial conditions to determine $x(t)$ for times in the future.
{\bf If one believes that only the initial position, $x_0$, and initial velocity,
$v_0$, are necessary to determine $x(t)$ for future times, then there can
be at most second derivatives of $x(t)$ in the equations of motion}. Under
the requirement of only two initial conditions, one only needs to consider the
acceleration of a particle in formulating the Laws of Motion.
It turns out that differential equations are very useful
in describing nature because the laws of physics often
take on a simple form when expressed in terms of infinitesimal changes. Thus, we will
try and understand physics using second order differential equations.
This is one of the important ideas that Newton demonstrated.
The three basic interactions (gravity, electromagnetism, weak and strong) can be understood
without introducing the concept of force. For example, in Phy132, you will learn that the law
of universal gravity between two "point" objects equals
\begin{eqnarray*}
m_1 {{d\vec{v}_1} \over {dt}} & = & G {{m_1 m_2} \over r^2} \hat{r}_{12} \\
m_1 \vec{a}_1 & = & G {{m_1 m_2} \over r^2} \hat{r}_{12}
\end{eqnarray*}
\noindent Note that there is no mention of force in this equation. The same is true of
the other fundamental interactions. One reason force is not needed because these
basic interactions occur without the particles actually touching each other. However, it is
useful to consider "force". We can pull and push things around, and in
practical engineering problems objects experience pushes and pulls from ropes, surfaces, etc. We will
refer to forces that actually touch an object as {\bf contact forces}. Since a push or pull
changes a particles velocity, we are motivated to define and quantify the amount
of force an object experiences by how it changes the object's velocity.
We start by doing simple experiments in an inertial reference frame, that is, a
reference frame floating freely in space. Let's examine what happens
when an object is subject to a {\bf constant contact force}.
\bigskip
\noindent {\bf Experiments with Contact Forces}
\bigskip
One can ask: how do we determine that a contact force acting on an object is constant?
One way might be to use a "perfect" spring and pull on the object such that the
spring's extention is constant. I think we can agree that if a
spring is streched by a {\it fixed amount} the contact force it exerts will not change.
By a "perfect" spring, we mean just that, that the spring does not weaken
over time but keeps the same constant force. Let's observe what
happens when an object is subject to a {\bf constant contact force}. What kind of
motion will result? Most likely not constant velocity, since this is the case if there
are no net forces. The object will accelerate, but will the acceleration change
in time? We need to do the experiment, and in lecture I will pull a cart on an air track
with a spring keeping the stretch of the string constant. If I do it correctly,
we will see that:\\
\bigskip
{\bf Experiment 1}: If an object is subject to a {\it constant force}, the motion
is one of {\it constant acceleration}.\\
\bigskip
Wow, this is a very nice result! Nature didn't have to be this simple. The
acceleration might have changed in time with our perception of constant force, but
within the limits of the experiment it doesn't. Experiments show that this
result is true for any object subject to any constant force. I should note that
this experimental result is only valid within the realm of non-relativistic
mechanics. If the velocities are large and/or the measurements very
very accurate, relativistic mechanics are needed to understand the
data.
We need another experiment determine how an object's acceleration is related to
it's mass. Take two identical objects, each of the same substance, and apply the constant force
$F_0$ to one object. The acceleration will be constant, call it $a_1$. Now connect the
two identical objects together and apply the same force $F_0$ to both. The acceleration
is constant, but how large is it? We will do a similar experiment in
lecture. The result from the experiment is:\\
\bigskip
{\bf Experiment 2}: If a constant force is applied to two identical objects which are
connected to each other, the measured acceleration is {it 1/2} the
acceleration of one of the objects if it is subject
to the same constant force.\\
\bigskip
This is also an amazing result, and nature didn't have be be so simple! It shows that if
the amount of material (or "$\rho V$ mass") is doubled,
the acceleration is cut in half for the same constant force. Experiment will also
show that $n$ times the amount of "$\rho V$ mass" results in an acceleration equal to
$1/n$ times $a_1$. Therefore, {\bf inertial mass is proportional to the amount of
material (or "$\rho V$ mass")} for any object made of a single substance.
This experimental result gives us a method to quantify inertial mass for objects
made of different materials. Decide on a reference mass $m_0$ (e.g. one kilogram). Use a constant
spring extension to produce a constant force. The constant force
applied to the reference mass, causes a constant acceleration ($a_0$). Now apply the same
constant force is to the (unknown) mass being measured. The acceleration will be constant,
$a$. The inertial mass of the unknown is $m = m_0 (a_0/a)$.
As far as we know, inertial mass, is an intrinsic property of an object,
proportional to the objects "$\rho V$ mass". There is no loss of generality taking
the proportionality constant equal to one, equating inertial mass to mass,
and using the same units for both. In
the MKS system, the unit is the kilogram (Kg). If $a$ is in units of $M/s^2$, then
force will have units of $Kg M/s^2$, and is a derived quantity. One $Kg M/s^2$
is called a Newton (N). If a constant force of one Newton is applied to an object whose
mass is 1 Kg, the object have a constant acceleration of 1 $M/s^2$.
For any single constant force, $F$, acting on an object of mass $m$, the acceleration is $a=F/m$
More force produces more acceleration, more mass results in less acceleration.
We need to one more experiment to see if the "dynamic" definition of force add like
vectors. Suppose an object is subject to two constant forces at once. Call them
$F_1$ and $F_2$. Since we are still experimenting in one dimension, the forces
can act towards the right (+) or left (-) direction. Forces to the right will
be considered positive and to the left negative. When both forces are applied
at the same time, here is what the experiment shows:
\bigskip
\noindent {\bf Experiment 3}: If a constant force $F_1$ produces an acceleration
$a_1$ on an object and a constant force $F_2$ produces an acceleration $a_2$
on the same object, then if the force $F_1 + F_2$ acts on the object
the measured acceleration is $a_1 +a_2$.\\
\bigskip
\noindent Wow, we should be grateful that nature behaves in such a
simple way. The experiment indicates that if a 5 Newton force to the
right and a 3 Newton force to the left are both applied to an object,
the resulting {\bf motion} is the same as if a 2 Newton force to the right
were applied. Forces applied in one dimension add up like real numbers.
(As discussed in the next section, in two and/or three dimensions
experiment shows they add up like vectors). We refer to the
"sum" of all the forces on an object as the net force, and give it
the label $\vec{F}_{net}$.
\bigskip
\centerline{\bf Two and Three Dimensional Frictionless Environment}
\bigskip
The extension to two and three dimensions is greatly facilitated by
using the mathematics of vectors. Displacement is a vector,
since displacements have the properties
that vectors need to have. A displacement 40 units east plus a
displacement 30 units north is the same as one displacement
50 units at an angle of $36.869...^\circ$ N of E. Similarly,
relative velocity is a vector, since it is the time
derivative of displacement. The extension of force and motion to two and three
dimensions follows:
\begin{equation}
\vec{F}_{net} \equiv m \vec{a}
\end{equation}
\noindent for a single particle. This equation is refered to as {\bf Newton's Second
Law of motion}. Although force is not needed to describe the fundamental interactions
of nature, force is a useful quantity in statics, and when
dealing with contact interactions (i.e. contact forces and friction).
In this mechanics class, we will deal with the contact forces of surfaces,
ropes, human pushes or pulls, and the non-contact force of gravity near a
planet's surface. The general
approach is to determine all the forces involved. Once the forces are known,
the acceleration of the objects are known. Once the accelerations are determined,
the position function $x(t)$ can be found by integration. Doing appropriate
experiments we can test our understanding of the "physics" behind the interactions.
Before we proceed with investigating different forces, we present Newton's Third Law.
\bigskip
\noindent {\bf Newton's Third Law: Symmetry in Interactions}
\bigskip
Let's consider what happens when two objects, "1" and "2", interact with each other. By interact,
we mean that object "1" causes a push or pull on object "2" for a certain amount of time. Likewise
object "2" can cause a push or pull on object "1". We'll do the following experiment in
lecture. Two carts will be held together with a spring between them attached
to cart "1". They will be released and
the spring will push them apart. The carts can roll on the horizontal surface of the table.
Is it possible for only one cart to move away, and the other one to stay still? As we shall see,
no it is not. If cart "1" moves off to the right, then cart "2" must move off to the left. If
cart "2" feels a push due to cart "1", then cart "1" also feels a push.
After they fly apart, and the
spring is no longer pushing them, they will travel with a constant velocity. Let's first
only examine the final velocities of the cars, and not be concerned with what is happening during
the pushing.
The experiment will show: if the carts are identical, they end up moving with the same
speed, but in opposite directions. We will also see that more "massive" carts end up moving slower
after "pushing away" a less massive cart. We can pure substances for the carts and
define mass as we did inititally,
{\bf as the amount of matter an object has}.
For cart "1", we have $m_1 = \rho V_1$, and for cart "2", $m_2 = \rho V_2$, where
the $V_i$ are the volumes of the carts. Now we can carry out some experiments.
The experiments will show that if $m_1 = 2 m_2$, the final speed of cart "2" will be twice that of
cart "1". If $m_1 = 3 m_2$, the final speed of cart "2" will be twice that of cart "1". Etc...
WOW!, this is a very nice result. The mass times speed
of cart "1" is always equal to the mass times speed of cart "2". Once again, the experiments
demonstrate that inertial mass is proportional to "$\rho V$" mass.
There is something special about the
product (mass)(speed). It is so special that is given a special name, {\bf momentum}. Since
velocity better describes the motion, we use velocity instead of speed
and define the momentum $\vec{p}$ of an object as
\begin{equation}
\vec{p} \equiv m \vec{v}
\end{equation}
\noindent Note that we have written momentum as a vector. We haven't
yet shown that momenta combine according
to the mathematics of vector addition, but we will check it out later.
Now, returning to the cart experiment.
Let $\vec{v}_1$ be the final velocity of cart "1", and $\vec{v}_2$ be the
final velocity of cart "2".
Then we see that $m_1 |\vec{v}_1| = m_2 |\vec{v}_2|$, or in vector notation:
\begin{eqnarray*}
m_1 \vec{v}_1 & = & - m_2 \vec{v}_2 \\
\vec{p}_1 & = & - \vec{p}_2
\end{eqnarray*}
\noindent Since cart "1" started off at rest with no momentum, $\vec{p}_1$ is the change in the
momentum of cart "1". Likewise, $\vec{p}_2$ is the change in the momentum of cart "2"
resulting from its interaction with cart "1". Newton postulated that this symmetry holds
whenever any two particles interact with each other:
\bigskip
\begin{center}
{\bf If object 1 interacts with object 2,\\
then the change in momentum of object 2 caused by object 1 is equal but opposite
to the change in momentum of object 1 due to object 2.}\\
\end{center}
\begin{equation}
\vec{\Delta p}_1 = - \vec{\Delta p}_2
\end{equation}
\noindent Since the forces act on each cart at the same time, we have
\begin{eqnarray*}
{{\vec{\Delta p}_1} \over {\Delta t}} & = & - {{\vec{\Delta p}_2} \over {\Delta t}} \\
m_1{{\vec{\Delta v}_1} \over {\Delta t}} & = & - m_2 {{\vec{\Delta v}_2} \over {\Delta t}} \\
m_1 a_1 & = & - m_2 a_2 \\
\vec{F}_{1net} & = & - \vec{F}_{2net}
\end{eqnarray*}
\noindent since the net force equals $ m d \vec{v} / dt = d \vec{p} / dt$.
we see that the magnitude of the force that each object experiences is the
same but opposite in direction. Let $\vec{F}_{12}$ be the force
that object "1" exerts on object "2", and
$\vec{F}_{21}$ be the force that object "2" exerts on object "1". Then,
\bigskip
\begin{center}
{\bf If object 1 exerts a force on object 2,\\
then object 2 exerts an equal but opposite force on object 1.}\\
\end{center}
\begin{equation}
\vec{F}_{12} = - \vec{F}_{21}
\end{equation}
This equality of interacting
forces that we showed in our cart experiments is called "Newton's Third Law".
It was a great insight of Newton to realize there was certain symmetry
in every interaction. He probabily reasoned that something must be
the same for each object. Interacting objects clearly can have different masses and
different accelerations. The only quantity left are the forces that
each object "feels". Nature is fair when it comes to interacting
particles, object 2 is not preferred to object 1 when it comes to the
force that each feels. Forces always come in pairs. Whenever there is
a force on a particle, there must be another force acting on another particle.
Newton's Third Law applies (in some form) to every type of interaction. It is true
if the objects are moving {\bf or if they are not}. It is true even in the static
case. As simple as it may seem, it is often miss-understood. Consider the example of a book
resting on a table in a room. The book feels a graviational force due to the
earth, which is it's weight. What is the paired force for the book's weight?
Most students answer is "the force of the table on the book". The table does exert
a force on the book equal to it's weight, but it is not the "reaction" force
to the book's weight. The "reaction" force to the book's weight is the force
on the earth due to the book. To determine the two forces that are "paired",
just replace "object 1" with one object and "object 2" with the other in the
statement above. "If the {\bf earth} exerts force on the {\bf book}, then the {\bf book}
exerts an equal but opposite force on the {\bf earth}."
\bigskip
Newton's third law and its connection to the conservation of momentum is an
example of how a symmetry in nature leads to a conservation law. We will
return to the conservation of momentum in the next section when we investigate
other conserved quantities.
\bigskip
\noindent {\bf Summary of Newton's Laws of Motion}
\bigskip
The laws of motion apply in an inertial reference frame.
\begin{enumerate}
\item Newton's "First Law": 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. This law enables one to quantify time, i.e. equal time
intervals.
\item Newton's "Second Law": $\vec{F}_{net} = m \vec{a}$.
\item Newton's "Third Law": $\vec{F}_{21}=-\vec{F}_{12}$.
\end{enumerate}
\bigskip
\centerline {\bf Some Simple Forces}
\bigskip
Newton's laws of motion give us a method for determining the resulting motion
when an object is subject to forces. One proceeds by first identifying all the forces
acting on the object. Then, one adds the forces via vector addition to find
$\vec{F}_{net}$. The object's acceleration is $\vec{a} = \vec{F}_{net}/m$.
Once the acceleration
is known at all times (or all positions) then the motion is determined. The
beauty of this approach is that the forces take on a simple form. That is,
the quantity that affects the acceleration of an object (the thing we are calling
a force) turns out to be a simple expression of position, velocity, etc. Here, we consider three
types of forces: contact forces, frictional forces, and weight (gravity near the
surface of a planet). In future courses we will consider the "universal" gravitational
force, spring forces, electric and magnetic forces, and atomic/nuclear/subatomic particle
interactions.
\bigskip
\noindent{\it Contact Force Example:}
As previously stated, by a contact force we mean the pushing or pulling caused by the touching (or
physical contact) of one object on another. Someone's hand pushing on, or a rope pulling
on an object are some examples. From Newton's third law, object 2 feels the same
contact force from object 1 that object 1 feels from the contact force from object 2.\\
\noindent Consider the following example shown in the figure:\\
\begin{figure}
\includegraphics[width=14cm]{notes1a.png}
\end{figure}
Two masses in an inertial reference frame are connected by a rope. The inertial mass
of the mass on the left is $M_1$, the inertial mass of the mass on the right is
$M_2$, and the rope connecting them has a mass of $m$. Someone pulls on the
mass on the right with a force of $F$ Newtons. Question: find the
resulting motion, and all the contact forces.
The method of applying Newton's Laws of motion to a system of particles is:
first: find the forces on each object in the system, then: the acceleration
of each object is just $F_{Net}/(mass)$. For the example above, the
forces on the "right mass" are $F$ minus the force that the rope pulls to the left, which we
label as $c_2$. The rope feels the "reaction" force $c_2$ to the right
and a force $c_1$ from the left mass. Finally, the mass on
the left feels only the "reaction" force from the rope which is $c_1$ to the left.\\
Summarizing:\\
\begin{center}
\begin{tabular}{ccc}
\underline{Object} & \underline{Net Force} & \underline{Equation of motion} \\
left mass & $c_1$ & $c_1 = M_1 a$ \\
rope & $c_2 - c_1$ & $c_2 - c_1 = m a$ \\
right mass & $F - c_2$ & $F - c_2 = M_2 a$ \\
\end{tabular}
\end{center}
\noindent Adding up the equations of motion for the various masses gives
$a=F/(M_1 + M_2 + m)$. Solving for the contact forces gives:
$c_1 = M_1 F/(M_1 + M_2 + m)$, and $c_2 = (M_1 + m)F/(M_1 + M_2 +m)$.
If the mass $m$ is very small compared to $M_1$ and $M_2$, then
the contact forces are approximately equal $c_1 \approx M_1 F/(M_1 + M_2)$,
and $c_2 \approx M_1 F/(M_1 + M_2)$, giving $c_1 \approx c_2$. This force
is called the tension in the rope, and is the same throughout for
a massless rope.\\
\bigskip
\noindent{\it Weight}
Weight is a force. In the Newtonian picture of gravity, the {\bf weight}
$W$ of an object is the gravitational force on the object due to all the
other matter in the universe. We will be considering the weight
of objects on or near the surface of a planet (neglecting the rotation of
the planet). In this case, the strongest
force the object experiences is due to the planet.
A remarkable property of nature is that the {\it motion} of all objects
due to the graviational force is not dependent on the
objects mass! This is true for objects falling in the classroom,
satelites orbiting the earth, planets orbiting a star, etc. We will demonstrate
this property in lecture by dropping two different objects with different masses.
They both fall with the same acceleration, which we label as $g$.
Since $F = ma$, the graviational force on an object must be $W = mg$,
where $g$ depends only on the location of the object and $m$ is
the inertial mass of the object. {\bf An objects weight is proportional
to it's inertial mass}. Since $g=W/m$ is the same for all objects, if the mass is doubled, so
is the objects weight. Thus, experiments show that all three aspects of mass,
"$\rho V$", inertial mass, and gravitational mass, are proportional to each other.
With no loss of generality, we can set them equal to each other and use
"$\rho V$" in each case.
To measure an objects weight on a planet, one can use a scale which keeps the
object at rest. Since the object is not accelerating relative to
the planet, the force the
scale exerts on the object equals its weight.
An object's mass is an intrinsic property and is the same everywhere.
An objects weight $W$ depends on its location (i.e. which planet it is
on or near).
The Einstein picture of gravity is somewhat different. Being in a
free falling elevator near the earth's surface "feels" the same as if
you were floating in free space or in the space shuttle. In
each case you are weightless. Thus, you can be near the surface of
the earth and be "weightless". Likewise, if you are in a rocket
ship in free space (outer space far away from any other objects)
that is accelerating at $9.8 M/s^2$ you feel
the same as if the rocket were at rest on the earth. Thus,
if you sit on a scale in a rocket that is accelerating in
free space, the scale will give you a "weight" reading eventhough
there are no "gravitational forces". Weight therefore is
a relative quantity, and depends on the reference frame. In an inertial reference frame,
everything is weightless. In a non-inertial reference frame,
the force needed to keep the object at rest relative to the
frame is the weight (or apparent weight) of the object. Mass (more specifically rest mass), on the other hand,
is an absolute intrinsic quantity and is the same everywhere. Students
interested in these philosophical topics should major in Physics.
In this introductory class, we will take the Newtonian point of
view for objects near the surface of a planet in which case the weight $W$ is:
\begin{equation}
W = m g
\end{equation}
Next quarter you will discover what Newton discovered, that the gravitational
force between two point objects of mass $m_1$ and $m_2$ that are
separated by a distance $r$ is: $F_{gravity} = G m_1 m_2 / r^2$.
You will also show that the acceleration near the surface of a
spherically symmetric planet is approximately
$g = G m_{planet}/R^2$, where $R$ is the planet's radius and
$G \approx 6.67 x 10^{-11} N M^2/Kg^2$.
\bigskip
\noindent{\it Projectile Motion}
\vspace{3mm}
Projectile motion is often used as an example in textbooks, and is the
motion of an object "flying through the air" near the surface of the
earth (or any planet). The approximations that are made are 1) that
the object is near enough to the surface to consider the surface as flat,
2) the acceleration due to gravity is constant (does not change with height),
and 3) air friction is neglected. Usually the $+\hat{j}$ direction is taken
as up, and the $\hat{i}$ direction parallel to the surface of the earth such that
the object travels in the $x-y$ plane. The objects acceleration is constant
and given by $a_0 = - g \hat{j}$ for all objects. Letting $\vec{v}(t)$ represent the objects
velocity vector and $\vec{r}(t)$ be the objects position vector we have:
\begin{equation}
\vec{v}(t) = \vec{v}_0 - gt \hat{j}
\end{equation}
and
\begin{equation}
\vec{r}(t) = -{{gt^2} \over {2}} \hat{j} + \vec{v}_0 t + \vec{r}_0
\end{equation}
\noindent where $\vec{v}_0$ is the initial velocity and $\vec{r}_0$ is the
initial position. If $\vec{r}_0 = 0$ and $\vec{v}_0 = v_0 cos(\theta) \hat{i}
+ v_0 sin(\theta) \hat{j}$ one has:
\begin{equation}
\vec{v}(t) = v_0 cos(\theta) \hat{i} + (v_0 sin(\theta) - gt) \hat{j}
\end{equation}
and for the position vector, one has:
\begin{equation}
\vec{r}(t) = v_0 cos(\theta) t \hat{i} + (v_0 sin(\theta) t - {{gt^2} \over 2} ) \hat{j}
\end{equation}
\noindent It is nice that the horizontal and vertical motions can be treated separately.
This is because force is a vector and the only force acting on the particle is gravity
which is in the vertical direction. The seemly complicated two-dimensional motion is
actually two simple one-dimensional motions.
In this example of projectile motion, two quantities remain constant: the acceleration
($-g \hat{j}$) and the x-component of the velocity. The x-component of the velocity
is constant since there is no force in the x-direction and consequently no acceleration
in the x-direction. Also note that vectors $\vec{r}$, $\vec{v}$, and $\vec{a}_0$
can (and usually) point in different directions.
On the earth, there is air friction which needs to be considered for an accurate
calculation. Also, even in the absence of air friction, the parabolic solution
above is not exactly correct. In the absence of friction, the path of
a projectile near a spherical planet is elliptical.
\bigskip
\noindent{\it Frictional Forces}
\vspace{3mm}
In this class we consider "contact" frictional forces. When two surfaces
touch each other, one surface exerts a force on the other (and visa-versa
by Newton's third law). It is convienient to "break-up" this force into
a component {\it perpendicular} to the surfaces (Normal Force) and a component
{\it parallel} to the surfaces (Frictional Force). We consider two cases
for the frictional force: 1) the two surfaces slide across each other (kinetic
friction) and 2) the two surfaces do not slide (static friction).
\bigskip
\noindent {\it Kinetic Friction}\\
\noindent If two surfaces slide across each other, the frictional force depends primarily
on two things: how much the surfaces are pushing against each other (normal force N)
and the type of material that makeup the surfaces. We will show in class that the
kinetic frictional force is roughly proportional to the force pushing the
surfaces together (normal force N), or $F_{kinetic\: friction} \propto N$.
We can change the proportional sign to an equal sign by introducing a
constant: $F_{kinetic\: friction} \approx \mu_K N$. The coefficient $\mu_K$
is called the coefficient of kinetic friction and depends on the material(s)
of the surfaces.
\bigskip
\noindent{\it Static Friction}\\
\noindent If the surfaces do not slide across each other, the
frictional force (parallel to the surfaces) is called static friction.
The static frictional force will have a magnitude necessary to keep
the surfaces from sliding. If the force necessary to keep the surfaces
from sliding is too great for the frictional force, then the surfaces
will slip. Thus, there is a maximum value $F_{Max}$ for the static friction:
$F_{static\: friction} \leq F_{Max}$. As in the case of sliding friction,
$F_{Max}$ will depend primarily on two things: the normal force $N$ and
the type of materials that make-up the surfaces. We will also show in
class the $F_{Max}$ is roughly proportional to the normal force,
$F_{Max} \propto N$. Introducing the coefficient of static friction,
$\mu_{S}$, we have: $F_{Max} = \mu_S N$. The static friction force
will only be equal to $F_{Max}$ just before the surfaces start slipping.
If the surfaces do not slip, $F_{static\: friction}$ will be just the
right amount to keep the surfaces from slipping. Thus, one usually
writes that $F_{static\: friction} \leq \mu_S N$.
Summarizing we have:
\begin{equation}
F_{kinetic\: friction} = \mu_K N
\end{equation}
\noindent and for static friction
\begin{equation}
F_{static\: friction} \leq \mu_S N
\end{equation}
\noindent We remind the reader that the above equations are not fundamental "Laws
of Nature", but rather models that approximate the forces of contact friction. The
fundamental forces involved in contact friction are the electromagnetic interactions
between the atoms and electrons in the two surfaces. To determine the frictional
forces from these fundamental forces is complicated, and we revert to the phenomenological
models described above.
It is interesting to note that in the case of kinetic friction, the net force that the
surface experiences must lie on a cone. The angle that the cone makes
with the normal is always $tan^{-1}(\mu_K)$. For the case of static friction, the
net force that the surface can experience must lie within a cone which makes an angle
with the normal of $tan^{-1}(\mu_S)$
\bigskip
\noindent {\it Uniform Circular Motion}
\vspace{3mm}
If an object travels with constant speed in a circle, we call
the motion uniform circular motion. The uniform meaning
constant speed. This motion is described by two parameters:
the radius of the circle, $R$, and the speed of the object, $v$.
The speed of the object is constant, but the direction of the velocity
is always changing. Thus, the object does have an acceleration.
We can determine the acceleration by differentiating the position
{\bf vector} twice with respect to time. For uniform circular motion,
the position vector is given by:
\begin{equation}
\vec{r}(t) = R (cos({{vt} \over {R}}) \hat{i} +
sin({{vt} \over {R}}) \hat{j})
\end{equation}
\noindent where $\hat{i}$ points along the +x-direction and
$\hat{j}$ points along the +y-direction. It is also convenient
to define a unit vector $\hat{r}$ which points from the origin to
the particle:
\begin{equation}
\hat{r}(t) = (cos({{vt} \over {R}}) \hat{i} +
sin({{vt} \over {R}}) \hat{j} )
\end{equation}
\noindent In terms of $\hat{r}$, the position vector $\vec{r}$ can be written as:
\begin{equation}
\vec{r}(t) = R \hat{r}(t)
\end{equation}
To find the velocity vector, we just differentiate the
{\bf vector} $\vec{r}(t)$ with respect to t:
\begin{eqnarray}
\vec{v}(t) & = & {{d \vec{r}} \over {dt}}\\
& = & R {v \over R} (-sin({{vt} \over {R}}) \hat{i}
+ cos({{vt} \over {R}}) \hat{j})\\
\vec{v}(t) & = & v (-sin({{vt} \over {R}}) \hat{i} +
cos({{vt} \over {R}}) \hat{j})
\end{eqnarray}
\noindent To find the acceleration of an object moving in uniform
circular motion one needs to differentiate the velocity
{\bf vector} $\vec{v}(t)$ with respect to $t$:
\begin{eqnarray}
\vec{a}(t) & = & {{d \vec{v}} \over {dt}}\\
& = & v {v \over R} (-cos({{vt} \over {R}}) \hat{i} -
sin({{vt} \over {R}}) \hat{j})\\
\vec{a}(t) & = & - {{v^2} \over {R}} \hat{r}
\end{eqnarray}
\noindent Thus for an object moving in uniform circular motion, the
magnitude of the acceleration is $|\vec{a}| = v^2/R$, and the direction
of the acceleration is towards the center of the circle.
In an inertial reference frame, net force equals mass times acceleration. Thus, if an object
is moving in a circle of radius $R$ with a constant speed of $v$, the net force on
the object must point towards the center and have a magnitude of $mv^2/R$.
Uniform circular motion is another case in which the three vectors $\vec{r}$, $\vec{v}$,
and $\vec{a}$ do not point in the same direction. In this case $\vec{a}$ points in the
opposite direction from $\vec{r}$, and $\vec{v}$ is perpendicular to both $\vec{r}$ and
$\vec{a}$.
\end{document}