Download Kinematics

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Period ________
Kinematics
We will begin our study of Physics by looking at the motion of objects and trying to
measure and describe what we observe. The study of motion and the use of mathematics
to describe different types of motion is known as Kinematics.
In order to clearly and completely describe motion, we have to define some key
quantities that we can either measure, or eventually, calculate from other measurements.
The two key quantities that we can measure easily are time (t) and distance (x). From
these two simple measurements, we will be able to derive some other quantities, allowing
us to give a clear and complete description of the motion of an object.
First, some definitions:

Distance (x): Basically, distance is the difference in position between any two points.
There are many units that can be used to measure a distance, but we will use the
meter (m). Distance is always positive. It is imperative to note that distance is also a
cumulative quantity. The more you walk, regardless of what direction you are
moving, the more distance you travel. For example… my home is 8.9 miles away
from school, but I drive a greater distance (~12 miles) when I take the scenic route.

Displacement (x): Displacement is a distance combined with a direction. It is also
the shortest distance between two points. Again, we will measure Displacement (x)
in meters. Unlike distance, your displacement depends on how far you walk and in
what direction. For example… my home is 8.9 miles away from school. As I drive
to school, regardless of what route I take and what distance I cover, my displacement
from when I start to when I finish is 8.9 miles. It is also best to include a direction
with your displacement; this allows us to choose a direction to call “negative” and
“positive”, making it easy to combine different motions to determine a displacement.
This indicates that a person could walk for a great distance, but end up with zero (or
negative!) displacement!

Time (t): We will always discuss the change in time between two events, and will
generally consider the first event to occur at a time t = 0. We will measure time in
seconds (s).

Speed (v): Speed is most easily defined as how fast you are traveling, but there are
two kinds of speeds. One of them is like the speed measured by the aptly-named
speedometer (or as I like to say, speed-o-meter) in your car; this is known as your
instantaneous speed. It tells you what your speed is at that exact point in time. You
can also discuss an object’s average speed, which is defined as the distance traveled
divided by time it takes to cover that distance. For example… as I drive to school, it
takes 12 minutes to travel 8.9 miles. From this information, I could determine my
average speed. However, at one point in time I may be moving faster or slower than
my average speed, and my speedometer will give my instantaneous speed at any point
in time. Speed will be measured in units of meters per second (m/s).
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Period ________

Velocity (v): Velocity is a speed combined with a direction. It can be defined as an
object’s displacement divided by the time it takes to move from the initial to the final
position. Velocity is also measured in units of meters per second (m/s). Again, we
can discuss both an instantaneous and average velocity. It is possible for someone to
move and still end up with zero (or negative) velocity. For example… imagine you
are watching a NASCAR race. As the cars travel around the track, often at a speed in
excess of 200 mph, the entire race ends up with a total displacement, and an average
velocity, of zero!

Acceleration (a): Acceleration is loosely defined as an object speeding up or slowing
down. More specifically, acceleration is best defined as the change in an object’s
velocity divided by the time taken for that velocity to change. Because an object’s
velocity can increase or decrease, acceleration can be positive or negative. However,
one must use caution when describing acceleration. Believe it or not, it is possible for
an object to speed up and have a negative acceleration! Acceleration is measured in a
unit of meters per second per second, or meters per second squared (m/s2).
Now that we have these terms defined, we need to learn how they are
another. To do this, we will build what I like to call a “Kinematics
contains four basic algebraic relationships that can tell us anything and
need to know about an object’s motion. (We are assuming here that the
constant.)
related to one
Toolbox” that
everything we
acceleration is
vf = vo + (a)t
x = (vo)t + ½ (a)t2
vf2 = vo2 + 2(a)x
x = ½ (vf + vo)t
Notice that each of these equations contains four different variables. As long as we know
at least 3 separate pieces of information, we can determine any other piece that we want.
What we also want to be able to do is interpret a graph to understand the motion that is
occurring. To do this, please consult the following graphic device.
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Bradshaw 05-06
Name _________________________________________________
Period ________
What this shows is how to interpret a graph of displacement vs. time, velocity vs. time,
and acceleration vs. time. What the above device shows is that, when looking at a graph
of x vs. t, the velocity at some time is found by looking at the slope of the graph at that
point. For instance, if I want to know the velocity of an object at t = 5 s, I would
determine the slope of the displacement vs. time graph at t = 5 s. The acceleration is
given by the slope of a v vs. t graph.
Similarly, velocity can be determined from the graph of a vs. t by analyzing the area
underneath the curve. Displacement can be found from the area under the v vs. t graph.
The four equations listed above can be derived by carefully analyzing these graphs. We
will explore these relationships in lab and in some homework problems and activities.
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Period ________
Vectors
A vector is a mathematical term that indicates a quantity with a certain size (we call this
magnitude) that is oriented in a given direction. For our purposes, any quantity that has
magnitude and direction is a vector. Some common examples we’ve already discussed
are displacement, velocity, and acceleration.
Vectors can often be tricky to deal with because of the directional aspect of them. One
exercise we will face with vectors is the combination of two different vectors of the same
type (i.e., two different velocities). This is a very simple exercise when the two vectors
point in the exact same (or exact opposite) direction; it is often challenging to try and
combine two similar vectors if they point in different directions. It can also difficult to
deal with a vector that moves in more than one direction at a time. See below for
examples of what I am referring to here…
In order to simplify this process as much as possible, we have developed a number of
techniques that reduce a complicated problem to a much simpler one. One of these will
be a change of notation, which will allow use to clearly indicate the exact magnitude and
direction of any vector, even if it points in more than one direction at a time. Our main
tool for working with vectors will be right-triangle Trigonometry (insert groaning noise
here).
Before we delve into the nitty-gritty of dealing with vectors, we need to first identify the
aspects of right triangles that we need to know and use. These are SOH CAH TOA, and
the Pythagorean Theorem.
SOH CAH TOA teaches us the relationships between the sides and angles of a right
triangle. Please consider the following right triangle:
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Name _________________________________________________
Period ________
SOH: Sine of  = Opposite side divided by Hypotenuse.
CAH: Cosine of  = Adjacent side divided by Hypotenuse.
TOA: Tangent of  = Opposite side divided by Adjacent side.
This will be especially useful for dealing with vectors that point in more than one
direction. We can use these definitions to express a single, multi-directional vector as the
sum of two distinct vectors, one of which points only in a horizontal (or x) direction, and
one which points only in a vertical (or y) direction. These two new one-directional
vectors are called vector components. So, after analyzing a multi-directional vector with
SOH CAH TOA, we will then be able to write this vector as the sum of its x-component
(*this ALWAYS goes with the Cosine of the angle*) and its y-component (*this
ALWAYS goes with the Sine of the angle*). As you will soon see, this makes
combining two (or more) vectors a much simpler endeavor.
The Pythagorean Theorem shows the relationships between the three sides of a right
triangle. Please consider the following diagram:
a2 + b2 = c2
This is most useful when we are trying to express a vector not by its components, but
rather in terms of its magnitude (*the hypotenuse of the right triangle will always be the
magnitude of the vector, unless the vector points purely in the x- or y-direction*).
In a general sense, we will use an angular measure to indicate the direction of a vector.
These angles are ALWAYS measured counter-clockwise from the +x-axis. (I.e., the +yaxis is 90 deg., the – x-axis is 180 deg., the – y-axis is 270 deg., and the +x-axis is 0 or
360 deg.)
So, our vectors will be expressed as a magnitude at some angle. For instance, take the
following two vectors: A = 12 @ 30 deg; B = 9 @ 135 deg.
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Name _________________________________________________
Period ________
Assume we wanted to determine a vector C: the sum of A + B. In order to do this, we
must follow a number of relatively simple steps:
1. First, we would need to first determine the x- and y-components of both A and B. To
do this, we will multiply the magnitude (ex: A = 12) of the vector by the cosine of
the angle (ex: for A,  = 30 degrees) for the x-component, and by the sine of the
angle for the y-component. This will yield 4 vector components: Ax, Ay, Bx, and By.
Ax = A Cos
Ay = A Sin
2. Next, we would combine the x-components of A and B; this will yield the xcomponent of the sum, vector C. We would do the same for the y-components.
Cx = Ax + Bx
Cy = Ay + By
3. At this point, we have two new x- and y-components, for the vector C. We must then
determine the magnitude and direction for C. To find the magnitude, we use the
Pythagorean Theorem, with the x- and y-components being the two sides, and the
magnitude being the hypotenuse.
C2 = Cx2 + Cy2
4. Determining the direction (or angle) for C is a more challenging process. To help me
in this process, I would make a sketch of what A + B should look like. I do this by
placing the tail of one vector at the head (or point) of the other vector. It does not
matter in which direction I do this, but generally we start with the first vector (A) and
finish with the last (B). This method is referred to as the Tail-to-Head method, for
obvious reasons. To finish, the sum (vector C) starts at the tail of the first vector and
ends at the head of the last vector. This is an important step, because it shows me in
what quadrant (and in what range of angle measure) the sum lies.
5. The simplest way to determine the angular direction of the vector C is to use an arctrig function (i.e., arcsin, arccos, arctan) which looks like Sin-1, Cos-1, or Tan-1 on
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Period ________
your calculator. I prefer Tan-1, but they will all work. To use Tan-1, you type into
your calculator the following expression; the resulting value is the angle giving the
direction of vector C.

 = Tan-1(Cy/Cx)
6. Once you have this result, we now have to think carefully about where this angle is.
The thing about calculators is that they aren’t very smart; it will always try to give
you a result that is between 0 and 90 degrees. For instance, if your calculator yields a
positive result, this indicates that the angle is either in the 1st or 3rd quadrant. You can
tell which quadrant it is in by the signs of the x- and y-components. If the x- and ycomponents are both positive, the angle is in the 1st quadrant, and you don’t need to
make any corrections. However, if the x- and y-components are both negative, the
angle is in the 3rd quadrant, and you must add 180 to this value to achieve the proper
result.
7. If your calculator yields a negative result, this means the angle is in the 2nd or 4th
quadrant. Again, we determine this by the signs of the components. If the xcomponent is negative, we must add this negative number to 180. If the y-component
is negative, we must add this negative number to 360.
8. At this point, we can now state this vector by its magnitude and direction.
So, now we can take a vector which is oriented in any direction and express it as two
individual vectors; one purely in the x-direction and one purely in the y-direction. This
provides a simple method of combining two vectors: add the x-components to get the xcomponent of the sum, add the y-components to get the y-component of the sum; use the
Pythagorean Theorem to get the magnitude of the sum; and use an arc-trig function (I like
arctan) to determine the direction of the vector sum.
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Bradshaw 05-06
Name _________________________________________________
Period ________
Free-Fall Motion
Continuing on our study of motion, we next investigate motion of objects as they are
moving in a vertical plane, such as being thrown straight upwards and subsequently
falling down towards the ground. Thankfully, the equations for free-fall are identical to
those for kinematics, with a couple of key differences. (N.B.: the term “free-fall”
indicates that we are ignoring the inherent resistance that air provides against moving
objects.)
 For displacement, we will use the letter (y) as opposed to the (x) that we used in
kinematics. We do this because the vertical axis in math is generally referred to
as the ‘y’ axis. Also, with the other variables, we will add a subscript ‘ y’ in order
to indicate we are dealing with a vertical motion (I.e.: v0y, ay, vfy). In all
equations, we can substitute the letter (y) for the letter (x).
 The acceleration (ay) is ALWAYS equal to -9.8 m/s2.
 The maximum vertical displacement is referred to as
ymax.
A key to dealing
with free-fall motion is the concept that, at ymax, the vertical velocity vy = 0.
This is required so that the object can cease its upwards motion and start moving
back downwards.
 When the vertical displacement y = 0, indicating that the object has returned to
the point from which it started, the object is moving downwards with the same
speed at which it was initially moving upwards. To say that in a different manner
that is more useful,
vfy = -v0y
where the negative sign indicates the downwards direction. However, the
magnitude (or numeric value) of these velocities are the same. In other words, it
is moving just as fast downwards at the bottom as it was moving upwards in the
very beginning.
 Furthermore, in the case where y = 0, the object will require ½ the total time to
reach its highest point ymax, and will require an equal amount of time to return to
the position from where it started.
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Name _________________________________________________
Period ________
Projectile Motion
To this point, we have learned how to deal with three distinct but apparently unrelated
topics:
Kinematics
Free-fall Motion
Vectors
What we have actually done is learned all the tools necessary to study what would appear
to be a very complicated situation, where an object is moving both horizontally (which
we studied in Kinematics) and vertically (which we studied in Free-fall) at the same time.
This can be very challenging to work with if you treat this two-dimensional motion as a
singular motion. In order to avoid confusion and make things as easy as possible, what
we do instead is not consider this motion as a single motion in two directions, but instead
as two separate, wholly unrelated motions (one of which is purely horizontal, the other
which is purely vertical) that occur at the same time.
Now, what’s all this crazy talk about “two separate, wholly unrelated motions”? It’s
clearly one thing moving in two directions at the same time. And that’s true; it is an
object moving in two directions at the same time. But, it’s also true that these two
motions are really not related to one another at all. If there was no horizontal motion, the
vertical motion would be the same; if there was no vertical motion, the horizontal motion
would be the same. The two motions are not related. The vertical motion has no bearing
on the horizontal motion; the horizontal motion has no bearing on the vertical motion.
So, when we start a projectile motion problem, we will treat it as two separate motions
that occur at the same time: one motion is horizontal, and one motion is vertical.
The only thing between these two motions that is related is the time.
The time for both motions is the same; everything else is totally
unrelated.
What this means, then, is that when we solve a projectile motion problem, we will
be trying to solve two separate problems at once: a free-fall problem as the object
is moving vertically through the air, and a horizontal kinematics problem as the
object moves horizontally above the ground.
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Name _________________________________________________
Period ________
Free-fall motion problems come in essentially three distinct flavors:
An object launched from
a height with initial
horizontal velocity, and
zero vertical velocity.
In this case, vox
An object launched from
the ground with an
initial velocity at some
angle that lands with
zero vertical
displacement.
= vo,
An object launched with
an initial velocity at an
angle that achieves some
vertical displacement.
In this case, we must
determine vox and voy
using SOH CAH TOA.
Now, the vertical
displacement = y.
In this case, we must
determine vox and voy
using SOH CAH TOA.
Also, the vertical
displacement is zero, so
vfy = -voy.
and voy = 0. This
indicates the vertical
displacement = –y.
(Note: the vertical
displacement can be
positive or negative.)
In each case, a key to successfully solving one of these problems is to keep organized. It is of
vital importance that you do not mix and match information. We deal with each motion
separately, so we must keep information for the horizontal motion together, and we must keep
information for the vertical motion together. We cannot mix and match x- and y-stuff; as the
Offspring would have said in the mid to late 90’s, “You Gotta Keep ‘Em Separated.” In order to
accomplish this, I advocate employing the following table:
X
?
Y
?
V0
V0Cos
V0Sin
Vf
V0Cos
?
a
0
- 9.8 m/s2
t
?
?
So, in each box, you would place the information that you are given, or those pieces of
information that we already know for free. For instance, the vertical acceleration (the ‘a’ box
under the ‘y’ column) would be filled with -9.8 m/s2. For the initial velocities, we would use
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Bradshaw 05-06
Name _________________________________________________
Period ________
what we know about vectors to determine the x- and y-components of our initial velocity (i.e.,
vox = voCos, voy = voSin

But what do we do for the boxes we can’t fill? Some of it we will obviously be trying to solve
for; but what about the horizontal acceleration (ax)? Well, this is also something that we will
know for free: ax = 0. This is because of a simple assumption that we will always make:
Ignore air resistance.
This assumption means that there is nothing pushing on the object as it moves horizontally, so
there is nothing that will change its velocity in that direction. This indicates that vfx = vox.
So, our job will always be to fill in the empty boxes in the table. This is accomplished by
solving kinematics problems (for the x-motion) or free-fall problems (for the y-motion). A key
idea here is that, even though we will treat each motion separately, the time is the same for each
motion. So, once we find that time (almost always done using the vertical free-fall motion), it is
the same for the x- and y-motion.
As we set out to solve these problems, we must still consider those things that we know about
free-fall (e.g., at ymax the vertical velocity vy =0). Oftentimes, these facts will make solving such
problems a bit easier. We will use the same equations as we’ve always used, but with a little
twist:
Horizontal Motion
Vertical Motion
x = voxt
y = voyt + ½ ayt2
(We only need one of the
tools here, because the
acceleration is zero, so
the others are useless)
vfy = voy + ayt
vfy2 = voy2 + 2ayy
y = ½ (vfy + voy)t
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Name _________________________________________________
Period ________
In the vast majority of problems, we will use the vertical motion to determine time. This
is because the time for every problem is the time the object remains in the air, which is
determined solely by the vertical motion. In rare cases, the time will be a given quantity;
however, the only time we can use the horizontal motion to find time is if we are
given both x and vox.
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Ed. 6/19/2017
Name _________________________________________________
Period ________
Newton’s Laws and Applications
Isaac Newton is often referred to as the father of modern science due to his numerous
important achievements that either changed the way that science was done, or introduced
ideas that greatly altered out ability to study, describe and explain the universe around us.
His contributions were numerous and fundamentally changed a number of different fields
of study. Some of his most important achievements were in the study of motion.
Through careful observation, critical thinking and meticulous mathematical derivation,
Newton laid down three simple, basic laws that govern the motion that we have studied.
1. Law of Inertia – An object will maintain its current state of motion, either at rest
or moving in a straight line with constant speed, unless a net external force acts on
the object.
2. Net Force – The vector sum of all forces acting on an object is equal to the
product of the object’s mass and acceleration. I.e.,
F = ma
3. Action/Reaction – For every action there exists an equal and opposite reaction.
(In other words, when object A exerts a force on object B, object B exerts the
same force F on object A, but in the opposite direction.)
These laws govern nearly all the motions that we observe, and while each law has its
merits, it is primarily the 2nd law that we are most interested in. The troubling issue with
the 2nd law is its sheer simplicity: a direct, linear relationship between two variables, with
mass being constant. Such a clear and basic concept; yet, the application of this law is
where the complication begins.
When applying the 2nd law, it is important to remember that Forces are vector quantities;
hence, we must always be careful to deal with the correct components of our forces. (If a
problem takes place on level ground, then we will want force components in the x- and ydirection. When we are on an inclined surface, we will want our force components to be
parallel to the surface, and perpendicular to the surface.)
Forces
There are certain forces that we must always be aware of in a given problem. These are
forces that are always present, and oftentimes do not have a great effect on the motion of
an object. However, that does not mean that we can forget about them. Some key forces
to remember:
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Ed. 6/19/2017
Name _________________________________________________

Period ________
Gravity – The force of gravity on an object near the Earth’s surface (often referred
to as the object’s weight) is equal to the object’s mass times the acceleration due
to gravity (g = 9.8 m/s2).
Fg = mg

Normal, FN – The normal force is a force exerted by a surface on an object. The
normal force is always perpendicular to the surface (the term normal is used in
math to refer to something that is perpendicular). The normal force always serves
to balance out any other forces that are perpendicular to a surface. This is
sometimes simply an object’s weight, although other forces or force components
can change the value of FN. To find the normal force, it is necessary to apply the
2nd law in a direction perpendicular to the surface, and to set that net force in that
direction equal to zero.

Tension, T – Tension is just a force that acts through something that is generally
very flimsy, such as a piece of string. It is important to remember that tension is a
force that can only pull, and it pulls equally hard in both directions. There is no
equation for Tension.

Friction, f – Friction is a force that opposes motion. When an object is at rest, it
is the force of static friction that must be overcome for an object to begin moving.
When an object is in motion, the force of kinetic friction will always try to bring
that object to rest. Static friction and kinetic friction have different strengths.
Static friction is always larger than kinetic; if you don’t believe me, go try and
push a car that is in neutral. You will find that it is much easier to keep it in
motion than it is to get it to start moving. The force of friction is determined by
two factors: the roughness of the two surfaces in contact (related by the
coefficient of friction, ; [note: the coefficient for static friction is always larger
than the coefficient for kinetic friction], and the normal force that acts on the
object. So,
f = FN
So, how do we know when to apply the 2nd law? Well, if we are asked to find
acceleration and we don’t have enough information to use kinematics, or, if we are asked
to determine a given force. These are generally the times we will want to apply the 2 nd
law. When applying the 2nd law, there are some strategies or techniques that we can use
to make the process a little easier.

Systems – In a situation where we have multiple objects and at least one tension
(see diagram below), this is a case where the idea of a system will simplify the
process. Basically the approach is as follows.
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Bradshaw 04-05
Ed. 6/19/2017
Name _________________________________________________
Period ________
o We must choose a direction to be positive or negative. We must be
consistent in this regard throughout the rest of the problem.
o We place everything in a large bag that encloses the mass of the system,
msys. This is equal to the total mass of all the objects enclosed in the bag.
Then, we determine what the net force is on the system that will cause it to
move. Then, we determine the acceleration of the system, asys.
o Once we know asys, we can then “open the bag” knowing that each part
of the system will have the same acceleration. Then, by carefully
choosing the amount of mass that the tension causes to accelerate, we can
determine T.
o Ex:
o In the special case of an Atwood’s machine (see below), we already know
what the acceleration of the system is:
a = (M - m) g
(M + m)
o Note: the above relationship assumes that M > m.
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Name _________________________________________________

Period ________
Inclined Planes – In these problems, there are a few basic rules that we will live
by.
1. The acceleration is parallel to the plane. Friction is always parallel to the
surface. Applied forces, unless stated otherwise, are parallel to the surface.
We choose either up the plane or down the plane to be positive, and then stick
with it.
2. The force of gravity must be broken up into components parallel to and
perpendicular to the surface.
3. The parallel component of Fg is mgSin.
4. The perpendicular component of Fg is mgCos. [This is often (but not
always) what the normal force is equal to.]
5. Carefully apply the 2nd law in each direction.
We can also use these techniques and apply them together for certain situations. For
example, a heavy object is pulled up a rough ramp by using a counterweight tied to a
cable that passes over a pulley. This will require all the techniques that we have
discussed and includes all the forces that have been mentioned above. There are
numerous other examples that will serve as good practice as well.
So, now all that remains is to try some different examples. I encourage you to try as
many or as few as you like.
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Bradshaw 04-05
Ed. 6/19/2017
Name _________________________________________________
Period ________
Friction
Friction is a force that always opposes motion. It tries to prevent an object from moving
if it is at rest, or it tries to stop an object’s motion when it is already moving. It is a
critical force for cell function and mobility, so it’s an important thing that we learn
something about.
The direction of the frictional force is ALWAYS parallel to a surface. The strength of
the force of friction is due to two different factors: the roughness of the two surfaces in
contact (this is given by a number called the coefficient of friction) and the normal force
that acts on an object. When we compute the frictional force f, we use
f = FN
where  is the coefficient of friction. Incidentally,  is a number between 0 and 1, is
always positive, and has no units. It will either be given in a problem, or you will be
provided with enough information to determine its value.
When an object is at rest, if you want to make it move, you have to overcome what is
referred to as Static Friction (fs). This means that if a force greater than static friction is
not acting on the object, the object will remain stationary. Once a force exceeding static
friction acts on the object, it will begin to move. Static friction is less than or equal to the
coefficient of static friction times the normal force. In other words,
fs ≤ FN
When an object is moving, friction tries to cause it to stop. This type of friction is
referred to as Kinetic Friction (fk). Much like static friction, kinetic friction is equal to
the coefficient of kinetic friction times the normal force, or
fk = FN
*Note: static friction is ALWAYS greater than kinetic friction; this is because the
coefficient of static friction is always greater than the coefficient of kinetic friction.
When solving problems that include friction, it is imperative that the frictional force be
directed opposite the direction that an object would tend to move or is moving as a result
of the other forces or force components that are horizontal to the surface.
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Name _________________________________________________
Period ________
Universal Law of Gravity and Kepler’s Laws
In the 17th Century, numerous achievements were made in the study of the planets and
satellite motion by some of the greatest minds in scientific history. One key discovery
was made by Isaac Newton; some of the others were set down by Johannes Kepler.
Universal Law of Gravity
While contemplating the orbit of the moon one day, Isaac Newton strolled through an
apple orchard on his family’s estate. While walking, he observed an apple as it fell from
a tree and plummeted to Earth, and he was struck with an epiphany.
The apple is pulled down towards the center of the Earth by an unseen
force. Perhaps the behavior of the moon is due to the same phenomena.
This realization led Newton in his development of the Universal Law of Gravity:

Any two massive objects, of mass m and M, whose centers are separated by a
distance r, will feel an equivalent attractive force of the same magnitude.
FG = GmM
r2
where FG is the gravitational force, and G is referred to as the Gravitational
Constant (G = 6.67 * 10-11 Nm2/kg2)
If one were so inclined, he or she could use the above expression to try and determine the
acceleration due to gravity for an object near another (say, for instance, a car of mass m
and Planet Earth® of mass M). We know from our studies of Newton’s Laws that an
object’s weight on Earth is equal to the strength with which the gravitational force pulls
on that mass. This gravitational force Fg = mg, where m is the mass of the object (say, a
car) and g is the acceleration due to gravity (recall: g = 9.8 m/s2). Now, we have a new
way to think about that gravitational force, as the mutual force exerted on both the car
and the Earth: FG. These two forces describe the same phenomena, and as such, they
should be equivalent. So,
Fg = FG
mg = GmM
r2
g = GM
r2
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where M is the mass of the Earth (5.98 * 1024 kg), r is the radius of the Earth
(6.37 * 106 m) and G is the Gravitational constant.
Kepler’s Laws
Shortly before Newton was solidifying his theories on Gravity, Johannes Kepler, a Dutch
astronomer and mathematician, was completing his study of the known solar system.
Working with the painstakingly collected observations of fellow astronomer Tycho
Brahe, Kepler formulated three distinct laws that describe the motion of the planets.
1. The orbits of all planets are ellipses, and the Sun lies at one focal point. (This was
a point of great contention, as astronomers had been assuming for thousands of
years that the motion of all celestial objects were perfectly circular.)
2. A certain object will sweep out identical areas of its elliptical orbit in equivalent
periods of time. (This serves as a justification as to why a planet will move more
quickly when it is near the sun, but more slowly when farther away.)
3. The ratio of orbital period squared (T2) to mean orbital radius cubed (R3) is equal
to a constant for all satellites orbiting the same object. (I.e., for each planet in our
solar system, the ratio of period squared to radius cubed is equal to the same
number.)
T2 = 42
R3
GM
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Work and Kinetic Energy
Work and Energy are very important concepts in Physics. They are related concepts, but
they are not technically the same thing. One of the great foundational tenets of all
science is the law of Conservation of Energy, a principle that is elegant in its simplicity
and practically unrivaled in its power or importance. However, before we can delve into
its depths and applications, we must first understand the concepts of Work and Energy.
Work
 In Physics, work is defined as the product of the component of force applied to an
object while it covers a given distance, and the distance it covers. In other words,
W = FxCos
where F is the force exerted on the object, x is the distance it covers, and  is the
angle between the direction of the force and the direction of motion.
o Note that this angle is often zero degrees (when the force and the motion are
parallel) but other times is given in the problem.

Work is measured in a unit called a Joule (J). 1 Joule equals 1 Newton times 1
meter. I.e., 1 J = 1 N * 1 m.

Work can be positive or negative, depending on the angle. When a force (or a
component of the force) opposes the object’s motion, then the work done by that
force is negative work. If a force (or a component of the force) is in the same
direction as the motion, then the work is positive.

It is also possible for an object to cover a given distance while experiencing a
force and have zero work done on the object. This is possible when the angle
between the force and the distance is 90 degrees.
o E.g., an object sliding over a horizontal table experiences a gravitational and
normal force; however, both of these forces are perpendicular to the direction
of motion, and hence, each force does zero work. (By definition, the normal
force never does any work.) Also, when an object travels in a circle, the
centripetal force is always perpendicular to the direction of motion, and so
there is no work done by the centripetal force.
Kinetic Energy
 This is the energy any object has due to its motion. From the slowest slug to the
fastest car, any object in motion has a certain amount of kinetic energy. This is
determined by two key factors: the object’s mass, and its speed.
KE = ½ mv2
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
Much like work, and all other forms of energy, KE is measured in Joules.

An object always has, at some time, a positive amount of KE. However, over a
period of time, an object can lose or gain KE, meaning that the change in KE can
be positive or negative. To determine the change in an object’s KE, we simply
compute the following:
KE = KEf – KE0 = ½ mvf2 – ½ mv02

Perceptive readers will note that Work and KE share the same unit. This is an
indication that there must be some connection between these two quantities.
Indeed, this is the case. The relationship between Work and KE is very simply
stated by the Work-Energy Theorem:
o The net work done on an object is equal to the change in that object’s
KE. (Note: the net work is the sum of all work done on an object; or, the net
work is the work done by the net force.)
W = KE
Potential Energy

Potential Energy (PE) is basically an energy that is associated with an object’s
position (Recall: Kinetic Energy was associated with an object’s motion). It is also
measured in joules. There are two main types of potential energy that we are going to
focus on.
1. Gravitational Potential Energy (PEg) – This is the energy an object has due to the
fact that it is some vertical height (h) above the ground (or some other chosen
point of reference). [Note: It is only the vertical position that matters here; the
horizontal location of an object has no effect on its gravitational potential energy.
This point will come in handy when we discuss the energy of an object that is
resting on an inclined surface; the amount of PE the object has still only depends
on its vertical height above the chosen point of reference.] The amount of
gravitational potential energy depends only on two variables: mass and height. A
simple way of writing the amount of gravitational potential energy any object has
is:
PEg = mgh
PEg = mgh
where g = 9.8 m/s2.
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The tricky bit about gravitational potential energy is that it is a relative quantity.
What I mean by that is, the amount of potential energy an object has is relative to
some point that is chosen as the “zero” level of PE. Typically, that point will be
the ground. So basically that means that an object sitting on the ground will have
zero gravitational potential energy, whereas an object some height off the ground
(say, a monkey in a tree) will have a certain amount of energy because it is up in
the tree. However, there are certain times when it is easier to choose a different
point than the ground to serve as the “zero” level of PE. In these cases though,
it’s still very simple, because then the amount of PE an object has only depends
on the difference in height from the chosen zero point.
For instance, let’s consider our friend Bubbles the monkey in the tree. Perhaps he
climbed up into the tree after a banana, which is some small height above the
branch where Bubbles is just chillin’. Both Bubbles and the banana have some
amount of PE relative to the ground, because they are both elevated some height
above the ground. It is likely that Bubbles has much more PE than the banana in
this case; why? (Answer: because a monkey has much more mass than a banana).
Let’s look at it from a couple of different perspectives now. Assume that we
choose the “zero” level of PE to be the point where Bubbles has reached in the
tree. Then, in this case, Bubbles actually has zero PE because he’s at the zero
level. The banana, above Bubbles’ head, would have some positive PE because it
is above the “zero” level that WE have chosen in this problem.
What if we choose the “zero” level of PE to be the location of the banana? Now,
our pal Bubbles is actually below the zero level of PE, and so Bubbles has a
negative amount of PE! Wacky!
So this raises an interesting point about PE: you can only accurately state the
amount of PE an object has if you also include the point of reference (the
chosen “zero” level of PE) for each situation.
Another simple example: y’all have zero PE with respect to the floor of our
classroom. Y’all also have zero potential energy relative to the floor of any other
classroom on the second floor of Upper House. However, you have some amount
of positive PE relative to the ground outside our windows. You also have some
negative amount of PE relative to the ceiling in our classroom.
So you can never say that a person, monkey, banana or any other object “has” an
amount of PE; it’s not like an absolute or anything. It is a relative value, and you
always have different amounts of PE, because it always depends on some
reference point. Once a reference point is stated, then you can accurately state the
value of an object’s PE. As one of my favorite high school teachers would always
say: “It’s all relative.”
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2. Spring Potential Energy (PEs) – This is the energy that a spring stores when it is
stretched or compressed some amount (x) from its normal relaxed state. It does
not matter if the spring is stretched or compressed; as long as the distance is the
same, the energy stored is equal. The amount of energy stored is determined by
two factors: the distance the spring is stretched or compressed, and the stiffness of
the spring (this is indicated by the spring constant, k). Like all other forms of
work and energy, this is measured in Joules (J). The amount of potential energy
stored in a spring is:
PEs = ½ kx2
PEs = ½ kxf2 – ½ kx02
where k is the spring constant (measured in N/m).
The spring constant indicates the stiffness of a spring. It is a constant for a given
spring (it is not the same value for every spring), and it tells the amount of force
that is required to stretch a given spring a certain distance. For example, a loose
spring would have a spring constant of, say, 200 N/m. This means that a 200 N
force would stretch this spring a distance of 1 meter. A stiff spring might have a
spring constant of 5,000 N/m, indicating that a 5,000 N force is required to stretch
the spring 1 meter.
The spring constant can be determined by Hooke’s Law. A brief historical
interlude: Robert Hooke lived in England at the same time as our friend the
Cookie Man, Sir Isaac Newton. Hooke and Newton were the two most famous
scientists in England at the time, and they were the most bitter of rivals. In fact, it
has been said that Isaac Newton refused to publish much of his work until Hooke
had died!
Robert Hooke developed Hooke’s Law, a simple but powerful idea. He stated
that when a spring is stressed (that is, stretched or compressed from its relaxed
state) it will respond by exerting a force that is proportional to the distance of the
stretch or compression, and is opposite the direction of the stretch or compression.
What this means is:
Fs = -kx
Where k is the spring constant, x is the compression/stretch of the spring, Fs is the
force exerted by the spring, and the minus sign just indicates the fact that Fs and x
are in opposite directions.
Many times, you will have to use Hooke’s Law to determine the spring constant,
where you will be given a force and a distance. In these cases, to determine the
spring constant, you will divide the Force (which sometimes will be the force of
gravity on an object) by the distance.
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Conservation of Energy
Conservation of Energy is one of the most powerful overriding concepts in the entire
field of Physics, and really throughout science in general. The great beauty of this
concept is how simple it is to both use and understand.
Essentially this theory can be boiled down to the following:
The total energy contained in a closed, isolated system* can never
change. The amount of energy is a constant. Energy is allowed to
change forms (Kinetic to Potential, Potential to Kinetic, Kinetic to
Thermal, etc.) but energy can not be lost or gained. The change in the
total energy in a closed, isolated system must be equal to zero.
[*Note: A ‘closed, isolated system’ is one in which there is no external force doing work
on the system. In the case where there does exist an external force that is doing work on
the system, that amount of work (and hence, energy) is added to or removed from the
system. In other words, if work is done on a system, the change in the total energy of the
system is equal to the work done on the system. We will spend most of our time working
with closed, isolated systems.]
In a mathematical sense, we can use this idea by considering the total energy ET as the
sum of KE, PEg, and PEs. Furthermore, we can use the idea that the change in ET is
equal to the sum of the changes in those three quantities, and this should be equal to zero:
ET = KE + PEg + PEs
ET = KE + PEg + PEs = 0
2
½ mv0 + mgh0+ ½ kx02 = ½ mvf2 + mghf+ ½ kxf2
This simple concept can be used in numerous different types of scenarios and situations.
In most problems, one type of energy will not be used. For instance, we will deal with
problems where there are no springs. Alternately, we will look at cases where an object
does not change its vertical position. However, we will investigate examples where all
three types of energy will be needed as it is transformed from one form to another and
another.
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Momentum
Introduction to Momentum

Momentum is a concept that is fundamental to the study of Physics, especially
the study of motion, which happens to be very closely related to some of the
concepts we have studied already in class (Isaac Newton referred to Momentum
as “the quantity of motion.”). It is one of the very basic concepts that govern our
universe on both the small scale and the large.
Momentum is a term that is very commonly used in the vernacular; one of the
most common usages deals with the arena of sport. It is very common to hear
announcers extol the importance and value of a team’s or a given player’s
momentum. Generally what these announcers are inferring is that the team or
player in question has been doing well and is “on a roll” or is particularly hard to
stop. We should all have some inkling as to what this Momentum means.
(Indeed, it would not be a stretch to think of Momentum as a different way of
looking at the Law of Inertia.)
It will be useful to think of Momentum as being due to an object’s motion (much
like KE is the energy we associate with an object’s motion). If an object is
stationary, it has no momentum; if an object is moving, it will have a certain
amount of momentum. How much, exactly? Well, I’m glad you asked!
An object’s Momentum is simply the product of its mass with its
velocity:
p = mv
where p represents Momentum (I know it’s silly and doesn’t make
any sense, but that’s the way it is, so let’s just get over it and deal.
Gosh!)
Momentum is measured in units of kg*m/s. This doesn’t have a cool
name (like the Newton, which is a grouping of other units renamed so it’s
easier to deal with; same applies for the Joule).

One of the things that we need to keep in mind about Momentum is the fact that
it is a vector quantity. That essentially tells us two main things:
1. There is a directional component to an object’s Momentum; and
2. Momentum can be either positive or negative, based on our sign
conventions.

Now I’m sure you’re saying to yourself, “Self, this Momentum thing, p = mv, is
a very simple idea. What is Bradshaw all about when he keeps saying it’s so
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important?” And that’s a fair question; yes, this is a simple idea (or at least it
appears so; remember Newton’s 2nd Law, anyone?) that is not mathematically
complicated. But it’s a running theme throughout all of Physics that often the
“biggest” and most important ideas appear to be deceptively simple and are
mathematically quite elegant.

So how does Momentum get to be complex? Well, the simple fact that it is a
vector can introduce certain complications, such as vector components, angles,
magnitudes and so forth. (Although, at this point, none of that should be a great
challenge to you.) No, where Momentum becomes difficult is when we do one
(or both) of the following things.
1. Investigate what happens when an object’s Momentum changes.
2. Investigate the Momenta of two (or more) objects simultaneously.
A Change in Momentum

An object’s Momentum will change anytime its velocity changes; notice that
doesn’t say speed? That’s because of the vector nature of Momentum. Even a
change in the direction of motion will cause a change in Momentum.
We can compute this change rather easily, provided that we have knowledge of
the object’s mass, initial velocity and final velocity. With these three pieces of
info, we can quickly determine the change in Momentum:
p = pf – p0 = mvf – mv0 = mv

But hold on a second… Momentum will change when velocity changes, fine, I
can see that. But, in order for velocity to change, there must be… an
acceleration! (Recall: a = v/t) And, in order for any object to accelerate, it
must experience… a net force!! (Recall: F = ma) So, what this means, is that
there is a way that we can relate the change in an object’s Momentum to the net
force that it experiences!
Alright, so let’s think about this. When an object feels a net force, it accelerates.
That means, over a period of time, its velocity will change; if its velocity changes,
then its Momentum will change. So, basically…, well, here, just check out the
math:
p = mv
but
a = v/t
so
v = at
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and then
p = mat
but
ma = F
so then
p = Ft
or
F = p/t
So this gives us, essentially, the way that Isaac Newton really stated his second
law: An object experiences a net force anytime it experiences a change in its
Momentum over a given period of time. Or, in order for an object to experience
a change to its momentum, a net force must act on for a certain period of time.
The change in an object’s Momentum can be referred to as a quantity called
Impulse, J. (Yes, I know this doesn’t make sense either; let’s just go with it and
not get caught up on symbolism, shall we?)

J = p = Ft
Impulse has the same units as Momentum. Like Momentum, it is also a vector
quantity.
Two (or more) Objects

When dealing with multiple objects, we are generally concerned with the
Momentum of the system, and what happens to it as a result of some sort of
interaction between the objects. To deal with this situation, we are going to take a
page from our study of energy, and use the concept of conservation again. In
other words, when we observe an interaction between two or more objects, we
will say that the Momentum for the system is conserved. In other words,
p = 
or
p0 = p f

We can then expand this expression for the two separate objects that we are
concerned about:
m1v01 + m2v02 = m1vf1 + m2vf2

So, this says that we cannot gain or lose Momentum; it can merely be transferred
from one object to another. Another way of approaching this conservation idea is
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to state that the total Momentum of the system before the interaction occurs must
equal the total Momentum after that interaction takes place.

This interaction is generally referred to as a Collision, even if the objects
technically don’t collide. (Imagine a single object that separates into multiple
parts; this is technically referred to as a Collision.) It is always assumed, and
often not stated, that Momentum is conserved during a collision. Collisions of
many varieties will be investigated, but they will all fit into one of three main
categories:
o Perfectly Elastic:
 In this type of a collision, Kinetic Energy is also conserved; i.e.,
there is an equal amount of KE before the collision as after.
 These collisions are very rare; the only true collision of this type is
one in which objects never actually touch, such as the interaction
between charged subatomic particles (where repulsive electrostatic
forces prevent them from every coming into contact). However, in
practice, the collision between billiard balls is close enough to
Perfectly Elastic that we will assume it is so. Also, when dealing
with the PASCO© Dynamics carts, anytime the magnetic sides are
adjacent, we will assume such a “collision” is Perfectly Elastic.
o Inelastic:
 This is the most common type of a collision. Only Momentum is
conserved. Nothing special here.
o Completely Inelastic
 This differs from the former in one key area. After the collision, if
the two objects become stuck together and henceforth move as one
single object (i.e., both objects have the same final velocity after
the collision takes place) then the collision is Completely Inelastic.
An example here is when a football player tackles a man, and the
two move after the collision as one new, more massive object.

Any collision problem will begin with Conservation of Momentum. Based on
the type of collision, any special scenarios will be explored (such as the case of
Perfectly Elastic, or Completely Inelastic).
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Rotational Motion
When we consider rotational motion, what we must think of is an object that is moving
such that its mass is circling some certain axis of rotation. A few simple examples of
rotation are the hands on a clock, the gears in a car’s transmission, or even the hard drive
of a computer.
In these cases, the object is not necessarily moving from one location to another, but is
merely spinning about some certain axis. As such, this is a new form of motion for us to
consider, and so we will follow a similar approach to learn about rotational motion as we
did for translational motion.
First, we will start with basic kinematics, allowing us to describe the rotational motion of
an object.
In Rotational Kinematics, we are not concerned with the horizontal or vertical motion
that an object goes through; what we need to think about is the angular distance that an
object covers. In other words, we need to measure what angle the object will sweep out
in its rotational motion.
Because we are going to be dealing with angular measurements, we need to introduce
new variables to describe these motions.

We define the angular displacement of an object in rotation as , which is
measured in radians (rad). This angular position is considered positive when
measured counterclockwise from the positive x-axis. Hence, as a general rule, we
consider counter-clockwise rotation to be a positive rotation, while clockwise is
considered the negative direction of rotation.

We can now define an angular velocity for an object that is rotating as , which
is measured in radians per second (rad/s). This essentially tells me how fast an
object is rotating; how many radians of angle will the object sweep through in a
certain period of time t. So, we can say that

 t

Furthermore, we can now define an angular acceleration , which will be
measured in radians per second squared (rad/s2). This tells me how the angular
velocity of the object is changing with respect to time. In other words,

t
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At this point, you may be saying to yourself that all this sounds very familiar. And it
should… because this is exactly how we began kinematics before. In fact, we will be
doing exactly the same things that we did before. We will even be using the same
equations as we did in the past, only we are exchanging variables.
We can say that  will replace x,  will replace v, and  will replace a. There is even
a way that we can relate these variables to one another:
x = R
v = R
a = R
where R is the radius of the rotation (in other words, it is how far the object is from the
rotational axis).
So, we now can set up for ourselves a little rotational kinematics toolbox:
f = o + t
 = ot + ½ t2
f2 = o2 + 2 
= ½ (f + o)t
Note: often information will be given in terms of the number of revolutions that an object
makes. When this is the case, these revolutions must be converted to radians, by the
relationship that 1 rev = 2 rad.
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Rotational Dynamics
As we saw earlier on in the year, when an object accelerates, this is due to the application
of a net force (recall Newton’s 2nd Law: F = ma). In the case of rotational motion, we
can think in the same manner. When an object’s rotational motion changes (i.e., when it
has an angular acceleration), this must be caused by a rotational force. This “rotational
force” is commonly referred to as a ‘Torque’.
A Torque (symbolized by the Greek letter Tau; say “Taw” or “Toww” {rhymes with
‘Ow’}] ) is created when a force F is applied to an object a distance r from a pivot point.
The distance r is often given the fancy name of a “lever arm” or “moment arm”; suffice
to say, it is the distance from the point of application of the force and the pivot point of
the rotation. Specifically, the torque is equal to the product of the lever arm and the
component of the force F that is perpendicular to the lever arm. Mathematically, we can
represent this as follows:
 = rFSin
where  is the smaller angle between the direction of the Force and the lever arm
distance. The unit for torque is N*m. A similar, non-SI unit is the foot-lb. [Note: this is
not Joules!]
In order to visualize the concept of a torque, imagine opening a door. You exert a force F
on one edge of the door; the hinge is on the opposite side a distance r from the point of
application of the force. To open the door most easily, you want your force F to be
perpendicular to the plane of the door (in other words, to the lever arm). If your force is
exerted at an angle, the door opens less easily (in essence, you are creating a smaller
torque). It requires more force the closer you get to the hinge (i.e., the pivot point)
because r is smaller, so  is smaller. Note that if you exert your force F parallel to the
door, the door will not open, because you are creating, essentially, no torque [if the angle
is 0 degrees, the sin is 0, so the torque is 0].
Now, how is it that the torque is related to the angular acceleration? Well, as was the
case with Newton’s 2nd Law, the net force was related to the acceleration by the mass. In
rotation, we have a similar relationship; however, it relies on more than the mass.
In rotation, an object will have a certain reluctance to change its angular velocity. This is
quantified by the introduction of Rotational Inertia (also known as moment of inertia).
An object’s rotational Inertia (we symbolize this with the letter I) is dependent on two
factors. One is mass; the more massive an object is, the more difficult it is to cause its
rotational motion to change; this is quite obvious. The other factor is more subtle. It
turns out that what also matters is how that mass is distributed. To treat this concept fully
requires the application of calculus; suffice to say, that the rotational depends on the
distribution of an object’s mass. The unit for I is kg*m2 (this will be explained more
explicitly shortly). So, what we essentially can state now is often referred to as Newton’s
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Name _________________________________________________
Period ________
2nd Law for rotational motion. The statement reads that “the net Torque on an object will
be equal to the product of its rotational inertia times its angular acceleration.” To put it
more precisely,

Kinetic Energy of Rotation
Consider a flat, uniform disk of radius R and mass M that is rotating with angular speed
 about an axis that passes through its center (think of a CD rotating about the hole in the
middle). All individual little massive particles m that make up the disk have the same
angular velocity; however, each of them has a different linear velocity because they are
all a different distance from the axis of rotation. So, each individual massive particle m
has a certain amount of kinetic energy. We can add all these together to get the total
kinetic energy for the whole disk:
KE =  ½ mv2
But each v
= R, so we can re-write the above equation as
KErot =  ½ m(R)2
KErot =  ½ mR2
However, for each of the objects,  is constant, so then we can say
KErot = ½ 2(mR2)
The expression in parentheses,
mR2, is what is referred to as the rotational inertia
(I). This form is the rotational inertia for a collection of individual massive particles
rotating about a given axis, where m is the mass of each particle, R is the distance of that
particle from the axis, and the  indicates that you will add each of the products (mR2)
together. For certain particular solid shapes, rotational inertia is given for certain axes.
So, we can now say, for an object that is rotating with an angular velocity  about some
axis that
KErot = ½ I 2
The unit for this rotational kinetic energy is, hopefully not surprisingly, Joules (J).
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Angular Momentum
When an object is in linear motion, it has a linear momentum due to this motion that is
equal to a product of its mass with its velocity. Similarly, when an object is rotating, that
motion creates a “rotational” momentum; we refer to as its Angular Momentum. Much
like linear momentum, Angular Momentum is equal to a product of a measure of inertia
times a measure of velocity. Specifically, we say that
L = I
where L is the Angular Momentum, I is the Rotational Inertia, and  is the angular
velocity. The unit for Angular Momentum is kg*m2/s.
Much like linear momentum, Angular Momentum is a vector; its direction is determined
by the direction of its rotation. However, the direction of L is not a rotational direction
(i.e., the direction is not clockwise or counter-clockwise). To determine the direction of
L, we have a (somewhat silly) rule known as “The Right-Hand Rule”:
With the fingers of your right hand, allow them to “curl” in the direction of the rotation
(if the rotation is clockwise, your fingers should curl clockwise). As they do, your thumb
on your right hand will point in the direction of the angular momentum (e.g., if the
rotation is clockwise, the angular momentum should be directed downwards).
This yields an interesting consequence, in that the angular momentum is always in a
direction that perpendicular to the plane of the rotation; in other words, angular
momentum is a three-dimensional concept.
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