Download AP Physics ----Administrative Chores

AP Physics ----Administrative Chores
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Pass out Syllabus
Comments on contact methods, web page, textbook, lectures, quizzes, homework.
Grading policy, grading scale, labs.
Course outline -- complete version on Thursday.
Calculators.
Study:
o Establish and keep to a regular study schedule.
o Read the assigned sections before lecture.
o Attempt the assigned problems before quiz section.
o Bring questions to lecture, quiz sections, or office hours.
o It can be helpful to work problems with friends -- but avoid the temptation of letting your friends do
all the work for you.
o Physics is like sports: if you want to be good at it you must practice.
Chapter 1
What Is Physics?
From Webster's Dictionary:
From Greek, physikos, meaning of nature,
1 : a science that deals with matter and energy and their interactions
2 a : the physical processes and phenomena of a particular system b : the physical properties and
composition of something
We will be most concerned with (1). So, physics is the science that deals with matter and energy and their
interactions. What isn't covered by this definition? Nearly everything around us is either matter or energy, so this
definition is rather broad. In this course we will be concerned with the topics of mechanics, thermodynamics, and
optics. We begin with mechanics.
Dimensions: Length, Mass, and Time
Mechanics deals with the motion of objects. To describe the motion of something, we need to know where it is at
some instant in time, i.e. we need to know distances (length) and times. Additionally, when looking at what causes
or controls motion, we will need to refer to the mass of objects -- the same cause of motion will have different
effects on objects of different mass.
As a convenience, we will use international standards for length, mass, and time: the meter, second, and kilogram.
Definitions for length, mass and time are given in the text.
Tables showing the range of values for masses, lengths, and times. Notice that these quantities can have enormous
variations from small to large.
Prefixes for Metric Units ------ See table in the text on page 5 .
Building Blocks of Matter
Matter is composed of molecules which are made of atoms. Atoms are composed of a positive nucleus surrounded
by a cloude of negatively charged electrons. The nucleus is composed of protons and neutrons which are in turn
composed of quarks. And it is now hypothesized that electrons and quarks are actually small vibrating strings that
exist in an 11 dimensional universe!
Dimensional Analysis or "You Can't Add Apples and Oranges"
We can treat the dimensions of quantities almost like variables. A length (L) divided by another length (L/L), is
unit-less, since the two lengths cancel. Length divided by length squared (L/L2) is one over length. The example
from the text demonstrates an application, where, without knowing what acceleration is, we can check that at least
the dimensions work out properly.
Chapter 1, cont'd
55mi/hr = 24.6m/s
Coordinate Systems
It is often useful to use coordinate systems to define where things are. For
instance, to specify the location of a chair in a room (a rectangular room),
we can give the (perpendicular) distances of the chair from two adjacent
walls
as shown to the right. Here I have labeled one of the distances X and the
other Y, and we write the location of the chair as (X,Y). If the chair is 2m
from the left wall and 5m from the bottom wall, then it's location is (2m,
5m). This type of coordinate system is known as a rectangular or
Cartesian coordinate system. Notice that this type of coordinate system
needs:
1. an origin, in this case, the corner of the room from which the
distances are measured,
2. 2 (or 3) coordinate axes, in the case of the room the walls serve as
the axes,
3. and a distance scale in which to measure along the axes.
In this example, distances are measured in meters.
Show example with "standard" x-y coordinate system, and mark some
positions on the plane.
Another way to specify a location is to give its distance from the origin, and its angle from a reference line, usually
taken to be the x-axis. This is commonly called an r-theta, or polar coordinate. Show some examples of locating
points with polar coordinates.
Trigonometry
We will have many occasions to make use of the properties of
right triangles.
Recall that for a right triangle, such as the one pictured here,
Pythagoran's theorem states that
a2 + b2 = c2
And, from trigonometry, we have the relations:
sin = a / c
cos = b / c
tan = a / b
We will have many occasions to use these relations to:
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find the distance between two points
decompose a vector (velocity, acceleration, or force) into x and y components
find the length and direction of a vector when we have the x and y components.
Problem Solving Strategies
Students should read through this now. I will return to this topic when I work examples.
Chapter 2: Motion in One Dimension
Mechanics is divided into two areas: dynamics, a cause and effect analysis of motion; and kinematics, the
description of motion. We begin with kinematics of motion in one dimension.
Displacement
To descripe motion, we need to keep track of how the position of an object changes. A change in position is called
a displacement, and is defined as the difference between the final and initial positions:
x = xf - xi
This calculation requires that first you establish a coordinate axes for measuring the position x.
Sketch a displacement, and show an example of how to calculate a positive and negative displacement.
Displacement is a vector. A vector is a quantity that has both a magnitude (value) and a direction. In the one
dimensional case, the direction is rather easy, it is either positive or negative (right or left). In 2 dimensions, you
must be more precise to indicate the direction.
Average Velocity
We define the average velocity of an object to be its displacement divided by the time interval for that
displacement:
vavg = x / t = (xf - xi) / (tf - ti)
Example: P2.2
A swimmer crosses a 50m pool in 20s, and returns to her starting point in an additional 22s. (a) What is her average
velocity for the first half of the swim?
1.
2.
3.
4.
5.
6.
Read the problem.
Make a sketch of the problem.
Identify the data: length of pool is the displacement, is 50m, and the time interval is 20s
Choose equation: vavg equation given above
Solve the equation: vavg = 50m/20s = 2.5m/s
Check answer.
(b) What is her average velocity for the second half of the swim?
Follow the same steps as above, but now the displacement is in the opposite direction, so let's call it -50m, and the
time interval is 22s. Now vavg = 2.3m/s.
(c) What is her average velocity for the round trip?
Again, follow the same steps as above. Since she returns to her starting point, xf = xi and the displacement is zero.
The time interval is 20s+22s = 42s. Note a curious thing about our definition about the average velocity, even
though there is motion during a time interval, if the ending point is the same as the starting point, then the
displacement is zero, and the average velocity is zero. This is the case here. Her speed is never zero, it is on average
100m/42s = 2.4m/s. But due to the formal, Physics definition, her average velocity is zero!
Graphical Interpretation of Velocity
It may be helpful to visualize the above problem on a graph of position versus time.
On such a graph, the average velocity between two points is simply the slope of the line connecting the two points.
The points represent the position of an object at a particular time. Note that even though the object moves in a
straight line (along x), the line on the graph doesn't have to be straight. This is because the line doesn't represent the
direction of motion, but shows the position as a function of time.
Instantaneous Velocity
In the above graph, the velocity changes from instant to instant. We can see this by choosing different final points,
and noting that the average velocity (the slope of the straight line connecting the initial and final points) is not
constant. What would the graph look like if the object were moving at a constant velocity? (ans. a straight line)
The instantaneous velocity, v, is defined as the limit of the average velocity as the time interval t becomes
infinitesimally short:
v = limit as t goes to 0 [x/t]
Graphically, the instantaneous velocity at a point is given by the slope of the line tangent to the position versus time
curve at that point.
The instantaneous speed is the magnitude of the instantaneous velocity.
Acceleration
Acceleration is probably a familiar concept to everyone. Most people are familiar with the feeling of acceleration in
a car or bus. In Physics, acceleration has a formal definition that may sometimes differ from common usage. The
average acceleration is defined as the change in velocity during an interval of time:
aavg = v/t = (vf - vi) / (tf - ti)
Note that acceleration can be positive (increasing velocity) or negative (decreasing velocity, also called
deceleration). And when we work problems in more than one dimension, we can have acceleration even though the
speed is constant -- the direction of the velocity changes, but the magnitude remains the same.
Example: P2.16
A car is travelling initially at +7.0m/s. It accelerates at a rate of +0.80m/s2 for 2.0s. What is the final velocity?
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Read the problem.
Make a sketch.
Identify the data: vi = 7.0m/s, a = 0.80m/s2, t = 2.0s.
We will use the equation a = (vf - vi) / t
Rewrite the equation to solve for vf = vi + at = 7.0m/s + (0.80m/s2)(2.0s) = 8.6m/s
Check the answer.
Recall:
1. Displacement: x = xf - xi
2. Average velocity: vavg = x / t
3. Instantaneous velocity: the average velocity in the limit the time interval goes to zero, also the slope on the
position versus time graph.
4. Average acceleration: aavg = v / t = (vf-vi)/(tf-ti). Note that vf and vi are instantaneous velocities.
Example: P2.16
A car is travelling initially at +7.0m/s. It accelerates at a rate of +0.80m/s2 for 2.0s. What is the final velocity?
1. Read the problem.
2. Make a sketch.
3. Identify the data: vi = 7.0m/s, a = 0.80m/s2, t = 2.0s.
4. We will use the equation a = (vf - vi) / t
5. Rewrite the equation to solve for vf = vi + at = 7.0m/s + (0.80m/s2)(2.0s) = 8.6m/s
6. Check the answer.
Motion Diagram
1-D Motion with Constant Acceleration
Free Fall
Chapter 3: Vectors and 2-D Motion --all the following topics are from last year
Vectors and Scalars
Some Properties of Vectors
Components of a Vector
Velocity in two dimensions
Projectile motion
Relative velocity
Recall: Kinematics
Displacement vector: r = rf - ri
Velocity vector: vavg = r/t
Instantaneous velocity: take the limit as t goes to zero.
Average acceleration: aavg = v/t
Example: P3-21
A car is on an incline that makes an angle of 24 degrees to the horizontal. The car rolls from rest down the incline
with a constant acceleration of 4.00m/s2 for a distance of 50.0m to the edge of the cliff. The cliff is 30.0m above the
ocean. Find (a) the car's position relative to the base of the cliff when the car lands in the ocean, and (b) the length
of time the car is in the air.
1. Reread the problem.
2. Draw a sketch.
3. Identify the data. For this problem, think in terms of (1) when the car is rolling down the incline, and (2)
when the car is falling to the ocean.
4. Choose the equations. For (1) vf=vi+at.
5. Solve the equations.
6. Check the answer.
Chapter 4: The Laws of Motion
We move from kinematics (description of motion) to dynamics (understanding the causes of motion, and the
resulting effects). The causes of motion we call "forces". Some types of forces are familiar, like if you pull on a
string or throw a ball. Other forces are just as real but less familiar, for instance, when you sit on a chair, the chair
pushes upward, against your weight. Otherwise, you would fall through the floor!
Types of Forces
Forces can be classified as contact or field forces. Contact forces involve the contact of two objects, such as in the
examples of someone pulling a string, throwing a ball, or sitting in a chair. Field forces are a bit more abstract. You
may have heard the phrase "action at a distance" to describe gravity. Initially, most of the forces we will use are
either contact forces or gravity.
Newton's Laws of Motion
The ideas embodied in Newton's three laws of motion were known to many. Newton gets credit primarily for
sythesising what was then known into three laws, and demonstrating the power of his formulation by tackling the
problem of planetary motion and gravitation.
Newton's First Law
What is the natural state of matter? Aristotle believed it is for matter to be at rest. But anybody who's driven on icy
Michigan roadways can attest to the fallacy of that blief. Galileo proposed that the natural state is to resist
acceleration, and this idea is formalized in the first law:
An object at rest remainst at rest, and an object in motion continues in motion with constant velocity (that is,
constant speed in a straight line), unless it experiences a net external force.
Stated differently, if the sum of all external forces on an object is zero, then its acceleration is zero: F=0 then a=0.
Example:
What are the forces acting on a person seated on a chair?
Newton's Second Law
The second law summarizes the observations about how an object accelerates when the net force is not zero:
The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its
mass.
We express this relation mathematically as: F= ma. This is a vector equation, being satisfied for each component
separately: Fx=max Fy=may Fz=maz
Units of Force and Mass
Recall that the SI unit of mass is the kilogram. Then, by dimensional analysis, we can see that the unit of force, call
the newton, is 1N = 1kg m/s2. Although we often equate the unit of pound to kilograms, one is a weight (a force)
while the other is a mass. The pound is related to the newton, 1N = 0.225lb. The weight of an object is due to the
force of gravity on the mass of the object. In projectile motion, we saw that freely falling objects near the earth's
surface experience an acceleration g, acting vertically downward. Since w = F = ma = mg, we see that the weight is
mg. Because g varies little over the surface of the earth, it is easy to confuse weight and mass. In fact, we often
measure the mass of objects by weighing them on spring scales. This difference may become more well understood
as we enter the era of the space station.
Example: P4-4
A freight train has a mass of 1.5x107kg. If the locomotive can exert a constant pull of 7.5x105N, how long does it
take to increase the speed of the train from rest to 80km/h?
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5.
Reread the problem.
Draw a diagram.
Identify the data. vi=0, vf=80km/hr=22m/s, m=1.5x107kg, F = 7.5x105N.
Choose equations. F=ma, and vf=vi+at.
Solve equations. a=F/m, t=(vf-vi)/a= (vf-vi)m/F = (22m/s)(1.5x107kg)/7.5x105N = 440 kg m/sN = 440 s or
about 7 and 1/3 minutes.
6. Check the answer. Assuming that the values given are reasonable, 7 minutes is a reasonable result for a
train.
Newton's Third Law
In nature, forces always come in pairs.
If two objects interact, the force exerted on object 1 by object 2 is equal in magnitude but opposite in direction th
the force exerted on object 2 by object 1.
This law is commonly stated as "for every action, there is an equal and opposite reaction". This
statement is fine, as long as you remember that action means the force exerted by 1 on 2, and
reaction is the force exerted by 2 back on 1. Mathematically, we will write this as F1 on 2= -F2 on 1.
Recall: Newton's Laws of Motion
1. If a body experiences no net external force, then its acceleration is zero: if F=0 then a=0.
2. F=ma
3. "For every action there is an equal and opposite reaction": F1 on 2 = -F2 on 1.
Demonstration: Office chairs.
Some Questions for discussion:
If a small car collides head-on with a massive truck, which vehicle experiences a greater inpact force? Which
vehicle experiences a greater acceleration? What is wrong with the following argument? A horse attempts to pull a
wagon from rest. If the horse pulls forward on the wagon, then Newton's third law holds that the wagon pulls back
equally hard on the horse. Thus the forces cancel and nothing happens. Is it possible for an object to move if no net
force acts on it?
Free-body diagram
When applying Newton's laws to solve a problem, the first thing that normally needs to be done is to identify the
forces acting on the object(s). For instance, a crate being pulled by a rope experiences the tension in the rope
(tension is a type of force), the pull of gravity on the center of mass of the crate, and the normal force of the floor
against the crate. These forces act on the crate. The reaction forces to each of those listed act on other objects, so
they are not listed. We can summarize these forces with a free-body diagram, a drawing of the crate and the forces
acting on it. (diagram)
Solving Problems
When tackling problems involving something in equilibrium (sometimes called statics problems), it is best to use
Newton's first law in component form (and we'll normally need only the x and y components, though an extension
to z should be obvious): Fx = 0
Fy = 0
Example: P4-15
A block of mass m=2.0kg is held in equilibrium on an incline of angle =60degrees by the horizontal force F, as in
the drawing. (a) Determine the value of F. (b) Determine the normal force exerted by the incline on the block
(ignore friction). 3 forces on the block: F, Fg, and n. Note that n is perpendicular to the surface of the incline! Let's
make a table with a row for each force, and columns for the x and y components:
x component y component
F F
0
Fg 0
-mg=-19.6N
n -n sin(60)
n cos(60)
The sum of the x components is: F- n sin(60) = 0, so F=n sin(60). The sum of the y components is: -19.6N + n
cos(60) = 0, so n=19.6N/cos(60) = 39N (the answer to part a). Putting this answer back into the equation from the x
components, we find for the answer to part b, F=(39N)sin(60) = 34N. NOTE: For problems involving ropes, chains,
etc., the tension in the rope is equal to the force it transmits to either end. If the object in a problem is not static, but
is accelerating, then Newton's second law is needed. Again, it is convenient to work with the component equations:
Fx = max
Fy = may
Example: P4-21
Assume that the three blocks in the figure move on a frictionless surface and that a 42N force acts as shown on the
3.0kg block. Determine (a) the acceleration given this system, (b) the tension in the cord connecting the 3.0kg and
the 1.0kg blocks, and © the force exerted on the 2.0kg block by the 1.0kg block. In the y direction, the weight of
the blocks is balanced by the normal force of the frictionless surface, such that Fy = 0. No motion occurs in the y
direction. Let's turn our attention to the x direction. To determine the acceleration of the entire system (all 3
blocks), apply the second law, F = ma, treating the 3 blocks as one object. Then m is the mass of the 3 blocks
(plus rope), m=1+2+3kg=6.0kg. F is the total exernal force on the blocks. The force of one block on another is an
internal force, so it can be ignored for now. The only remaining force is the 42N applied force. The remaining
quantity in the equation is the acceleration, a. Solving for a: a = F/m = (42N)/(6.0kg) = 7.0m/s2. Now, regarding
the 3 blocks as 2 separate groups, one being pulled right by the cord, and the other being pulled right by 42N and
pulled left by the cord. The tension in the cord must be sufficient to supply the first pair with an acceleration of
7.0m/s2 (since the whole accelerates at this rate, then each part must also accelerate at this rate). The mass of the
pair of blocks is 3.0kg, so the applied force from the tension in the cord must be F=ma=(3.0kg)(7.0m/s2) = 21N.
Finally, to accelerate the 2.0kg block, the 1.0kg block must exert a force on it of F=ma=(2.0kg)(7.0m/s2)= 14N.
Friction
To finish this chapter on the laws of motion, we will discuss the force of friction. Friction arises when two surfaces
slide past one another, like a book sliding across a table. Experiments reveal that friction has the following basic
properties:
1. The frictional force is independent of the contact are.
2. The frictional force is proportional to the normal force on the object. (NOTE: this is not necessarily equal to
the weight.) The constant of proportionality is called .
3. There is a static friction and kinetic (sliding) friction. Kinetic friction is constant, equals kn, and is directed
opposite the motion. Static friction is <sn, and directed to balance the net applied force.
Demonstration with weights and spring scale. Question: Describe a few examples in which the force of friction
exerted on an object is in the direction of motion of the object.
Example: P4-41
Find the acceleration experienced by each of the two masses shown if the coefficient of kinetic friction between the
7.00kg mass and the plane is 0.250. Draw a free body diagram for both masses. Find the normal force for the 7kg
mass, then, assuming it is moving up the plane, find the frictional force. This step is a bit subtle, since you will still
get an answer if you choose the wrong direction for the frictional force, but afterward, a check of the answer will
reveal that the force is pointing in the same direction as the motion.
Web access to book site.
The book has a web site at http://www.saunderscollege.com/physics/college . To login to the student site you will
need the username, cpstudent35, and password, phyisfun. WARNING: I looked at the site, went to the chapter 2
quiz problems, and found that the answer to the first problem is wrong. I haven't had time to check other things on
the site, but this is not a good sign!
Chapter 5: Work and Energy
Energy is an extremely important concept in physics, however energy is an abstract idea. You can't simply look at a
piece of coal and see it's energy! We begin to understand energy by relating it to mechanical quantities, the first of
which is work.
Work
Work has a distinct definition in physics. Work is done on an object when a force is applied to move an object a
distance s in the direction of the force, or an object is moved a distance s, with a component of the force F along the
motion.
W = (F cos)s
The cos factor takes care of any difference in direction between F and s. The unit of work is the joule (J). From the
definition, 1J = (1N)(1m) = 1Nm.
This definition differs considerably from everyday usage of the word "work". If you stand, holding 20kg of books,
you will eventually feel more tired, and probably think it is because it's alot of work to hold 20kg of books. But,
since the books don't move, no work is done according to the physics definition!!
Example:
How much work is done to lift (slowly) 20kg of books to a height of 2m?
W = F cos s. The force on the books must be just larger than, and opposite to gravity, so the angle between the
force and the motion is 0 degrees. Thus W = Fs = (20kg)(9.8m/s2)(2m)= 392kg m2/s2 = 392J.
Work is done to lift the books, since a force must be applied to overcome the force of gravity on the books, and
negative work is done when the books are lowered. Negative work means that we get energy out of the books, but
more about this when we discuss energy.
Kinetic Energy
When an object moves with a constant acceleration, then W = Fs = (ma)s. But we also know that v2 = v02 + 2as, or
as = ( v2 - v02 )/2. Using this in the expression for work gives:
W = m(v2-v02)/2 = (1/2)mv2 - (1/2)mv02.
We call the quantity (1/2)mv2 the kinetic energy, and write KE = (1/2)mv2. Thus we can say that W = KEf - KEi, or
in words, the work done on an object equals the change in its kinetic energy.
Example:
A car traveling with speed v skids a distance d when its brakes lock. If the car is traveling with speed 2v, and the
brakes apply the same stopping force as before, how far will it skid?
Since W=Fs=KE, and F remains constant, we see that s = KE/F. If the speed doubles, then the KE quadruples
(and also KE), so the stopping distance also quadruples. So the car will skid a distance 4d.
Potential Energy
The books of the earlier example didn't have KE after they were lifted, yet work was done. Work was done to
counteract the force of gravity, and change the position of the books in the Earth's gravitational field. If the force
supporting the books is removed, the books will free fall, gaining KE equal to the work done to lift them. Because
the books have the potential to do a certain amount of work, we say that they have gained "potential energy" or PE.
If the books were allowed to drop, then the work done by the gravitational force is W = mgs = mgyi - mgyf. We call
the quantity mgy the PE, PE=mgy. So the work done by gravity on the books is W = PEi - PEf.
Reference Levels
Since the work done depends only on the difference in PE, the zero point of PE is arbitrary. That is, you are free to
choose the zero level for any problem. Sometimes the appropriate zero will be the ground, but at other times it
might be some other location (bottom of a well, top of a platform, top of a desk).
Example: P5-21
A 40N chile is in swing with 2.0m ropes. Find the gravitational potential energy of the child relative to her lowest
position (a) when the ropes are horizontal, (b) when the ropes make a 30 degree angle with vertical, and c) at the
bottom of the swing.
Sketch. a) PE = mgh = (40N)(2.0m) = 80J. b) PE = mgh = (40N)(2.0m)(1-cos(30)) = 11J. c) PE = mgh = 0J.
Conservative and Non-Conservative Forces
A force is conservative if the work it does on an object moving between two points is independent of the path the
object takes between the points. If the path is closed, so that the final position is the same as the starting postion,
then the work done is zero if the force is conservative.
A force is non-conservative if it leads to a dissipation of mechanical energy.
Force
Type
Gravity
Conservative
normal
Conservative (acts perp. to motion)
Friction
Non-conservative
Spring force Conservative
Conservation of Mechanical Energy
If a particular system involves only conservative forces, then W = PEi-PEf = KEf-KEi, and we can write
KEi+PEi = KEf+PEf = constant
This is a statement of conservation of mechanical energy. Conservation of mechanical energy holds when there is
no mechanism for adding additional energy (no unaccounted for external forces) or removing energy from the
system (no non-conservative forces).
Example: P5-26
Tarzan swings on a 30.0m long vine initially inclined at an angle of 37.0 deg. What is his speed at the bottom of the
swing if he starts from rest?
Initially he has only PE. Make PE=0 at the bottom of the swing, then at the bottom he has only KE, and his KE at
the bottom will equal his PE at 37.0deg. PE = mgh = mg(30.0m)(1-cos(37)) = 6.04mg.
KE = 1/2 mv2 = PE = 6.04mg
v = [2(6.04)g]1/2 = 10.9m/s
Potential Energy Stored in a Spring
It is found that springs exhibit a force that is proportional to their compression or elongation, F=kx, where x is the
distance from the equilibrium (unstretched) position. This is known as Hooke's law. This force is conservative (as
long as the spring is not overstretched). The potential energy stored in a spring that is a distance x from equilibrium
is PE=(1/2)kx2.
Non-conservative Forces and Work-Energy Theorem
If non-conservative forces are present, then the sum of KE and PE is not constant. The difference between the final
and initial energies equals the work done by the non-conservative forces. Wnc = ΔKE
Recall:
Work = W = Fs cos
Kinetic Energy = KE = (1/2)mv2
Gravitational Potential Energy = PEg = mgh
Work done on an object by conservative forces = change in kinetic energy plus change in potential energy.
Using (conservation of) energy often allows a problem to be solved much more easily than by using forces and
Newton's laws.
Potential Energy Stored in a Spring
It is found that springs exhibit a force that is proportional to their compression or elongation, F=kx, where x is the
distance from the equilibrium (unstretched) position, and k is the spring constant with units of N/m. This is known
as Hooke's law. This force is conservative (as long as the spring is not overstretched). The potential energy stored
in a spring that is a distance x from equilibrium is PEs=(1/2)kx2.
Example: P5.29
When a spring of unknown spring constant, k, is compressed 0.120m, it can launch a 20.0g projectile to a height of
20.0m above its starting point. Neglect resistive forces. What is k?
Use conservation of energy. Initially, there is only potential energy of the compressed spring: PEi = 1/2 kx2 =
k(0.0072). At the top of the motion, all of the energy is gravitational potential energy: PEf = mgh =
(0.0200kg)(9.8m/s2)(20.0m) = 3.92J. Since all the forces are conservative, the inital and final energies must be
equal, or: (0.0072)k = 3.92J, such that k=544N/m.
Non-conservative Forces and Work-Energy Theorem
If non-conservative forces are present, then the sum of KE and PE is not constant. The difference between the final
and initial energies equals the work done by the non-conservative forces.
Wnc = (KEf+PEf)-(KEi+PEi)
Example: P5.32
An 80.0N box is pulled 20.0m up a 30 degree incline by an applied force of 100N parallel to the incline. If the
coefficient of kinetic friction between box and incline is 0.220, calculate the change in kinetic energy of the box.
Use Wnet = KEf-KEi, and the fact that Wnc = Fs cos. We assume that initially the box is not moving, and assign the
zero for gravitational potential energy to the bottom of the incline. Wnc=-kmg cos30 s = (0.220)(80.0N)(20.0m)cos30 = -305J. Wg = -PEf = -mgh = -mgs sin = -(80.0N)(20.0m)sin30 = -800J. WF = Fs
cos = (100N)(20.0m) = 2000J. KEf = WF + Wnc + Wg = 2000-305-800 = 895J.
As a general rule, it is found that energy is conserved in the universe. Energy comes in many forms, and can
generally be converted from one form to another, but there no evidence of energy non-conservation has yet been
found. This observation is now so fundamental, that when a situation arises where it seems that energy is not
conserved, other possible explanations are always sought first.
Recall
Work = W = Fs cos
Power
Power is defined as work divided by the time in which the work is accomplished, P = W/t. Power is measured in
watts, W. For engines and motors, it is still traditional to specify power in the unit horsepower, 1hp = 746W. Note
that if we use W=Fs, then P=Fs/ t = Fv. This last form is an easy way to calculate the instantaneous power of
an object that is being accelerated, like a car.
Example: P5.46
A 650kg elevator starts from rest. It moves upward for 3.00s with constant acceleration until it reaches its cruising
speed of 1.75m/s. (a) What is the average power delivered by the elevator motor during the acceleration? (b) What
is the instantaneous power at t=3.00s?
The acceleration is a=v/t=(1.75m/s)/(3.00s) = 0.583m/s2. The net force on the elevator must be Fnet=ma =
(650kg)(0.583) = 379N. The net force is the force from the motor minus the force of gravity, so Fmotor = Fnet + mg =
379N + (650kg)(9.8) = 6750N. In 3s the elevator moves a distance x=1/2 at2 = (.583)(3.00s)2/2 = 2.62m. The work
done by the motor is then Wmotor= Fmotorx = (6750N)(2.62m) = 17,700J. The average power of the motor is P=W/t =
17700J/3.00s = 5900W
At t=3.00s, the instantaneous power is P=Fv = (6750N)(1.75m/s) = 11,800 W.
Chapter 6: Momentum and Collisions
Momentum
The linear momentum of an object is defined as the product of its mass times its velocity, p=mv, a vector
relationship. There is no special unit for momentum, its units are simply kg m/s.
The component equations for momentum are: px = mvx and py = mvy.
Newton actually expressed his second law not as F=ma, but as F = p/t. Written this way, the second law
basically says that the time rate of change of momentum of an object is equal to the net force acting on it.
Rewriting the above expression we get the impulse-momentum theorem: Ft = p = mvf - mvi.
The quantity Ft is called the impulse. The above expression says that the impulse acting on an object equals the
change in momentum of the object. It is often not possible to know exactly what force is acting on an object. For
instance, when a ball bounces off the floor, the force changes rapidly with time, from nearly zero when the ball first
touches the floor to a large value as the velocity of the ball comes to zero, and reverses, then back to a small value
as the ball begins to move up from the floor. But by using the impulse, we don't need to know exactly how the force
varies of this short time, we can use the average force exerted on the ball during the time it is in contact with the
floor.
Example: like P6.13
A 1.3kg basketball bounced off the floor with a speed of 10.0m/s rebounds with a velocity of 9.8m/s. (a) What is
the impulse delivered to the ball? (b) What is the average force on the ball if it is in contact with the floor for
2.6x10-3s?
(a) Impulse = Ft = pf-pi = mvf - mvi = (1.3kg)(-10.0 - 9.8m/s) = 26kg m/s.
(b) Favg = Ft/t = (26kg m/s)/(2.6x10-3s) = 104N.
Conservation of Momentum
Conservation of momentum is as fundamental in physics as conservation of energy, but better. Momentum is
conserved in all isolated systems. A system is isolated if there is no outside influence (forces) acting on it.
Examples of isolated systems are:






a collision between two atoms of a gas,
a rocket in space, plus the exhaust from its thrusters,
a girl standing on a plank lying on a frictionless surface,
the collision of "glider cars" on an air track,
a satellite flying past a planet,
or the collision of 2 vehicles on a roadway.
You should notice the repeated use of the word collision. The interaction of two objects is generally treated as a
collision.
Consider a collision between two objects, 1 and 2. (Drawing of 2 objects before and after collision, indicating
masses and velocities.) The impulse on object 1 is F1t = m1v1f - m1v1i. Likewise for object 2, F2t = m2v2f - m2v2i.
Recall Newton's third law: F2 on 1 = -F1 on 2. In terms of the forces given above this means that F1 = -F2. (Although
the impulse equation involves an average force, since F1 equals -F2 at every instant in time, then the same result
holds for the averages.) Therefore, F1t = -F2t, or
m1v1f - m1v1i = -(m2v2f - m2v2i)
Putting the initial quantities on the left and the final quantities on the right yields:
m1v1i + m2v2i = m1v1f + m2v2f
This result is called conservation of momentum. It says that the initial momentum equals the final momentum. The
key to this result is Newton's third law, and the fact that there are no other forces acting on the system. In this
derivation, we used a system with two objects, but the result holds for systems with any number of objects.
Example: P6.15
A 730N man stands in the middle of a frozen pond of radius 5.0m. He is unable to get to the side because of a lack
of friction between his shoes and the ice. To overcome this difficulty (since he just learned about conservation of
momentum) he throws his 1.2kg physics textbook horizontally toward the north shore, at a speed of 5.0m/s. How
long does it take him to reach the south shore?
Apply conservation of momentum. When he throws the textbook, he will get a push from the reaction force of the
textbook on him.
mbvb = -mmvm
vm = -mbvb/mm = -(1.2kg)(5.0m/s)/((730N)/(9.8m/s2)) = -0.081m/s. (minus sign means the directions are opposite,
I'll ignore it from here.) The time to move to the edge is t=(5.0m)/(0.081m/s) = 62s.
Collisions
Interactions between two objects often fall into the category of collisions -- even many things that don't seem to be
related to collisions at all, for example, air pressure or glaucoma testing (see text). While momentum is strictly
conserved in collisions (emphasizing again that there are no external forces), energy need not be conserved. Quick
example: When a ball is dropped and bounces off the floor, it doesn't return to the same height it was dropped from.
Energy is lost in the collision. Momentum is conserved, but you must consider the system of the ball and the Earth!
We call a collision where momentum is conserved but energy is not an inelastic collision. When the two objects
stick together, like two train cars that hit and couple together, we say that the collision is perfectly inelastic.
When both momentum and energy are conserved, we say that the collision is elastic. Examples of truly elastic
collisions are collisions between subatomic particles, or "collisions" between stellar objects. In the latter, the
objects should not make physical contact, but we can treat their gravitational interaction as a collision.
Example: Inelastic Collision P6.25
A 0.030kg bullet is fired vertically at 200m/s into a 0.15kg baseball that is initially at rest. How high does the
combination rise after the collision, assuming the bullet embeds in the ball.
Since the bullet embeds (sticks) in the ball, this is a perfectly inelastic collision. Initially we have 2 objects, and
afterward we have one object with a mass that is the sum of the original masses.
(Drawing) m1v1+m2v2 = (m1+m2)vf.
Thus vf = (m1v1+m2v2) / (m1+m2) = (0.030kg)(200m/s)/(0.030kg+0.15kg) = 33.3m/s.
So the ball will rise to a height y = v2/2g = (33.3m/s)2/2(9.8m/s2) = 57m.
Example: Elastic Collision P6.30
A 10.0g object moving to the right at 20.0cm/s makes an elastic head-on collision with a 15.0g object moving in the
opposite direction at 30.0cm/s. Find the velocity of each object after the collision.
Since this is an elastic collision, momentum and energy are conserved, so we have 2 equations:
m1v1i + m2v2i = m1v1f + m2v2f, and
1/2 m1v1i2 + 1/2 m2v2i2 = 1/2 m1v1f2 + 1/2 m2v2f2
From the statement of the problem we know m1 = 0.0100kg, m2 = 0.0150kg, v1i = 0.200m/s, and v2i = -0.300m/s.
We are left to find v1f, v2f. We have 2 equations and 2 unknowns. m1v1i + m2v2i = 0.00650 kg m/s. 1/2 m1v1i2 + 1/2
m2v2i2 = 0.000875J. Begin by eliminating v2f = (m1v1i + m2v2i - m1v1f) / m2 = 0.433 - 0.667 v1f from the energy
equation.
0.000875J = 1/2 m1v1f2 + 1/2 m2(0.433 - 0.667 v1f)2 = 0.00500 v1f2 + 0.00141 - 0.0100 v1f + 0.00334 v1f2
This results in a quadratic equation:
0.00834 v1f2 - 0.0100 v1f + 0.000535 = 0
The solutions are ________________
Glancing Collisions
Recall that in collisions, momentum is always conserved. Energy is conserved is the collision is elastic, but not
conserved if the collision is inelastic. All the examples I showed last week had initial and final velocities oriented
along one axis. But clearly this is not required, and the initial and final velocities can be at angles to each other. (As
long as there are only 2 objects involved in the collision, it can be shown that all the velocities must lie in a plane.)
In glancing collisions, the initial and final velocities can be at angles to each other. In this case, the x and y
components of momentum are separately conserved. The additional equation (conservation of y momentum)
compensates for the additional angle variables.
The following example should be added to your list of homework problems.
Example: P6.37
A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves
at 4.33 m/s at an angle of 30 degrees with respect to the original line of motion. (a) Find the velocity (magnitude
and direction) of the second ball after the collision. (b) Was this an inesastic collision or an elastic collision?
Call the initial velocity vi = 5.00 m/s, and the final speeds v1 = 4.33 m/s and v2. Both balls have the same mass, m.
Call the unknown angle .
(Drawing) For part a, apply conservation of momentum. Call the incoming direction of the first ball the x direction.
Then the inital momentum is mvi in the x direction, the initial y component is zero. After the collision, the x
component of momentum is mv1cos30 + mv2cos. Cancel the common m factor and set final and inital equal to get
v2cos = 5.00 m/s - 4.33cos30 = 1.25 m/s.
After the collision, the y component of momentum is mv1sin30 - mv2sin. Cancel the common mass factor and set
final equal to inital to get v2sin = 4.33sin30 = 2.16 m/s.
This means that v2 = Sqrt(1.252 + 2.162) = 2.50 m/s, and  = Asin(2.16/2.50) = 60 degrees opposite to the first ball.
For part (b), let's just calculate the initial and final kinetic energies, since we know the speeds. Notice that the mass,
although unknown, will be a common factor to all the kinetic energy terms, so I will divide it out. KEi/m = (5.00
m/s)2/2 = 12.5 J/kg. KEf/m = ((4.33 m/s)2 + (2.50 m/s)2)/2 = 12.5 J/kg. Since the initial and final kinetic energies
are equal, the collision is elastic.
Chapter 7: Circular Motion and the Law of Gravity
Measuring Angles in Radians
In the equations we will be dealing with in this and the next chapter, angles are measured in radians instead of
degrees. By using radians, the equations are substantially simplified. Radians are defined such that 2 radians = 360
degrees.
An easy way to visualize what this means is to look at a circle of radius 1 (Drawing). An angle of 360 degrees
makes one complete turn around the circle, covering the entire circumference of 2r = 2. The length of the
perimeter on the r=1 circle equals the angle in radians. If we use an angle of less than 360 degrees, then we sweep
out less than the full circumference. For instance, an angle of 180 degrees will sweep around half the perimeter of
the circle, or 2/2 = . And for an angle d, the fraction of the perimeter it will sweep around is d/360, or a length
of 2d/360 = d/180.
This gives us the transformation from degrees to radians:
r = d/180
where r is the angle in radians, and d is the angle in degrees.
Angular Speed and Acceleration
When discussing linear motion, we began by defining displacement. We will now define an analogous
displacement for circular motion. Imagine a point P at a radius r from the center of a circle, and make a reference
line that passes through P. If the circle rotates by an angle r, (the reference line doesn't move), the point P moves
an arc length s = (r/2) 2r = r/r. That is the fraction of the full circumference intercepted by the angle.
If P moves from angle 1 to 2 in a time t, then it moves with an average angular speed avg = (2 - 1) / (t2 - t1) =
 / t. As was the case with linear velocity, taking the limit as t goes to zero gives the instantaneous angular
speed. The units of angular speed are radians per second, rad/s. NOTE: Radians are not a dimension in the same
sense as meters or seconds, and are often omitted from an answer. However, keeping radians in place is often
helpful to see that things cancel appropriately.
Example: What's the angular speed of the earth rotating on its axis?
The earth makes 1 revolution in 24 hours. Therefore  = (1 rev/24 hr)(2 rad/rev)(1 hr/3600 s) = 7.27x10-5 rad/s.
Recall that in chapter 2, after defining velocity we defined acceleration. We'll do the same thing now for the case of
circular motion. We define angular acceleration as the change in angular speed divided by time, avg = (2 - 1)/(t2
- t1) = /t. The Greek letter  is used for angular acceleration. The units of angular acceleration are rad/s2.
Instantaneous angular acceleration is obtained by taking the limit as t goes to zero.
Note that when a rigid object is rotating, every point on the object has the same angular speed and the same angular
acceleration. The linear speed and accelerations change, depending on the radius from the axis of rotation, but the
angular quantities are the same everywhere.
Motion with Constant Angular Acceleration
Note the similarities between the equations for linear velocity and acceleration, and angular speed and acceleration.
Rotational Motion About a
Fixed Axis with  Constant
Linear Motion with "a"
Constant
t
v = v0 + at
0t + (1/2)t2
x = v0t + (1/2)at2

v2 = v02 + 2ax
By measuring angles in radians, and using angular speed and acceleration, the equations for rotational motion are
very similar to the familiar equations for linear motion. Therefore, techniques similar to those you have used for
solving linear motion problems can be used to solve rotational motion problems. Let's work an example to
demonstrate.
Example: P7.10
The tub of a washer goes into it’s spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0
s. At this point the person doing the laundry opens the lid, and a safety switch stops the spinning. The tub slows to
rest in 12.0 s. Through how many revolutions does the tub turn during this 20.0 s interval? Assume constant angular
acceleration.
NOTE: In solving this problem we will determine the angle through which the tub rotates, and this angle will be
larger than 2. An angle greater than 2 simply means that the object rotated by more than one revolution. For
instance, if the object makes 3 revolutions then it rotates through an angle of 3x2 = 6.
First, convert the angular speed to rad/s:  = (5.0 rev/s)(2 rad/rev) = 10 rad/s = 31.4 rad/s. Next, we need the
angular accelerations: i =  / t = (31.4 rad/s)/(8.0 s) = 3.9 rad/s2, and f = - / t = (-31.4 rad/s)/(12.0 s) = -2.6
rad/s2.
Now we calculate the angular displacements: 1 = 0t + (1/2)1t2 = (1/2)(3.9 rad/s2)(8.0 s)2 = 125 rad, and 2 =
(31.4 rad/s)(12.0 s) + (1/2)(-2.6 rad/s2)(12.0 s)2 = 190 rad. The total angular displacement is tot = 1 + 2 = 125 +
190 rad = 315 rad. Or, in terms of revolutions, tot = (315 rad)(1 rev/2 rad) = 50 rev.
Relations Between Angular and Linear Quantities
Consider a rotating object with a point P a radius r from the axis of rotation. When the object makes an angular
displacement , then from the definition the radian, we know that the point P moves along the arc of a circle or
radius r, and the arc length is s = r. We know that /t = , and s / t = v, in the limit that t goes to zero.
Upon dividing the previous equation on both sides by t, we get s/t = r /t, or v = r. This linear speed, v, is
the instantaneous speed of point P. The direction of v is tangent to the circular path, and therefore this speed is
often referred to as the tangential speed, or vt = r.
While every point on the object has the same angular speed, , only points at the same radius from the axis have
the same tangential speed, vt.
Here is my first real chance to point out that radians are not a "real" unit, in the sense of meters or seconds. Let's
dimensionally analyze the expression vt = r. [vt] = L/T. [r] = L angle/T. The only way these two can be the same
is if angle is not a unit. This is true if (and only if) the angle is measured in radians, since then  = s/r, which has
dimensions of L/L = no unit. If the angle is measured in degrees, these equations do not work. Remember to
convert angles from degrees to radians!
Remember to add problem 6.37 to your list of homework problems
Recall: Quantities for Rotational Motion




Angular displacement,  = s/r, measured in radians,  radians = 180 degrees.
Angular speed,  = /t
Angular acceleration,  = /t
Tangential speed, vt = r
Tangential Acceleration
If the angular speed of an object changes from 1 to 2 in a time t, then its angular acceleration is  = /t. The
point P will initially have tangential speed vt1 = r1, and end with tangential speed vt2 = r2. Therefore vt = r.
Dividing both sides by t gives: vt/t = r/t. In the limit that t goes to zero, vt/t is the tangential
acceleration of P, at. at = r.
A company called Sea Launch Corporation launched its first satellite into orbit on Sunday. The company launches
rockets from a converted offshore oil platform which can be moved to a location on the equator. Why do they
launch from the equator?
Because the earth is spinning, every point on the surface has a tangential speed proportional to its distance from the
earth's axis. Rocket launchers can take advantage of this tangential speed to give their rockets a head start on their
journey into orbit, and the greatest tangential speed is obtained at locations furthest from the earth's axis, i.e. on the
equator.
Example:
A floppy disk in a computer rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s. The radius of
the disk is 4.45 cm. What is the angular acceleration of the disk, the tangential acceleration of a point on the edge of
the disk, and the final linear speed of this point.
Use  = /t = (31.4 rad/s)/(0.892s) = 35.2 rad/s2. Then from the relationship between linear and angular
acceleration we have at = r = (0.0445 m)(35.2 rad/s2) = 1.57 m/s2. Finally, the relation between linear and angular
speed gives vt = r = (0.0445 m)(31.4 rad/s) = 1.40 m/s.
Centripetal Acceleration
To any change in velocity there is a corresponding acceleration. Since velocity is a vector quantity, the changes can
be in magnitude (what we were primarily concerned with in the preceding chapters) or direction. Circular motion
involves constant changes in direction. A simple demonstration will easily verify that there is a force, and therefore
an acceleration, for an object in circular motion, even when the speed is constant. Imagine a small weight attached
to the end of a string, and spin it above your head. There is tension in the string, and thus a force on the weight. The
force is directed inward, toward the center of rotation. Furthermore, if you release the string, the force will no
longer be present, and instead of moving in a circle, the weight will fly off in a straight (neglecting gravity) line.
So, an object in circular motion experiences acceleration towards the center of the circle. This acceleration is called
centripetal (center seeking) acceleration, and has a magnitude of ac = vt2/r = r2. A derivation of this equation can
be found on p.186 of the text.
A rotating object always has centripetal acceleration. It may also have tangential acceleration if the rate of rotation
(angular speed) is changing. Centripetal and tangential acceleration are always perpendicular. Neither of them has a
fixed direction -- the direction changes as the object moves around in its rotation -- but they are always oriented
perpendicularly to each other. Therefore, we find the total acceleration using Pythagorean's theorem: atot = Sqrt{ac2
+ at2}.
Example: P7.20
A race car starts from rest on a circular track of radius 400 m. The car's speed increases at the constant rate of 0.500
m/s2. At the point at which the magnitudes of the centripetal and tangential accelerations are equal, determine (a)
the speed of the race car, (b) the distance traveled, and © the elapsed time.
at = r = 0.500 m/s2.  = at/r = 0.500 m/s2/400 m = 0.00125 rad/s2. (a) at = ac = v2/r, so v = Sqrt{r at} = Sqrt{(400
m)(0.500 m/s2} = 14.1 m/s. We can also find  = v/r = 0.0354 rad/s.
(b) Using 2 = 02 + 2, we see that  = 2/2 = (0.0354 rad/s)2 / 2(0.00125 rad/s2) = 0.50 rad = 0.080 rev. In
terms of meters, this is a distance of s = r = 200 m.
© Use  = 0t + (1/2)t2 to find the time. t = Sqrt{2/} = Sqrt{2(0.50 rad)/(0.00125 rad/s2)} = 28.3 s.
This can also be worked using tangential speed and acceleration.
Centripetal Force
Consider a mass m tied to a string of length r, being swung in a horizontal circle. For the mass to remain in a circle,
it experiences a centripetal acceleration, ac = vt2/r. To every acceleration there is associated a force, in this case it is
called the centripetal force, Fc = mac = mvt2/r. If the centripetal force is removed, for instance by letting go of the
string tied to the mass, then, following Newton's first law, the mass moves in a straight line.
It is the centripetal force, acting perpendicular to the instantaneous motion of the object that keeps it moving in a
circular path. The force must act perpendicular to the motion, since any component parallel to the motion will tend
to speed up or slow down the object. Remember from chapter 5, an impulse Ft causes a change in momentum
mv.
When working problems involving a centripetal force, it is not convenient to think in terms of x and y components,
since the direction of the centripetal force changes constantly. It is normally much more convenient to work with
the components of the forces that are parallel (tangential) and perpendicular to the circular path.
Example: P7.53
In a popular amusement park ride, a rotating cylinder of radius 3.00 m is set in rotation at an angular speed of 5.00
rad/s. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What
minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping?
fs,max = μn, and in this case the normal force is the centripetal force exerted on the rider to keep her moving in a
circle. That centripetal force is provided by the wall of the cylinder, and is Fc = mv2/r = m2r. We are told that  =
5.00 rad/s, and r = 3.00 m. Since we don't know the mass, let's write Fc/m = 2r = (5.00 rad/s)2(3.00 m) = 75.0 m/s².
The force of static friction must balance the gravitational force on each rider, fs = μFc > mg. Therefore, μ > mg/Fc =
(9.8 m/s²)/(75.0 m/s²) = 0.13. Since μ must be greater than or equal to 0.13, the minimum value is 0.13.
Describing Motion of a Rotating System
What happens if you're in a car that turns to the left at high speed? Assuming the car stays upright, you feel pulled
to the right side of the car. This "force" that you feel is a fictitous force, this one being called a centrifugal force. It
arises because for you to turn along with the car, you must be acted upon by a centripetal force, and this force
comes from the car seat, steering wheel (if you are holding on to it), or whatever part of the car you are touching.
But this is not a real force in the sense we have been using. A person standing outside the car doesn't "see" this
force, but instead sees you trying to continue moving in a straight line while the car turns to the left. You feel the
force because your local environment (the car) is accelerating (a centripetal acceleration). Newton's laws don't work
as is in an accelerating environment, additional corrections must be included to compensate for the acceleration.
This is not something we will be studying in this course.
A Rollercoaster Loop
Consider a rollercoaster car going around a loop of radius R. What must be the speed at the bottom, as it enters the
loop, so that it just makes it over the top of the loop without falling?
What we want is that at the top of the loop, the centripetal acceleration needed to keep the car moving on the circle
just equals the gravitational force. This means that the rollercoaster track exerts no force on the car at the top of the
loop, the car feels its own weight only. The we have mv2top/R = mg, or vtop = Sqrt{gR}.
Let us take the zero point of potential energy at the bottom of the loop. Then, at the top of the loop, the energy of
the car is both kinetic and potential, while at the bottom it is all kinetic, and since we have only conservative forces
(we neglect friction), the energy at the top equals the energy at the bottom.
What is the kinetic energy of something moving in a circle? It is still ½mv2, where v is the total velocity of the
object. In the case of the rollercoaster car, v is just the tangential velocity, as we calculated above for the top of the
loop. So, Etop = KEtop + PEtop = ½mv2top + mg2R = ½mgR + 2mgR = 2.5mgR. At the bottom of the loop, Ebot =
KEbot = ½mv2bot. Using conservation of energy to equate this gives ½mv2bot = 2.5mgR, or vbot = Sqrt{5mgR}.
To have this amount of energy, the car must start at a height of at least 2.5R above the bottom of the loop. Note that
this is more than 2R, the height of the loop. The additional starting height (meaning additional energy) is needed to
have enough speed to stay on the loop as the car goes over the top.
Newton's Universal Law of Gravitation
Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of
their masses and inversely proportional to the square of the distance between them.
This is Newton's Universal Law of Gravitation in words. In mathematical notation, let the two masses be m1 and
m2, and let the distance between their centers be designated r. Then we write
F = G m1m2/r2
where G is a universal constant called the gravitational constant, with the value G = 6.673 × 10-11 Nm²/kg².
Notice some features of this force. First, it is proportional to 1/r2. We say that this is an inverse square law force,
that is, the force is "inversely proportional to the square of the distance". If the distance between two objects is
doubled, the force decreases by a factor of 4 (is 1/4 the original), or if the distance between two objects is halved,
the force increases by a factor of 4.
Next, the force of m1 on m2, written F21, is equal in magnitude to the force of m2 on m1, written F12. The
gravitational force is always attractive, so the directions of F21 and F12 are opposite. Written mathematically, F21 = F12 as required by Newton's third law.
If one or both masses are extended, spherical objects (the usual case for our work), then the same law holds. This is
true as long as the distance between the objects is measured between their centers, and the objects don't overlap.
Pay attention to objects overlapping when, for instance, calculating what happens to something "launched" from the
surface of the earth into space.
Example: P7.32
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the
Sun. (a) what force is exerted on the Moon by the Sun? (b) What force is exerted on the Moon by the Earth? ©
What force is exerted on the Earth by the Sun?
From problem 31, the distance from the Earth to the Moon is 384,000 km = 3.84x108 m, and from table 7.3, the
distance from the Earth to the Sun is 1.496x1011 m. During a solar eclipse, the distance from the Moon to the Sun is
the difference of the above distances, or 1.492x1011 m. The mass of the Sun is 1.991x1030 kg, the Earth is 5.98x1024
kg, and the Moon is 7.36x1022 kg.
(a) Fmoon-sun = (6.673x10-11 Nm²/kg²)(1.991x1030 kg)(7.36x1022 kg)/(1.492x1011 m)2 = 4.39x1020 N
(b) Fmoon-earth = 1.99x1020 N
© Fearth-sun = 3.55x1022 N
Historically, there was one small difficulty with the law of gravitation applied to planets and stars: for n
gravitational relations, there were n+1 unknown quantities, namely G, plus n masses. This problem was finally
solved in 1798 by Cavendish (112 years after Newton published his law in 1687). He arranged an apparatus to
determine G, using known masses in his laboratory. See the text for a description.
Gravitational Potential Energy Revisited
In chapter 5 we introduced gravitational potential energy, and found that in a uniform gravitational field, like near
the surface of the earth, PE = mgh. For objects far from the surface of the earth, this is no longer valid, and an
alternate expression must be used:
PE = -GMEm/r
ME = 5.98x1024 kg is the mass of the earth. This equation "fixes" the zero point of the potential energy at an infinite
distance from the earth. And the above expression does reduce to mgh for objects near the surface of the earth, with
g = GME/RE² = (6.67x10-11 N m²/kg²)(5.98x1024 kg)/(6.38x106 m)2 = 9.84 m/s².
Example: Escape Speed from the Earth
Let's apply the general gravitational potential energy formula to determine the minimum speed necessary for an
object to escape Earth's gravity. First, what is the meaning of "escape earth's gravity", given the fact that the
gravitational force reaches to infinity? We will say that an object can escape earth's gravity if it has sufficient speed
to (eventually) travel an infinite distance from Earth. When given the minimum speed, the object will arrive at
infinity with exactly zero speed.
We will use conservation of mechanical energy. When the object reaches infinity with zero speed, its kinetic energy
is zero, and at infinity, the gravitational potential energy is also zero, so the final mechanical energy is zero.
The initial mechanical energy is Ei = KEi + PEi = ½mvi2 - GMEm/RE. We use RE, since the object starts at the
surface of the earth, a distance RE from the center.
Setting Ei = Ef = 0 gives:
½mvi2 - GMEm/RE
or
vi = Sqrt{2GME/RE} = Sqrt{2(6.67x10-11 N m²/kg²)(5.98x1024 kg)/(6.38x106 m)} = 1.12x104 m/s = 11.2 km/s =
40,000 km/h = 25,000 mi/h.
Kepler's Laws
Before Newton wrote down the law of gravitation, Johannes Kepler deduced three "laws" about the orbits of
planets around the sun. The three laws are:
1. All planets move in elliptical orbits with the Sun at one of the focal points.
2. A line drawn from the Sun to any plant sweeps out equal areas in equal time intervals.
3. The square of the orbital period of any planet is proportional to the cube of the average distance from the
planet to the Sun.
The last one is of most interest as regards homework problems. The statement transforms into the equation:
T2 = (4²/GMs) r3
where T is the orbital period of the planet, Ms is the mass of the sun, and r is the average distance from the planet to
the Sun. This relation can be applied to the period of a satellite orbiting the earth by replacing the mass of the sun
with the mass of the earth, and using the average distance from the satellite to the center of the earth.
Example: Low Orbit Objects
All objects (satellites, space shuttles, space capsules, and space stations) in low earth orbit circle the earth
approximately once every 90 minutes. Why?
Objects in low earth orbit are only a few hundred kilometers above the surface. This distance is small compared to
the radius of the earth (the distance from the center of the earth to the surface), 6380 km. Let's apply Kepler's third
law, using r = 6380 km = 6.38x106 m.
T = 2 Sqrt{Re3/GME} = 2 Sqrt{(6.38x106 m)3/(6.67x10-11 N m²/kg²)(5.98x1024kg)} = 5070 s = 84.5 min.
This period is independent of the mass of the object, and the exact shape of the orbit.
Chapter 8: Rotational Equilibrium and Rotational Dynamics
We will complete our discussion of rotational motion by investigating rotational dynamics -- how forces produce
rotational motion -- and equilibrium for objects that can rotate. In the process, we will come across the concepts of
torque, angular momentum, angular kinetic energy, and conservation of angular momentum.
Torque
Torque is the measure of the ability of an applied force to cause rotation. Torque is denoted by the greek letter tau,
, and is equal to the applied force times the "lever arm", d,
 = Fd.
The units of torque are force times distance, or Nm. (People may also be familiar with the English unit ft lb.) The
lever arm is the perpendicular distance from the axis of rotation to a line drawn parallel to the force. Alternatively,
we can take the lever arm as the distance from the axis of rotation to the point where the force is applied, but then
we must use only the component of the force perpendicular to the lever arm.
Examples:
Some examples to help understand how to calculate torque.
1. A wrench: See Fig. 8.2 on page 218 in the text.
2. A door.
3. A barbell supported off center, as drawn to the right. The triangle represents the point of support. There are
two torques in this example, one from the weight on the left which tends to rotate the barbell
counterclockwise, and one from the weight on the right which tends to rotate it clockwise. As you might
imagine, we assign torques that tend to cause a counterclockwise (CCW) rotation a positive value, and those
that tend to cause a clockwise (CW) rotattion a negative value. Therefore, the torque from the left mass is
left = +Mgx, and the torque from the right mass is right = -Mg(d-x). The total torque is a sum of the two, tot
= left + right = Mgx - Mg(d-x) = Mg(2x-d). Notice that if the support is in the middle, x=d/2, then the total
torque is zero. Also, note that the support must supply a normal force, n = 2Mg, which supports the whole
barbell, but since this force acts at the point of support, it's torque is zero.
Recall:


Torque =  = Fd, where F is the applied force and d is the lever arm.
Remember, a positive torque tends to cause a CCW rotation, and a negative torque tends to cause a CW
rotation. This is our convention.
Torque and the Second Condition for Equilibrium
The first condition for equilibrium, from Chapter 4, is that the net force acting on an object must be zero. In fact,
we can write this down for the x and y components separately:
Fx =0, and Fy =0
Now consider the barbell example given above. All the applied forces, gravity and the normal force from the
support, act in the y direction, and they sum to zero. Yet, unless the support is exactly in the middle of the barbell,
there will be a net torque, and the barbell will tend to rotate. We cannot say it is in equilibrium!
But this does lead to a second condition for equilibrium, mainly that the net torque must also be zero. To the above
force equations we should add:
 = 0.
The primary implication is that not only must we pay attention to the size and direction of the applied forces, but
also their point of application on an object.
Example: P8.13
The chewing muscle, the masseter, is one of the strongest in the human body. It is attached to the mandible (lower
jawbone), as shown in Figure P8.13a. The jawbone is pivoted about a socket just in front of the auditory canal. The
forces acting on the jawbone are equivalent to those acting on the curved bar in Figure P8.13b: C is the force
exerted against the jawbone by the food being chewed, T is the tension in the masseter, and R is the force exerted
on the mandible by the socket. Find T and R if you bite down on a piece of steak with a force of 50.0N.
We must apply the two conditions for equilibrium (this of course assumes that the jaw doesn't move):
T - C - R = 0, and
C(11 cm) - T(3.5 cm) + R(0 cm) = 0.
From the second equation, we find T = C(11 cm)/(3.5 cm) = (50.0N)(11cm)/(3.5cm) = 157 N. Then from the first
equation we find that R = T - C = 157N - 50.0N = 107N.
If the object is in equilibrium, it doesn't matter where you put the axis of rotation for calculating the net torque; the
location of the axis is completely arbitrary.
Center of Gravity
How do we handle the force of gravity for an extended object? Up to now we've simply taken gravity as acting on
an object, with no real concern for where it acts. But now, to handle torques, we must concern outselves with where
the force of gravity acts on an object.
For instance, consider the barbell from the previous lecture. I calculated the
gravitational force for each weight separately, and used the two forces to
determine the torque. But, there is another approach.
What if we take the total mass of the barbell, 2M, and say that the gravitational
force due to the total mass acts at the center of the barbell. Then the torque is  =
2Mg(x-d/2) = Mg(2x-d), the same result arrived at previously. We call the point
at the center of the barbell its center of gravity. We can consider the net gravitational force as acting on (or at) the
center of gravity.
We define center of gravity (a.k.a. center of mass) of an object as the point (not necessarily on the object) through
which the total gravitational force acts, regardless of the orientation of the object.
The barbell is a rather symmetric object, and finding its center of gravity is easy. The center of gravity of a
homogeneous, symmetric object lies on the "axis of symmetry". Notice that we already implicitly assumed that the
force of gravity acted at the center of the spherical weights of the barbell.
For an object without symmetry, we need a mathematical expression:
xcg = (m1x1 + m2x2 + m3x3 + ...)/(m1 + m2 + m3 + ...) = mixi/mi
To apply the above expression, break the object up into n symmetrical or point objects, each with mass mi and c.g.
located at xi. Then the calculation will yield the x coordinate of the center of gravity of the composite object. To
determine the y coordinate of the center of gravity, apply the same technique, exchanging y for x in the expression.
Example: P8.7
A water molecule consists of an oxygen atom with two hydrogen atoms bound to it. The bonds are 0.100 nm in
length and the angle between the two bonds is 106°. Determine the location of the center of gravity of the molecule.
Consider the mass of an oxygen atom to be 16 times the mass of a hydrogen atom. See the figure in the text.
There is symmetry about the x-axis. If we flip the water molecule around the x-axis, we get back the same water
molecule (all H atoms are identical and indistinguishable). Therefore, the c.g. must lie on the x-axis, giving ycg = 0.
We can find xcg using the equation given above. Let's call the center of the oxygen atom x=0. Then xcg =
((0)(16mH) + 2(0.100 nm)cos53° mH)/(16mH + 2mH) = 0.00669 nm from the oxygen atom.
So the center of mass of the water molecule is located at (xcg, ycg) = (0.00669, 0) nm.
Example: Conceptual Question 6
Why does holding a long pole help a tightrope walker stay balanced?
With a long pole, the tightrope walker can lower his center of gravity. (If the pole is curved, he can even get his
center of gravity below the rope.) A lower center of gravity means that the torque exerted on him, due to
misalignment of the cg with the rope, is reduced.
Example:
When a car speeds up, the front rises, and when it brakes, the front drops. Why?
The force that accelerates or decelerates a car comes ultimately from the contact of the tires with the road. The
body of a car is attached to the wheels through a "suspension" which allows for independent motion to help smooth
over bumps. The suspension serves as an axis of rotation, and because it is located above the road surface, there is a
lever arm. The force applied to speed up a car will produce a torque which tends to rotate the front of the car
upward, and a braking force produces a torque which tends to rotate the front downward.
Some Example Equilibrium Problems
Example: P8.17
A 500 N uniform rectangular sign 4.00 m wide and 3.00 m high is suspended from a horizontal, 6.00 m long,
uniform, 100 N rod, as drawn. The left end of the rod is supported by a hinge and the right end is supported is
supported by a thin cable making a 30.0° angle with the vertical. (a) Find the tension, T, in the cable. (b) Find the
horizontal and vertical components of force exerted on the left end of the rod by the hinge (or wall).
We can determine the tension in the cable by determining the force required on the end of the rod to keep it in
equilibrium; the tension is this force. Begin by drawing a force diagram for the rod. First, consider the rod and sign
as one object. Their combined center of gravity is at xcg = [(3.00 m)(100 N) + (4.00 m)(500 N)]/(100 N + 500 N) =
3.83 m. Therefore, the weight of the rod and sign, 600 N, acts at a point 3.83 m from the wall. In addition to the
weight, there is the force from the tension in the cable, which acts in the direction of the cable, and the force of the
hinge, which acts in an undetermined direction. For convenience, let's call the x and y components of the hinge
force Fx and Fy.
Next, apply the first condition for equilibrium, that the x and y components of the forces on the rod must sum to
zero. For the x direction, we get:
T sin30° = Fx
And for the y direction:
T cos30° + Fy = 600 N
Finally, apply the second condition for equilibrium, that the sum of the torques is zero. As long as the rod is
stationary, we can calculate the torque about any convenient axis. The usual approach is to pick one of the three
locations where forces are applied; as this will remove one of the forces from the torque calculation (the lever arm
will be zero). I will choose the axis of rotation at the hinge, and the resulting torque equation is:
(6.00 m)T cos30° - (3.83 m)(600 N) = 0
(a) From this last equation we find T = (3.83 m)(600 N)/(6.00 m)cos30° = 442 N.
(b) Use this result with the first 2 equilibrium equations to find the x and y components of the hinge force.
Fx = T sin30° = (442 N)sin30° = 221 N, and
Fy = 600 N - T cos30° = 600 N - (442 N)cos30° = 217 N.
Y will be obtain the same result if you work the problem with a different axis of rotation. Try it and see.
Example: P8.20
A 15.0 m, 500 N uniform ladder rests against a frictionless wall, making an angle of 60.0° with the horizontal, as
shown. (a) Find the horizontal and vertical forces exerted on the base of the ladder by the Earth when an 800 N fire
fighter is 4.00 m from the bottom. (b) If the ladder is just on the verge of slipping when the fire fighter is 9.00 m up,
what is the coefficient of static friction between ladder and ground?
Again, start by drawing a force diagram for the ladder, being careful to note the location where forces are applied.
Since the ladder is uniform, its center of gravity is right in the middle, 7.5 m from either end. The first condition for
equilibrium gives:
nwall - Fx = 0
Fy - 500 N - 800 N = 0 or Fy = 1300 N
Since the wall is frictionless, there is no vertical force from the wall. The second condition for equilibrium,
choosing the base of the ladder as the axis of rotation, gives:
(500 N)(7.50 m)cos60° + (800 N)(4.00m)cos60° - nwall(15.0 m)sin60° = 0
The last equation gives
nwall = [(500 N)(7.50 m)cos60° + (800 N)(4.00m)cos60°]/(15.0 m) sin60° = 268 N
Thus Fx = nwall = 268 N, and Fy = 1300 N for part (a).
(b) To determine the coefficient of static friction, we need the x component of the force on the base of the ladder,
equal to the friction force, and the y component, equal to the normal force. The coefficient of friction is the ratio
Fx/Fy. When the fire fighter is 9.00 m up the ladder, the equilibrium equations read:
nwall - Fx = 0
Fy - 500 N - 800 N = 0 or Fy = 1300 N
(500 N)(7.50 m)cos60° + (800 N)(9.00m)cos60° - nwall(15.0 m)sin60° = 0
We find that Fx = nwall = 421 N, and again Fy = 1300 N. This give s = Fx/Fy = 421 N/1300 N = 0.324.
Relationship Between Torque and Angular Acceleration
Consider a mass, m, fixed to a central point by a string of length r, and moving about that point in a circle on a
frictionless surface. If a tangential force is applied to the mass, it will experience a tangential acceleration given by
Ft = mat. Multiply both sides of this expression by the radius, so that the left hand side is the torque, to get Ftr =  =
mrat. Now use the relation at = r and the expression becomes  = mr2. This relation says that the torque is
proportional to (or produces) the angular acceleration, times a factor of mr2. The factor mr2 is called the moment of
inertia.
What if the rotating object is not a small mass at the end of a string, but a larger object, like a wheel, a cylinder, or a
sphere? In this case, consider the extended object as being composed of lots of little pieces, each piece feeling a
torque. Then the total torque is the sum of the torques on each piece:
tot =  = (mr2).
Since the angular acceleration, , is the same for all the pieces of the object, it can be factored out of the sum in the
last expression to yield:
tot = (mr2) .
We call I = mr2 the moment of inertia of the object, and then write tot = I, the rotational analog of F = ma.
Relationship Between Torque and Angular Acceleration
First, recall that angular speed =  = /t,  = /t, vt = r, at = r, and torque =  = Fd.
Consider a mass, m, fixed to a central point by a string of length r, and moving about that point in a circle on a
frictionless surface. If a tangential force is applied to the mass, it will experience a tangential acceleration given by
Ft = mat. Multiply both sides of this expression by the radius, so that the left hand side is the torque, to get Ftr =  =
mrat. Now use the relation at = r and the expression becomes  = mr2. This relation says that the torque is
proportional to (or produces) the angular acceleration, times a factor of mr2. The factor mr2 is called the moment of
inertia.
What if the rotating object is not a small mass at the end of a string, but a larger object, like a wheel, a cylinder, or a
sphere? In this case, consider the extended object as being composed of lots of little pieces, each piece feeling a
torque. Then the total torque is the sum of the torques on each piece:
tot =  = (mr2).
Since the angular acceleration, , is the same for all the pieces of the object, it can be factored out of the sum in the
last expression to yield:
tot = (mr2) .
We call I = mr2 the moment of inertia of the object, and then write tot = I, the rotational analog of F = ma.
Moments of Inertia
The moment of inertia is an important part of the above result. It plays the same role as mass in Newton's second
law, but has some special properties. While an object has only one mass, it can have different moments of inertia
which depend on the axis of rotation, and the distribution of mass (applicable for people and object with moving
parts). Table 8.1 lists the moments of inertia for a number of shapes and axes of rotation.
Rotational Kinetic Energy
If an object is (purely) rotating, then each particle of the object has kinetic energy. The sum of all the kinetic
energies is the kinetic energy of rotation, KErot = (½mivi2) = ½(miri2)2 = ½I2. Notice the similarity of the
expressions KE = ½mv2 and KErot = ½I2.
Rotational kinetic energy is a useful concept, especially when an object is both rotating and translating, like a wheel
rolling down an incline. The total kinetic energy of the wheel is the sum of the kinetic energies of each particle in
the wheel. Due to the combined rotation and translation, the speed of each particle depends on its angle on the
wheel, and the computation becomes quite complex. But we can consider the translational and rotational motions
separately.
The total kinetic energy of a rolling object is the sum of the translational kinetic energy (mass of the object, speed
of the axis of rotation), and the rotational kinetic energy:
KEtot = KEtrans + KErot = ½mv2cg + ½I2.
Example: Which rolls faster down an incline, a solid disk, or a ring?
If we ignored the rotational kinetic energy, then we would predict that both will move down an inclined plane at the
same speed, since after moving downward a vertical distance h, the kinetic energies would be ½mv2 = mgh, or v =
Sqrt{2gh}. Since these objects are rolling, we must include the rotational kinetic energy in the conservation of
energy equation. Then we find that:
½mv2 + ½I2 = mgh.
The moment of inertia of a solid disk is Idisk = ½mr2, and the moment of inertia of a ring is Iring = mr2. Inserting
these into the conservation of energy expression, yields:
½mv2 + ¼mr22 = mgh for the disk, and
½mv2 + ½mr22 = mgh for the ring.
Divide through by the mass, and substitute v = r, resulting in: ½v2 + ¼v2 = gh for the disk, and
½v2 + ½v2 = gh for the ring.
Thus we find that the disk has a speed of vdisk = Sqrt{4gh/3}, and the ring has a speed of vring = Sqrt{gh}. The disk
is moving faster by a factor of Sqrt{4/3} for the same distance traveled. Demonstration.
The disk has a smaller moment of inertia than the ring. Therefore a smaller fraction of its kinetic energy goes to
rotational kinetic energy, and a larger fraction goes to translational kinetic energy, and its the translational kinetic
energy that gets the object down the incline.
Angular Momentum
Continuing to draw parallels between rotational and translational motion, we can define an angular momentum.
Linear momentum was defined such that F = p/t. Similarly, we can write  = I = I/t = (I)/t. Now it is
easy to see that the rotational analog of p=mv is angular momentum = L = I.
We can write  = L/t, which states that "the torque acting on an object is equal to the time rate of change of the
object's angular momentum." If the net torque on an object is zero, then the angular momentum is conserved.
Li = Lf or Iii = Iff.
We applied the principle of conservation of linear momentum to collisions involving two bodies. Since angular
momentum depends on the moment of inertia and angular speed, there are a number of interesting applications that
involve only one object, but an object whose moment of inertia can change. For instance, a figure skater will begin
a spin with her arms extended, then bring her arms in close to her body, seemingly spinning faster as she does so.
To a reasonable approximation, the ice is frictionless, and her angular momentum is conserved. The moment of
inertia of the skater with her arms extended is larger than with her arms close to her body. To conserve angular
momentum, her angular speed must increase when she brings her arms in, so she doesn't seem to spin faster, she
does spin faster!
Demonstration with rotating turntable.
Example: P8.47
The Physics 2130 instructor is standing on a turntable holding two 3.0 kg masses. When his arms are extended, the
masses are 1.0 m from the axis of rotation, and he rotates with an angular speed of 0.75 rad/s. The moment of
inertia of the instructor and turntable is 3.0 kg m² and is assumed to be constant. the instructor then pulls the masses
to 0.30 m from the axis of rotation. (a) Find the new angular speed of the instructor. (b) Find the kinetic energy of
the instructor before and after the masses are pulled in.
The initial moment of inertia of the two masses is Imi = 2mr2 = 2(3.0 kg)(1.0 m)2 = 6.0 kg m². The final moment of
inertia of the two masses is Imf = 2(3.0 kg)(0.30 m)2 = 0.54 kg m². Including the moment of inertia of the instructor
and turntable, the initial moment of inertia is Ii = 9.0 kg m², and the final moment of inertia is If = 3.5 kg m².
(a) Using conservation of momentum, Iii = Iff, or
f =  iIi/If = (0.75 rad/s)(9.0 kg m²)/(3.5 kg m²) = 1.9 rad/s.
(b) All kinetic energies are purely rotational.
KEi = ½Iii2 = ½(9.0 kg m²)(0.75 rad/s)2 = 2.5 J
KEf = ½Iff2 = ½(3.5 kg m²)(1.9 rad/s)2 = 6.3 J
Notice that the kinetic energy increased! Work (negative) was done by the centripetal force when the masses were
pulled in, resulting in the increase in kinetic energy.
Chapter 9: Solids and Fluids
Since we covered sections 1 and 2 in previous classes, I will be covering primarily sections 9.3 to 9.6.
States of Matter
You are all familiar with the 3 common (on earth) states of matter: solid, liquid, and gas. Physicists recognize that
there is a fourth, and just recently, a fifth state of matter. The fourth is called a plasma, basically a collection of
highly ionized atoms, as found in the sun and other stars. The fifth is a Bose-Einstein condensate, a state of matter
only recently created in laboratories by cooling atoms to ultra cold temperatures using laser beams. For our
discussions, we will concentrate on solids, liquids, and gases.
We can think of a solid as a material where the position of one atom is basically fixed with respect to another. I say
basically, because if we compress or stretch a solid, the atoms will move slightly with respect to one another, that
is, we say that solids are elastic. (Remember Hooke's law for springs.)
The molecules in liquids and gases are basically free to move about. From a physics perspective, the difference
between a liquid and a gas is primarily the space between molecules: the space is about 10 times larger in a gas than
in a liquid.
Density and Pressure
"The density of a substance of uniform composition is defined as its mass per unit volume."
Density =  = M/V, and has units of kg/m³ in SI units.
The specific gravity of a substance is the ratio of its density to the density of water at 4°C, which is 1.0x103 kg/m³
A fluid exerts a force on any object submerged in it. This force tends to compress the object, and is perpendicular to
the surface of the object. We define the pressure exerted by the fluid as the force divided by the area that it acts on,
P = F/A. The units of pressure are newtons per square meter (N/m²) also called pascals (Pa).
To determine pressure, we need to know the force, and the area that this force is applied to. This idea explains the
sharp edges of knives, and the sharp point of an ice pick or a push pin (to concentrate force and produce a large
pressure) and the idea behind snow shoes (to spread out force and produce a low pressure). The pressure inside a
tire produces a force on the road equal to the weight supported by the tire.
Example: Snowshoes
How much pressure is applied to snow by a 700 N man standing in boots with an area of 0.060 m² (2 boots of 10 by
30 cm), or using snowshoes with an area of 0.20 m² (2 shoes of 20 by 50 cm).
In boots the pressure is P=F/A = 700 N/0.060 m² = 12,000 Pa, and in snow shoes P = 700 N/0.20 m² = 3,500 Pa.
all:


Density =  = mass/volume = M/V, and is measured in kg/m³ in SI units.
Pressure = P = force/area = F/A, and is measured in N/m² = Pa.
Variation of Pressure with Depth
"If a fluid is at rest in a container, all portions of the fluid must be in static equilibrium." The situations we will
consider involve fluids in static equilibrium, at least at the initial and final instances. To be in equilibrium, "all
points at the same depth must be at the same pressure", otherwise parts of the fluid would feel a net force, and
would tend to move, contrary to the assumption of static equilibrium.
Note: Although individual molecules are free to move about in a fluid, this is not contrary to static equilibrium. The
individual molecules move independently, and more or less randomly. On average, for every molecule that moves
to the left, another moves to the right, and the net sum of all the motions is zero. The fluid motion mentioned above
is a bulk motion, or a flow, where an entire collection of molecules has some net motion.
To understand how pressure varies with depth, let's imagine a container of fluid, and consider the forces acting on a
column of fluid with area A, and extending from the top to a depth h. In equilibrium, there is no net horizontal force
on this column. The force of air pressure acts on the fluid, exerting a downward force of F0 = P0A. The weight of
the fluid also exerts a downward force of Fw = Mg, where M = V = hA. These downward forces must be
balanced by an upward force. This upward force comes from the pressure of the fluid at depth h, so calling this
pressure P, the upward force on the column of fluid is Fup = PA.
These are all the forces, and they must sum to zero, otherwise the system is not in equilibrium. PA - Mg - P0A = 0,
or, inserting the expression for M in terms of , and canceling the common factor of A,
P = P0 + gh.
Normally, P0 stands for the atmospheric pressure, which has a standard (average) value of 1.01x105Pa. As one goes
deeper into a fluid, the pressure increases as gh.
Notice that if P0 is equal to atmospheric pressure plus some additional pressure, then the additional pressure shows
up everywhere in the fluid. That is, the above expression for P depends on P0 and the depth only. This fact is known
as Pascal's principle:
Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the
containing vessel.
Applications of this principle include hydraulic systems like brakes, and garage lifts.
Pressure Measurements
A manometer is a simple device for measuring pressure. It consists of a tube containing a fluid, one end of which is
open to the atmosphere, while the other end is sealed a fixed volume of trapped gas. (Drawing) Points A and B are
at the same "depth" relative to the top of the fluid at the open end of the container. Therefore, the pressure at A will
equal the pressure at B, and the pressure at B is P = P0 + gh, where h is the distance to the top of the fluid.
P is called the absolute pressure (meaning relative to absolute zero pressure, which is a perfect vacuum), and P - P0
is called gauge pressure (meaning relative to atmospheric pressure, as is measured by many gauges). The
manometer measures gauge pressure, since P - P0 = gh.
A barometer is another simple device for measuring pressure. A barometer is a long (>0.76 m) tube, filled with
mercury (traditionally), then inverted into a dish of mercury, such that there is no gas inside the tube. The mercury
will pull away from the closed end of the tube, leaving a vacuum -- P = 0 at the top of the mercury in the tube.
Using our equation for P versus depth, putting a negative sign in front of h since it is higher, not lower than the
open surface of the mercury, we have 0 = P0 - gh, or P0 = gh. (It is traditional to keep h a positive number.) So
the barometer measures absolute atmospheric pressure.
At a standard one atmosphere of pressure, the height of mercury is 0.76 m = 760mm = 760 torr. The density of
mercury is 13.595x103 kg/m³ which means that standard atmospheric pressure is
P0,standard = gh = (13.595x103 kg/m³)(9.8m/s²)(0.76 m) = 1.013x105Pa.
Bouyant Forces and Archimedes' Principle
"Any body completely or partially submerged in a fluid is buoyed up by a force whose magnitude is equal to the
weight of the fluid displaced by the body."
This upward force is called the bouyant force. The bouyant force is equal to the weight of the displaced fluid, and
acts at the center of gravity of the displaced fluid. An object dropped into water is acted on by two net forces, its
weight acting downward and the bouyant force acting upward. If the weight exceeds the bouyant force, the object
sinks; if not, it floats. Since the weight is given by the average density of the object, avg, times the volume and
g,avgVg, and the bouyant force is given by waterVg, we see that the object sinks if its density is greater than the
density of water.
So how do fish and submersibles remain in equilibrium at fixed depths in water? They are each able to adjust their
density to match the surrounding water density. By increasing or decreasing their density, they are able to change
their depth up or down.
Example: P9.26
An object weighing 300 N in air is immersed in water after being tied to a string connected to a spring scale. The
scale now reads 265 N. Immersed in oil, the object weighs 275 N. Find (a) the density of the object and (b) the
density of the oil.
We will use P = P0 + gh. Since the information supplied in the problem is not pressure but force, we will multiply
the above equation by A, and use F = PA to get:
F = F0 + fgV
F is the force measured on the scale. The subscript f on  is to remind us that we use the density of the fluid the
object is immersed in.
(a) To determine the density of the object, we need its mass and volume. The mass we get by dividing the weight
(in air) by g. The volume we get by considering the volume of water displaced to cause the change in the scale
reading. That is, V = (F - F0)/waterg = (300 - 265 N)/(1.0x103 kg/m³)(9.8 m/s²) = 3.6x10-3 m³. The density is then 
= M/V = F/gV = (300 N)/(9.8m/s²)(3.6x10-3 m³) = 8.6x103 kg/m³.
(b) Now that we know the volume of oil the object will displace, we can determine the density of the oil.
oil = (F - F0)/gV = (300 - 275 N)/(9.8 m/s²)(3.6x10-3 m³) = 0.71x103 kh/m³.