physics 120-01 class notes thru 012314 lecture
TRANSCRIPT
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Physics and Technological Progress The laws of the physics of motion were developed by Galileo
(15641642) and Newton (16421727)
These laws form the basis for: Understanding the motion of objects and fluids on earth and in space
Steam and gasoline engines, modern transportation
Aerodynamics and the development of the airplane
Rocketry and space travel
The laws of electricity and magnetism were assimilated by
Maxwell (18311879)
These laws form the basis for:
Electrical power generation and transmission
Electronics and computer industry Electrical machinery
Lasers and Lighting
In PHY 120 you will learn the laws of physics of motion, conservation of energy and
momentum, thermodynamics and heat transfer and how to apply these laws
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Basic Approach to Understanding Physical
Phenomena Physicists establish hypotheses(theories to be tested)
explaining observations
These hypotheses are tested against observations over an extended
period of time (say, decades). They must be stated in such a way that
they can be proven wrong.
If the hypotheses prove to accurately predict observed phenomena overthis extended period of time they are called theories and if the
theories become widely accepted over a century or two they are called
laws
The observations that stand the test of time are called facts
If a theory is found incorrect by more accurate observations it issuperseded by a new set of hypotheses
If the new hypotheses prove accurate over an extended period of time
they are accepted as theories (may become laws) and the supporting
observations are called facts. and the cycle continues2
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Measurements Play a Key Role in Physics
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The way in which theories are tested is by observations made in
experiments designed by scientists to test the theories. Theseobservations must be accurately measured to determine the
degree to which the theories explain the experimental findings.
Among the key measurements that physicists make are:
1.The time interval between two events.2.The dimensions of an object
3.The mass of an object
4.The electric current through an electrical circuit.
5.The temperature of an object.6.The luminous intensity of a light source.
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xper menta servat ons are a e yMeasurements in the International (SI) Units
The Scientific Method and Experiments Hypotheses must be supported by experiments made with precise
measurements
This requires precise instrumentation and diagnostic equipment, as well as
precisely defined standard measures for mass, length, time, electric current,
temperature and luminosity
The International System (SI) Units are the accepted standard
today, the units we will use in this course are:
Unit Standard Basis for Standard
Length Meter Distance between two marks on a Pt/Ir bar kept at 0 degrees C
Mass Kilogram Mass of a standard cylinder of Pt/Ir
Time Second Time for 9,192,631,770 wave cycles of Cs atomic clock to occur
In this course we will be concerned primarily with the units of length (meters),
time (seconds) and mass (kilograms) 4
A i l B li f h M i f Obj
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Aristotles Beliefs on the Motion of Objects
Held Sway for 2000 Years
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Aristotle (~350 B.C.) believed that there were two types of motion.
1. Natural Motion. Light things like smoke rise and heavy things likedropped boulders fall.
2. Violent Motion Motions that resulted from a push or a pull.
Although Aristotles beliefs were ultimately proven wrong by Galileo and
Newton and there were Greeks who lived at the same time as Aristotle thathad different beliefs, Aristotle was so highly respected for his intellect that his
beliefs were accepted without challenge.
As a result of his belief in Natural Motion he thought that a heavy object
would fall to earth faster than a lighter one.
Aristotle and Ptolemy (~ 150 A.D.) also believed that the sun, planets and
stars revolved around the earth (geocentric hypothesis) rather than the sun
(heliocentric hypothesis)
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With the advent of telescopes, experiments designed to test hypotheses and methods
of measuring time durations Galileo was able to disprove Aristotles hypotheses
Aristotles (incorrect) hypotheses were
thought correct for 2000 Years
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Aristotles thoughts held the upper hand for 2000 years until Galileo and Kepler
made more accurate observations in the 16thcentury A.D.1. With the advent of the telescope Galileo was able to make more accurate
observations and show that not all motion was geocentrica. He observed that Jupiter had moons which orbited around Jupiter, not
around Earth
2. Galileo was also able to show that any two objects fell to earth at the same
speed, independent of their weight, opposed to Aristotles hypotheses, and he
performed accurate experiments on motion of objects using pendulums and
water clocks.
3. Keplers observations on the motion of the planets superseded Ptolemys
predictions
Galileo and Keplers observations were the foundation of Newtons theories ofmotion and gravitation. With Newtons theories in hand there was a theoretical
basis for the observations of Galileo and Kepler and the ideas of Aristotle and
Ptolemy on motion were finally discarded.
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Galileos Experiment Proved Aristotles
Hypothesis Wrong
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A key to the scientific method is making accurate measurements.
A scientific assertion (hypothesis) can be disproven by a well-
designed experiment which allows for accurate observations.
Aristotle (384 B.C.322 B.C.) believed that an object falls at a
speed proportional to its weight. He was so highly respected his
assertion was held to be true for 2000 years.
According to historians, in 1589 Galileo dropped two balls from the
leaning tower of Pisa simultaneously and found they both
impacted the ground at the same time, disproving Aristotles
assertion.
Why do you suppose Aristotle was wrong and so many peoplebelieved Aristotle for so long?
Would the same result have occurred if Galileo had
dropped a feather and a heavier rock at the same time in a
second experiment?
What causes the difference in the two experiments?
Galileo was the first true physicist because he developed well-designed
experiments and made precise measurements of his observations
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Units of Measure
We will be primarily using the SI (International)System of Units of measures in this course The key units for studying motion and the familiar English
unit equivalent are: 1 Meter = 39.37 inches = 3.281 feet
1 Kilogram = 2.205 pounds (mass equivalent)
The acceleration of gravity on earth in SI units The mass of the earth creates a gravitational field that
exerts a force directed toward the center of the earth
g = 9.80 meters/second/second
A body free-falling under the force of gravity will gainspeed at the rate of 9.80 meters per second
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Mass and Weight
Mass is a measure of the inertia (resistance to changein motion) of an object It is related to the quantity of matter in an object
Objects which are subjected to a (net) force will accelerate in
inverse proportion to their mass Weight and Mass are different
The weight of an object is the force with which it is pulled tothe center of the earth (or any other astronomical body) bythe force of gravity
Weight = mass x acceleration of gravity
Mass is an inherent quantity, weight varies with thegravitational field (e.g., objects weigh less on the moon thanon earth, whereas mass is same)
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Converting from one system of units to
another
Write down the units explicitly For example 65 (miles/hour) (convert to feet/second)
Find the conversion factor from one system
to the other 1 mile = 5280 feet, one hour = 3600 seconds
Set up as multiplicative factors so that only
the new units appear in the result 65 miles/hour x 5280 feet/mile x 1 hour/3600 secs
= 95.33 feet/sec
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Check Dimensions
For motion studies the fundamental units are Mass, Length and Time
The answer to a motion problem must have the correctdimensions (check numerator and denominator)
For example Distance is always going to be in [L], (meters, feet, miles)
Speed is always going to be in [L]/[T] (meters/sec, feet/sec,miles/hour)
Acceleration is always going to be in [L]/[T]2
(meter/sec2
,feet/sec2, miles/hour2)
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Common Physics Units (Motion)
Physical Parameter Units Example (SI Units)
Displacement (x) [L] meter, (m)
Time (t) [T] second, (s)
Mass (m) [M] kilogram, (kg)
Speed (v) [L]/ [T] meter per second, (m/s)
Acceleration (a) [L]/ [T]2 meter/sec2, m/s2
Force (F) [M] [L]/ [T]2 newton, kg m/s2
Energy (E) [M] [L]2/ [T]2 joule, kg m2/s2
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Check Dimensions: Example
Check for consistency of dimensions the eq. v = at2
The units of v are [L]/[T], the units of a are [L]/[T]2 , and the
units of time are [T]
We ask the question do the dimensions on the right = thedimensions on the left?
The dimension on the rhs of the equation are:
[L]/[T]2x [T]2= [L], (length)
The dimension of velocity are length/time, [L]/[T], so the two
dont match and the equation is invalid
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Terminology of a right triangle
90
h = hypotenuse
h0 = length of sideopposite the angle
ha = length of side
adjacent to the angle
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Trigonometric Relationships
90
h h0= h sin
ha= h cos
h2= h02 + ha
2; sin = h0/h, cos = ha/h, tan = h0/ha
= sin-1 (h0/h) = cos-1(ha/h) = tan
-1(h0/ha)
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Additional Info on Trig/Geom.
hh0
ha
90 + + = 180
+ = 90 (angles are complementary)
= 90 -
h cos = h sin = ha
h sin = h cos = h0
tan = h0/ha
tan = ha/h0
h
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Scalars and Vectors
A scalar quantity is one that can be described with a singlenumber specifying the size or magnitude of a measurement
The mass of a rock is 30 kg
The height of a tree is 20 m
A vector quantity is one where both the magnitude and the
direction are essential characteristics
The acceleration of gravity is 9.80 m/sec2directed toward the center
of the earth
A boat is headed due north at 1 meter/second (velocity)
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Vector Addition (two vectors @ 90)
Start
N
A
B
R
R = A + B, R ={(275m)2 + (125m)2}1/2= 302m is the magnitude of the
vector (Pythagorean Theorem).
We know that tan = 125m/275m =125/275= 0.454, so = tan-1(0.454) =
24.4, so that the vector points 24.4 north of due east.
275 m due E
125 mdue N
90
Travel due east for 275 m (A) and travel due north for 125 m (B). What isthe resultant vector?
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Any Vector can be resolved into
components related to a co-ordinate system
X
Y
V
Vx
VY
V = VX+ VY= 1X VX+ 1Y VYwhere 1Xand 1Yare unit vectors along the X and Yaxes respectively and VXand VYare the magnitudes of the respective vectors.
VX= V cos , VY= V sin , where V is the magnitude of the vector. This gives us a
method of adding together many vectors, by resolving them into the X and Y
components and simply adding the magnitudes of the X and Y components andaligning them along the X-axis and Y-axis respectively
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Add multiple vectors together by adding their
x and y components (Ref. X-Y Origin)
Y
X
V1V2
V1XV2X
V2YV1Y
V2XV1X+
V2Y V1Y+ V1+ V2
Resultant Vector Magnitude = {(V1x+ V2x)2+ (V1Y+ V2Y)
2}1/2,
Resultant Vector Direction = tan-1{(V1Y+ V2Y)/(V1X+ V2X)}
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Add multiple vectors together by adding their x
and y components (Head to Tail)
Y
X
V1
V2
V1XV2X
V2Y
V1Y
V2XV1X+
V2Y V1Y+ V1+ V2
Resultant Vector Magnitude = {(V1x+ V2x)2+ (V1Y+ V2Y)
2}1/2,
Resultant Vector Direction = tan-1{(V1Y+ V2Y)/(V1X+ V2X)}
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Positive Vector and its Negative
V
- V
-VX= -V cos
VX= V cos
VY= V sin
- VY= - V sin
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Problem 1-9
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Determine the number of milliliters in one ounce given:1 gallon = 128 ounce; 1 gallon = 3.785 x 10-3 m3 and
1mL = 10-6m3
From the first equation: 1 ounce = 1 gallon/128;
From the second equation: 1 gallon = 3.785 x 10-3
m3
So, 1 ounce = 3.785 x 10-3 m3 /128 = 2.96 x 10-5m3
In order to get this into mL, from the third equation we have 1 mL = 10-6 m3or
1m3= 106mL
So, 1 ounce = 2.96 x 10-5m3 = 2.96 x 10-5 m3x 106mL/m3=
2.96 x 101 mL = 29.6 mL
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Problem 1-16
1Line of
Sight
2
85.0 meters
1= 35.0 = line of sight to top of building
2 = 38.0 = line of sight to top of
antenna
What is height of
Antenna onBuilding?
ha
hb
hb= height of buildingha= height of antenna
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Problem 1-16
1Line of
Sight
2
85.0 meters
1= 35.0
2 = 38.0
What is height of
Antenna onBuilding?
ha
hb
Tan (1) = hb/85, hb= 85 tan (35.0)
Tan (2) =(ha+ hb)/85, ha+ hb = 85 tan (38.0)
ha= (ha+ hb)hb= 85 {tan (38.0)tan (35.0)} =
85.0(0.7810.700)
ha = 6.89 meters
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Problem 1-23(a)
N
B = 325 N
A = 445 N
325
R
Rmagnitude = (3252+ 4452)1/2 = 551 N
tan = (325/445) = 0.730
= tan-1 (325/445) = tan-1 (0.730) =
36.1 north of due west
Two forces on a crateOne force is 445 N due west
The other is 325 N due north
What is resultant R?
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P bl 1 23 (b)
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Problem 1-23 (b)
N
B = 325 N due north
A = 445 Ndue west
325
1
R1
R1mag = (3252+ 4452)1/2 = 551 N
With B pointing due north
1= tan-1 (325/445) = tan-1 (0.730) = 36.1north of due west
If B is reversed, it points due south
2= 36.1 degrees south of due west
Direction of resultant vector R2 is changed, but
the magnitude (551N) is the same.
-B = 325 N due south2
R2
325
What if B is reversed?
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P bl 1 36
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Problem 1 - 36
Soccer
Net
A
X
Y
AY
AX
Player #1
Player #2
30.0
N
The direction of AYis due south
The direction of AX is due east
The magnitude of A is 8.6 meters, what are the
magnitudes of AXand AY?
so the magnitude of AX = A sin (30) = 8.6 x 0.5
= 4.3 meters. Its direction is due east
and the magnitude of AY= A cos (30) = 8.6 x
0.866 = 7.4 meters. Its direction is due south.
90
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Problem 147 (Prev. Edition)
35.0
5.0 m
15.0 m
18.0 m
R = A + B + C
N
A football player runs 5.0 m due N, then 15.0 m due E, and then
18.0 m 35south of due east, what is the resultant vector(magnitude and direction)?
A
B
C
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Problem 147 (components of vector A)
5.0 m
N
To solve the problem we will find the x and y components
of the vectors A, B and C individually. We will then add the x-components together for each vector to find the x-component
of the resultant vector R and the y-components of each vector
to find the y-component of the resultant vector R. That is,
Rx= Ax+ Bx+ Cx, Ry= Ay+ By+ Cy , tan = Ry/ Rx
= tan-1 (Ry/ Rx) R = (Rx2+ Ry
2)1/2
Starting with the vector A Football player runs 5.0 m in a
direction due N
The components of A are Axand Ay
The value of Ax= 0 (there is no x component because the
vector is pointing North (+y) direction)
The value of Ay= +5.0 m
A
+Y
+X
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Problem 1 - 47 (Components of Vector B)
15.0 m
N
Continuing with the vector B. Football player runs 15.0 m in a
direction due E. The vector B then has a component in the +xdirection, but none in the +y direction.
Bx= + 15.0 m. and By= 0
B
+Y
+X
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Problem 147 (Components of Vector C)
35.0
18.0 m
N
Continuing with the vector C. Player runs 18.0 m in a direction
35south of due eastCx= + 18.0 m cos (35) Cy= - 18.0 m sin (35)
Cx= + 14.7 m, Cy= - 10.3 m
C
Cx= + 18.0 m cos (35)
Cy=-
18.0ms
in(35)
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