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Transcript
GOALS
1. Distinguish between instantaneous and average speeds.
2. Use the formula v = d/t to solve problems that involve distance, time, and speed.
3. Distinguish between scalar and vector quantities and give several examples of each.
4. Use the Pythagorean theorem to add two vector quantities of the same kind that act as right
angles to each other.
5. Define acceleration and find the acceleration of an object whose speed is changing.
1 2 6. Use the formula v = v + at to solve problems that involve speed, acceleration, and time.
1 7. Use the formula d = v t + ½ at to solve problems that involve distance, time, speed, and 2
acceleration.
8. Explain what is meant by the acceleration of gravity.
9. Describe the effect of air resistance on falling objects.
10. Separate the velocity of an object into vertical and horizontal components in order to
determine its motion.
11. Define force and indicate its relationship to the first law of motion.
12. Discuss the significance of the second law of motion, F = ma.
13. Distinguish between mass and weight and find the weight of an object of given mass.
14. Use the third law of motion to relate action and reaction forces.
15. Explain the significance of centripetal force in motion along a curved path.
16. Relate the centripetal force on an object moving in a circle to its mass, speed, and the radius
of the circle.
17. State Newton's law of gravity and describe how gravitational forces vary with distance.
18. Account for the ability of a satellite to orbit the earth without either falling to the ground or
flying off into space.
1
19. Define escape speed.
Motion



Everything in the universe is in nonstop
motion
The laws of motion that govern the behavior
of atoms and stars apply just as well to the
objects of everyday life
Motion is the ability of an object to move
from one place to another

The first step in analyzing motion is to describe how
fast or slow the motion is and establish a reference
frame
2
What is a Day?
Earth
Sun
• Time from 1 to 2 is a sidereal day ( depends only on Earths rotation rate )
• Time from 1 to 3 is a solar day Varies from 21 sec early to 29 sec late depending
on Earth’s orbital speed which varies
•Currently the length of a mean solar day is about 86,400.002 seconds.
24.0000006 hrs = Mean Solar day
•The sidereal day is 23 h 56 m 4.1 s,
Motion

SPEED


Rate at which
something covers
distance
Speed is distance
divided by time
 v = speed
 d = distance
 t = time
V is the average speed over the time t
d
v
t
4
Speed Example 2.1

A car going 40
miles/hour in a 6 hour
period travels how far?

v=


t=


40 mi/hr
6 hours
d=vxt
240 miles
This is an average speed.
The speedometer of a car
shows an instantaneous
speed.
5
Vector Quantities

Vectors have direction as well as magnitude



They tell us “which way” as well as “how much”
velocity (v) and force (F) are examples
An arrowed line that represents the magnitude
and direction of a quantity is called a vector
F
6
Vector Quantities have
Direction And Magnitude
Vector v is 40 km/s to the right
Tam2s6_3
7
Figure 2-4
Vectors can be Added
Tam2s6_4
Figure 2-5
8
Adding Vectors


Besides manually
measuring the length of
“C” as in the previous
slide, vectors can be
added using the
Pythagorean theorem
when you have a right
angle triangle
More accurate than
using a scale drawing
2
a
+
2
b
=
2
c
9
Using the Pythagorean theorem

Example 2.3 – use the
Pythagorean theorem
to find the
displacement of a car
that goes north for 5
km and then east for 3
km, as in Fig. 2-5
c  a b
2
2
C  5km  3km  5.83km
2
2
10
Scalar Quantities

Need only a number and a unit; Not direction





mass (m)
speed (v)
Pressure ( pascals)
Kinetic Energy (joules)
Density ( kg/ vol )
11
Vector and Scalar Quantities
Vector quantities are printed in boldface type
Scalar quantities are printed in italic type.
12
Acceleration

A change in velocity or direction (Curved Path)




Acceleration, in general, is a vector quantity
For our purposes, we will be calculating straight
line motion, where acceleration is the rate of
change of speed (a scalar quantity)
SI units are m/s2
It is customary to write (m/s)/s as just m/s2 since
m/s
m
m

 2
s
( s )(s ) s
13
Acceleration
v2  v1
a
t
change in speed
Acceleration 
time interval
a=acceleration
v2=final speed
v1=initial speed
t=time
14
Acceleration Example 2.4
The speed of a car changes from 15 m/s to 25 m/s in 20 s when its gas
pedal is pressed hard. Find its acceleration.
Figure 2-8
Tam2s6_6
15
Acceleration Example 2.4
vi 15 m / s
v f  25 m / s
t  20 s
25 m / s 15 m / s
a
20 s
v2  v1
a
t
change in speed
Acceleration 
time interval
= 0.5m/s2
16
Free Fall

Galileo Galilei
(1564-1642) –
Italian physicist
who discovered
that the higher a
stone is when
dropped, the
greater its speed
when it reached the
ground
17
Free Fall



This means the stone is
accelerated
Acceleration is the same for all
stones, big or small
Galileo’s experiments showed
that if there were NO AIR for
them to push their way through,
all falling objects near the
Earth’s surface would have the
same acceleration of 9.8 m/s2
18
Free Fall

In general, the speed of an
object falling from rest
can be calculated as

Vdownward = g t
Note that speed if free fall is linear proportional to
time of fall. Meaning speed after 4 seconds is twice
that is is at 2 seconds and four times what it is at 1
second



v=downward speed
g=acceleration due to
gravity
t=time
19
How Far does a Falling Object Fall?

To find out
how far, h,
the object
has fallen in
the time, t,
use

h=½ g t 2
t2 factor means the “h” increases much faster
than the object’s speed “v”
20
Distance Traveled when Accelerating



When starting from rest it is D= ½ at2
If tossed vertically h = ½ gt2
When starting with an Initial Velocity v1

D = V1 t + ½ at2

If falling due to gravity, a becomes g ( 9.8m/s2 )
21
Example 2.7

A stone dropped from a
bridge strikes the water
2.2 s later. How high
is the bridge above the
water?
1 2
h  gt
2
1
(9.8m / s 2 )(2.2s ) 2  24m
2
22
Trajectory in two dimensions
B
V
VY
H max
ɵ Vx
X
A
C
range
V = 1000 m /s
ɵ = 45 degrees
Statement of the problem:
Given the initial Velocity and its angle ɵ
Determine H max , Time to Impact, and total
distance X ( ie, range)
Trajectory in two dimensions
B
V
VY
H max
ɵ Vx
X
A
C
range
The initial Velocity V can be expressed as a
vector having
a vertical component VY and a horizontal
component Vx
ɵ
The relative values of VY and Vx
the angle ɵ
depend on
Trajectory in two dimensions
(1)
The time to reach H max will equal VY / g
= t tot / 2
B
V
Parabolic Path
Vy
H max
Vx
A
X
Vx is constant along the range
(2)
t tot = 2 ( Vy / g)
(3)
Total Range is Vx x t tot
(4)
H max = 1/2 g (t tot / 2)2
C
Air Resistance




Keeps falling things from developing the full g
In air, a stone falls faster than a feather because air
resistance affects the stone less
In a vacuum, the stone and feather fall with the same
acceleration (9.8m/s2)
Air resistance increases with speed squared until the
object cannot go any faster, when drag = weight.

It then continues to fall at a constant terminal speed that
depends upon the size, shape, and weight
26
Free Fall at terminal Velocity

If a sky diver is falling at a constant velocity
i.e, her terminal velocity, and weighs 400N,

What is the force exerted by the AIR on the
skydiver?
If the sky diver rolls into a ball, would the
force be different? Would terminal velocity be
different?

27
Vacuum vs. Air Resistance
tam2s6_12
Figure 2-15
28
Force and Motion




What makes something at rest begin to move?
Why do some objects move faster than
others?
Why are some accelerated and others not?
Questions like these led Isaac Newton to
formulate three principles that summarize
behavior of moving bodies
29
Force and Motion

These are known as
Newton’s Three
Laws of Motion
30
Force and Motion

Newton’s First Law of Motion



If no net force acts on it, an object at
rest remains at rest and an object in
motion remains in motion at constant
velocity (that is, at constant speed in
a straight line)
An object at rest never begins to
move all by itself- a force is needed
to start it off
If an object is moving, the object will
continue going at constant velocity
unless a force acts to slow it down,
speed it up, or change its direction
Only a Net Force Can Accelerate
an Object
31
Force, Inertia, Mass

Force is defined as any influence that can change the
speed or direction of motion of an object



SI unit of force is the Newton (N)
Inertia is the reluctance of an object to change its
state of rest or of uniform motion in a straight line
Mass is the property of matter that shows itself as
inertia


The more mass something has, the greater its inertia and
the more matter it contains
SI unit of mass is the kilogram (kg)
32
Inertia
Figure 2-19
Tam2s6_15
33
Force and Motion

Newton’s Second
Law of Motion


Force is equal to the
product of mass and
acceleration of an
object
The direction of the
force is the same as
that of the acceleration
F  ma
F = Force (SI unit is the Newton which
is a (kg)(m / s2)
m=mass (SI unit is the kilogram (kg))
a=acceleration (SI unit m/s2)
34
Newton’s Second Law of Motion
Twice the force means
twice the acceleration
When the same force acts
On different masses, the
Greater mass receives the
Smaller acceleration
Tam2s6_17
Figure 2-22
35
Mass and Weight

Weight



Force with which something is attracted by the
earth’s gravitational pull.
Weight is a force; if you weigh 150 lbs, the Earth
is pulling you down with a FORCE of 150 lbs
Although weight and mass are very closely
related, weight is different from mass

Mass refers to how much matter something contains
36
Calculating Weight



Newton’s second law of motion says F=ma
In the case of an object at the Earth’s surface, the
force gravity exerts on it is its own weight
So, substituting w for F and g for a in the F=ma
equation leads to:

w = mg




weight = (mass) (acceleration of gravity)
w = weight
m = mass
g = acceleration of gravity
37
Weight

Weight and mass are always proportional to
one another ( Weight = Mass x Gravity )

Mass rather than weight is normally specified
in the SI system as kilo grams
SI system unit for weight is the Newton, since
weight is a force

38
Weight

The weight of 5 kg of potatoes is:


Mass is a more basic property than weight since the
pull of gravity varies from place to place



w=mg = (5kg)(9.8m/s2)=49 (kg)(m/s2)=49 N
Pull of gravity is less on a mountaintop than at sea level,
less at the equator than near the poles (earth bulges at
equator)
Mass remains the same no matter where you are in
the universe
Weight changes depending where you are in the
universe, since g changes
39
Force and Motion

Newton’s Third Law of Motion


When one object exerts a force on a second
object, the second object exerts an equal force in
the opposite direction on the first object
Another way to state this is for every action force
there is an equal and opposite reaction force



This law is responsible for walking-as you move
forward the Earth is moving backward
Recoil of a canon
Action of a rocket
40
Newton’s Third Law of Motion
Figure 2-27
Tam2s6_21
41
Circular Motion

Before we consider how gravity works, we
must look into exactly how curved paths
come about


Planets orbit the sun in curved paths, not straight
lines
A curved path requires an inward pull known
as centripetal force
42
Centripetal Force
The velocity vector v is always tangent to the circle
and the centripetal force vector Fc is always directed
toward the center of the circle
Figure 2-31
Tam2s6_23
43
Centripetal Force

Centripetal force
(Fc) is calculated
using the following
equation where:

m = mass
v = velocity

r = radius of circle

mv
Fc 
r
2
44
What Happens to Centripetal Force when
Mass, Velocity, and Radius are changed?
Figure 2-32
45
Tam2s6_24
Centripetal Force Example 2.10


Find the centripetal force
needed by a 1000-kg car
moving at 5 m/s to go
around a curve 30 m in
radius
m=



1000 kg
v=

mv
Fc 
r
2
5 m/s
833 N
r=

30 m
46
Centripetal Force Example
Fc must be balanced
by friction of tires
on ground
Figure 2-33
Tam2s6_25
47
The slaying of the giant





David was told there was a giant needing slain
He selected his 1.0 m sling and his best
smooth flat river rock weighing .35 kg
He knew his accuracy suffered if the
centripetal force on the sling was more than
80 pounds ( made his arm tired )
How fast can he launch the stone at the giant?
What was the kinetic energy of the stone
48
Newton’s Law of Gravity


Newton arrived at the Law of Gravity that
bears his name using Kepler’s 1st and 2nd laws
and Galileo’s work on falling bodies
Every object in the universe attracts every
other object with a force proportional to both
of their masses and inversely proportional to
the square of the distance between them
49
Newton’s Law of Gravity
Gm1m2
Gravitatio nal Force - F 
2
R
•R – distance between the objects
•G – Gravitational Constant of nature, the same
number everywhere in the universe =
6.670 X 10-11 N m2/kg
m - mass of objects
50
Newton’s Law of Gravity
The inverse square – 1/R2 – means the gravitational
force drops off rapidly with increasing R
Figure 2-35
Tam2s6_26
51
Circular Orbital Velocity





All objects in orbit have a perfect balance of
centripetal force and Gravitational Force
Fc = mv2/ r Centripetal Force
F = GMm/ r2 Gravitational Force
Solving for v
v= (GM/r)0.5
52
Weight and Distance Relationship
Weight or Force falls off with Inverse Square of Distance
Surface
2x distance
1/4 Weight
3x distance
1/9 weight
Figure 2-37
100x distance
1/10,000 weight
The End
54
EXAMPLES
Ex. 2.1
Ex. 2.2
Ex. 2.3
Ex. 2.4
Ex. 2.5
Ex. 2.6
Ex. 2.7
Ex. 2.8
Ex. 2.9
Ex. 2.10
Ex. 2.11
Ex. 2.12
Ex. 2.13
Ex. 2.14
Ex. 2.15