Transcript HNRS 227 Lecture #2 Chapters 2 and 3

HNRS 227 Lecture #2
Chapter 2
Motion
presented by Prof. Geller
Recall from Chapter 1
Units of length, mass and time
Metric Prefixes
Density and its units
The Scientific Method
Main Concepts - Chapter 2
Speed vs. Velocity
Acceleration
Force
Falling Objects
Newton’s Laws of Motion
Momentum
Circular Motion
Universal Law of Gravity
Speed and Velocity
Speed
distance traveled in a unit of time
a scalar quantity
Velocity
speed and direction
a vector quantity
Questions for Thought
What is the difference between speed and
velocity?
Speed is a scalar, and velocity is a vector. Speed is the
magnitude of the velocity vector.
Questions for Thought
What is acceleration?
Acceleration is the ratio of the change in velocity per
change in time. Acceleration can also be viewed as
the derivative (remember calculus?) of the velocity.
Questions for Thought
An insect inside a bus flies from the back
toward the front at 5.0 miles/hour. The bus is
moving in a straight line at 50 miles/hour.
What is the speed of the insect?
The speed of the insect relative to the ground is the
50.0 mi/hr of the bus plus the 5.0 mi/hr of the
insect relative to the bus for a total of 55 mi/hr.
Relative to the bus alone the speed of the insect is 5.0
mi/hr.
Force
Definition of force
something that causes a change in the
motion of an object
a push or pull
an electric, magnetic, gravitational effect
a vector quantity
Net force - Resultant Force
Inertia
Defining Inertia
tendency of an object to remain in its current
state of motion
the more massive the more inertia
think of stopping a car vs. truck
Acceleration Due to Gravity
Direction of acceleration due to gravity
directed to center of Earth
Think: scalar or vector?
a vector quantity
Why?
has magnitude and direction
Generalized Motion
Motion can be viewed as a combination of
movements
vertical component
typically gravitational acceleration
horizontal component
some force from muscle, gunpowder, etc.
Question for Thought
What happens to the velocity and acceleration
of an object in free fall?
Assuming a short free fall distance near the
earth’s surface, the velocity increases
downward as the acceleration remains
constant.
Question for Thought
In the equation d=1/2*a*t2, if a is 9.8 meters
per second per second and t is in seconds,
what is the unit of d?
Meters.
Question for Thought
What is inertia?
Inertia is the property of matter that an object
will remain in unchanging motion or at rest in
the absence of an unbalanced force.
Question for Thought
Where does the unit s2 (or concept of “square
second”) come from?
 Acceleration is change in velocity per change in time, with
units of (m/s)/s. When the fraction is simplified, you get
meters per second squared. The “seconds squared” indicates
that something that changes in time is changing in time, that is,
the ratio of change in distance per unit of time is changing in
time.
Question for Thought
Neglecting air resistance, what are the forces
acting on a bullet that has left the barrel of a
rifle?
After it leaves the rifle barrel, the force of
gravity acting straight down is the only force
acting on the bullet.
Question for Thought
How does the force of gravity on a ball change
as a ball is thrown straight up in the air?
The force of gravity on the ball remains constant
because the force of gravity is independent of the
motion of the object near the surface of the earth.
Sample Question
An object falls from a bridge and hits the
water 2.5 seconds later.
A) With what velocity did it strike the water?
B) What is the average velocity during the
fall?
C) How high is the bridge?
Sample Answer
23. These three questions are easily answered by using the three sets of relationships, or
equations, that were presented in this chapter:
(a)
vf
vf
 at  vi
m
 m
 9.8 2 

2.50
s


0
s 
s
m
 9.8  2.50 2  s
s
 25 m / s
(b)
v

(c)
v

d
vf  vi
2

25 m / s  0
2
d
 d  vt
t
 m
 13 2.50 s
 s 
 13  2.50
 33 m
m
s
s
 13
m
s
Newton’s Laws of Motion
Newton’s First Law of Motion
body at rest tends to stay at rest and body in
uniform motion will stay in straight line
uniform motion unless acted upon by an
outside force
Newton’s Second Law of Motion
the acceleration of a body is proportional to
the force being applied
•F = m*a
Newton’s Laws of Motion
Newton’s Third Law of Motion
for every force there is an equal and
opposite force (action and reaction)
Question for Thought
How can there ever be an unbalanced
force on an object if every action has an
equal and opposite reaction?
The action and reaction forces are between two
objects that are interacting. An unbalanced
force occurs on a single object as the result of
one or more interactions with other objects.
Question B 4 (page 50)
What force would an asphalt road have to
be to give a 6,000 kilogram truck in order
to accelerate it at 2.2 meters per second
per second?
Question B 4 (page 50)
2.
F = ma
m
4
F  6,000 kg  2.2 2  1.3  10 N
s
Momentum
By definition momentum is the product of

mass and velocity
Conservation of momentum
total momentum of a closed
system remains constant
Question for Thought
Is it possible for a small car to have the
same momentum as a large truck?
Explain.
 Yes, the small car would have to be moving with a much
higher velocity, but it can have the same momentum
since momentum is mass times velocity.
Question B 9 (p. 50)
A 30.0 kilogram shell is fired from a 2,000
kilogram cannon with a velocity of 500
meters per second. What will be the
velocity of the cannon?
15. Listing the known and unknown quantities:
Shell m = 30.0 kg
Shell v = 500 m/s
Cannon m = 2,000 kg
Cannon v = ? m/s
This is a conservation of momentum question, where the shell and cannon can be
considered as a system of interacting objects:
Shell momentum
B9
(p. 50)
= - Cannon momentum
mvs = - mvc
mvs - mvc = 0

30.0 kg500
m 
 2,000 kgvc  0
s 
m 

15,000
kg

 2,000 kg vc   0

s 
m 

15,000 kg s  2,000 kg vc 
m
15,000 kg 
s
vc 
2,000 kg

15,000 kg
1
m


2,000 1
kg s
 7.5
m
s
The cannon recoils with a velocity of 7.5 m/s.
Circular Motion
Centripetal Force
usually gravity, but can be any force (such as
the pull of a string) that forces an object into
circular motion
Centrifugal Force
a fictitious force, an apparent outward force
felt by an object in circular motion
Acceleration in circular motion is a = v2 / r
therefore in circular motion F = mv2 / r
Sample Question
What is the maximum speed at which a
1000 kilogram car can move around a
curve with a radius of 30.0 meters if the
tires provide a maximum frictional force of
2700 Newtons? (1 Newton is a unit of
force needed to accelerate a 1 kilogram
mass 1 meter per second per second.)
28.
Sample Question

v2
v 2 

F  ma and a 
 F  m
 v 


r
r
v 

rF
m
30.0m   2700.0 N
 9.00
1,000.0 kg
m
km
or 32.4
s
hr
rF
m
Universal Law of Gravity
Newton’s Universal Law of Gravitational
Attraction
every object is attracted to every other
object
the force is proportional to masses and
inversely proportional to the distance squared
–F = (G*m*M) / r2
A Problem of Very Little
Attraction
What is the gravitational force between
two 100 kilogram people separated by 1
meter?
Really just a little
Use the formula for the force of gravity:
F = G*m*M / r2
where G = 6.67 x 10-11 Nm2/kg2
and m = 100 kg
and M = 100 kg
and r = 1
=======================
Then
F = 6.67 x 10-11 Nm2/kg2 *100 kg*100 kg / (1 meter) 2
F = 6.67 x 10-7 N