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Analysis of Motion
&
Newton’s Laws
Class Objective
Learn and apply Newton’s First, Second, and Third Laws Unidirectional Multidirectional
Learn the relationship between position, velocity, and acceleration
RAT 10.1
Some Definitions (1D)
Position - location on a straight line
-3 -2 -1 0 1 2 3 4
Displacement - change in location on a straight line
-3 -2 -1 0 1 2 3 4
x
x
x = x2 - x1 = 4 - (-2) = 6
Some Definitions (1D)
Average Velocity - rate of position change with time
TimeElapsed
ntDisplaceme
t
xvave
dt
dx
t
x
tv
0
lim
Instantaneous Velocity
Average and Instantaneous Velocity (1D)
Position
Timet1 t2
x1
x2
Slope = = Average Velocity x2 - x1
t2 - t1
Some Definitions (2D)
Position -- a location usually described by a graphic on a map or by a coordinate system
54321-3 -2 -1
4
3
2
1
00
-1
-2
-3
-4(-2, -3)
(4, 3)
Some Definitions (2D)
Displacement -- change in position,
where
r
4
3
2
1
1 2 3 4
00
-1-1
-2
-2
-3
-3
-4(-2, -3)
(4, 3)
51r
r
Δ
2r
12 rrrΔ
Some Definitions (2D)
Average velocity
Instantaneous velocity
Speed - the magnitude of instantaneous velocity (scalar)
scalarTimeElapsed
vectorntDisplaceme
t
rvave
dt
rd
t
r
tv
0
lim
vspeed
Some Definitions (1D)
Average Acceleration - rate of velocity change with time
t
v
tt
vva
12ave
12
Instantaneous Acceleration
dt
dv
t
v
0t
lima
Average and Instantaneous Acceleration (1D)
Velocity
Timet1 t2
v1
v2
Slope = = Average Acceleration
v2 - v1
t2 - t1
Some Definitions (2D)
Average Acceleration
Instantaneous Acceleration
t
v
tt
vva
12ave
12
dt
vd
t
v
0t
lima
ExampleOne-dimensional motion
Sp
eed
,
mil e
s p
er
h
ou
r
3210
010
20
Time, hours
Paired ExerciseWhat is the distance traveled?
What is the acceleration at 1.25 hours? S
pee
d,
mil e
s p
er
h
ou
r
3210
010
20
Time, hours
For constant acceleration...
if acceleration is constant
integrating both sides
v0 is the original value at the beginning of the time interval
dt
vda
dta vd o
dtavd o
tavv oo
(Definition)
Constant Acceleration
substituting the velocity equation from the previous page
integrating both sides
yields
dt
xdv
dtv xd
dttavxd oo
dttavxd oo
2ooo ta
2
1tvxx
(Definition)
Equations of Motion (Constant Acceleration)
Velocity
Position (in terms of x)
tavv oo
2ooo ta
2
1tvxx
Multiple DirectionsEquations of motion can be written for each direction independently.
Velocity
Position
tavvo0 xxx
tavvoo yyy
2xxo ta
2
1tvxx
oo
2yyo ta
2
1tvyy
oo
Distance, Velocity, and Acceleration
Suppose a dragster has constant acceleration.
If a dragster starts from rest and accelerates to 60 mph in 10 seconds. How far did it travel?
Plot Speed vs time
60 mph
10 secondstime
spee
d
What does the area under the line represent?
(1/6 min)
(1 mi/min)
Distances….
Area = distance?
Sure: Right?
So:
xdxdtdt
dxvdt
mile12
1min
6
1
min
mile1
2
1
x
Your Turn:
RAT 10.2
Momentum
v
m
p = m v
momentum
Newton’s 1st Law:
“Every body persists in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by forces impressed upon it.”
“In the absence of a net force applied to an
object, momentum stays constant.”
Newton
Holtzapple
Newton’s Second Law
The time-rate-of-change of momentum is proportional to the net force on the object.
If mass is constant...
netFdt
vmd
dt
pd
)(
netFamdt
vdm
dt
vmd
)(
Newton’s Third Law
“To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal and directed in contrary parts.”
Newton
Newton’s Third Law
Other statements
Forces always exist by the interaction of two (or more) bodies
The force on one body is equal and opposite to the force on another body
It is impossible to have a single isolated force acting in one direction
The designation of an “action force” and a “reaction force” is arbitrary because there is mutual interaction between the bodies
Newton’s Third Law
A Consequence
The earth and the moon orbit about a common point about 1000 miles below the surface of the earth because the earth pulls on the moon and the moon pulls on the earth.
Example: Newton’s 3rd Law
Consider a rocket with constant exhaust gas velocity:
The mass changes (obviously) as the fuel is burned and the gas is ejected.
fuelve
mv
Positive
Example: Newton’s 3rd Law
The magnitude of the net force acting on the rocket can be determined by observing its acceleration
where m is the instantaneous mass of the rocket and dv is the instantaneous change in rocket velocity.
netFdt
vdmam
Example: Newton’s 3rd Law
The magnitude of the net force acting on the ejected gas is
where ve is the velocity of ejected gas and dm/dt is the rate mass is ejected from the rocket.(Note: The origin of this equation will become more clear when we do Accounting for Momentum.)
dt
dmvF enet
Example: Newton’s 3rd Law
From Newton’s 3rd Law, these two forces must be opposite and equal to each other, so:
or,
mdv
dtv
dm
dte
mdv v dme
Example: 3rd Law
Using calculus, this can be solved to yield:
where m0 is the initial mass of rocket including fuel
m m e v ve 0
/
Why Newton’s Laws?
Engineers use models to predict things such as motion, fluid flow, lift on an airplane wing, movement of neutrons in a nuclear reactor, deflection of beams or columns, etc.
Newton’s laws are widely used and a good first example of engineering models.
More on Models
Question: If I toss a piece of chalk at a sleeping student, does its path follow a parabola?
Answer: Not exactly, because air resistance affects the motion. Also, we should consider the effect of the spinning earth as it moves around the sun in an ellipse. However, for most practical work, a parabola is close enough.
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