Download Foundation of Newtonian Mechanics

The Foundation of Newtonian Mechanics
Abstract
Due to the large number of equations involved in AP Physics 1, I (DCW) was curious to
know what equations and key concepts were the absolute minimum starting information
that the students need in order to have a chance to be successful in this subject. For
Newtonian mechanics only, I went through the key concepts and derivations, and
concluded that there is a very small set of starting information that the students build
upon. This information consists of the four kinematic equations and Newton’s three
laws. Because this information is so fundamental to student success, I recommend that
students be pop quizzed frequently on this subject matter, to the extent that they
memorize this information.
Absolutely KEY definitions and concepts
A proper understanding of Newtonian Mechanics absolutely requires a minimal amount
of “working memory” (i.e., memorization) and understanding in the following areas:
 Definition of delta and the recognition of the appropriate symbol for delta
 Mathematical definition of average
 Definition of velocity
 Definition of acceleration
 A thorough understanding of the standard nomenclature
 A thorough understanding of the role of units and dimensional analysis
 A thorough understanding of the need for subscripts, and what those subscripts
stand for
 A thorough understanding of algebra, separation of an unknown variable, etc.
 A thorough understanding of the four kinematic equations
 A thorough understanding of Newton’s laws
CHAPTER 2
Displacement
Displacement is defined as final position – initial position. In equation form, this
becomes
x  x f  xi
(1)
Velocity
Velocity is defined as displacement per unit time, or
x
v
t
Acceleration
Acceleration is defined as change in velocity per unit time, or
v
a
t
(2)
(3)
Kinematics
1st Kinematic Equation
Equation (2) can be solved for Δx.
x  vt
Since this displacement occurs for the whole time interval Δt, any changes in velocity
that caused the resultant displacement are averaged out over the whole time interval.
This leads to the following brief derivation:
x  vavg t
(4)
If acceleration for time interval Δt is constant, the average velocity in equation (4) is
defined by its mathematical equivalent of
1
vavg  ( vi  v f )
(5)
2
A substitution of equation (5) into equation (4) leads to the first kinematic equation.
1
x  (vi  vf )t
(6)
2
2nd Kinematic Equation
Equation (3) can easily be solved for change in velocity, again assuming a constant
acceleration:
 v  a t
(7)
The delta-v term may be expanded, and equation (7) can be easily solved for final
velocity when initial velocity, acceleration, and time interval are known.
v f  v i  a t
v f  v i  a t
(8)
3rd Kinematic Equation
Under some conditions, the final velocity is unknown, so equation (6) cannot be solved
without additional work. Inspection of equation (8) indicates that it can be substituted
into equation (6) to eliminate the need to know final velocity, resulting in the following:
1
x  ( vi  ( vi  at )) t
2
Collection of like terms inside the parentheses yields:
1
x  ( 2vi  at ) t
2
Multiplying everything in parentheses by ½ yields:
1
1
x  ( ( 2vi )  ( at )) t
2
2
Simplifying the above equation leads to the 3rd kinematic equation.
1
x  vit  a(t ) 2
2
(9)
4th Kinematic Equation
Under some conditions, the time interval is unknown, so equation (8) cannot be solved
without additional work. Inspection of equation (6) indicates that it can be solved for
delta-t, and the result can be substituted into equation (8), resulting in the following:
From equation (6)
2x  ( vi  vf )t
2x
 t
vi  v f
(10)
Equation (10) can now be substituted into equation (8)
 2 x 

v f  vi  a 
v  v 
f 
 i
 2x
v f  vi  a 
v  v
f
 i
 2 a x
v f  vi  
v  v
i
 f
v
f








 vi v f  vi   2ax
v 2f  vi2  2ax
v 2f  vi2  2ax
(11)
Note that only the 1st and 2nd kinematic equations are mathematically independent. The
other two kinematic equations must be derived via substitutions from the first two
kinematic equations.
For motion close to the earth’s surface, it is seen that vertical motion is accelerated while
horizontal motion is constant. Due to this, the kinematic equations can be applied
independently in the horizontal and vertical directions, leading to the following additional
topics:


One dimensional motion
Projectile motion
o The range equation (no change in elevation)
o Changes in elevation
o Path of a projectile
Various projectile motion problem variable specifications are important for the fact that
students must apply much of the algebra that they have previously learned in order to
work those problems.
CHAPTER 4
Newton’s Laws
1st Law
Law of inertia – objects in motion stay in motion in a straight line until a net force acts on
them.
2nd Law
The acceleration of an object is equal to the net force on it divided by its mass. In
equation form,
(12)
 F  ma
Newton originally wrote this law differently. A slightly different derivation of Newton’s
2nd law, recognizing that acceleration is the change in velocity per unit of time, leads to
the following:
v
F m
t
Expanding the delta-v term in the numerator leads to
m( v f  vi )
F
t
F
mv f  mvi
t
The product of mass and velocity is momentum, denoted by the symbol “p”. Since the
final momentum minus the initial momentum is the change in momentum,
p
F
(13)
t
Equation (13) leads to the following, which is the impulse-momentum theorem.
Ft  p
(14)
Note that momentum will be discussed and developed in following work. Also note that
Newton’s 2nd law is seen in very many different contexts in physics.
3rd Law
For every action, there is an equal and opposite reaction.
Combinations
Force x displacement
The product of force and displacement, where the force is applied parallel to the
displacement, is known as work. Work has units of Joules, and is an energy unit.
CHAPTER 5
An object that is dropped from rest, accelerates under the influence of gravity, and gains
velocity as it loses height. This means that gravity is applying a force to that object, and
since the object’s displacement is parallel to the force of gravity, gravity is doing work on
the object. This leads to the following:
From equation (11)
v 2f  vi2  2 gy ,
where “g” is the acceleration due to gravity, delta-y indicates displacement in the vertical
direction, and “down” is taken as the positive direction. Since the initial velocity is zero
(the object was dropped from rest),
v 2f  2 gy
v 2f
2
 g y
Multiplication of both sides of the above equation by the object’s mass yields
mv 2f
 mgy
2
1
mv 2f  mgy
2
(15)
The left hand side of equation (15) is known as kinetic energy (energy of motion), while
the right hand side of equation (15) is known as gravitational potential energy. Equation
(15) states that energy is conserved, and it says that as a falling object loses gravitational
potential energy, it gains an equal amount of kinetic energy.
The net work done on an object will accelerate it, based on Newton’s 2nd law. This leads
to the following derivation:
W  Fx
Substitution of mass times acceleration for F yields
W  max
(16)
Another substitution from equation (11) (the 4th kinematic equation) leads to the
following:
v 2f  vi2  2ax
v 2f  vi2  2ax
v 2f  vi2
2
 a x
Substitution into equation (16) yields
 v 2f  vi2 

W  m
 2 


 mv 2f  mvi2 

W 


2


W 
1
1
mv 2f  mvi2
2
2
W  KE
(17)
Equation (17) is known as the work/kinetic-energy theorem.
Force x velocity
The product of force and velocity leads to the following:
x Fx W
Fv  F


P
t
t
t
(18)
The product of force and displacement is work. Work per unit time is power.
CHAPTER 6
Conservation of momentum
Equation (14), the impulse momentum theorem, states that the product of force and the
time interval over which that force was applied, causes a change in momentum for the
object that the force was applied to. When two objects interact, Newton’s 3rd law states
that the forces between them are equal in magnitude and opposite in direction. This leads
to the conclusion that the change in momentum that one object causes to a second object
is equal and opposite to the change in momentum that the first object experiences. Thus,
due to Newton’s 2nd and 3rd laws, momentum is conserved.
Conclusion 1
Based on the definitions of displacement, velocity and acceleration, kinematic equations
1 and 2 are derived. Substitutions from each of these equations into the other equation
leads to kinematic equations 3 and 4. The combination of the four kinematic equations
and Newton’s 3 laws leads to development of the following subject areas in Newtonian
mechanics:
 One and two dimensional motion
 The impulse momentum theorem
 The definition of kinetic energy
 Conservation of energy
 Work and the work/KE theorem
 The definition of power
 Conservation of momentum
These areas cover chapters 2-6 in the Holt Physics textbook (pre-AP physics, 203 pages
of material). Thus, it is VITALLY important that physics students become intimately
familiar with the kinematic equations and Newton’s laws.
CHAPTER 7
Rotation
There is a rotational equivalent of displacement, velocity, and acceleration. When these
equivalent variables are substituted into the kinematic equations, the rotational kinematic
equations result, as shown on the next page.
Equivalent variables: x and θ; v and ω; a and ά
Kinematics
x
v
t
a
v
t
Rotational Kinematics


t


t
(20)
1
(vi  vf )t
2
1
x  vit  a(t ) 2
2
 
vf  vi  at
f  i  t
(23)
vf 2  vi 2  2ax
f 2  i 2  2
(25)
x 
1
(i  f )t
2
1
  it   (t ) 2
2
(19)
(21)
(22)
Memorization of the rotational kinematic equations is made MUCH easier if students are
thoroughly familiar with the linear kinematic equations.
In addition to the rotational kinematic equations, there are rotational equivalents for the
variables associated with Newton’s laws:
 Torque is equivalent to force, and is equal to the product of force and lever-arm
distance, where the force must be perpendicular to the lever arm.
 Moment of inertia, which is the rotational equivalent of mass, can be derived
directly from Newton’s 2nd law
F  ma
Fr  mar
a  r
Fr  mr r
Fr    mr 2 
  I
Thus, it is readily seen that moment of inertia, or I, is equal to mr2.
A mathematical treatment of rotational variables in a manner exactly analogous to what
was done above leads to development of rotational kinetic energy, rotational work,
rotational power, and conservation of angular momentum.
CHAPTER 8
Simple Harmonic Motion
For a round object of radius R that is rotating at a constant angular velocity, one
occasionally wants to know the position of a particular point on the object. Assuming
that the horizontal axis is denoted as zero radians, it is obvious that the Cartesian
coordinates of any given point are
x  R cos 
(26)
y  R sin 
(27)
Since the object is rotating at a constant rate, ωi equals ωf, which equals ω. When ω is
substituted into the rotational kinematic equation that describes this situation (e.g. eq
(21)), the result is
1
  (   ) t
2
   t
Since θi is equal to zero radians, Δθ is equal to θ. Also, the data for any related physics
experiment normally starts at time zero, so Δt is equal to t. Finally, when the given
rotating point gets to the vertical axis, it is as far from the horizontal axis as it is going to
get. For the purpose of simple harmonic motion, this is known as its amplitude, denoted
by the symbol “A”. Thus, substitutions into equations (26) and (27) lead to the following
simple harmonic motion equations:
x  A cos  t
y  A sin  t
Conclusion 2
Rotational relationships, while seemingly complex and abstract, can be developed in a
manner that is functionally equivalent to the development of the linear equations outlined
above. In effect, the rotational equations have an exact duplicate in the functional form
of the linear equations, and the linear equations should be treated as a spring-board into
the rotational equations. Thus, the linear kinematic equations and Newton’s laws in the
linear world, are also vitally important for learning about rotation. Such a connection
between linear and rotational equations leads to the following subject areas:
 Rotational motion
 Rotational kinetic energy
 Rotational work
 Rotational power
 Conservation of angular momentum
 Simple harmonic motion
These areas cover chapters 7-8 in the Holt Physics textbook (73 pages of material). Thus,
it is once again VITALLY important that physics students become intimately familiar
with the kinematic equations and Newton’s laws, as these four equations and three laws
are the key material behind almost 300 pages of the Holt Physics textbook.
Physics Binder
The physics binder should have several dividers in it, as shown immediately below.





Standard nomenclature, with subscripts, superscripts, and units
Definitions/Laws
o Delta
o Average
o Displacement
o Velocity
o Acceleration
o Vector resolution
o Vector addition
o Newton’s 3 laws
o Rotational equivalent of linear variables
Derivations and equations
o 4 kinematic equations
o Range equation
o Definition of kinetic energy
o Impulse/momentum theorem
o Force x displacement and work
o Work/KE theorem
o Force x velocity and power
o 4 rotational kinematic equations
o Torque
o Rotational kinetic energy, work, and power
o Simple harmonic motion
Key concepts
o 1 D motion
o Projectile motion
o Conservation of energy
o Conservation of momentum
o Circular motion
o Moment of inertia
o Conservation of angular momentum
o Relationship between circular motion and simple harmonic motion
Standard physics models
Kinematical Models
Constant velocity
Constant acceleration
Simple harmonic oscillator
Uniform circular motion
Collision
Causal Models
Free particle: net force = 0
Constant force: net force = constant
Linear binding force: net force = -kΔx
“Center seeking force”, and force = constant
Impulsive force