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Chapter 3: The Causes of Motion: Dynamics
Chapter 3 Goals:
• To introduce the concept of inertia and its relationship
to changes in motion
• To show how the net force is the physical cause of
changes in motion
• To present and understand Newton’s First and Second
laws of motion
• To introduce the idea of momentum
•As important examples: to discuss uniformly
accelerated motion, and simple harmonic motion
• To introduce the ideas of work and energy
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Newton’s First Law
Also called the law of inertia
“An object in motion remains in motion, and an object at
rest remains at rest, unless a net force acts on the object”
Not a quantitatively useful result, but it captures an
incredibly insightful observation about motion
Until Isaac Newton, most people assumed that motion
would cease unless the force continued to act, but this
was because ‘friction’ is ubiquitous
If an observer sees motion that obeys N1 (again, with
no forces present), the observer is said to be in an
‘inertial frame of reference’
Example of non-inertial reference frame: the bed of an
accelerating pickup truck
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Some of the Forces in Nature
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Relationship of Mass to Weight
Weight force W is the gravity force on a body of mass m
Mass m is the amount of matter in a body [m] = kg, slug
W  mg and the direction is toward the center of the Earth
g = 9.81 m/s2, so a 1 kg mass has a weight of 9.8 N
g = 32.2 ft/s2, so a 1 slug mass has a weight of 32.2 lb
• If an object ONLY feels the gravity force, then it is in
free-fall, which means that it is accelerating downward at
g (whether moving up or down!).
•If the object is YOU, you would say you were
‘weightLESS’ but in fact what you lack is a force to
counteract W.
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Weight Force and Tension Force
• Lamp is object
W
• There are two forces acting on
the object:
• Tension force (from chain) is
UP and therefore POSITIVE
• Gravity force (from Earth) is
DOWN and therefore
NEGATIVE
• By N1, if the two forces add to
zero, the lamp is either
motionless or moving at
constant velocity (its
acceleration is zero!)
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Weight Force and Normal Force
N
W
• Monitor is object
• There are two forces acting on
the object:
• Normal force (from desk) is
UP and therefore POSITIVE
• Gravity force (from Earth) is
DOWN and therefore
NEGATIVE
• By N1, if the two forces add to
zero, the monitor is either
motionless or moving at
constant velocity (its
acceleration is zero!)
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Newton’s Second Law
Serves to define the mass of the object
“When a net force acts on a body, the body accelerates in
the direction of the net force. The acceleration is directly
proportional to the force, and inversely proportional to
the body’s mass”
Fnet :
F
Fnet
a
or
Fnet  ma
m
SI: [a] = m/s2 [m] = kg  [F] = kg-m/s2 := N(ewton)
A 1 N force causes a 1 kg mass to accelerate at 1 m/s2
USA: [a] = ft/s2 [m] = slug  [F] = slug-ft/s2 := lb
(‘pound’)
A 1 lb force causes a 1 slug mass to accelerate at 1 ft/s2
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What is a force?
“When a net force acts on a body, the body accelerates in
the direction of the net force. The acceleration is directly
proportional to the force, and inversely proportional to
the body’s mass”
Often you literally feel them: when an object presses on
you, your biology can sense it
Usually: but from the moment you came to life you have
felt gravity, and it acts at all points in your body, so you
are NOT aware of it until it goes away
What is a force? A force is a push or a pull.
Enuf said? NO, but that definition has survived and
works really well. It is a tautology: anything that is a
push or a pull must be a force
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An Alternative Formulation of N2,
and a Mention of Newton’s Third Law
Momentum of a moving object is p := mv
dm
Assume the mass is constant: then
0
dt
dp d
dv
dm
 mv   m  v
 ma
dt dt
dt
dt
dp
F 
Therefore N2 takes the elegant form net dt
The real power of the momentum concept will
emerge once we consider systems of bodies, and
Newton’s Third Law:
“To every action (force) on a body there is an equal
but opposite reaction (force) (on a different body!)”
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Remarks on T and N
• The direction of N is by definition always ‘normal’
to the surface
• If a surface exerts a non-normally directed force on
a body, it is usually treated as some kind of friction—
or GLUE!
• Often, the normal force magically adjusts itself to
suit what the kinematics is doing!
• If a weight is supported by a surface, the normal
force is equal to the weight in magnitude—but ONLY
if the weight is not accelerating!
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Remarks on T and N
• T lives without change in all parts of the string or
cable that is ‘under’ tension
• T can go around corners by the use of pulleys
• If a weight is tied to a string and it hangs, the
tension is the weight—but ONLY if the weight is
not accelerating!
• Often, the tension magically adjusts itself to suit
what the kinematics is doing!
Example: a 2 kg hockey skate is accelerating sideways
at 4 m/s2 on an ice rink because a child is pulling on its
lace. Find the tension.
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Some Scenarios Combining T, N and W
Let the weight W of the body be 12 N (so what is mass?)
At the beginning, the body is on a horizontal surface
Tension T is provided by a hand pulling up on the rope
 The body may or may not be in equilibrium (that is, a
may or may not be zero), and there may or may not be a
third force N
T=10 N
T=12 N
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T=15 N
The Spring Force
There does NOT need to be a block, yet!!!
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The Wonders of the Spring Force
• It is much like tension or compression, depending
whether the spring is stretched longer than
equilibrium position, or squeezed shorter
• There is a very elegant expression for the force
exerted BY the spring: Hooke’s Law F = – k (x-x0)
• Here, k is the spring constant
• x is the location of the movable end of the spring
• x0 is a constant too: the location of the movable
end when the spring is neither stretched nor
squeezed
• Often, one takes x0 to be zero
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A graph of Hooke’s law F(x)
Stiff spring
F exerted BY spring
Limp spring
x0
• Slope of the graph is –k
• [k] = N/m
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x
What kind of acceleration occurs to a mass?
• At this instant, the spring is squeezed (x < 0) so its
force is to the right (F > 0)
• Therefore, acceleration of mass is to the right
• If it is moving to the right, it will be speeding up
and fly right through equilibrium, since a = 0 there
• If it is moving to the left, it will be slowing down
and eventually stop, since a is growing and opposed
to the motion
{show Active Figure AF_1502}
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What kind of acceleration occurs to a mass?
• The only horizontal force on the mass is the spring
(ignore up/down forces)
• Fnet = – kx = ma  a = – kx/m
• So the acceleration is in the opposite direction to
the position, and proportional to it!!
d 2x k
 x  0 a differential equation for x(t )
2
m
dt
{show Active Figure AF_1501}
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What kind of acceleration occurs to a mass?
2
d x
k
 x0
2
m
dt
a differential equation for x(t )
• What functions of time look an awful lot like their
own second derivatives, to the point where if the
function is added to its second derivative the answer
could be zero?
• NOT powers of t . . . Thinking . . .
• AHA! Trig functions!!
x(t )  A sin ( t   )
a trial solution for x(t )!!
• What are the meanings of A,  and ? Let’s plug
in and see what happens…
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Plugging the trial solution into the DE
x(t )  A sin  t     [ A]  m [ ]  rad [ ]  rad/s
dx(t )
d

: v(t )   Acos t     t   
dt
 dt

  A cos t   
dv(t ) d 2 x(t )
d


: a(t )  A sin  t     t   
2
dt
dt
 dt

   2 A sin  t    so a does indeed look like x!!
Inserting this stuff into the differential equation, we get
d 2 x(t )
dt 2
k
k
2
 x(t )    A sin  t     A sin  t     0
m
m
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Plugging the trial solution into the DE
d 2 x(t )
k
k
2
 x(t )    A sin  t     A sin  t     0
2
m
m
dt
• It works great, if we take  = √(k/m)
• No information about A or  at this stage
x(t )  A sin  t   
Cosine works just as well!!
–
• A is amplitude
• T is the period and T =
2p/ 2p √(m/k)
•  is the phase constant
{show Active Figure AF_1506}
Here,  is roughly +150° = 2.6 rad
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Relationship of Simple Harmonic Motion to
Circular Motion
• This kind of motion: x = A sin t or A cos t ,
with perhaps a phase constant  in there, is called
Simple Harmonic Motion
• One can think of it as a ‘projection’ of circular
motion at the same angular speed/frequency
{show
Active
Figure
AF_1513}
y
A
A sin t + 
A cos t + 
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t + 
x
Relationship of v(t) and a(t) to x(t)
x(t )  A sin  t   
Position amplitude  A
phase constant  = + p/2
A
–A
v(t )   A cos t   
Velocity amplitude  A
A
v(t ) is 90  ahead of x(t )
– A
a(t )   2 A sin  t   
Acceleration amplitude   2 A
a(t ) is 180  ahead of x(t )
2A
– 2A
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Problem 3.7
A body is constrained to move on a straight path so
that its acceleration is proportional to its
displacement from equilibrium, that is a = -g x
where g = 2.5 s-2
a) Determine the period of the oscillations.
b) If the amplitude of the oscillations is 0.40 m,
determine the speed of the body as it passes
through the centre of oscillation.
c) [added] If the body is at 0.2m and moving to
the right at t = 0.0 s, sketch x(t) and v(t) [hint:
you must deal with the phase constant].
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The definition of Work
• Consider a force that varies with a body’s position F(x)
• If the body moves a small displacement dx, so small that
F(x) doesn’t change much, in the same (or opposite)
direction as the force, the small bit of work done by the
force is
dW := F(x) dx
• This is of course the (signed) area of the thin strip of
width dx and height F(x) on a graph of F versus x
• For a non-small displacement (from xA to xB), the work
done by the force is the integral of dW (area under F(x)):
xB

W AB : F ( x) dx
xA
NOTE: this is the work done
by any force!! If it is the net
force, see the next slide!!
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The definition of Kinetic Energy,
and the Work-Energy TheoremxB
• If the force is the net force we get W AB,net :
 Fnet ( x) dx
xA
• Since Fnet = ma = m dv/dt we have (watch closely!!)
W AB,net
xB
xB
xB
2 B
xA
xA
xA
A
dv
dx
v
 m dx  m dv
 m v dv  m
dt
dt
2



1 2
• Define Kinetic Energy of a body K : mv
2
• Thus we have the Work-Energy Theorem: the work
done on a body by the net force that acts in some
process is equal to the change in kinetic energy of the
body during that process, or WAB,net = DK
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Units of Work and Energy
• [W] = N-m = kg-m2/s2 = Joule (J)
• If a force of 1 N acts on a body, and the body moves in
the direction of the force a distance of 1 m, the force does
1 J of work (pretty small…)
• If a force of 1 dyne (= 1 g-cm/s2) acts, and the body
moves1 cm, the force does 1 erg of work (really small…)
• If a force of 1 lb… moves 1 ft… 1 foot-pound of work
• 1 calorie (cal) = 4.186 J [mechanical equivalent of heat]
• 1 British Thermal Unit (BTU) = amount of heat needed
to raise 1 lb of water 1 °F = 1055 J
• 1 kcal = 1 Cal = 1 food calorie
• 1 electron-Volt (ev) = 1.602x10-19 J
• 1 kilowatt-hr = another common energy unit…
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The Potential Energy
associated with an x-dependent force, and the
Mechanical Energy
xB
• Work W AB : F ( x) dx  f ( xB )  f x A   Df AB

xA
• Here, f(x) is called the antiderivative of F(x)
• If F is not a function of x, [“F ≠ F(x)”] you can’t do this!
• Potential Energy Change of the body-force
combination is DUAB := – DfAB
• So we argue that when the K of a body grows, due to
forces acting, those forces lose U (and vice versa)
• Therefore, in such a process, the total mechanical
energy does not change: DE=DK + DU = 0
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Mechanical Energy and Energy conservation
• Mechanical Energy of a situation E:=K+U
• Mechanical energy change in a process is
DE=DK + DU +ELOST
• K is owned by the body (in m and v)
• U is owned by the body’s location relative to
the source (s) of the force(s), in some way
• There are many forces in nature for which one
cannot even define a potential energy function!!
• Examples include FRICTION, TENSION,
NORMAL…
• If all the forces are ‘derivable’ from potentials,
and if any non-potential-owning forces do no
work, Mechanical Energy is Conserved: DE = 0
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Potential energy facts
• Suppose the process is in the presence of F(x)
• Let us move from a starting point x0 to a variable
endpoint x. Introduce the ‘dummy’ variable x’:
x

U ( x)  U ( x0 )   F ( x' )dx' 
x0
 (signed area under F ( x) graph between x0 and x)
• U(x0) is called the reference value of the potential
energy and if you are clever you can make it 0
• Since U is the integral of F, F is the derivative of U
dU ( x)
F ( x)  
dx
 (slope of tangent to U ( x) at x)
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Potential energy example: Gravity
• Almost trivial example: the weight force W = – mg
y
y


y0
y0
U ( y )  U ( y0 )   F ( y ' )dy'  (mg ) dy'  mgy  mgy0
• Reference potential energy U(y0) is evidently mgy0
• So just take your initial height to BE zero and the
reference potential energy to be zero too. Freedom!!
• Conclusion: potential energy of gravity is U ( y)  mgy
• Check: what is the force of gravity, from the potential?
dU ( y)
d
F ( y)  
  mgy  mgyo   mg  0  mg
dy
dy
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Potential energy example: Spring
• Important example: the spring force F = – k(x-x0)
x
x


x0
x0
U ( x)  U ( x0 )   F ( x' )dx'  (k ) ( x' x0 )dx'
2
2
kx
1 2x
kx
x
 kx'
 kx0 x' x 
 0  kx0 x  kx02
0
x0
2
2
2
k 2
k
2
 x  x0  2 xx0  x  x0 2
2
2


• Reference potential energy U(x0) is buried in there
• But you can just take potential energy to be zero if x
= x0 and so U(x0) is taken to be zero: thus one says
1
dU ( x)
2
U ( x)  k x  x0  and F ( x)  
 k ( x  x0 )
2
dx
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Potential energy example: Spring
• Checking the integral as an area
• For simplicity, take x0 = 0
• The end of the spring
moves from xmax < 0 to 0
• The force of the spring
is positive, and the area
under the curve is too
• Area = ½ (base)(height)
= ½ (–xmax)((– k xmax)
1 2
Area  kxmax
2
1 2
So DU   kxmax
2
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Potential energy functions U(x)
dU ( x)
• What information is encoded in U(x)? F ( x)  
dx
• For x > 0, slope of U is
positive, so F is to left
• For x < 0, slope of U is
negative, so F is to right
• A restoring force
• For x = 0, slope of U is
zero, so F is zero
• Stable equilibrium point
• Unstable equilibrium point:
object tends to ‘fly away’
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Power: the rate of energy consumption
• Power is the time rate of energy consumption, or
production, or the time rate of doing work
dW Formal definition and fairly useless
 P :
dt
F dx
 since dW  F dx, P 
 Fv Cool!!
dt
• [P] = J/s = Watt (W)
• so 1 J = 1 W-s and 1 kW-hr = 3.6x106 J
• a kW-hr of energy costs about 12 cents
• 1 horse-power (hp) = 746 W
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Energetics of Simple Harmonic Motion
 x(t )  A sin t    [ x0  0]
1 2 A2
 U  kx 
sin 2 t   
2
2
sin and sin2
 v(t )  A cost   
2
1 2 m 2 A2
kA
 K  mv 
cos 2 t    
cos 2 t   
2
2
2


2
kA2
kA
E  K U 
sin 2 t     cos 2 t    
constant!!
2
2
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