Transcript Chapter 6

Uniform Circular Motion,
Acceleration

A particle moves with a constant speed in a
circular path of radius r with an acceleration:
v2
ac 
r


The centripetal acceleration, ac is directed toward
the center of the circle
The centripetal acceleration is always
perpendicular to the velocity
Uniform Circular Motion, Force


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A force, Fr , is
associated with the
centripetal acceleration
The force is also
directed toward the
center of the circle
Applying Newton’s
Second Law along the
radial direction gives
v2
 F  mac  m r
Uniform Circular Motion, cont
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A force causing a centripetal
acceleration acts toward the
center of the circle
It causes a change in the
direction of the velocity vector
If the force vanishes, the
object would move in a
straight-line path tangent to
the circle

See various release points in
the active figure
Motion in a Horizontal Circle
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The speed at which the object moves
depends on the mass of the object and the
tension in the cord
The centripetal force is supplied by the
tension
Tr
T=mv2/r hence
v
m
Motion in Accelerated Frames

A fictitious force results from an accelerated
frame of reference

A fictitious force appears to act on an object in the
same way as a real force, but you cannot identify
a second object for the fictitious force

Remember that real forces are always interactions
between two objects
“Centrifugal” Force
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From the frame of the passenger (b), a
force appears to push her toward the door
From the frame of the Earth, the car
applies a leftward force on the passenger
The outward force is often called a
centrifugal force

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It is a fictitious force due to the centripetal
acceleration associated with the car’s
change in direction
In actuality, friction supplies the force
to allow the passenger to move with
the car

If the frictional force is not large enough,
the passenger continues on her initial
path according to Newton’s First Law
“Coriolis Force”

This is an apparent
force caused by
changing the radial
position of an object in
a rotating coordinate
system
The result of the rotation is the curved path of object
Ball in figure to the right, winds, rivers and currents on
earth. For winds we get the prevailing wind pattern below.
Fictitious Forces, examples
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Although fictitious forces are not real forces,
they can have real effects
Examples:
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Objects in the car do slide
You feel pushed to the outside of a rotating
platform
The Coriolis force is responsible for the rotation of
weather systems, including hurricanes, and ocean
currents
Introduction to Energy
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The concept of energy is one of the most
important topics in science and engineering
Every physical process that occurs in the
Universe involves energy and energy
transfers or transformations
Energy is not easily defined
Work
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The work, W, done on a system by an agent
exerting a constant force on the system is the
product of the magnitude F of the force, the
magnitude Dr of the displacement of the point
of application of the force, and cos q, where q
is the angle between the force and the
displacement vectors
Work, cont.

W = F Dr cos q  F. Dr

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The displacement is that
of the point of application
of the force
A force does no work on
the object if the force
does not move through a
displacement
The work done by a force
on a moving object is
zero when the force
applied is perpendicular
to the displacement of its
point of application
Work Example

The normal force and
the gravitational force
do no work on the
object

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cos q = cos 90° = 0
The force F is the only
force that does work on
the object
Units of Work
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Work is a scalar quantity
The unit of work is a joule (J)
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1 joule = 1 newton . 1 meter
J = N · m ( Fr)
The sign of the work depends on the direction of the
force relative to the displacement
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Work is positive when projection of onto
is in the
same direction as the displacement
Work is negative when the projection is in the opposite
direction
Work Done by a Varying Force
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Assume that during a very
small displacement, Dx, F
is constant
For that displacement,
W ~ F Dx
For all of the intervals,
xf
W   Fx Dx
xi
Work Done by a Varying Force,
cont

lim
Dx 0
xf
 F Dx  
x
xi


xf
xi
Fx dx
Therefore,W 

xf
xi
Fx dx
The work done is equal
to the area under the
curve between xi and xf
Work Done By A Spring
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A model of a common
physical system for
which the force varies
with position
The block is on a
horizontal, frictionless
surface
Observe the motion of
the block with various
values of the spring
constant
Hooke’s Law

The force exerted by the spring is
Fs = - kx
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x is the position of the block with respect to the equilibrium
position (x = 0)
k is called the spring constant or force constant and measures
the stiffness of the spring
This is called Hooke’s Law
Hooke’s Law, cont.
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When x is positive
(spring is stretched), F
is negative
When x is 0 (at the
equilibrium position), F
is 0
When x is negative
(spring is compressed),
F is positive
Hooke’s Law, final
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The force exerted by the spring is always
directed opposite to the displacement from
equilibrium
The spring force is sometimes called the
restoring force
If the block is released it will oscillate back
and forth between –x and x
Hooke’s Law consider the spring
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When x is positive (spring is
stretched), Fs is negative
When x is 0 (at the
equilibrium position), Fs is 0
When x is negative (spring is
compressed), Fs is positive
Hence the restoring force
Fs =Fs = -kx
Work Done by a Spring
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Identify the block as the system and see figure below
The work as the block moves from xi = - xmax to xf = 0 is ½ kx2
Note: The total work done by the spring as the block moves
from –xmax to xmax is zero see figure also
Ie. From the General definition
W W
net
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Or
  ( F)  dr
0
xf
Ws   Fs  dr   (kxi dxi   (kx)dx
xi
- xmax
1 2

kx
dx

kxmax
 
 xmax
2
xf
Ws   Fx dx  
xi
n
0
ie. x dx  x
n 1
/(n  1)
Work Done by a Spring,in
general
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Assume the block undergoes an arbitrary
displacement from x = xi to x = xf
The work done by the spring on the block is
Ws  
xf
xi
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1 2 1 2
 kx  dx  kxi  kxf
2
2
If the motion ends where it begins, W = 0
NOTE the work is a change in the expression
1/2kx2 We say a change in elastic potential
energy..in general a energy expression is defined
for various forces and the work done changes that
energy.
Kinetic Energy and WorkKinetic Energy Theorem
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Kinetic Energy is the energy of a particle due
to its motion
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K = ½ mv2
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K is the kinetic energy
m is the mass of the particle
v is the speed of the particle
A change in kinetic energy is one possible
result of doing work to transfer energy into a
system
Kinetic Energy

Calculating the work:
W 
xf
xi
vf
 F dx  
xi
W   mv dv
vi
xf
ma dx
IE. a=dv/dt
adx=dv/dt dx
=dv dx/dt=vdv
1 2 1
 W  2mv f  2 mv i2
Wnet  K f  K i  DK
The
Work-Kinetic Energy Theorem states SW = Kf – Ki = DK
Hence K=1/2 mv2 is a a natural for energy expression..
And the last equation is called the Work-Kinetic Energy Theorem
Again we note that the work done changes an energy expression …
in this case a change in Kinetic energy
The speed of the system increases if the work done on it is positive
The speed of the system decreases if the net work is negative
Also valid for changes in rotational speed
Potential Energy in general
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Potential energy is energy related to the
configuration of a system in which the
components of the system interact by forces
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The forces are internal to the system
Can be associated with only specific types of
forces acting between members of a system
Gravitational Potential Energy
NEAR SURFACE OF EARTH ONLY
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The system is the Earth
and the book
Do work on the book by
lifting it slowly through a
vertical displacement
Dr  Dyˆj
The work done on the
system must appear as
an increase in the
energy of the system
Gravitational Potential Energy,
cont
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There is no change in kinetic energy since
the book starts and ends at rest
Gravitational potential energy is the energy
associated with an object at a given location
above the surface of the Earth
 
W  Fapp  D r
W  (mgˆj)   y f  y i  ˆj
W  mgy f  mgy i
Gravitational Potential Energy,
final
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The quantity mgy is identified as the gravitational
potential energy, Ug
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Ug = mgy
THIS IS ONLY NEAR THE EARTH’s surface
……………WHY???????
Units are joules (J)
Is a scalar
Work may change the gravitational potential energy
of the system
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Wnet = DUg
Conservative Forces and
Potential Energy
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Define a potential energy function, U, such
that the work done by a conservative force
equals the decrease in the potential energy of
the system
The work done by such a force, F, is
xf
WC   Fx dx  DU
xi

DU is negative when F and x are in the same
direction
Conservative Forces and
Potential Energy
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The conservative force is related to the potential
energy function through
dU
Fx  
dx

The x component of a conservative force acting on
an object within a system equals the negative of the
potential energy of the system with respect to x
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Can be extended to three dimensions
Conservative Forces and
Potential Energy – Check

Look at the case of a deformed spring
dUs
d 1 2
Fs  
   kx   kx
dx
dx  2

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This is Hooke’s Law and confirms the equation for
U
U is an important function because a
conservative force can be derived from it
Energy Diagrams and
Equilibrium
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Motion in a system can be observed in terms of a graph of its
position and energy
In a spring-mass system example, the block oscillates
between the turning points, x = ±xmax
The block will always accelerate back toward x = 0
Energy Diagrams and Stable
Equilibrium
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The x = 0 position is one of
stable equilibrium
Configurations of stable
equilibrium correspond to
those for which U(x) is a
minimum
x = xmax and x = -xmax are
called the turning points
Energy Diagrams and Unstable
Equilibrium
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Fx = 0 at x = 0, so the
particle is in equilibrium
For any other value of x, the
particle moves away from
the equilibrium position
This is an example of
unstable equilibrium
Configurations of unstable
equilibrium correspond to
those for which U(x) is a
maximum
Neutral Equilibrium
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Neutral equilibrium occurs in a configuration
when U is constant over some region
A small displacement from a position in this
region will produce neither restoring nor
disrupting forces
Ways to Transfer Energy Into
or Out of A System
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Work – transfers by applying a force and
causing a displacement of the point of
application of the force
Mechanical Waves – allow a disturbance to
propagate through a medium
Heat – is driven by a temperature difference
between two regions in space
A word from our sponsors:
CONDUCTION, CONVECTION, RADIATION
More Ways to Transfer Energy
Into or Out of A System
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Matter Transfer – matter physically crosses
the boundary of the system, carrying energy
with it
Electrical Transmission – transfer is by
electric current
Electromagnetic Radiation – energy is
transferred by electromagnetic waves
Two New important Potential Energies
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In the universe at large Gravitational force as
defined by Newton prevails
Ie.. F = -Gm1m2 /r2 m the masses G a universal
constant and r distance between the masses
(negative is attractive force)
In the atomic world the electric force dominates
defined as F=kq1q2 /r2 here r is the distance
between the electric charges represented by q
and k a universal constant
Charges can be + or The Constant values.G,k depend upon units
used
Gravitational and Electric
Potential energies (3D)
xf
WC   Fx dx  DU
xi
With r replacing x we get and using the gravitational and electric forces
equations and S for integration from point initial to final
W = S FGdr = - Gm1m2 = S 1/r2 dr = -Gm1m2 (1/rf -1/ri)
W = S Fedr = kq1q2 =S 1/r2 dr = kq1q2 (1/rf -1/ri)
Or potential energies for these forces go as 1/r
Note from above that F = -dU/dr with
UG = Gm1m2 /r Ue = kq1q2 /r we get back the 1/r2 forces
Conservation of Energy
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Energy is conserved
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This means that energy cannot be created nor
destroyed
If the total amount of energy in a system changes,
it can only be due to the fact that energy has
crossed the boundary of the system by some
method of energy transfer!
Isolated System
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For an isolated system, DEmech = 0
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Remember Emech = K + U
This is conservation of energy for an isolated system with
no nonconservative forces acting
If nonconservative forces are acting, some energy is
transformed into internal energy
Conservation of Energy becomes DEsystem = 0
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Esystem is all kinetic, potential, and internal energies
This is the most general statement of the isolated system
model
Isolated System, cont
 (example
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book falling)
The changes in energy DEsystem = 0
Or DK +DU=0
DK=-DU
Ie. Kf - Ki = -(Uf –Ui)
can be written out and rearranged
Kf + Uf = Ki + Ui Remember, this applies only to a
system in which conservative forces act
Or 1/2mvf2 +mghf =1/2mgvi2+mghi
Example – Free Fall
example 8-1
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Determine the speed of
the ball at y above the
ground
Conceptualize


Use energy instead of
motion
Categorize


System is isolated
Only force is gravitational
which is conservative
Example – Free Fall, cont

Analyze


Apply Conservation of Energy
Kf + Ugf = Ki + Ugi

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Ki = 0, the ball is dropped
Solving for vf
vf  v i2  2g  h  y 

Finalize

The equation for vf is consistent with the results
obtained from kinematics
For the electric force
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Total energy
Is K+U=1/2mv2 +kq1q2 /r
Specifically in a hydrogen atom using charge
units e (CALLED ESU we get rid of K) and
the proton and electron both have the same
charge =e
Or total energy for electron in orbit
=1/2mv2 +e2 /r we will use this in chapter 3
Instantaneous Power


Power is the time rate of energy transfer
The instantaneous power is defined as
dE

dt

Using work as the energy transfer method,
this can also be written as
avg
W

Dt
Power


The time rate of energy transfer is called
power
The average power is given by




W
P
Dt
when the method of energy transfer is work
Units of power:
what is a Joule/sec called ?
Answer WATT! 1 watt=1joule/sec
Instantaneous Power and
Average Power

The instantaneous power is the limiting value
of the average power as Dt approaches zero
dW
dr
lim W
  Dt 0

 F
 F v
Dt
dt
dt

The power is valid for any means of energy
transfer
NOTE: only part of F adds to power ?

Units of Power

The SI unit of power is called the watt


A unit of power in the US Customary system
is horsepower


1 watt = 1 joule / second = 1 kg . m2 / s2
1 hp = 746 W
Units of power can also be used to express
units of work or energy

1 kWh = (1000 W)(3600 s) = 3.6 x106 J
T
Example 8.10 melev =1600kg passengers =200kg
A constant retarding force =4000 N
How much power to lift at constant rate of 3m/s
How much power to lift at speed v with a=1.00 m/ss
f
W
USE SF =0 in first part and =ma in second then use
Next equation

lim
Dt 0
W dW
dr

 F
 F v
Dt
dt
dt