Transcript PHYS 1443 – Section 501 Lecture #1

PHYS 1443 – Section 003
Lecture #20
Monday, Nov. 25, 2002
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
6.
Simple Harmonic and Uniform Circular Motions
Damped Oscillation
Newton’s Law of Universal Gravitation
Free Fall Acceleration and Gravitational Force
Kepler’s Laws
Gravitation Field and Potential Energy
Today’s homework is homework #20 due 12:00pm, Monday, Dec. 2!!
Monday, Nov. 25, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
1
Announcements
• Class on Wednesday
• Remember the Term Exam on Monday, Dec. 9 in the
class
– Covers chapters 11 – 15
– Review on Wednesday, Dec. 4
Monday, Nov. 25, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
2
Simple Harmonic and Uniform Circular Motions
Uniform circular motion can be understood as a
superposition of two simple harmonic motions in x and y axis.
y
y
A
P
y
f
O
x
P
y
w
A
q
O x Q
y
q
A a
q
O
Q
A
O vx Q
x
t=t
t=0
q=wt+f
When the particle rotates at a uniform angular
speed w, x and y coordinate position become
Since the linear velocity in a uniform circular
motion is Aw, the velocity components are
Since the radial acceleration in a uniform circular
motion is v2/A=w2A, the components are
Monday, Nov. 25, 2002
v
P
x
ax P
x  A cosq  Acoswt  f 
y  A sin q  Asin wt  f 
vx  v sin q   Aw sin wt  f 
v y  v cosq  Aw coswt  f 
a x  a cosq   Aw 2 coswt  f 
a y  a sin q   Aw 2 sin wt  f 
PHYS 1443-003, Fall 2002
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3
x
Example 13.7
A particle rotates counterclockwise in a circle of radius 3.00m with a constant angular
speed of 8.00 rad/s. At t=0, the particle has an x coordinate of 2.00m and is moving to
the right. A) Determine the x coordinate as a function of time.
Since the radius is 3.00m, the amplitude of oscillation in x direction is 3.00m. And the
angular frequency is 8.00rad/s. Therefore the equation of motion in x direction is
x  A cosq  3.00mcos8.00t  f 
Since x=2.00, when t=0
 2.00 

  48.2
 3.00 
1
2.00  3.00mcos f f  cos 
However, since the particle was
moving to the right f=-48.2o,

x  3.00m cos 8.00t  48.2

Find the x components of the particle’s velocity and acceleration at any time t.


 

dx
Using the
vx 
 3.00  8.00sin 8.00t  48.2   24.0m / s sin 8.00t  48.2
displacement
dt
Likewise,
dv
2

a







24
.
0

8
.
00
cos
8
.
00
t

48
.
2


192
m
/
s
cos
8
.
00
t

48
.
2
x
from velocity
dt

Monday, Nov. 25, 2002
PHYS 1443-003, Fall 2002
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Damped Oscillation
More realistic oscillation where an oscillating object loses its mechanical
energy in time by a retarding force such as friction or air resistance.
F
Let’s consider a system whose retarding force
is air resistance R=-bv (b is called damping
coefficient) and restoration force is -kx
The solution for the above 2nd order
differential equation is
The angular frequency w
for this motion is
w
x
 kx  bv  max
dx
d 2x
 kx  b
m 2
dt
dt
x
k  b 


m  2m 
e
2

b
t
2m
Acoswt  f 
Damping Term
This equation of motion tells us that when the retarding force is much smaller than restoration
force, the system oscillates but the amplitude decreases, and ultimately, the oscillation stops.
We express the
angular frequency as
Monday, Nov. 25, 2002
w
 b 
 w 2  

 2m 
2
Where the natural
w0 
frequency w0
PHYS 1443-003, Fall 2002
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k
m
5
More on Damped Oscillation
The motion is called Underdamped when the magnitude of
the maximum retarding force Rmax = bvmax <kA
How do you think the damping motion would change as
retarding force changes?
 bvmax  kA
As the retarding force becomes larger, the amplitude reduces
more rapidly, eventually stopping at its equilibrium position
Under what condition this system
does not oscillate?
The system is Critically damped
What do you think happen?
w
 0 w 
b
2m
b  2mw  2 mk
Once released from non-equilibrium position, the object
would return to its equilibrium position and stops.
If the retarding force is larger
Rmax  bvmax  kA The system is Overdamped
than restoration force
Once released from non-equilibrium position, the object would return
Monday, Nov. 25, 2002 to its equilibrium
PHYSposition
1443-003,and
Fall 2002
stops, but a lot slower than before 6
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Newton’s Law of Universal Gravitation
People have been very curious about the stars in the sky, making
observations for a long time. But the data people collected have not been
explained until Newton has discovered the law of gravitation.
Every particle in the Universe attracts every other particle with a
force that is directly proportional to the product of their masses and
inversely proportional to the square of the distance between them.
How would you write this
principle mathematically?
G is the universal gravitational
constant, and its value is
Fg 
m1m2
r122
With G
G  6.673 10
11
Fg  G
Unit?
m1m2
r122
N  m 2 / kg 2
This constant is not given by the theory but must be measured by experiment.
This form of forces is known as an inverse-square law, because the magnitude of the
force is inversely proportional to the square of the distances between the objects.
Monday, Nov. 25, 2002
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More on Law of Universal Gravitation
Consider two particles exerting gravitational forces to each other.
m1
r̂12
F21
r
m2
Two objects exert gravitational force on each other
following Newton’s 3rd law.
F12
Taking r̂12 as the unit vector, we can
write the force m2 experiences as
What do you think the
negative sign mean?
F 12
m1m2
 G 2 r̂12
r
It means that the force exerted on the particle 2 by
particle 1 is attractive force, pulling #2 toward #1.
Gravitational force is a field force: Forces act on object without physical contact
between the objects at all times, independent of medium between them.
How do you think the
The gravitational force exerted by a finite size,
gravitational force on the
spherically symmetric mass distribution on a particle
outside the distribution is the same as if the entire mass surface of the earth look?
of the distributions was concentrated at the center.
M Em
Fg  G 2
RE
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Dr. Jaehoon Yu
Free Fall Acceleration & Gravitational Force
Weight of an object with mass m is
mg. Using the force exerting on a
particle of mass m on the surface of
the Earth, one can get
What would the gravitational
acceleration be if the object is at
an altitude h above the surface of
the Earth?
mg
g
M Em
RE2
ME
G
RE2
G
M Em  G M Em
Fg  mg '  G
2
2


R

h
r
E
ME
g'  G
RE  h 2
What do these tell us about the gravitational acceleration?
•The gravitational acceleration is independent of the mass of the object
•The gravitational acceleration decreases as the altitude increases
•If the distance from the surface of the Earth gets infinitely large, the weight of the
object approaches 0.
Monday, Nov. 25, 2002
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Example 14.2
The international space station is designed to operate at an altitude of 350km. When
completed, it will have a weight (measured on the surface of the Earth) of 4.22x106N.
What is its weight when in its orbit?
The total weight of the station on the surface of the Earth is
FGE  mg
ME
M Em
6
G
2  4.22  10 N
RE
Since the orbit is at 350km above the surface of the Earth,
the gravitational force at that height is
FO
M Em
RE2
FGE
 mg'  G R  h 2 
2
RE  h 
E
Therefore the weight in the orbit is
FO


2
RE2
6.37 106

FGE 
2
RE  h 
6.37 106  3.50 105

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6

4
.
22

10

3
.
80

10
N
2

10
Example 14.3
Using the fact that g=9.80m/s2 at the Earth’s surface, find the average density of the Earth.
Since the gravitational acceleration is
g
So the mass of the Earth is
Therefore the density of the
Earth is
G
ME
11 M E

6
.
67

10
2
2
RE
RE
2
R g
ME  E
G
2


ME
VE
RE g
3g
G


4GRE
4
3
RE

3  9.80
3
3


5
.
50

10
kg
/
m
4  6.67 10 11  6.37 106
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Kepler’s Laws & Ellipse
a
b
F1
c
F2
Ellipses have two different axis, major (long) and
minor (short) axis, and two focal points, F1 & F2
a is the length of a semi-major axis
b is the length of a semi-minor axis
Kepler lived in Germany and discovered the law’s governing planets’
movement some 70 years before Newton, by analyzing data.
•All planets move in elliptical orbits with the Sun at one focal point.
•The radius vector drawn from the Sun to a planet sweeps out equal
area in equal time intervals. (Angular momentum conservation)
•The square of the orbital period of any planet is proportional to the
cube of the semi-major axis of the elliptical orbit.
Newton’s laws explain the cause of the above laws. Kepler’s third law is
the direct consequence of law of gravitation being inverse square law.
Monday, Nov. 25, 2002
PHYS 1443-003, Fall 2002
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The Law of Gravity and the Motion of Planets
•Newton assumed that the law of gravitation applies the same
whether it is on the Moon or the apple on the surface of the Earth.
•The interacting bodies are assumed to be point like particles.
Apple g
RE
aM
Moon
Newton predicted that the ratio of the Moon’s
acceleration aM to the apple’s acceleration g would be
2
aM

1 / rM 
 RE   6.37 106 
4


  


2
.
75

10
2  
g
1 / RE   rM   3.84 108 
2
v
2
Therefore the centripetal acceleration of the Moon, aM, is
aM  2.75 104  9.80  2.70 103 m / s 2
Newton also calculated the Moon’s orbital acceleration aM from the knowledge of its distance
from the Earth and its orbital period, T=27.32 days=2.36x106s
9.80
4 2  3.84 108
v 2 2rM / T 2 4 2 rM
3
2



2
.
72

10
m
/
s



aM r
2
6 2
2


60
r
2
.
36

10
T
M
M


This means that the Moon’s distance is about 60 times that of the Earth’s radius, its acceleration
is reduced
by 25,
the2002
square of the ratio. PHYS
This 1443-003,
proves that
the inverse square law is valid.
Monday, Nov.
Fall 2002
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Dr. Jaehoon Yu
Kepler’s Third Law
It is crucial to show that Keper’s third law can be predicted from the
inverse square law for circular orbits.
v
r
Since the gravitational force exerted by the Sun is radially
directed toward the Sun to keep the planet circle, we can
apply Newton’s second law
GM s M P M p v 2

r2
r
Ms
2r
Since the orbital speed, v, of the planet with period T is v  T
2
GM s M P
M p 2r / T 
The above can be written

2
r
Solving for T one
can obtain
T
2
2  4 r 3  K r 3 and


s
 GM s 
 4 2
K s   GM
s

r

  2.97 10 19 s 2 / m3

This is Keper’s third law. It’s also valid for ellipse for r being the length of the
semi-major axis. The constant Ks is independent of mass of the planet.
Monday, Nov. 25, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
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Example 14.4
Calculate the mass of the Sun using the fact that the period of the Earth’s orbit around the
Sun is 3.16x107s, and its distance from the Sun is 1.496x1011m.
Using Kepler’s third law.
The mass of the Sun, Ms, is
2


4

3
3
2 

T  GM s r  K s r
 4 2  3
r
M s  
 GT 


4 2
11




1
.
496

10
11
7 
6
.
67

10

3
.
16

10



 1.99 10 kg
30
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PHYS 1443-003, Fall 2002
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
3
Kepler’s Second Law and Angular Momentum Conservation
Consider a planet of mass Mp moving around the Sun in an elliptical orbit.
D
S
C
r
Since the gravitational force acting on the planet is
A always toward radial direction, it is a central force
dr
B Therefore the torque acting on the planet by this
force is always 0.
  r  F  r  Frˆ  0
Since torque is the time rate change of angular
momentum L, the angular momentum is constant.


dL
0
dt
L  const
Because the gravitational force exerted on a
 r  p  r  M p v  M p r  v  const
planet by the Sun results in no torque, the
angular momentum L of the planet is constant.
Since the area swept by the
L
1
1
dA  L

dt
 const
dA  r  d r  r  vdt
motion of the planet is
2M p
2M p
2
2
dt
L
This is Keper’s second law which states that the radius vector from the Sun
to a planet sweeps our equal areas in equal time intervals.
Monday, Nov. 25, 2002
PHYS 1443-003, Fall 2002
Dr. Jaehoon Yu
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