Transcript lecture06

6. Centripetal force
mv2
 2 
2
F
 m r  m  r
r
T 
2
F = ma
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Example: A stone of mass m sits at the bottom of a bucket. A string is
attached to the bucket and the whole thing is made to move in circles.
What is the minimum speed that the bucket needs to have at the highest
point of the trajectory in order to keep the stone inside the bucket?
m v2
m v2
m v2
F
 N  mg 
N
 mg
r
r
r
N min  0
vmin  gr
2
Examples (centripetal force)
(1)
N
mg
(2)
N
F = ma
2
m v2
F
r
2
mv
N  mg
r
mv
N
 mg
r
m v2
N  mg
r
m v2
N
 mg
r
mg
(3)
N
2
mg
mv
m g N 
r
m v2
N  m g
r
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Question 1: The ball whirls around a pole.
In what direction does the net force on the ball point?
1) toward the top of the pole
2) toward the ground
3) along the horizontal component of the tension force
4) along the vertical component of the tension force
5) tangential to the circle
The vertical component of the tension balances the weight.
The horizontal component of tension provides the
centripetal force that points toward the center of the circle.
Question 2: A particle moves at a constant
speed along the presented trajectory.
Compare the magnitude of the acceleration A
of the particle at points A and B.
T
mg
B
Question 3: A ball is going around in a circle at
constant speed. What is the angle between the
acceleration vector and the velocity vector of the ball?
A) 0
B) 45
C) 90
D)180
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Newton’s Law of Gravitation
Gm1m2
Fg 
r2
m1
m2
r
F 2 on 1
F 1on 2
Gravitational constant: G  6.671011 N  m2 / kg 2
Gravitational force:
• one of the fundamental forces of nature
• always attractive
• exist between any two objects and always act along the line joining
the two objects
• one of the fundamental forces acting in our galaxy and the main force
of interaction between the sun and planets including Earth
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1. Determining the value of G
For determining of G, Henry Cavendish in 1798 used an instrument
called a torsion balance. A modern version of the Cavendish torsion
balance is shown below.
G
Fg r 2
m1m2
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2. Weight and Law of Gravitation
The weight of a body is the total gravitational force acting on that body
Consider an object near
the surface of the earth:
Mm
Fg  G 2  mg
r
M
g G 2
r
Fg
What happened if
the object will move
far from the earth?
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Question 1: The mass of the earth is 81 times the mass of the moon.
The magnitude of the gravitational force of Earth on the Moon is __ times
the magnitude of the gravitational force of the Moon on the Earth.
A. 81
B. 812
C. 1
D. 1/81
E. (1/81)2
1.
2.
3.
4.
5.
Question 2: The direction of the net gravitational
force on m due to the two masses M is ___.
up
down
to the right
to the left
There is no
net force on m
Question 3: The planet Saturn has 100 times the mass of Earth. Saturn is 10
times further from the Sun than Earth is. The magnitude of the acceleration
of Earth in its orbit around the Sun is ___times the acceleration of Saturn.
A. 100
B. 10
C. 1
D. 1/10
E. 1/100
Question 4:
RMoon REarth  0.273
M Moon M Earth  0.0123
g Moon g Earth  ?
A.
B.
C.
D.
0.06
0.17
0.39
0.62
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3. Orbits of planets and satellites
(Fundamentally important application of Newton’s Mechanics)
•Copernicus in 1543 proposed that the sun was the center of the Solar
System with the planets moving in circular orbits.
•In 1619 Kepler showed that planets followed elliptical orbits using huge
amount of high quality data gathered by Tycho Brahe by naked eye
astronomy.
•Kepler characterized planetary orbits using “Kepler’s Three Laws”.
•In 1683 Newton showed that Kepler’s laws follow from his “Law of Gravity”
and his “Three Laws of Mechanics”.
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3a. Circular Orbits
mE m m v2
G 2 
r
r
Velocity:
Period:
GmE
v
r
2r
2r 3 / 2
T

v
(GmE )1/ 2
Example: We want to place 2000-kg satellite into a circular orbit 400 km
above the earth’s surface. For the earth: RE = 6380 km, mE = 5.971024kg.
v
Gm E

r
(6.67  10 11 N  m 2 / kg 2 )( 5.97  10 24 kg )
 7664 m / s
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6
(6.38  10  0.4  10 )m
2r 2 (6.78106 m)
T

 5556s  92.6 min
v
7664m / s
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3b. Satellites Orbits
vcircular  GmE / r
vescape  2GmE / r
vcircular  7.9 km/s
If r = RE = 6380 km then
vescape  11.2km/s
Trajectories of a projectile launched from point A in the direction AB with
different speeds:
7) hyperbolic orbit: v > vescape
6) parabolic orbit: v = vescape
5) elliptical orbit: vescape>v >vcircular
4) circular orbit:
1,2,3) elliptical orbit:
v = vcircular
v < vcircular
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Question: The Moon does not crash into Earth because:
1) it’s in Earth’s gravitational field
2) the net force on it is zero
3) it is beyond the main pull of Earth’s gravity
4) it’s being pulled by the Sun as well as by Earth
5) none of the above
The Moon does not crash into Earth because of its high speed. If it
stopped moving, it would, of course, fall directly into Earth. With its high
speed, the Moon would fly off into space if it weren’t for gravity providing
the centripetal force.
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