Download Chapter 6 - Santa Rosa Junior College

Chapter 6
Circular
Motion, Orbits,
and Gravity
© 2010 Pearson Education, Inc.
PowerPoint® Lectures for
College Physics: A Strategic Approach, Second Edition
© 2010 Pearson Education, Inc.
Slide 6-3
© 2010 Pearson Education, Inc.
Slide 6-5
Circular Motion
Milky Way Galaxy
Orbital Speed of Solar System: 220 km/s
Orbital Period: 225 Million Years
Mercury: 48 km/s
Venus: 35 km/s
Earth: 30 km/s
Mars: 24 km/s
Jupiter: 13 km/s
Neptune: 5 km/s
Differential Solar Rotation
Entangled Magentic Field
Lines makes Sun Spots
Equatorial Rotational Period: 27.5 days
464m/s
Precession
causes the
position of the
North Pole to
change over a
period of
26,000 years.
Orbital Speed of Earth: ~ 30 km/s
Although the
Moon is always
lit from the
Sun, we see
different
amounts of the
lit portion from
Earth
depending on
where the
Moon is
located in its
month-long
orbit.
Orbital Speed of Moon: ~ 1 km/s
155mph~70m/s
RADAR: RAdio Detecting And Ranging
200mph~90m/s
Electrons in Bohr Orbit: 2 million m/s
(Speed of Light: 200 million m/s)
Maximal Kerr Black Hole
Rotational Speed = Speed of Light
Uniform Circular Motion
The velocity vector is tangent to the path
The change in velocity vector is due to the change
in direction.
∆v
The centripetal acceleration
a
=
changes the direction of motion:

∆t
Angular Position: Rotation Angle
The rotation angle is the ratio of arc length to radius of curvature.
For a given angle, the greater the radius, the greater the arc length.
∆s
∆θ =
r
∆θ = rotation angle
∆s = arc length
r = radius
Angular & Tangential Velocity
ω : angular velocity (rad/s)
v: tangential velocity (m/s)
∆θ v
=
ω =
∆t r
v = ωr
ω is the same for every point & v varies with r.
Period & Frequenccy of
Rotation
What is the period of the Earth? Moon?
Centripetal Acceleration
• The acceleration is always
perpendicular to the path of the motion
• The acceleration always points toward
the center of the circle of motion
• This acceleration is called the
centripetal acceleration
2
v
• Magnitude is given by:
aC =
r
Circular Motion: centripetal acceleration
There is an acceleration
because the velocity is
changing direction.
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Slide 3-43
v2
2π r
aC =
, v
=
r
Τ
Space Station Rotation
A space station of diameter 80 m is turning about
its axis at a constant rate. If the acceleration of the
outer rim of the station is 2.5 m/s2, what is the period
of revolution of the space station?
a.
22 s
b.
19 s
c.
25 s
d.
28 s
e.
40 s
Centripetal and Centrifugal
Center SEEKING and Center FLEEING
Factual = Centripetal Force
Ffictitious = Centrifugal Force
In Physics, we use
ONLY
CentriPETAL acceleration
NOT
CentriFUGAL acceleration!
The Earth rotates once per day around its axis as shown.
Assuming the Earth is a sphere, is the rotational speed at Santa
Rosa greater or less than the speed at the equator?
366 m/s
464 m/s
Is the centripetal acceleration
greater at the Equator or at Santa
Rosa?
2
v
ac =
r
0.027 m / s 2
0.034m / s 2
The Earth rotates once per day around its axis. Assuming the Earth is a sphere
with radius 6.38 x 106m, find the tangential speed of a person at the equator
and at 38 degrees latitude (Santa Rosa!) and their centripetal accelerations.
At the equator, r = 6.38 x 106m:
2π r 2π (6.38 x106 m)
v =
=
= 464m / s
86, 400 s
∆t
464m / s )
(
v
2
a=
=
=
.034
m
/
s
c
r 6.38 x106 m
2
2
At Santa Rosa, r = 6.38 x 106m cos38:
2π r 2π (6.38 x106 m) cos 38
v =
=
= 366m / s
86, 400 s
∆t
366m / s )
(
v
2
=
ac =
=
.027
m
/
s
r cos 38 6.38 x106 m cos 38
2
2
ac
ac
What is the total acceleration
acting on a person in Santa Rosa?
The vector sum.
ac
g
Is your apparent weight as
measured on a spring scale more
at the Equator or at Santa Rosa?
ac
g
Since you are standing on the Earth
(and not in the can) the centrifugal force tends
to throw you off the Earth. You weigh less
where the centripetal force is greatest because
that is also where the centrifugal force is
greatest – the force that tends to throw you
out of a rotating reference frame.
Centripetal & Centrifugal Force
Depends on Your Reference Frame
Outside Observer
(non-rotating frame) sees
Centripetal Force pulling
can in a circle.
Center-seeking
Center-fleeing
Inside Observer
(rotating reference frame)
feels Centrifugal Force
pushing them against the
can.
Centrifugal Force is Fictitious?
The centrifugal force is a real
effect. Objects in a rotating frame
feel a centrifugal force acting on
them, trying to push them out. This
is due to your inertia – the fact that
your mass does not want to go in a
circle. The centrifugal force is
called ‘fictitious’ because it isn’t
due to any real force – it is only due
to the fact that you are rotating.
The centripetal force is ‘real’
because it is due to something
acting on you like a string or a car.
Artificial Gravity
How fast would the space station segments A and B have to rotate in
order to produce an artificial gravity of 1 g?
v A = 56m / s ~ 115mph
vB = 104m / s ~ 210mph
Important: Inside vs Outside the
Rotating Frame
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Horizontal (Flat) Curve
 The force of static
friction supplies the
centripetal force
mv
∑ F=c f= r
2
 The maximum speed at
which the car can
negotiate the curve is
v = µ gr
 Note, this does not
depend on the mass of
the car
© 2010 Pearson Education, Inc.
∑F
y
=N − mg =0
f = µ mg
Example Problem
A level curve on a
country road has a
radius of 150 m. What
is the maximum speed
at which this curve can
be safely negotiated on
a rainy day when the
coefficient of friction
between the tires on a
car and the road is
0.40?
© 2010 Pearson Education, Inc.
Slide 6-28
Horizontal (Flat) Curve
A highway curve has a radius of 0.14 km and is
unbanked. A car weighing 12 kN goes around the
curve at a speed of 24 m/s without slipping. What
is the magnitude of the horizontal force of the
road on the car? What is μ? Draw FBD.
a.
12 kN
b.
17 kN
c.
13 kN
d.
5.0 kN
e.
49 kN
© 2010 Pearson Education, Inc.
Banked Curve
These are designed with friction
equaling zero - there is a
component of the normal force
that supplies the centripetal force
that keeps the car moving in a
circle.
2
mv
=
sin θ
∑ Fr n=
r
∑ F=y n cosθ − mg= 0
Dividing:
2
v
tan θ =
rg
Banked Curve
A race car travels 40 m/s around a banked (45° with
the horizontal) circular (radius = 0.20 km) track.
What is the magnitude of the resultant force on the
80-kg driver of this car?
a.
0.68 kN
b.
0.64 kN
c.
0.72 kN
d.
0.76 kN
e.
0.52 kN
Vertical Circle has Non-Uniform Speed
Where is the speed Max? Min?
Where is the Tension Max? Min?
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Example Problem: Loop-the-Loop
A roller coaster car goes through a vertical loop at a constant
speed. For positions A to E, rank order the:
• centripetal acceleration
• normal force
• apparent weight
© 2010 Pearson Education, Inc.
Slide 6-32
QuickCheck 8.10
Humps in the Road: Outside the Vertical Loop
A car that’s out of gas coasts
over the top of a hill at a steady
20 m/s. Assume air resistance
is negligible. Which free-body
diagram describes the car at
this instant?
© 2010 Pearson Education, Inc.
Slide 8-80
QuickCheck 8.10
Humps in the Road: Outside the Vertical Loop
A car that’s out of gas coasts
over the top of a hill at a steady
20 m/s. Assume air resistance
is negligible. Which free-body
diagram describes the car at
this instant?
© 2010 Pearson Education, Inc.
Now the centripetal
acceleration points down.
Slide 8-81
QuickCheck 8.11
Loop d’ Loops: Inside the Vertical Loop
A roller coaster car does a loopthe-loop. Which of the free-body
diagrams shows the forces on
the car at the top of the loop?
Rolling friction can be neglected.
© 2010 Pearson Education, Inc.
Slide 8-82
Humps in the Road
Outside the Vertical Loop
A roller-coaster car has a mass of 500 kg when fully
loaded with passengers. The car passes over a hill of
radius 15 m, as shown. At the top of the hill, the car has
a speed of 8.0 m/s. What is the force of the track on the
car at the top of the hill?
a.
7.0 kN up
b.
7.0 kN down
c.
2.8 kN down
d.
2.8 kN up
e.
5.6 kN down
© 2010 Pearson Education, Inc.
Maximum Speed for Vertical
Circular Motion
Humps in the Road
What is the maximum speed the car can have as it
passes this highest point without losing contact with
the road?
:
Max speed without losing contact MEANS:
n
Take : n = 0
Therefore:
mv
mg =
r
2
v = gr
mg
Maximum Speed to not loose contact with road only depends on R!
ROOT GRRRRRRRR
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What is the maximum speed the vehicle
can have at B and still remain on the
track?
© 2010 Pearson Education, Inc.
QuickCheck 8.11
Loop d’ Loops: Inside the Vertical Loop
A roller coaster car does a loopthe-loop. Which of the free-body
diagrams shows the forces on
the car at the top of the loop?
Rolling friction can be neglected.
The track is above the car, so
the normal force of the track
pushes down.
© 2010 Pearson Education, Inc.
Slide 8-83
Loop d’ Loops: Inside the Vertical Loop
Minimum Speed to get to the Top.
What is the minimum speed so that
the car barely make it around the
loop the riders are upside down and
feel weightless ? R = 10.0m
© 2010 Pearson Education, Inc.
Loop d’ Loops: Inside the Vertical Loop
A roller-coaster car has a mass of 500 kg when fully loaded with passengers. At
the bottom of a circular dip of radius 40 m (as shown in the figure) the car has a
speed of 16 m/s. What is the magnitude of the force of the track on the car at the
bottom of the dip?
a.
3.2 kN
b.
8.1 kN
c.
4.9 kN
d.
1.7 kN
e.
5.3 kN
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Universal Law of Gravity
G = 6.67 x10
d
m
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GmM
F=
2
d
−11
2
Nm
2
kg
M
Gravity: Inverse Square Law
GmM
F=
2
d
© 2010 Pearson Education, Inc.
Gravitational Force INSIDE the Earth
Inside the Earth the Gravitational Force is Linear.
Acceleration decreases as you fall to the center (where your
speed is the greatest) and then the acceleration increases
but in the opposite direction, slowing you down to a stop at
the other end…but then you would fall back in again,
bouncing back and forth forever!
© 2010 Pearson Education, Inc.
Finding little g
Calculate the acceleration of gravity acting on you at the
surface of the Earth. What is g?
F=
Source of the Force
This is your
WEIGHT!
Gm you M E
RE
F = m you a
2
Gm you M E
RE
2
= m you a
GM E
a=
RE2
© 2010 Pearson Education, Inc.
Response to the Force
Independent
of your mass!
This is why a rock and
feather fall with the
same acceleration!
Finding little g
Calculate the acceleration of gravity acting on you at the
surface of the Earth. What is g?
F=
Gm you M E
F = m you a
2
E
R
GM E
a=
RE2
Source of the Force
6.673 x10
(
a=
−11
Reaction to the Force
Nm 2 / kg 2 )( 5.98 x1024 kg )
( 6.38x10 m )
6
2
a = 9.81 m / s 2
© 2010 Pearson Education, Inc.
= g!
Earth-Moon Gravity
Calculate the magnitude of the force of gravity between the
Earth and the Moon. The distance between the Earth and Moon
centers is 3.84x108m
FEM
FEM =
GmM
=
2
d
−11
2
2
24
22
6.673
x
10
Nm
/
kg
5.98
x
10
kg
7.35
x
10
kg )
(
)(
)(
( 3.84 x10 m )
8
2
FEM = 2.01x10 N
20
Earth-Moon Gravity
Calculate the acceleration of the Earth due to the Earth-Moon
gravitational interaction.
FEM
aE =
mE
20
2.01x10 N
=
24
5.98 x10 kg
−5
aE = 3.33 x10 m / s
2
Earth-Moon Gravity
Calculate the acceleration of the Moon due to the Earth-Moon
gravitational interaction.
FEM
aM =
mM
2.01x1020 N
=
22
7.35 x10 kg
−3
aM = 2.73 x10 m / s
2
Earth-Moon Gravity
The acceleration of gravity at the Moon due to the Earth is:
−3
aM = 2.73 x10 m / s
2
The acceleration of gravity at the Earth due to the moon is:
−5
aE = 3.33 x10 m / s
2
Why the difference?
FORCE is the same. Acceleration is NOT!!!
BECAUSE MASSES ARE DIFFERENT!
Force is not Acceleration!
FEarth on Moon = − FMoon on Earth
The forces are equal but the accelerations are not!
Checking Understanding:
Gravity on Other Worlds
A 60 kg person stands on each of the following planets. Rank
order her weight on the three bodies, from highest to lowest.
A.
B.
C.
D.
E.
A>B>C
B>A>C
B>C>A
C>B>A
C>A> B
© 2010 Pearson Education, Inc.
Slide 6-38
Answer
A 60 kg person stands on each of the following planets. Rank
order her weight on the three bodies, from highest to lowest.
A.
B.
C.
D.
E.
A>B>C
B>A>C
B>C>A
C>B>A
C>A> B
© 2010 Pearson Education, Inc.
Slide 6-39
Recall: Curvature of Earth
Curvature of the Earth: Every 8000 m, the Earth curves by 5 meters!
If you threw the ball at 8000 m/s off the surface of the Earth
(and there were no buildings or mountains in the way)
how far would it travel in the vertical direction in 1 second?
=
∆y
1 2
gt ~ 5=
m / s 2 ⋅ (1s 2 ) 5m
2
The ball will achieve orbit.
Orbital Velocity
If you can throw a ball at 8000m/s, the Earth curves away
from it so that the ball continually falls in free fall around the
Earth – it is in orbit around the Earth!
Ignoring
air
resistance.
Above the atmosphere
Projectile Motion/Orbital Motion
Projectile Motion is Orbital motion that hits the Earth!
Orbital Motion| & Escape Velocity
8km/s: Circular orbit
Between 8 & 11.2 km/s: Elliptical orbit
11.2 km/s: Escape Earth
42.5 km/s: Escape Solar System!
Orbit Question
Find the orbital speed of a satellite 200 km above the Earth.
Assume a circular
orbit. M E 5.97
=
=
x1024 kg , RE 6.38 x106 m
ms M E G
F=
2
r
v2
a=
r
F = ms a
2
ms M E G
v
=
F =
ms
2
r
r
v=
−11
24
M EG
RE + h
2
What is this?
2
(5.97 x10 kg )(6.67 x10 Nm / kg )
v=
6
6.58 x10 m
v = 7.78 x10 m / s
3
Notice that this
is less 8km/s!
Orbit Question
What is the period of a satellite orbiting 200 km above the Earth?
Assume a circular=
orbit. M E 5.97
=
x1024 kg , RE 6.38 x106 m
2π r
v=
Τ
If you don’t know the velocity:
GMm
v2
=
F =
m
2
r
r
2π r
Τ=
v
(2π r / Τ) 2
=m
r
2π (6.58 x10 m)
=
3
7.78 x10 m / s
6
=
Τ 5314
=
s 88 min
2
4
π
r3
Τ2 =
GM
Kepler’s 3rd!
Period increases with r!
g and v Above the Earth’s Surface
If an object is some
distance h above the
Earth’s surface, r
becomes RE + h
F=
g=
GMm
r2
GM E
( RE + h )
2
v2
=g
r
The tangential speed of an
object is its orbital speed
and is given by the
centripetal acceleration, g:
GM E
v2
=
RE + h ( RE + h )2
Orbital speed decreases with
increasing altitude!
GM E
v=
( RE + h)
Additional Questions
A satellite orbits the earth. A Space Shuttle crew is sent to boost the
satellite into a higher orbit. Which of these quantities increases?
A.
B.
C.
D.
E.
Speed
Angular speed
Period
Centripetal acceleration
Gravitational force of the earth
© 2010 Pearson Education, Inc.
Slide 6-43
Answer
A satellite orbits the earth. A Space Shuttle crew is sent to boost the
satellite into a higher orbit. Which of these quantities increases?
A.
B.
C.
D.
E.
Speed
Angular speed
Period
Centripetal acceleration
Gravitational force of the earth
© 2010 Pearson Education, Inc.
Slide 6-44