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© 2010 Pearson Education, Inc.
PowerPoint® Lectures for College Physics: A Strategic Approach, Second Edition
Chapter 6
Circular Motion, Orbits, and Gravity
© 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
Equatorial Rotational Period: 27.5 days
Differential Solar Rotation Entangled Magentic Field Lines makes Sun Spots
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. The centripetal acceleration changes the direction of motion:
∆=∆
vat
Angular Position: Rotation Angle
∆θ = rotation angle
∆s = arc length
r = radius
sr
θ ∆∆ =
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.
: angular velocity (rad/s)v: tangential velocity (m/s)ω
vt rθω ∆
= =∆
v rω=
Angular & Tangential Velocity
is the same for every point & v varies with r. ω
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
• Magnitude is given by:
2
Cvar
=
© 2010 Pearson Education, Inc.
Circular Motion: centripetal acceleration
There is an acceleration because the velocity is changing direction.
Slide 3-43
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
2 2, Cv ra vr
π= =
Τ
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
cvar
= 20.034 /m s
20.027 /m s
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: 62 2 (6.38 10 ) 464 /
86,400r x mv m st sπ π
= = =∆
( )222
6
464 /.034 /
6.38 10= = =c
m sva m sr x m
At Santa Rosa, r = 6.38 x 106m cos38: 62 2 (6.38 10 )cos38 366 /
86,400r x mv m st sπ π
= = =∆
( )222
6
366 /.027 /
cos38 6.38 10 cos38= = =c
m sva m sr x m
ca
ca
What is the total acceleration acting on a person in Santa Rosa?
cag
The vector sum.
Is your apparent weight as measured on a spring scale more at the Equator or at Santa Rosa?
cag
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
Inside Observer (rotating reference frame) feels Centrifugal Force pushing them against the can.
Outside Observer (non-rotating frame) sees Centripetal Force pulling can in a circle.
Center-seeking
Center-fleeing
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?
56 / ~ 115Av m s mph=
104 / ~ 210Bv m s mph=
© 2010 Pearson Education, Inc.
Important: Inside vs Outside the Rotating Frame
© 2010 Pearson Education, Inc.
Horizontal (Flat) Curve
The force of static friction supplies the centripetal force
The maximum speed at which the car can negotiate the curve is
Note, this does not depend on the mass of the car
v grµ=
2
= =∑ cmvF f
r
0= − =∑ yF N mgµ=f mg
© 2010 Pearson Education, Inc.
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?
Slide 6-28
© 2010 Pearson Education, Inc.
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
Horizontal (Flat) Curve
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
tan vrg
θ =
2
sinθ= =∑ rmvF n
rcos 0θ= − =∑ yF n mg
Dividing:
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
Banked Curve
© 2010 Pearson Education, Inc.
Vertical Circle has Non-Uniform Speed
Where is the speed Max? Min? Where is the Tension Max? Min?
© 2010 Pearson Education, Inc.
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
Slide 6-32
© 2010 Pearson Education, Inc.
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?
QuickCheck 8.10
Slide 8-80
Humps in the Road: Outside the Vertical Loop
© 2010 Pearson Education, Inc.
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?
QuickCheck 8.10
Now the centripetal acceleration points down.
Slide 8-81
Humps in the Road: Outside the Vertical Loop
© 2010 Pearson Education, Inc.
A roller coaster car does a loop-the-loop. Which of the free-body diagrams shows the forces on the car at the top of the loop? Rolling friction can be neglected.
QuickCheck 8.11
Slide 8-82
Loop d’ Loops: Inside the Vertical Loop
© 2010 Pearson Education, Inc.
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.
:
n
mg
Humps in the Road
: 0=Take n
=v gr
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:
2mvmgr
=Therefore:
Maximum Speed for Vertical Circular Motion
Maximum Speed to not loose contact with road only depends on R! ROOT GRRRRRRRR
© 2010 Pearson Education, Inc.
What is the maximum speed the vehicle can have at B and still remain on the track?
© 2010 Pearson Education, Inc.
A roller coaster car does a loop-the-loop. Which of the free-body diagrams shows the forces on the car at the top of the loop? Rolling friction can be neglected.
QuickCheck 8.11
The track is above the car, so the normal force of the track pushes down.
Slide 8-83
Loop d’ Loops: Inside the Vertical Loop
© 2010 Pearson Education, Inc.
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
Loop d’ Loops: Inside the Vertical Loop Minimum Speed to get to the Top.
© 2010 Pearson Education, Inc.
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
Loop d’ Loops: Inside the Vertical Loop
© 2010 Pearson Education, Inc.
Universal Law of Gravity
d M
m
211
26.67 10 NmG xkg
−=
2
GmMFd
=
© 2010 Pearson Education, Inc.
Gravity: Inverse Square Law
2
GmMFd
=
© 2010 Pearson Education, Inc.
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!
Gravitational Force INSIDE the Earth
© 2010 Pearson Education, Inc.
Finding little g
Calculate the acceleration of gravity acting on you at the surface of the Earth. What is g?
2you E
E
Gm MF
R= youF m a=
2you E
youE
Gm Mm a
R=
Response to the Force
2E
E
GMaR
=Independent of your mass! This is why a rock and feather fall with the same acceleration!
Source of the Force
This is your WEIGHT!
© 2010 Pearson Education, Inc.
Finding little g
2E
E
GMaR
=
Calculate the acceleration of gravity acting on you at the surface of the Earth. What is g?
2you E
E
Gm MF
R= youF m a=
( )( )( )
11 2 2 24
26
6.673 10 / 5.98 10
6.38 10
−
=x Nm kg x kg
ax m
29.81 /a m s=
Source of the Force Reaction to the Force
= g!
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
2EMGmMF
d=
( )( )( )( )
11 2 2 24 22
28
6.673 10 / 5.98 10 7.35 10
3.84 10
−
=EM
x Nm kg x kg x kgF
x m
202.01 10EMF x N=
Earth-Moon Gravity
Calculate the acceleration of the Earth due to the Earth-Moon gravitational interaction.
EME
E
Fam
=
20
24
2.01 105.98 10
x Nx kg
=
5 23.33 10 /Ea x m s−=
Earth-Moon Gravity
EMM
M
Fam
=
20
22
2.01 107.35 10
x Nx kg
=
3 22.73 10 /Ma x m s−=
Calculate the acceleration of the Moon due to the Earth-Moon gravitational interaction.
Earth-Moon Gravity
The acceleration of gravity at the Moon due to the Earth is:
3 22.73 10 /Ma x m s−=
5 23.33 10 /Ea x m s−=
The acceleration of gravity at the Earth due to the moon is:
Why the difference?
FORCE is the same. Acceleration is NOT!!!
BECAUSE MASSES ARE DIFFERENT!
Earth-Moon Gravity
Force is not Acceleration!
The forces are equal but the accelerations are not!
Earth on Moon Moon on EarthF F= −
© 2010 Pearson Education, Inc.
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. A > B > C B. B > A > C C. B > C > A D. C > B > A E. C > A > B
Slide 6-38
© 2010 Pearson Education, Inc.
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. A > B > C B. B > A > C C. B > C > A D. C > B > A E. C > A > B
Slide 6-39
Recall: Curvature of Earth
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?
Curvature of the Earth: Every 8000 m, the Earth curves by 5 meters!
The ball will achieve orbit.
2 2 21 ~ 5 / (1 ) 52
y gt m s s m∆ = ⋅ =
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!
Above the atmosphere
Ignoring air resistance.
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.
2var
=2
s Em M GFr
= sF m a=
2
2s E
sm M G vF m
r r= = E
E
M GvR h
=+
24 11 2 2
6
(5.97 10 )(6.67 10 / )6.58 10
x kg x Nm kgvx m
−
=
37.78 10 /v x m s=
24 65.97 10 , 6.38 10E EM x kg R x m= =
Notice that this is less 8km/s!
What is this?
Orbit Question
2 rvπ
Τ =
What is the period of a satellite orbiting 200 km above the Earth? Assume a circular orbit.
2 rv π=
Τ
5314 88minsΤ = =
24 65.97 10 , 6.38 10E EM x kg R x m= =
6
3
2 (6.58 10 )7.78 10 /
x mx m s
π=
2
2GMm vF m
rr= =
2(2 / )rmr
π Τ=
22 34 r
GMπ
Τ =
Kepler’s 3rd!
If you don’t know the velocity:
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
( )2E
E
GMgR h
=+
The tangential speed of an object is its orbital speed and is given by the centripetal acceleration, g:
2v gr=
2
E
vR h+
( )E
E
GMvR h
=+
( )2E
E
GMR h
=+
2GMmF
r=
Orbital speed decreases with increasing altitude!
© 2010 Pearson Education, Inc.
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. Speed B. Angular speed C. Period D. Centripetal acceleration E. Gravitational force of the earth
Slide 6-43
© 2010 Pearson Education, Inc.
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. Speed B. Angular speed C. Period D. Centripetal acceleration E. Gravitational force of the earth
Slide 6-44
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