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Two kinds of moving in circles
• Rotary (rotational)= axis of motion
at center of movement (sit and
spin)
• Circular motion=axis of rotation
outside of object
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Angular quantities
• Always use radians=r=57.3˚
• 1 revolution=360 ˚ = 2 π radians
–Counterclockwise is +
–Clockwise -
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Holt book p247 #4
• Θ(theta)= angular displacement in rad
• Θ =Δs/ r
• s= arc length in meters. R= radius
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Holt p 248 # 2
• ω(lower case omega)=angular
speed
• measured in rad/s or rev/s
ω = Θ/ Δ t
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Holt p 250 #2
• α ( lower case alpha)= angular
acceleration
• Rad/s/s
• Δ ω/ Δ t
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Chapter 7
Section 1 Circular Motion
Tangential Speed
• tangential speed (vt) =speed along
an imaginary line drawn tangent to
the circular path.
• Larger radius= greater vt
• uniform circular motion= speed is
constant.
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Holt p255 #4
• Vt= r ω
• Where Vt = tangential velocity
(linear velocity)
at=rα
Where at= tangential acceleration
(Linear acceleration)
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Chapter 7
Section 1 Circular Motion
Centripetal Acceleration
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Chapter 7
Section 1 Circular Motion
Centripetal Acceleration
• centripetal acceleration= acceleration
in a circular path at constant speed is
due to a change in direction.
CENTRIPETAL ACCELERATION
vt 2
ac 
r
2
(tangential speed)
centripetal acceleration =
radius of circular path
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Chapter 7
Section 1 Circular Motion
• (a.
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Chapter 7
Section 1 Circular Motion
• Centripetal force =name given to the
net force on an object in uniform
circular motion.
• Any type of force or combination of
forces can provide this net force.
– friction
– gravitational force.
– Tension
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Chapter 7
Section 1 Circular Motion
Centripetal Force,
Centripetal force is always directed
to the center of the circular path.
2
vt
ac 
r
mvt
Fc  mac 
r
2
mass  (tangential speed)2
centripetal force =
radius of circular path
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Chapter 7
Section 1 Circular Motion
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Objectives
• Explain how Newton’s law of universal gravitation
accounts for various phenomena, including satellite
and planetary orbits, falling objects, and the tides.
• Apply Newton’s law of universal gravitation to solve
problems.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Gravitational Force, continued
• The centripetal force that holds the planets in orbit
is the same force that pulls an apple toward the
ground—gravitational force.
• Gravitational force is the mutual force of attraction
between particles of matter.
• Gravitational force depends on the masses and on
the distance between them.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Gravitational Force, continued
Newton's Law of Universal Gravitation
m1m2
Fg  G 2
r
mass 1 mass 2
gravitational force  constant 
(distance between masses)2
• The constant G, called the constant of universal
gravitation, equals 6.673  10–11 N•m2/kg.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Gravitational Force, continued
• The gravitational forces always
equal and opposite.
•
• Action reaction pair.
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
Newton’s Law of Universal Gravitation
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Chapter 7
Section 2 Newton’s Law of
Universal Gravitation
• Newton’s law of gravitation
accounts for ocean tides.
• High and low tides are due to
the gravitational force exerted
on Earth by its moon and the
sun.
• Semi diurnal tidal pattern (two
high and low tides)
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Section 2 Newton’s Law of
Universal Gravitation
Chapter 7
Applying the Law of Gravitation, continued
• weight = mass  gravitational field strength
• Because it depends on gravitational field
strength, weight changes with location:
weight = mg
Fg
GmmE GmE
g

 2
2
m
mr
r
• On the surface of any planet, the value of g, as
well as your weight, will depend on the planet’s
mass and radius.
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Universal gravitation
• What is the gravitational force between you and the
earth given your mass of 70 kg and the earths mass
of 5.98 X1024 kg and a radius of 6.38 X10 6 m?
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Chapter 7
Section 3 Motion in Space
Objectives
• Describe Kepler’s laws of planetary motion.
• Relate Newton’s mathematical analysis of
gravitational force to the elliptical planetary orbits
proposed by Kepler.
• Solve problems involving orbital speed and period.
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Chapter 7
Section 3 Motion in Space
Kepler’s Laws
Kepler’s laws describe the motion of the planets.
• First Law planets travel in elliptical orbits around the
sun, the sun is at one of the focal points.
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Chapter 7
Section 3 Motion in Space
Keplers 2nd law
the planet
travels faster when it
is closer to the sun
and slower when it is farther away.
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Chapter 7
Section 3 Motion in Space
• Kepler’s third law states that T2  r3.
• T= period of planet (time for 1 cycle or
revolution.
• The constant of proportionality is 4p2/Gm,
where m is the mass of the object being
orbited.
2
 4p  3
• Thus
2
T 
r
 Gm 
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Section 3 Motion in Space
Chapter 7
Kepler’s Laws, continued
• Kepler’s third law the period of an object in a circular
orbit depends on the same factors as speed of an
object in a circular orbit :
3
r
T  2p
Gm
m
vt  G
r
• Also Vt=2πr/T
• hat m is the mass of the central object
• The mean radius (r) =distance between the centers of
the two bodies.
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Chapter 7
Section 3 Motion in Space
Planetary Data
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Chapter 7
Section 3 Motion in Space
Sample Problem
Period and Speed of an Orbiting Object
Magellan was the first planetary spacecraft to be
launched from a space shuttle. During the spacecraft’s
fifth orbit around Venus, Magellan traveled at a mean
altitude of 361km. If the orbit had been circular, what
would Magellan’s period and speed have been?
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Chapter 7
Section 3 Motion in Space
Sample Problem, continued
1. Define
Given:
r1 = 361 km = 3.61  105 m
Unknown:
T=?
vt = ?
2. Plan
Choose an equation or situation: Use the equations for
the period and speed of an object in a circular orbit.
r3
T  2p
Gm
vt 
Gm
r
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Chapter 7
Section 3 Motion in Space
Sample Problem, continued
Use Table 1 in the textbook to find the values for the
radius (r2) and mass (m) of Venus.
r2 = 6.05  106 m
m = 4.87  1024 kg
Find r by adding the distance between the spacecraft
and Venus’s surface (r1) to Venus’s radius (r2).
r = r1 + r2
r = 3.61  105 m + 6.05  106 m = 6.41  106 m
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Chapter 7
Section 3 Motion in Space
Sample Problem, continued
3. Calculate
r3
(6.41  10 6 m)3
T  2p
=2p
Gm
(6.673  10 –11 N•m 2 /kg 2 )(4.87  10 24 kg)
T  5.66  10 3 s
Gm
(6.673  10 –11 N•m 2 /kg 2 )(4.87  10 24 kg)
vt 

r
6.41  10 6 m
vt  7.12  10 3 m/s
4. Evaluate
Magellan takes (5.66  103 s)(1 min/60 s)  94 min to complete
one orbit.
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Chapter 7
Section 3 Motion in Space
Weight and Weightlessness
To learn about apparent weightlessness, imagine that
you are in an elevator:
– When the elevator is at rest, the magnitude of the
normal force acting on you equals your weight.
– If the elevator were to accelerate downward at 9.81
m/s2, you and the elevator would both be in free fall.
You have the same weight, but there is no normal
force acting on you.
– This situation is called apparent weightlessness.
– Astronauts in orbit experience apparent
weightlessness.
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Chapter 7
Section 3 Motion in Space
Weight and Weightlessness
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Chapter 7
Section 4 Torque and Simple
Machines
Objectives
• Distinguish between torque and force.
• Calculate the magnitude of a torque on an object.
• Identify the six types of simple machines.
• Calculate the mechanical advantage of a simple
machine.
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Chapter 7
Section 4 Torque and Simple
Machines
Rotational Motion
• Rotational and translational motion can be
analyzed separately.
– For example, when a bowling ball strikes the pins, the pins
may spin in the air as they fly backward.
– These pins have both rotational and translational motion.
• In this section, we will isolate rotational motion.
• In particular, we will explore how to measure the
ability of a force to rotate an object.
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Chapter 7
Section 4 Torque and Simple
Machines
The Magnitude of a Torque
• Torque is a quantity that measures the ability of a
force to rotate an object around some axis.
• How easily an object rotates on both how much
force is applied and on where the force is applied.
• The perpendicular distance from the axis of rotation
to a line drawn along the direction of the force is
equal to d sin q and is called the lever arm.
t = Fd sin q
torque = force  lever arm
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Chapter 7
Section 4 Torque and Simple
Machines
The Magnitude of a Torque, continued
• The applied force may
act at an angle.
• However, the direction of
the lever arm (d sin q) is
always perpendicular to
the direction of the
applied force, as shown
here.
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Chapter 7
Section 4 Torque and Simple
Machines
Torque
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Chapter 7
Section 4 Torque and Simple
Machines
Torque and the Lever Arm
In each example, the cat is pushing on the
door at the same distance from the axis.
To produce the same torque, the cat must
apply greater force for smaller angles.
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Chapter 7
Section 4 Torque and Simple
Machines
The Sign of a Torque
• Torque is a vector quantity. In this textbook, we will
assign each torque a positive or negative sign,
depending on the direction the force tends to rotate
an object.
• We will use the convention that the sign of the torque
is positive if the rotation is counterclockwise and
negative if the rotation is clockwise.
Tip: To determine the sign of a torque, imagine that the torque is
the only one acting on the object and that the object is free to
rotate. Visualize the direction that the object would rotate. If
more than one force is acting, treat each force separately.
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Chapter 7
Section 4 Torque and Simple
Machines
The Sign of a Torque
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Chapter 7
Section 4 Torque and Simple
Machines
Sample Problem
Torque
A basketball is being pushed by two players during tipoff. One player exerts an upward force of 15 N at a
perpendicular distance of 14 cm from the axis of
rotation.The second player applies a downward force
of 11 N at a distance of 7.0 cm from the axis of
rotation. Find the net torque acting on the ball about its
center of mass.
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Chapter 7
Section 4 Torque and Simple
Machines
Sample Problem, continued
1. Define
Given:
F1 = 15 N
d1 = 0.14 m
F2 = 11 N
d2 = 0.070 m
Unknown:
tnet = ?
Diagram:
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Chapter 7
Section 4 Torque and Simple
Machines
Sample Problem, continued
2. Plan
Choose an equation or situation: Apply the
definition of torque to each force,and add up the
individual torques.
t = Fd
tnet = t1 + t2 = F1d1 + F2d2
Tip: The factor sin q is not included in the torque
equation because each given distance is the
perpendicular distance from the axis of rotation to a
line drawn along the direction of the force. In other
words, each given distance is the lever arm.
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Chapter 7
Section 4 Torque and Simple
Machines
Sample Problem, continued
3. Calculate
Substitute the values into the equation and
solve: First,determine the torque produced by each
force.Use the standard convention for signs.
t1 = F1d1 = (15 N)(–0.14 m) = –2.1 N•m
t2 = F2d2 = (–11 N)(0.070 m) = –0.77 N•m
tnet = t1 + t2 = –2.1 N•m – 0.77 N•m
tnet = –2.9 N•m
4. Evaluate
The net torque is negative,so the ball rotates in a
clockwise direction.
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Chapter 7
Section 4 Torque and Simple
Machines
Simple Machines
• A machine is any device that transmits or modifies
force, usually by changing the force applied to an
object.
• All machines are combinations or modifications of six
fundamental types of machines, called simple
machines.
• These six simple machines are the lever, pulley,
inclined plane, wheel and axle, wedge, and screw.
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Chapter 7
Section 4 Torque and Simple
Machines
Simple Machines
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Chapter 7
Section 4 Torque and Simple
Machines
Simple Machines, continued
• Because the purpose of a simple machine is to
change the direction or magnitude of an input force,
a useful way of characterizing a simple machine is to
compare the output and input force.
• This ratio is called mechanical advantage.
• If friction is disregarded, mechanical advantage
can also be expressed in terms of input and output
distance.
Fout
din
MA 

Fin dout
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Chapter 7
Section 4 Torque and Simple
Machines
Simple Machines, continued
The diagrams show two examples of a trunk
being loaded onto a truck.
• In the first example, a force (F1) of
360 N moves the trunk through a
distance (d1) of 1.0 m. This requires
360 N•m of work.
• In the second example, a lesser
force (F2) of only 120 N would be
needed (ignoring friction), but the
trunk must be pushed a greater
distance (d2) of 3.0 m. This also
requires 360 N•m of work.
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Chapter 7
Section 4 Torque and Simple
Machines
Simple Machines, continued
• The simple machines we have considered so far are
ideal, frictionless machines.
• Real machines, however, are not frictionless. Some
of the input energy is dissipated as sound or heat.
• The efficiency of a machine is the ratio of useful
work output to work input.
Wout
eff 
Win
– The efficiency of an ideal
(frictionless) machine is 1, or 100
percent.
– The efficiency of real machines is
always less than 1.
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Chapter 7
Section 4 Torque and Simple
Machines
Mechanical Efficiency
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Chapter 7
Standardized Test Prep
Multiple Choice
1. An object moves in a circle at a constant speed.
Which of the following is not true of the object?
A. Its acceleration is constant.
B. Its tangential speed is constant.
C. Its velocity is constant.
D. A centripetal force acts on the object.
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Chapter 7
Standardized Test Prep
Multiple Choice
1. An object moves in a circle at a constant speed.
Which of the following is not true of the object?
A. Its acceleration is constant.
B. Its tangential speed is constant.
C. Its velocity is constant.
D. A centripetal force acts on the object.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 2–3.
A car traveling at 15 m/s on a flat surface turns in a
circle with a radius of 25 m.
2. What is the centripetal acceleration of the car?
F. 2.4  10-2 m/s2
G. 0.60 m/s2
H. 9.0 m/s2
J. zero
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 2–3.
A car traveling at 15 m/s on a flat surface turns in a
circle with a radius of 25 m.
2. What is the centripetal acceleration of the car?
F. 2.4  10-2 m/s2
G. 0.60 m/s2
H. 9.0 m/s2
J. zero
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 2–3.
A car traveling at 15 m/s on a flat surface turns in a
circle with a radius of 25 m.
3. What is the most direct cause of the car’s centripetal
acceleration?
A. the torque on the steering wheel
B. the torque on the tires of the car
C. the force of friction between the tires and the road
D. the normal force between the tires and the road
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the passage below to answer questions 2–3.
A car traveling at 15 m/s on a flat surface turns in a
circle with a radius of 25 m.
3. What is the most direct cause of the car’s centripetal
acceleration?
A. the torque on the steering wheel
B. the torque on the tires of the car
C. the force of friction between the tires and the road
D. the normal force between the tires and the road
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
4. Earth (m = 5.97  1024 kg) orbits the sun (m = 1.99 
1030 kg) at a mean distance of 1.50  1011 m. What is
the gravitational force of the sun on Earth? (G =
6.673  10-11 N•m2/kg2)
F. 5.29  1032 N
G. 3.52  1022 N
H. 5.90  10–2 N
J. 1.77  10–8 N
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
4. Earth (m = 5.97  1024 kg) orbits the sun (m = 1.99 
1030 kg) at a mean distance of 1.50  1011 m. What is
the gravitational force of the sun on Earth? (G =
6.673  10-11 N•m2/kg2)
F. 5.29  1032 N
G. 3.52  1022 N
H. 5.90  10–2 N
J. 1.77  10–8 N
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
5. Which of the following is a correct interpretation of
the expression ag  g  G mE ?
r2
A. Gravitational field strength changes with an
object’s distance from Earth.
B. Free-fall acceleration changes with an object’s
distance from Earth.
C. Free-fall acceleration is independent of the falling
object’s mass.
D. All of the above are correct interpretations.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
5. Which of the following is a correct interpretation of
the expression ag  g  G mE ?
r2
A. Gravitational field strength changes with an
object’s distance from Earth.
B. Free-fall acceleration changes with an object’s
distance from Earth.
C. Free-fall acceleration is independent of the falling
object’s mass.
D. All of the above are correct interpretations.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
6. What data do you need to calculate the orbital speed
of a satellite?
F. mass of satellite, mass of planet, radius of orbit
G. mass of satellite, radius of planet, area of orbit
H. mass of satellite and radius of orbit only
J. mass of planet and radius of orbit only
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
6. What data do you need to calculate the orbital speed
of a satellite?
F. mass of satellite, mass of planet, radius of orbit
G. mass of satellite, radius of planet, area of orbit
H. mass of satellite and radius of orbit only
J. mass of planet and radius of orbit only
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
7. Which of the following choices correctly describes the
orbital relationship between Earth and the sun?
A. The sun orbits Earth in a perfect circle.
B. Earth orbits the sun in a perfect circle.
C. The sun orbits Earth in an ellipse, with Earth
at one focus.
D. Earth orbits the sun in an ellipse, with the sun
at one focus.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
7. Which of the following choices correctly describes the
orbital relationship between Earth and the sun?
A. The sun orbits Earth in a perfect circle.
B. Earth orbits the sun in a perfect circle.
C. The sun orbits Earth in an ellipse, with Earth
at one focus.
D. Earth orbits the sun in an ellipse, with the sun
at one focus.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the diagram to answer
questions 8–9.
8. The three forces acting on
the wheel have equal
magnitudes. Which force will
produce the greatest torque on the wheel?
F. F1
G. F2
H. F3
J. Each force will produce the same torque.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the diagram to answer
questions 8–9.
8. The three forces acting on
the wheel have equal
magnitudes. Which force will
produce the greatest torque on the wheel?
F. F1
G. F2
H. F3
J. Each force will produce the same torque.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the diagram to answer
questions 8–9.
9. If each force is 6.0 N, the
angle between F1 and F2 is
60.0°, and the radius of the
wheel is 1.0 m, what is the
resultant torque on the wheel?
A. –18 N•m
B. –9.0 N•m
C. 9.0 N•m
D. 18 N•m
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
Use the diagram to answer
questions 8–9.
9. If each force is 6.0 N, the
angle between F1 and F2 is
60.0°, and the radius of the
wheel is 1.0 m, what is the
resultant torque on the wheel?
A. –18 N•m
B. –9.0 N•m
C. 9.0 N•m
D. 18 N•m
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
10. A force of 75 N is applied to a lever. This force lifts a
load weighing 225 N. What is the mechanical
advantage of the lever?
F. 1/3
G. 3
H. 150
J. 300
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
10. A force of 75 N is applied to a lever. This force lifts a
load weighing 225 N. What is the mechanical
advantage of the lever?
F. 1/3
G. 3
H. 150
J. 300
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
11. A pulley system has an efficiency of 87.5 percent.
How much work must you do to lift a desk weighing
1320 N to a height of 1.50 m?
A. 1510 J
B. 1730 J
C. 1980 J
D. 2260 J
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
11. A pulley system has an efficiency of 87.5 percent.
How much work must you do to lift a desk weighing
1320 N to a height of 1.50 m?
A. 1510 J
B. 1730 J
C. 1980 J
D. 2260 J
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
12. Which of the following statements is correct?
F. Mass and weight both vary with location.
G. Mass varies with location, but weight does not.
H. Weight varies with location, but mass does
not.
J. Neither mass nor weight varies with location.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
12. Which of the following statements is correct?
F. Mass and weight both vary with location.
G. Mass varies with location, but weight does not.
H. Weight varies with location, but mass does
not.
J. Neither mass nor weight varies with location.
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
13. Which astronomer discovered that planets travel in
elliptical rather than circular orbits?
A. Johannes Kepler
B. Nicolaus Copernicus
C. Tycho Brahe
D. Claudius Ptolemy
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Chapter 7
Standardized Test Prep
Multiple Choice, continued
13. Which astronomer discovered that planets travel in
elliptical rather than circular orbits?
A. Johannes Kepler
B. Nicolaus Copernicus
C. Tycho Brahe
D. Claudius Ptolemy
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Chapter 7
Standardized Test Prep
Short Response
14. Explain how it is possible for all the water to remain
in a pail that is whirled in a vertical path, as shown
below.
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Chapter 7
Standardized Test Prep
Short Response
14. Explain how it is possible for all the water to remain
in a pail that is whirled in a vertical path, as shown
below.
Answer: The water
remains in the pail even
when the pail is upside
down because the water
tends to move in a
straight path due to
inertia.
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Chapter 7
Standardized Test Prep
Short Response, continued
15. Explain why approximately two high tides take place
every day at a given location on Earth.
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Chapter 7
Standardized Test Prep
Short Response, continued
15. Explain why approximately two high tides take place
every day at a given location on Earth.
Answer: The moon’s tidal forces create two bulges on
Earth. As Earth rotates on its axis once per day, any
given point on Earth passes through both bulges.
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Chapter 7
Standardized Test Prep
Short Response, continued
16. If you used a machine to increase the output force,
what factor would have to be sacrificed? Give an
example.
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Chapter 7
Standardized Test Prep
Short Response, continued
16. If you used a machine to increase the output force,
what factor would have to be sacrificed? Give an
example.
Answer: You would have to apply the input force over
a greater distance. Examples may include any
machines that increase output force at the expense
of input distance.
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Chapter 7
Standardized Test Prep
Extended Response
17. Mars orbits the sun (m = 1.99  1030 kg) at a mean
distance of 2.28  1011 m. Calculate the length of
the Martian year in Earth days. Show all of your
work. (G = 6.673  10–11 N•m2/kg2)
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Chapter 7
Standardized Test Prep
Extended Response
17. Mars orbits the sun (m = 1.99  1030 kg) at a mean
distance of 2.28  1011 m. Calculate the length of
the Martian year in Earth days. Show all of your
work. (G = 6.673  10–11 N•m2/kg2)
Answer: 687 days
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Chapter 7
Section 1 Circular Motion
Centripetal Acceleration
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Chapter 7
Section 1 Circular Motion
Centripetal Force
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Chapter 7
Section 3 Motion in Space
Kepler’s Laws
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Chapter 7
Section 4 Torque and Simple
Machines
The Magnitude of a Torque
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Chapter 7
Section 4 Torque and Simple
Machines
Simple Machines
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