AP%20Physics%201%20%20%20%20%20Chapter%207%20Circular%20Motion%20and%20Gravitation

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AP Physics 1 Chapter 7Circular Motion
and Gravitation
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Chapter 7 Circular Motion and Gravitation

• Angular Measure
• Angular Speed and Velocity
• Uniform Circular Motion and Centripetal
  Acceleration
• Angular Acceleration
• Newtons Law of Gravitation
• Keplers Laws and Earth Satellites

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Homework for Chapter 7

• Read Chapter 7
• Chapter 7  s 15,16,17,21,23,25,26,27,29,51 
• 18,19,22,30,31,32,54,55,60,65,66,68,70,
• 62,63,66,68,70, 71,75
•  33,35,36,37,38,40
• 34,39,56,57,5941-47,53

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Angular MeasureAngular Speed and Velocity
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Angular Measure
The relationship between rectangular coordinates
and polar coordinates are x r cos ? y r
sin ?
rotation axis of rotation lies within the body
(example Earth rotates on its axis) revolution
axis of rotation lies outside the body
(example Earth revolves around the Sun)
Circular motion is conveniently described using
polar coordinates (r,?) because r is a constant
and only ? varies. ? is measured
counter-clockwise from the x axis.
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Angular Measure
Angular distance (?? ? ?0) may be measured in
either degrees or radians (rad). 1 rad 57.3
or 2?? rad 360
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Angular Measure
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Angular Measure
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Angular Measure
Example 7.1 When you are watching the NASCAR
Daytona 500, the 5.5 m long race car subtends and
angle of 0.31. What is the distance from the
race car to you?
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Angular Speed and Velocity
Linear analogy a ? v ? t
Linear analogy v ? x ? t
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Angular Speed and Velocity
The units of angular acceleration are rad/s2.
The way to remember this is the right-hand rule
When the fingers of the right hand are curled in
the direction of rotation, the extended thumb
points in the direction of the angular velocity
or angular acceleration vector.
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Angular Speed and Velocity

• Tangential and angular speeds are related by v
  r?, with ? in radians per second.
• Note, all of the particles rotating about a
  fixed axis travel in circles.
• All of the particles have the same angular speed
  (?).
• Particles at different distances from the axis
  of rotation have different tangential speeds.
• Sparks from a grinding wheel illustrate
  instantaneous tangential velocity.

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Angular Speed and Velocity
For every linear quantity or equation there is an
analogous angular quantity or equation. (Assume
x0 0, ?0 0, t0 0). Substitute ? ? x, ? ? v,
a ? a.
Quantity Linear / Tangential Angular
distance (arc length) s r?
tangential speed v r?
tangential acceleration a ra
displacement x v t ? ? t
average velocity v v v0 2 ? ? ?0 2
kinematics eqn. 1 v v0 at ? ?0 at
kinematics eqn. 2 x v0t ½ at2 ? ?0t ½ at2
kinematics eqn. 3 v2 v02 2ax ?2 ?02 2a?
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Angular Speed and Velocity
When angular speed and velocity are given in
units of rpm (revolutions per minute) you should
first convert them to rad/s before trying to
solve the problem.
Example 7.2a Convert 33 rpm to rad/s.
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Angular Speed and Velocity
f frequency T period ? angular speed
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Angular Speed and Velocity
The SI unit of frequency is 1/sec or hertz (Hz).
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Angular Speed and Velocity

• Example 7.2b A bicycle wheel rotates uniformly
  through 2.0 revolutions in 4.0 s.
• What is the average angular speed of the wheel?
• What is the tangential speed of a point 0.10 m
  from the center of the wheel?
• What is the period?
• What is the frequency?

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Check for Understanding
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Check for Understanding
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Check for Understanding
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Check for Understanding
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Uniform Circular Motion andCentripetal
AccelerationAngular Acceleration
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Uniform Circular Motion and Centripetal
Acceleration
Physics Warmup 35
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Uniform Circular Motion and Centripetal
Acceleration
Physics Warmup 35
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Uniform Circular Motion and Centripetal
Acceleration
uniform circular motion An object moves at a
constant speed in a circular path.
The speed of an object in uniform circular motion
is constant, but the objects velocity changes in
the direction of motion. Therefore, there is an
acceleration.
Fig. 7.8 p.218
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Uniform Circular Motion and Centripetal
Acceleration
centripetal acceleration center-seeking For
and object in uniform circular motion, the
centripetal acceleration is directed towards the
center. There is no acceleration component in
the tangential direction. If there were, the
magnitude of the velocity (tangential speed)
would change. ac v2 r?2 r
Fig. 7.10, p.219
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Uniform Circular Motion and Centripetal
Acceleration
From Newtons second law, Fnet ma. Therefore,
there must be a net force associated with
centripetal acceleration. In the case of
uniform circular motion, this force is called
centripetal force. It is always directed toward
the center of the circle since we know the net
force on an object is in the same direction as
acceleration. Fc mac mv2 mr?2
r Centripetal force
is not a separate or extra force. It is a net
force toward the center of the circle. A
centripetal force is always required for objects
to stay in a circular path. Without it, an object
will fly out along a tangent line due to inertia.

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Uniform Circular Motion and Centripetal
Acceleration
The time period T, the frequency of rotation f,
the radius of the circular path, and the speed of
the particle undergoing uniform circular motion
are related by T 2 p r 1 2 p
v f ? centrifugal
force center-fleeing force a fictitious force
something made up by nonphysicists the vector
equivalent of a unicorn Hint Do not label a
force as centripetal force on your free-body
diagram even if that force does act toward the
center of the circle. Rather, label the actual
source of the force i.e., tension, friction,
weight, electric force, etc. Question 1 What
provides the centripetal force when clothes move
around a dryer? (the inside of the
dryer) Question 2 What provides the centripetal
force upon a satellite orbiting the Earth?
(Earths gravity)
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Uniform Circular Motion and Centripetal
Acceleration
Example 7.a
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Uniform Circular Motion and Centripetal
Acceleration

• Example 7.3 A car of mass 1500 kg is negotiating
  a flat circular curve of radius 50 meters with a
  speed of 20 m/s.
• What is the source of centripetal force on the
  car?
• What is the magnitude of the centripetal
  acceleration of the car?
• What is the magnitude of the centripetal force on
  the car?

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Uniform Circular Motion and Centripetal
Acceleration
Example 7.3a A car approaches a level, circular
curve with a radius of 45.0 m. If the concrete
pavement is dry, what is the maximum speed at
which the car can negotiate the curve at a
constant speed?
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Uniform Circular Motion and Centripetal
Acceleration
Check for Understanding 1. In uniform circular
motion, there is a a. constant velocity b.
constant angular velocity c. zero
acceleration d. net tangential acceleration
Answer b
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Uniform Circular Motion and Centripetal
Acceleration
Check for Understanding 2. If the centripetal
force on a particle in uniform circular motion is
increased, a. the tangential speed will
remain constant b. the tangential speed will
decrease c. the radius of the circular path
will increase d. the tangential speed will
increase and/or the radius will decrease
Answer d Fc mv2
r
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Uniform Circular Motion and Centripetal
Acceleration
Check for Understanding 3. Explain why mud
flies off a fast-spinning wheel. Answer
Centripetal force is proportional to the square
of the speed. When there is insufficient
centripetal force (provided by friction and
adhesive forces), the mud cannot maintain the
circular path and it flies off along a tangent.
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Angular Acceleration

• Average angular acceleration (?) is
• ? ? ?
• ? t
• The SI unit of angular acceleration is rad/s2.
• The relationship between tangential and angular
  acceleration is
• at r ?
• (This is not to be confused with centripetal
  acceleration, ac).

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Angular Acceleration
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Angular Acceleration
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In uniform circular motion, there is centripetal
acceleration but no angular acceleration (a 0)
or tangential acceleration (at r a 0).
In nonuniform circular motion, there are angular
and tangential accelerations. at ?v ?(r?)
r?? ra ?t ?t ?t
Fig. 7.16, p.226
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7.4 Angular Acceleration
There is always centripetal acceleration no
matter whether the circular motion is uniform or
nonuniform. It is the tangential
acceleration that is zero in uniform circular
motion.

• Example 7.4 A wheel is rotating wit a constant
  angular acceleration of 3.5 rad/s2. If the
  initial angular velocity is 2.0 rad/s and is
  speeding up, find
• the angle the wheel rotates through in 2.0 s
• the angular speed at t 2.0 s

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Angular Acceleration
Example 7.5 The power on a medical centrifuge
rotating at 12,000 rpm is cut off. If the
magnitude of the maximum deceleration of the
centrifuge is 50 rad/s2, how many revolutions
does it rotate before coming to rest?
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Angular Acceleration
Check for Understanding 1. The angular
acceleration in circular motion a. is equal
in magnitude to the tangential acceleration
divided by the radius b. increases the
angular velocity if in the same direction c.
has units of rad/s2 d. all of the above
Answer d
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Angular Acceleration
Check for Understanding 2. Can you think of an
example of a car having both centripetal
acceleration and angular acceleration?
Answer Yes, when a car is changing its speed on
a curve.
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Angular Acceleration
Check for Understanding 3. Is it possible for a
car in circular motion to have angular
acceleration but not centripetal acceleration?
Answer No, this is not possible. Any car in
circular motion always has centripetal
acceleration.
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Newtons Law of GravitationKeplers Laws and
Earth Satellites
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Uniform Circular Motion and Centripetal
Acceleration
Physics Warmup 48
Solution It would decrease. You would have mass
below you pulling downward and mass above you
pulling upward. At the center of the earth, you
would weigh zero.
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Newtons Law of Gravitation
G is the universal gravitational constant.
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Newtons Law of Gravitation
F a 1 Inverse Square Law r2
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For homogeneous spheres, the masses may be
considered to be concentrated at their centers.
Any two particles, or point masses, are
gravitationally attracted to each other with a
force that has a magnitude given by Newtons
universal law of gravitation.
Fig. 7.17, p.228
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Newtons Law of Gravitation
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Newtons Law of Gravitation
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Newtons Law of Gravitation
We can find the acceleration due to gravity,
ag, by setting Newtons 2nd Law the Law of
Gravitation F mag GmM (m cancels
out) r2 so, ag GM This
is the acceleration due to gravity at a
r2 distance r from a planets center.
At the Earths surface agE g GME ME 6.0 x
1024 kg RE2 RE 6.4 x 106
m where ME and RE are the mass and radius of
the Earth. At an altitude h above the Earths
surface ag GME (RE
h)2
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Newtons Law of Gravitation
Example 7.7 Calculate the acceleration due to
gravity at the surface of the moon. The radius of
the moon is 1750 km and the mass of the moon is
7.4 x 1022 kg.
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Newtons Law of Gravitation
Note it is just r, not r2, in the denominator.
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Newtons Law of Gravitation
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Gravitational potential energy
U - Gm1m2 r
Note U mgh only applies to objects near the
surface of the earth.
Fig. 7.20, p. 231
On Earth, we are in a negative gravitational
potential energy well. Work must be done against
gravity to get higher in the well in other
words, U becomes less negative. The top of the
well is at infinity, where the gravitational
potential energy is chosen to be zero.
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Newtons Law of Gravitation

• Example 7.6 The hydrogen atom consists of a
  proton of mass 1.67 x 10-27 kg and an orbiting
  electron of mass 9.11 x 10-31 kg. In one of its
  orbits, the electron is 5.4 x 10-11 m from the
  proton and in another orbit, it is 10.6 x 10-11 m
  from the proton.
• What are the mutual attractive forces when the
  electron is in these orbits, respectively?
• If the electron jumps from the large orbit to
  the small one, what is the change in potential
  energy?

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Newtons Law of Gravitation Check for
Understanding

• The gravitational force is
• a linear function of distance
• an infinite-range force
• applicable only to our solar system
• sometimes repulsive

Answer b
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Newtons Law of Gravitation Check for
Understanding

• 2. The acceleration due to gravity on the Earths
  surface
• is a universal constant like G
• does not depend on the Earths mass
• is directly proportional to the Earths radius
• does not depend on the objects mass

Answer d
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Newtons Law of Gravitation Check for
Understanding
3. Astronauts in a spacecraft orbiting the Earth
or out for a spacewalk are seen to float in
midair. This is sometimes referred to as
weightlessness or zero gravity (zero g). Are
these terms correct? Explain why an astronaut
appears to float in or near an orbiting
spacecraft.
Answer No. Gravity acts on the astronauts and
the spacecraft, providing the necessary
centripetal force for the orbit, so g is not zero
and there is weight by definition (wmg). The
floating occurs because the spacecraft and
astronauts are falling (accelerating toward
Earth at the same rate).
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7.3 Uniform Circular Motion and Centripetal
Acceleration
Physics Warmup 33
Boeing 747
Freedom 7
Space Shuttle
ISS
Hubble
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Satellites

• Johannes Kepler (1571-1630)
• German astronomer and mathematician
• formulated three law of planetary motion
• The laws apply not only to planets, but to any
  system of a body revolving about a more massive
  body (such as the Moon, satellites, some comets)

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Satellites
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Satellites
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Satellites
When a planet is nearer to the sun, the radius of
orbit is shorter, and so its linear momentum must
be greater in magnitude (it orbits with greater
speed) for angular momentum to be conserved.
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Satellites
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Satellites
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Satellites
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Satellites

• We can find the tangential velocity of a planet
  or satellite where m is orbiting M.
• Set
• Centripetal Force Force of Gravity
• F mv2 GmM
• r r2
• Solve for v v GM
  tangential velocity
• r of an orbiting body
• Keplers third law can be derived from this
  expression. Since v 2 ?? r / T (circumference /
  period), and M is the mass of the Sun,
• 2 ?? r GM
• T r
• Squaring both sides and solving for T2 gives
• T2 4 ??2 r3 or T2 Kr3
  Keplers 3rd Law or
  GM Keplers Law of Periods

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Satellites
Example 7.8 The planet Saturn is 1.43 x 1012 m
from the Sun. How long does it take for Jupiter
to orbit once about the Sun?
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Satellites

• Example 11 A satellite is placed into a circular
  orbit 1000 km above the surface of the earth (r
  1000 km 6400 km 7400 km). Determine
• the time period (T) of the satellite
• the speed (v) of the satellite

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Satellites
escape speed the initial speed needed to escape
from the surface of a planet or moon. At the
top of a planets potential energy well, U 0.
An object projected to the top of the well would
have an initial velocity of vesc. At the top of
the well, its velocity would be close to zero.
From the conservation of energy, final equals
initial K0 U0 K U ½
mvesc2 GmM 0 0 r
vesc 2GM escape speed
r On Earth, since g
GME/ RE2, vesc 2gRE A tangential speed
less than the escape speed is required for a
satellite to orbit. Notice, escape speed does
not depend on the mass of the satellite.
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Satellites
Example 7.9 If a satellite were launched from
the surface of the Moon, at what initial speed
would it need to begin in order for it to escape
the gravitational attraction of the Moon?
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Satellites
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7.6 Satellites
Note a geosynchronous satellite orbits the earth
with a period of 24 hours so its motion is
synchronized with the earths rotation. Viewed by
an observer on earth, a geosynchronous satellite
appears to be stationary. All geosynchronous
satellites with circular orbits have the same
orbital radius (36,000 km above sea level for
Earth).
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Check for Understanding
A Space Shuttle orbits Earth 300 km above the
surface. Why cant the Shuttle orbit 10 km above
Earth? a) The Space Shuttle cannot go fast
enough to maintain such an orbit. b) Because r
appears in the denominator of Newtons law of
gravitation, the force of gravity is much larger
closer to the Earth this force is too strong to
allow such an orbit. c) The closer orbit would
likely crash into a large mountain such as
Everest because of its elliptical nature. d)
Much of the Shuttles kinetic energy would be
dissipated as heat in the atmosphere, degrading
the orbit.
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Check for Understanding
Answer d. A circular orbit is allowed at any
distance from a planet, as long as the satellite
moves fast enough. At 300 km above the surface
Earths atmosphere is practically nonexistent. At
10 km, though, the atmospheric friction would
quickly cause the shuttle to slow down.
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7.6 Satellites
Check for Understanding
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7.6 Satellites
Check for Understanding
The period of a satellite is given by the
formula T2 K r3. This means a specific period
maps onto a specific orbital radius. Therefore,
there is only one orbital radius for a
geosynchronous satellite with a circular
orbit.
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Check for Understanding
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Check for Understanding
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Internet Activity Put a satellite in
orbit http//www.lon-capa.org/mmp/kap7/orbiter/
orbit.htm