Transcript Chapter 1 Units and Problem Solving

AP Physics
Chapter 7
Circular Motion and Gravitation
Chapter 7: Circular Motion and
Gravitation
7.1 Angular Measure
7.2 Angular Speed and Velocity
7.3 Uniform Circular Motion and Centripetal
Acceleration
7.4 Angular Acceleration
7.5 Newton’s Law of Gravitation
7.6 Kepler’s Laws and Earth Satellites
Homework for Chapter 7
• Read Chapter 7
• HW 7.A: p.244: 5, 6, 8, 9, 19, 24-28, 30, 33.
• HW 7.B: p.245: 43-46, 48, 50, 51, 63, 64.
• HW 7.C: p.248: 72-76, 78 ,80 ,86-88.
7.1 Angular Measure
7.2 Angular Speed and Velocity
7.1 Angular Measure
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,Ө)
because r is a constant and only Ө
varies.
• Ө is measured counter-clockwise from
the +x axis.
The relationship between
rectangular coordinates
and polar coordinates
are:
x = r cos Ө
y = r sin Ө
7.1 Angular Measure
Angular distance (∆Ө = Ө – Ө0) may be measured in either degrees or radians (rad).
1 rad ≈ 57.3° or 2𝜋 rad = 360°
7.1 Angular Measure
7.1 Angular Measure
7.1 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?
7.2 Angular Speed and Velocity
Linear analogy:
v=∆x
∆t
Linear analogy:
a=∆v
∆t
7.2 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.
7.2 Angular Speed and
Velocity
a) Tangential and angular speeds are
related by v = rω, with ω 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 (ω).
Particles at different distances from
the axis of rotation have different
tangential speeds.
b) Sparks from a grinding wheel
illustrate instantaneous tangential
velocity.
7.2 Angular Speed and Velocity
For every linear quantity or equation there is an analogous angular quantity or
equation. (Assume x0 = 0, θ0 = 0, t0 = 0). Substitute θ → x, ω → v, α → a.
Quantity
Linear / Tangential
Angular
distance (arc length)
s
rθ
tangential speed
v
rω
tangential acceleration
a
rα
displacement
x=vt
θ=ωt
average velocity
v = v + v0
2
ω = ω + ω0
2
kinematics eqn. #1
v = v0 + at
ω = ω0 + αt
kinematics eqn. #2
x = v0t + ½ at2
θ = ω0t + ½ αt2
kinematics eqn. #3
v2 = v02 + 2ax
ω2 = ω02 + 2αθ
7.2 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.
7.2 Angular Speed and Velocity
f = frequency
T = period
ω = angular speed
7.2 Angular Speed and Velocity
•
The SI unit of frequency is 1/sec or hertz (Hz).
7.2 Angular Speed and Velocity
Example 7.2b: A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s.
a) What is the average angular speed of the wheel?
b) What is the tangential speed of a point 0.10 m from the center of the wheel?
c) What is the period?
d) What is the frequency?
Check for Understanding
Check for Understanding
Check for Understanding
Check for Understanding
Homework 7.A
Sections 7.1 & 7.2
• HW 7.A: p.244: 5, 6, 8, 9, 19, 24-28, 30, 33.
7.3 Uniform Circular Motion and
Centripetal Acceleration
7.4 Angular Acceleration
7.3 Uniform Circular Motion and Centripetal Acceleration
Physics Warmup # 35
7.3 Uniform Circular Motion and Centripetal Acceleration
Physics Warmup # 35
7.3 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
object’s velocity changes in the
direction of motion. Therefore, there is
an acceleration.
Fig. 7.8 p.218
7.3 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 = rω2
r
Fig. 7.10, p.219
7.3 Uniform Circular Motion and Centripetal Acceleration
• From Newton’s 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 = mrω2
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.
7.3 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=
2πr
v
=1
f
=2π
ω
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?
(Earth’s gravity)
7.3 Uniform Circular Motion and Centripetal Acceleration
Example 7.a:
7.3 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.
a) What is the source of centripetal force on the car?
b) What is the magnitude of the centripetal acceleration of the car?
c) What is the magnitude of the centripetal force on the car?
7.3 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?
7.3 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
7.3 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
7.3 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.
7.4 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).
7.4 Angular Acceleration
7.4 Angular Acceleration
In uniform circular motion, there is
centripetal acceleration but
no angular acceleration (α = 0) or
tangential acceleration (at = r α = 0).
In nonuniform circular motion, there are
angular and tangential accelerations.
at = ∆v = ∆(rω) = r∆ω = rα
∆t
∆t
∆t
Fig. 7.16, p.226
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
a) the angle the wheel rotates through in 2.0 s
b) the angular speed at t = 2.0 s
7.4 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?
7.4 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
7.4 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.
7.4 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.
Homework 7.B
Sections 7.3 & 7.4
• HW 7.B: p.245: 43-46, 48, 50, 51, 63, 64.
7.5 Newton’s Law of Gravitation
7.6 Kepler’s Laws and Earth
Satellites
7.3 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.
7.5 Newton’s Law of Gravitation
G is the universal gravitational constant.
7.5 Newton’s Law of Gravitation
F α 1
r2
Inverse Square Law
Any two particles, or point masses, are
gravitationally attracted to each other
with a force that has a magnitude given
by Newton’s universal law of gravitation.
For homogeneous spheres, the masses
may be considered to be concentrated at
their centers.
Fig. 7.17, p.228
7.5 Newton’s Law of Gravitation
7.5 Newton’s Law of Gravitation
7.5 Newton’s Law of Gravitation
• We can find the acceleration due to gravity, ag, by setting
Newton’s 2nd Law = the Law of Gravitation
F = mag = GmM
r2
so,
ag = GM
r2
(m cancels out)
This is the acceleration due to gravity at a
distance r from a planet’s center.
• At the Earth’s surface: agE = g = GME
RE2
ME = 6.0 x 1024 kg
RE = 6.4 x 106 m
where ME and RE are the mass and radius of the Earth.
• At an altitude h above the Earth’s surface: ag = GME
(RE + h)2
7.5 Newton’s 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.
7.5 Newton’s Law of Gravitation
Note: it is just r, not r2, in the
denominator.
7.5 Newton’s Law of Gravitation
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.
7.5 Newton’s 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.
a) What are the mutual attractive forces when the electron is in these orbits,
respectively?
a) If the electron jumps from the large orbit to the small one, what is the change in
potential energy?
7.5 Newton’s Law of Gravitation: Check for Understanding
1. The gravitational force is
a. a linear function of distance
b. an infinite-range force
c. applicable only to our solar system
d. sometimes repulsive
Answer: b
7.5 Newton’s Law of Gravitation: Check for Understanding
2. The acceleration due to gravity on the Earth’s surface
a. is a universal constant like G
b. does not depend on the Earth’s mass
c. is directly proportional to the Earth’s radius
d. does not depend on the object’s mass
Answer: d
7.5 Newton’s 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 (w=mg). The “floating” occurs because the spacecraft
and astronauts are “falling” (“accelerating” toward Earth at the same rate).
7.3 Uniform Circular Motion and Centripetal Acceleration
Boeing 747
Freedom 7
Space Shuttle
ISS
Hubble
Physics Warmup # 33
7.6 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)
7.6 Satellites
7.6 Satellites
7.6 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.
7.6 Satellites
7.6 Satellites
7.6 Satellites
7.6 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
r
tangential velocity
of an orbiting body
• Kepler’s third law can be derived from this expression. Since v = 2 𝜋 r / T
(circumference / period), and M is the mass of the Sun,
2 𝜋 r = GM
T
r
Squaring both sides and solving for T2 gives
T2 = 4 𝜋2 r3 or T2 = Kr3 Kepler’s 3rd Law or
GM
Kepler’s Law of Periods
For our solar system, K = 2.97 x 10-19 s2/m3
7.6 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?
7.6 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
a) the time period (T) of the satellite
b) the speed (v) of the satellite
7.6 Satellites
escape speed – the initial speed needed to escape from the surface of a planet
or moon.
• At the top of a planet’s 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
r
• On Earth, since g = GME/ RE2,
vesc =
escape speed
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.
7.6 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?
7.6 Satellites
7.6 Satellites
Note: a geosynchronous satellite orbits the earth with a period of 24 hours so
its motion is synchronized with the earth’s 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).
7.6 Check for Understanding
A Space Shuttle orbits Earth 300 km above the surface. Why can’t 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 Newton’s 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 Shuttle’s kinetic energy would be dissipated as heat in
the atmosphere, degrading the orbit.
7.6 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 Earth’s atmosphere
is practically nonexistent. At 10 km, though, the atmospheric friction would quickly
cause the shuttle to slow down.
7.6 Check
Satellites
for Understanding
7.6 Check
Satellites
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.
7.6 Check for Understanding
7.6 Check for Understanding
Internet Activity
Put a satellite in orbit:
http://www.lon-capa.org/~mmp/kap7/orbiter/orbit.htm
Homework 7.C
Sections 7.5 & 7.6
• HW 7.C: p.248: 72-76, 78, 80, 86-88.
Warmup: TSAR
Think (3 minutes)
quietly about the following question:
How does Newton’s Law of Gravitation relate to the motion of
satellites?
Write your answer using full sentences in paragraph form. You
may include formulas and sketches.
Share (3 minutes)
• Pair up. Read your paragraph word for word to your partner.
Explain any sketches.
• Your partner will listen silently and prepare to give quality
feedback. Your partner may not give feedback yet, but they
may ask for clarification.
• Now, switch roles.
Advise (3 minutes)
• Your partner will now give you advice on your answer.
Listen to your partner, ask questions to clarify, and evaluate
the advice.
• Now, switch roles.
Revise (3 minutes)
• Decide what advice was useful.
• Revise work in red pen.