Transcript Chapter 8: Motion in Circles

Chapter 8: Motion in Circles
 8.1 Circular Motion
 8.2 Centripetal Force
 8.3 Universal Gravitation and Orbital Motion
Chapter 8 Objectives
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Calculate angular speed in radians per second.
Calculate linear speed from angular speed and viceversa.
Describe and calculate centripetal forces and
accelerations.
Describe the relationship between the force of gravity and
the masses and distance between objects.
Calculate the force of gravity when given masses and
distance between two objects.
Describe why satellites remain in orbit around a planet.
Chapter 8 Vocabulary
 angular displacement
 linear speed
 angular speed
 orbit
 axis
 radian
 centrifugal force
 centripetal acceleration
 centripetal force
 circumference
 ellipse
 gravitational constant
 law of universal gravitation
 revolve
 rotate
 satellite
Inv 8.1 Motion in Circles
Investigation Key
Question:
How do we describe
circular motion?
8.1 Motion in Circles
 We say an object rotates
about its axis when the
axis is part of the moving
object.
 A child revolves on a
merry-go-round because
he is external to the merrygo-round's axis.
8.1 Motion in Circles
 Earth revolves around the
Sun once each year while
it rotates around its northsouth axis once each day.
8.1 Motion in Circles
 Angular speed is the rate at which an object
rotates or revolves.
 There are two ways to measure angular speed
 number of turns per unit of time (rotations/minute)
 change in angle per unit of time (deg/sec or rad/sec)
8.1 Circular Motion
 A wheel rolling along the ground has both a
linear speed and an angular speed.
 A point at the
edge of a wheel
moves one
circumference in
each turn of the
circle.
8.1 The relationship between linear
and angular speed
 The circumference is the distance around a
circle.
 The circumference depends on the radius of the
circle.
8.1 The relationship between linear
and angular speed
 The linear speed (v) of a point at the edge of a
turning circle is the circumference divided by
the time it takes to make one full turn.
 The linear speed of a point on a wheel depends
on the radius, r, which is the distance from the
center of rotation.
8.1 The relationship between linear
and angular speed
Circumference
(m)
C = 2π r
Radius (m)
Distance (m)
Speed
(m/sec)
v = d2π r
t
Time (sec)
8.1 The relationship between linear
and angular speed
Linear speed
(m/sec)
v=wr
Radius (m)
Angular speed
(rad/sec)
*Angular speed is represented with
a lowercase Greek omega (ω).
Calculate linear from angular
speed
Two children are spinning around on a merry-goround. Siv is standing 4 meters from the axis of
rotation and Holly is standing 2 meters from the
axis. Calculate each child’s linear speed when the
angular speed of the merry go-round is 1 rad/sec?
1.
2.
You are asked for the children’s linear speeds.
You are given the angular speed of the merry-go-round and radius to
each child.
3.
4.
Use v = ωr
Solve:
For Siv: v = (1 rad/s)(4 m) v = 4 m/s.
For Holly: v = (1 rad/s)(2 m) v = 2 m/s.
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8.1 The units of radians per second
 One radian is the angle
you get when you rotate
the radius of a circle a
distance on the
circumference equal to the
length of the radius.
 One radian is
approximately 57.3
degrees, so a radian is a
larger unit of angle
measure than a degree.
8.1 The units of radians per second
 Angular speed
naturally comes out
in units of radians
per second.
 For the purpose of
angular speed, the
radian is a better unit
for angles.
 Radians are better for angular speed because a radian is
a ratio of two lengths.
8.1 Angular Speed
Angular speed
(rad/sec)
w=q
t
Angle turned (rad)
Time taken (sec)
Calculating angular speed
in rad/s
A bicycle wheel makes six turns in
2 seconds. What is its angular speed in
radians per second?
1.
You are asked for the angular speed.
2.
You are given turns and time.
3.
There are 2π radians in one full turn. Use: ω = θ ÷ t
4.
Solve: ω = (6 × 2π) ÷ (2 s) = 18.8 rad/s
8.1 Relating angular speed, linear speed
and displacement
 As a wheel rotates, the point touching the
ground passes around its circumference.
 When the wheel has turned one full rotation, it
has moved forward a distance equal to its
circumference.
 Therefore, the linear speed of a wheel is its
angular speed multiplied by its radius.
Calculating angular speed
from linear speed
A bicycle has wheels that are 70 cm in diameter (35
cm radius). The bicycle is moving forward with a
linear speed of 11 m/s. Assume the bicycle wheels
are not slipping and calculate the angular speed of
the wheels in rpm.
1.
You are asked for the angular speed in rpm.
2.
You are given the linear speed and radius of the wheel.
3.
Use: v = ωr, 1 rotation = 2π radians
4.
Solve: ω = v ÷ r = (11 m/s) ÷ (0.35 m) = 31.4 rad/s.
 Convert to rpm: 31.4 rad x 60 s x 1 rotation = 300 rpm
1s
1 min 2 π rad
Chapter 8: Motion in Circles
 8.1 Circular Motion
 8.2 Centripetal Force
 8.3 Universal Gravitation and Orbital Motion
Inv 8.2 Centripetal Force
Investigation Key Question:
Why does a roller coaster stay on a track upside
down on a loop?
8.2 Centripetal Force
 We usually think of acceleration as a change
in speed.
 Because velocity includes both speed and
direction, acceleration can also be a change in
the direction of motion.
8.2 Centripetal Force
 Any force that causes an object to move in a
circle is called a centripetal force.
 A centripetal force is always perpendicular to
an object’s motion, toward the center of the
circle.
8.2 Calculating centripetal force
 The magnitude of the centripetal force needed to
move an object in a circle depends on the
object’s mass and speed, and on the radius of
the circle.
8.2 Centripetal Force
Mass (kg)
Centripetal
force (N)
Fc = mv2
r
Linear speed
(m/sec)
Radius of path
(m)
Calculating centripetal force
1.
A 50-kilogram passenger on an amusement
park ride stands with his back against the wall
of a cylindrical room with radius of 3 m. What
is the centripetal force of the wall pressing into
his back when the room spins and he is
moving at 6 m/sec?
You are asked to find the centripetal force.
2.
You are given the radius, mass, and linear speed.
3.
Use: Fc = mv2 ÷ r
4.
Solve: Fc = (50 kg)(6 m/s)2 ÷ (3 m) = 600 N
8.2 Centripetal Acceleration
 Acceleration is the rate at which an object’s
velocity changes as the result of a force.
 Centripetal acceleration is the acceleration of
an object moving in a circle due to the
centripetal force.
8.2 Centripetal Acceleration
Centripetal
acceleration (m/sec2)
ac = v 2
r
Speed
(m/sec)
Radius of path
(m)
Calculating centripetal acceleration
A motorcycle drives around a bend with a 50meter radius at 10 m/sec. Find the motor
cycle’s centripetal acceleration and compare
it with g, the acceleration of gravity.
1.
2.
3.
4.
5.
You are asked for centripetal acceleration and a
comparison with g (9.8 m/s2).
You are given the linear speed and radius of the motion.
Use: ac = v2 ÷ r
4. Solve: ac = (10 m/s)2 ÷ (50 m) = 2 m/s2
The centripetal acceleration is about 20%, or 1/5 that of
gravity.
8.2 Centrifugal Force
 We call an object’s tendency to
resist a change in its motion its
inertia.
 An object moving in a circle is
constantly changing its direction
of motion.
 Although the centripetal force pushes you toward the
center of the circular path...it seems as if there also is a
force pushing you to the outside.
 This “apparent” outward force is often incorrectly
identified as centrifugal force.
8.2 Centrifugal Force
 Centrifugal force is not a true
force exerted on your body.
 It is simply your tendency to
move in a straight line due to
inertia.
 This is easy to observe by twirling a small object at the
end of a string.
 When the string is released, the object flies off in a
straight line tangent to the circle.
Chapter 8: Motion in Circles
 8.1 Circular Motion
 8.2 Centripetal Force
 8.3 Universal Gravitation and Orbital Motion
Inv 8.3 Universal Gravitation and Orbital
Motion
Investigation Key Question:
How strong is gravity in other
places in the universe?
8.3 Universal Gravitation
and Orbital Motion
 Sir Isaac Newton first deduced that
the force responsible for making
objects fall on Earth is the same
force that keeps the moon in orbit.
 This idea is known as the law of
universal gravitation.
 Gravitational force exists between
all objects that have mass.
 The strength of the gravitational
force depends on the mass of the
objects and the distance between
them.
8.3 Law of Universal Gravitation
Force (N)
F = m1m2
r2
Mass 1
Mass 2
Distance between
masses (m)
Calculating the weight of a person
on the moon
The mass of the Moon is 7.36 × 1022 kg. The
radius of the moon is 1.74 × 106 m. Use the
equation of universal gravitation to calculate the
weight of a 90-kg astronaut on the Moon’s surface.
1.
You are asked to find a person’s weight on the Moon.
2.
You are given the radius and the masses.
3.
Use: Fg = Gm1m2 ÷ r 2
4.
Solve:
8.3 Orbital Motion
 A satellite is an object
that is bound by gravity
to another object such as
a planet or star.
 An orbit is the path
followed by a satellite.
The orbits of many
natural and man-made
satellites are circular, or
nearly circular.
8.3 Orbital Motion
 The motion of a satellite is
closely related to projectile
motion.
 If an object is launched above
Earth’s surface at a slow
speed, it will follow a
parabolic path and fall back to
Earth.
 At a launch speed of about 8 kilometers per second,
the curve of a projectile’s path matches the curvature
of the planet.
8.3 Satellite Motion
 The first artificial satellite, Sputnik I, which
translates as “traveling companion,” was
launched by the former Soviet Union on
October 4, 1957.
 For a satellite in a circular orbit, the force of
Earth’s gravity pulling on the satellite equals the
centripetal force required to keep it in its orbit.
8.3 Orbit Equation
 The relationship between a satellite’s orbital
radius, r, and its orbital velocity, v is found by
combining the equations for centripetal and
gravitational force.
8.3 Geostationary orbits
 To keep up with Earth’s rotation,
a geostationary satellite must
travel the entire circumference
of its orbit (2π r) in 24 hours, or
86,400 seconds.
 To stay in orbit, the satellite’s
radius and velocity must also
satisfy the orbit equation.
Use of HEO
 All geostationary satellites must orbit
directly above the equator.
 This means that the geostationary “belt”
is the prime real estate of the satellite
world.
 There have been international disputes
over the right to the prime
geostationary slots, and there have
even been cases where satellites in
adjacent slots have interfered with each
other.