Download Relative Motion in Two Dimensions

SECTION
Circular Motion
6.2
MAIN IDEA
An object in circular motion has an acceleration toward the
circle’s center due to an unbalanced force toward the
circle’s center.
Essential Questions
•
Why is an object moving in a circle at a constant speed
accelerating?
•
How does centripetal acceleration depend upon the
object’s speed and the radius of the circle?
•
What causes centripetal acceleration?
SECTION
Circular Motion
6.2
Review Vocabulary
• Average velocity the change in position divided by the
time during which the change occurred; the slope of an
object’s position-time graph.
New Vocabulary
• Uniform circular motion
• Centripetal acceleration
• Centripetal force
SECTION
6.2
Circular Motion
Describing Circular Motion
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SECTION
6.2
Circular Motion
Centripetal Acceleration (cont.)
• Solve the equation for acceleration and give it the
special symbol ac, for centripetal acceleration.
• Centripetal acceleration always points to the
center of the circle. Its magnitude is equal to the
square of the speed, divided by the radius of
motion.
SECTION
6.2
Circular Motion
Centripetal Acceleration (cont.)
• One way of measuring the speed of an object moving in a
circle is to measure its period, T, the time needed for the
object to make one complete revolution.
• During this time, the object travels a distance equal to the
circumference of the circle, 2πr. The object’s speed, then,
is represented by v = 2πr/T.
• Velocity is always tangent to the circular motion
SECTION
6.2
Circular Motion
Centripetal Acceleration (cont.)
• The acceleration of an object moving in a circle is
always in the direction of the net force acting on it,
there must be a net force toward the center of the
circle. This force can be provided by any number
of agents.
SECTION
6.2
Circular Motion
Centripetal Acceleration (cont.)
• When an Olympic hammer thrower swings the
hammer, the force is the tension in the chain
attached to the massive ball.
• However, when the thrower releases the hammer,
it will go in the direction of its velocity.
SECTION
6.2
Circular Motion
Centripetal Acceleration (cont.)
• When an object moves in a circle, the net force
toward the center of the circle is called the
centripetal force.
• To analyze centripetal acceleration situations
accurately, you must identify the agent of the force
that causes the acceleration. Then you can apply
Newton’s second law for the component in the
direction of the acceleration in the following way.
SECTION
6.2
Circular Motion
Centripetal Acceleration (cont.)
• Newton’s Second Law for Circular Motion
• The net centripetal force on an object moving in a
circle is equal to the object’s mass times the
centripetal acceleration.
SECTION
6.2
Circular Motion
Centripetal Acceleration (cont.)
• When solving problems, it is useful to choose a
coordinate system with one axis in the direction of
the acceleration.
• For circular motion, the direction of the
acceleration is always toward the center of the
circle.
SECTION
6.2
Circular Motion
Centrifugal “Force”
• According to Newton’s first law, you will continue
moving with the same velocity unless there is a net
force acting on you.
• The passenger in the
car would continue to
move straight ahead if it
were not for the force of
the car acting in the
direction of the
acceleration.
SECTION
6.2
Circular Motion
Centrifugal “Force” (cont.)
• The so-called centrifugal, or outward force, is a
fictitious, nonexistent force.
• You feel as if you are being
pushed only because you
are accelerating relative to
your surroundings. There is
no real force because there
is no agent exerting a force.
SECTION
6.2
Section Check
Explain why an object moving in a
circle at a constant speed is
accelerating.
SECTION
6.2
Section Check
Answer
Acceleration is the rate of change of velocity, the
object is accelerating due to its constant change in
the direction of its motion.
SECTION
6.2
Section Check
What is the direction of the velocity vector
of an accelerating object?
A. toward the center of the circle
B. away from the center of the circle
C. along the circular path
D. tangent to the circular path
SECTION
Section Check
6.2
Answer
Reason: While constantly changing, the velocity vector
for an object in uniform circular motion is
always tangent to the circle. Vectors are never
curved and therefore cannot be along a
circular path.
SECTION
Relative Velocity
6.3
MAIN IDEA
An object’s velocity depends on the reference frame
chosen.
Essential Questions
•
What is relative velocity?
•
How do you find the velocities of an object in different
reference frames?
SECTION
Relative Velocity
6.3
Review Vocabulary
• resultant a vector that results from the sum of two
other vectors.
New Vocabulary
• Reference frame – a coordinate system from which
motion is viewed is a frame of reference
SECTION
6.3
Relative Velocity
Relative Motion in One Dimension
• Suppose you are in a school bus
that is traveling at a velocity of
8 m/s in a positive direction. You
walk with a velocity of 1 m/s
toward the front of the bus.
• If a friend of yours is standing on
the side of the road watching the
bus go by, how fast would your
friend say that you are moving?
SECTION
6.3
Relative Velocity
Relative Motion in One Dimension (cont.)
• If the bus is traveling at 8 m/s,
this means that the velocity of
the bus is 8 m/s, as measured
by your friend in a coordinate
system fixed to the road.
• When you are standing still,
your velocity relative to the
road is also 8 m/s, but your
velocity relative to the bus is
zero.
SECTION
6.3
Relative Velocity
Relative Motion in One Dimension (cont.)
• In the previous example, your motion is viewed
from different coordinate systems.
• A coordinate system from which motion is
viewed is a reference frame.
SECTION
6.3
Relative Velocity
Relative Motion in One Dimension (cont.)
• Walking at 1m/s towards the front of the bus
means your velocity is measured in the
reference from of the bus.
• Your velocity in the road’s reference frame is
different
• You can rephrase the problem as follows: given
the velocity of the bus relative to the road and
your velocity relative to the bus, what is your
velocity relative to the road?
SECTION
6.3
Relative Velocity
Relative Motion in One Dimension (cont.)
• When a coordinate system is moving, two velocities are
added if both motions are in the same direction, and
one is subtracted from the other if the motions are in
opposite directions.
• In the given figure,
you will find that
your velocity
relative to the
street is 9 m/s, the
sum of 8 m/s and
1 m/s.
SECTION
6.3
Relative Velocity
Relative Motion in One Dimension (cont.)
• You can see that when the velocities are along the
same line, simple addition or subtraction can be
used to determine the relative velocity.
SECTION
6.3
Relative Velocity
Relative Motion in One Dimension (cont.)
• Mathematically, relative velocity is represented as
vy/b + vb/r = vy/r.
• The more general form of this equation is:
Relative Velocity va/b + vb/c + va/c
• The relative velocity of object a to object c is the
vector sum of object a’s velocity relative to object b
and object b’s velocity relative to object c.
SECTION
Relative Velocity
6.3
Relative Motion in One Dimension (cont.)
• Can a moving car be a frame of reference?
•
Yes
SECTION
6.3
Relative Velocity
Relative Motion in Two Dimensions
• The method for adding relative velocities also
applies to motion in two dimensions.
• As with one-dimensional motion, you first draw a
vector diagram to describe the motion and then
you solve the problem mathematically by
resolving vectors into x and y components and
finding the resultant magnitude and direction
SECTION
6.3
Relative Velocity
Relative Motion in Two Dimensions (cont.)
• For example, airline pilots
must take into account the
plane’s speed relative to
the air, and their direction
of flight relative to the air.
They also must consider
the velocity of the wind at
the altitude they are flying
relative to the ground.
SECTION
6.3
Relative Velocity
Relative Motion in Two Dimensions (cont.)
SECTION
6.3
Relative Velocity
Relative Motion in Two Dimensions (cont.)
How are vectors used to describe relative motion in two dimensions?
Resolve the velocity vectors into their x and y components. Each
component gives the speed in the corresponding direction relative to a
given reference frame.
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble
Ana and Sandra are riding on a ferry boat that is
traveling east at a speed of 4.0 m/s. Sandra rolls a
marble with a velocity of 0.75 m/s north, straight
across the deck of the boat to Ana. What is the
velocity of the marble relative to the water?
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
Step 1: Analyze and Sketch the Problem
• Establish a coordinate system.
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
• Draw vectors to represent the velocities of the boat
relative to the water and the marble relative to the
boat.
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
Identify known and unknown variables.
Known:
Unknown:
vb/w = 4.0 m/s
vm/w = ?
vm/b = 0.75 m/s
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
Step 2: Solve for the Unknown
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
Because the two velocities are at right angles, use
the Pythagorean theorem.
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
Substitute vb/w = 4.0 m/s, vm/b = 0.75 m/s
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
Find the angle of the marble’s motion.
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
Substitute vb/w = 4.0 m/s, vm/b = 0.75 m/s
= 11° north of east
The marble is traveling 4.1 m/s at 11° north of east.
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
Step 3: Evaluate the Answer
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
Are the units correct?
Dimensional analysis verifies that the velocity is in m/s.
Do the signs make sense?
The signs should all be positive.
Are the magnitudes realistic?
The resulting velocity is of the same order of magnitude
as the velocities given in the problem.
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
The steps covered were:
Step 1: Analyze and Sketch the Problem
Establish a coordinate system.
Draw vectors to represent the velocities of the
boat relative to the water and the marble relative
to the boat.
SECTION
6.3
Relative Velocity
Relative Velocity of a Marble (cont.)
The steps covered were:
Step 2: Solve for the Unknown
Use the Pythagorean theorem.
Step 3: Evaluate the Answer
SECTION
6.3
Section Check
Steven is walking on the top level of a double-decker bus
with a velocity of 2 m/s toward the rear end of the bus. The
bus is moving with a velocity of 10 m/s. What is the velocity
of Steven with respect to Anudja, who is sitting on the top
level of the bus and to Mark, who is standing on the street?
A.
The velocity of Steven with respect to Anudja is 2 m/s and is 12 m/s
with respect to Mark.
B.
The velocity of Steven with respect to Anudja is 2 m/s and is 8 m/s
with respect to Mark.
C.
The velocity of Steven with respect to Anudja is 10 m/s and is 12
m/s with respect to Mark.
D.
The velocity of Steven with respect to Anudja is 10 m/s and is 8 m/s
with respect to Mark.
SECTION
Section Check
6.3
Answer
Reason: The velocity of Steven with respect to Anudja is 2
m/s since Steven is moving with a velocity of 2
m/s with respect to the bus, and Anudja is at rest
with respect to the bus.
The velocity of Steven with respect to Mark can
be understood with the help of the following
vector representation.
SECTION
6.3
Section Check
Which of the following formulas correctly relates the
relative velocities of objects a, b, and c to each other?
A. va/b + va/c = vb/c
B. va/b  vb/c = va/c
C. va/b + vb/c = va/c
D. va/b  va/c = vb/c
SECTION
6.3
Section Check
Answer
Reason: The relative velocity equation is
va/b + vb/c = va/c.
The relative velocity of object a to object c
is the vector sum of object a’s velocity
relative to object b and object b’s velocity
relative to object c.
SECTION
6.3
Section Check
An airplane flies due south at 100 km/hr relative to the air.
Wind is blowing at 20 km/hr to the west relative to the
ground. What is the plane’s speed with respect to the
ground?
A. (100 + 20) km/hr
B. (100 − 20) km/hr
C.
D.
SECTION
Section Check
6.3
Answer
Reason: Since the two velocities are at right angles, we
can apply the Pythagorean theorem. By using
relative velocity law, we can write:
vp/a2 + va/g2 = vp/g2
CHAPTER
Motion in Two Dimensions
6
Resources
Physics Online
Study Guide
Chapter Assessment Questions
Standardized Test Practice
SECTION
Projectile Motion
6.1
Study Guide
•
The vertical and horizontal motions of a projectile are
independent. When there is no air resistance, the
horizontal motion component does not experience an
acceleration and has constant velocity; the vertical
motion component of a projectile experiences a
constant acceleration under these same conditions.
SECTION
Projectile Motion
6.1
Study Guide
•
The curved flight path a projectile follows is called a
trajectory and is a parabola. The height, time of flight,
initial velocity and horizontal distance of this path are
related by the equations of motion. The horizontal
distance a projectile travels before returning to its initial
height depends on the acceleration due to gravity an on
both components on the initial velocity.
SECTION
Circular Motion
6.2
Study Guide
•
An object moving in a circle at a constant speed has an
acceleration toward the center of the circle because the
direction of its velocity is constantly changing.
•
Acceleration toward the center of the circle is called
centripetal acceleration. It depends directly on the
square of the object’s speed and inversely on the radius
of the circle.
SECTION
Circular Motion
6.2
Study Guide
•
A net force must be exerted by external agents toward
the circle’s center to cause centripetal acceleration.
SECTION
Relative Velocity
6.3
Study Guide
•
A coordinate system from which you view motion is
called a reference frame. Relative velocity is the velocity
of an object observed in a different, moving reference
frame.
•
You can use vector addition to solve motion problems of
an object in a moving reference frame.
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
What is the range of a projectile?
A. the total trajectory that the projectile travels
B. the vertical distance that the projectile travels
C. the horizontal distance that the projectile
travels
D. twice the maximum height of the projectile
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
Reason: When a projectile is launched at an angle, the
straight (horizontal) distance the projectile
travels is known as the range of the projectile.
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
Define the flight time of a trajectory.
A. time taken by the projectile to reach the
maximum height
B. the maximum height reached by the projectile
divided by the magnitude of the vertical velocity
C. the total time the projectile was in the air
D. half the total time the projectile was in the air
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
Reason: The flight time of a trajectory is defined as
the total time the projectile was in the air.
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
What is centripetal force?
Answer: When an object moves in a circle, the net
force toward the center of the circle is called
centripetal force.
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
Donna is traveling in a train due north at 30 m/s.
What is the magnitude of the velocity of Donna
with respect to another train which is running
due south at 30 m/s?
A. 30 m/s + 30 m/s
B. 30 m/s − 30 m/s
C. 302 m/s + 302 m/s
D. 302 m/s − 302 m/s
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
Reason: The magnitude of the velocity of Donna
relative to the ground is 30 m/s and the
magnitude of velocity of the other train
relative to the ground is also 30 m/s.
Now, with this speed, if the trains move in
the same direction, then the relative
speed of one train to another will be zero.
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
Reason: In this case, since the two trains are
moving in opposite directions, the relative
speed (magnitude of the velocity) of Donna
relative to another train is 30 m/s + 30 m/s.
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
What is the relationship between the magnitude
of centripetal acceleration and the radius of a
circle?
A.
B.
C.
D.
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
Reason: Centripetal acceleration always points to
the center of the circle. Its magnitude is
equal to the square of the speed, divided
by the radius of motion. That is,
Therefore,
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
Which of the following formulas can be used to
calculate the period (T) of a rotating object if the
centripetal acceleration (ac) and radius (r) are
given?
A.
C.
B.
D.
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
Reason: We know that the centripetal acceleration always
points toward the center of the circle. Its
magnitude is equal to the square of the speed
divided by the radius of motion. That is, ac = v2/r.
To measure the speed of an object moving in a
circle, we measure the period, T, the time needed
for the object to make one complete revolution.
That is, v = 2πr/T, where 2πr is the
circumference of the circle.
CHAPTER
6
Motion in Two Dimensions
Chapter Assessment
Reason: Substituting for v in the equation
v = 2πr/T, we get,
CHAPTER
6
Motion in Two Dimensions
Standardized Test Practice
A 1.60-m tall girl throws a football at an angle of
41.0° from the horizontal and at an initial
velocity of 9.40 m/s. How far away from the girl
will it land?
A. 4.55 m
B. 5.90 m
C. 8.90 m
D. 10.5 m
CHAPTER
6
Motion in Two Dimensions
Standardized Test Practice
A dragonfly is sitting on a merry-go-round, 2.8
m from the center. If the tangential velocity of
the ride is 0.89 m/s, what is the centripetal
acceleration of the dragonfly?
A. 0.11 m/s2
B. 0.28 m/s2
C. 0.32 m/s2
D. 2.2 m/s2
CHAPTER
6
Motion in Two Dimensions
Standardized Test Practice
The centripetal force on a 0.82-kg object on the
end of a 2.0-m massless string being swung in a
horizontal circle is 4.0 N. What is the tangential
velocity of the object?
A. 2.8 m/s2
B. 3.1 m/s2
C. 4.9 m/s2
D. 9.8 m/s2
CHAPTER
6
Motion in Two Dimensions
Standardized Test Practice
A 1000-kg car enters an 80-m radius curve at 20
m/s. What centripetal force must be supplied by
friction so the car does not skid?
A. 5.0 N
B. 2.5×102 N
C. 5.0×103 N
D. 1.0×103 N
CHAPTER
6
Motion in Two Dimensions
Standardized Test Practice
A jogger on a riverside path sees a rowing team
coming toward him. If the jogger is moving at 10
km/h, and the boat is moving at 20 km/h, how
quickly does the jogger approach the boat?
A. 3 m/s
B. 8 m/s
C. 40 m/s
D. 100 m/s
CHAPTER
6
Motion in Two Dimensions
Standardized Test Practice
Test-Taking Tip
Practice Under Testlike Conditions
Answer all of the questions in the time provided
without referring to your book. Did you complete the
test? Could you have made better use of your time?
What topics do you need to review?
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
Relative Velocity
Another example of combined relative velocities is
the navigation of migrating neotropical songbirds.
In addition to knowing in which direction to fly, a
bird must account for its speed relative to the air
and its direction relative to the ground.
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
Relative Velocity
If a bird tries to fly over the Gulf of Mexico into a
headwind that is too strong, it will run out of energy
before it reaches the other shore and will perish.
Similarly, the bird must account for crosswinds or it
will not reach its destination.
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
The Flight of a Ball
A ball is launched at 4.5 m/s at 66° above the
horizontal. What are the maximum height and flight
time of the ball?
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
Relative Velocity of a Marble
Ana and Sandra are riding on a ferry boat that is
traveling east at a speed of 4.0 m/s. Sandra rolls a
marble with a velocity of 0.75 m/s north, straight
across the deck of the boat to Ana. What is the
velocity of the marble relative to the water?
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
Trajectories of Two Softballs
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
Motion Diagrams for Horizontal and
Vertical Motions
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
Projectiles Launched at an Angle
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
The Flight of a Ball
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
A Player Kicking a Football
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
The Displacement of an Object in Circular
Motion
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
Vectors at the Beginning and End of a Time
Interval
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
Uniform Circular Motion
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
A Nonexistent Force
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
Calculating Relative Velocity
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
The Plane’s Velocity Relative to the
Ground
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
Relative Velocity of a Marble
CHAPTER
6
Motion in Two Dimensions
Chapter Resources
A Hammer Thrower Swings a Hammer
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