Download Relative Motion in Two Dimensions

PHYSICS
Principles and Problems
Chapter 6: Motion in Two Dimensions
CHAPTER
6
Motion in Two Dimensions
BIG IDEA
You can use vectors and Newton’s laws to
describe projectile motion and circular motion.
CHAPTER
6
Table Of Contents
Section 6.1
Projectile Motion
Section 6.2
Circular Motion
Section 6.3
Relative Velocity
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SECTION
Projectile Motion
6.1
MAIN IDEA
A projectile’s horizontal motion is independent of its vertical
motion.
Essential Questions
•
How are the vertical and horizontal motions of a
projectile related?
•
What are the relationships between a projectile’s height,
time in the air, initial velocity and horizontal distance
traveled?
SECTION
Projectile Motion
6.1
Review Vocabulary
• Motion diagram a series of images showing the
positions of a moving object taken at regular time
intervals.
New Vocabulary
• Projectile
• Trajectory
SECTION
6.1
Projectile Motion
Path of a Projectile
• If you observed the movement of a golf ball being
hit from a tee, a frog hopping, or a free throw
being shot with a basketball, you would notice
that all of these objects move through the air
along similar paths, as do baseballs, and arrows.
• Each path rises and then falls, always curving
downward along a parabolic path.
SECTION
6.1
Projectile Motion
Path of a Projectile (cont.)
• An object shot through the air is called a
projectile.
• A projectile can be a football or a drop of water.
• You can draw a free-body diagram of a launched
projectile and identify all the forces that are acting
on it.
SECTION
6.1
Projectile Motion
Path of a Projectile (cont.)
• No matter what the object is, after a projectile has
been given an initial thrust, if you ignore air
resistance, it moves through the air only under
the force of gravity.
• The force of gravity is what causes the object to
curve downward in a parabolic flight path. Its path
through space is called its trajectory.
SECTION
6.1
Projectile Motion
Path of a Projectile (cont.)
• You can determine an object’s trajectory if you
know its initial velocity.
• Two types of projectile motion are horizontal and
angled.
SECTION
6.1
Projectile Motion
Independence of Motion in Two
Dimensions
Click image to view movie.
SECTION
6.1
Projectile Motion
Angled Launches
• When a projectile is launched at an angle, the
initial velocity has a vertical component as well as
a horizontal component.
• If the object is launched upward, like a ball tossed
straight up in the air, it rises with slowing speed,
reaches the top of its path, and descends with
increasing speed.
SECTION
6.1
Projectile Motion
Angled Launches (cont.)
• The adjoining figure
shows the separate
vertical- and
horizontal-motion
diagrams for the
trajectory of the ball.
SECTION
6.1
Projectile Motion
Angled Launches (cont.)
• At each point in the vertical
direction, the velocity of the
object as it is moving upward
has the same magnitude as
when it is moving downward.
• The only difference is that the
directions of the two velocities
are opposite.
SECTION
6.1
Projectile Motion
Angled Launches (cont.)
• The adjoining figure
defines two quantities
associated with a
trajectory.
SECTION
6.1
Projectile Motion
Angled Launches (cont.)
• One is the maximum
height, which is the
height of the projectile
when the vertical
velocity is zero and
the projectile has only
its horizontal-velocity
component.
SECTION
6.1
Projectile Motion
Angled Launches (cont.)
• The other quantity
depicted is the range, R,
which is the horizontal
distance that the projectile
travels when the initial and
final heights are the same.
• Not shown is the flight
time, which is how much
time the projectile is in the
air.
SECTION
6.1
Projectile Motion
Angled Launches (cont.)
• For football punts,
flight time is often
called hang time.
SECTION
6.1
Projectile Motion
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?
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Step 1: Analyze and Sketch the Problem
• Establish a coordinate system with the initial
position of the ball at the origin.
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
• Show the positions of the ball at the beginning, at
the maximum height, and at the end of the flight.
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
• Draw a motion diagram showing v, a, and Fnet.
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Identify the known and unknown variables.
Known:
Unknown:
yi = 0.0 m
ymax = ?
θi = 66°
t=?
vi = 4.5 m/s
ay = −9.8 m/s2
Vy, max = 0.0 m/s
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Step 2: Solve for the Unknown
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Find the y-component of vi.
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Substitute vi = 4.5 m/s, θi = 66°
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Use symmetry to find the y-component of vf
vyf = - vyi = - 4.1 m/s
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Solve for the maximum height.
Vy,max2 = vyi2 + 2a(ymax – yi)
(0.0 m/s2) = vyi + 2a(ymax – 0.0m/s)
ymax = - vyi / 2a
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Substitute vyi = 4.1 m/s, a = -9.8 m/s2
ymax = (4.1 m/s)2 = 0.86m
2(-9.8 m/s2)
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Solve for the time to return to the launching height.
Vyf = vyi + at
SECTION
Projectile Motion
6.1
The Flight of a Ball (cont.)
Solve for t.
t = vyf – vyi
a
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Substitute vyf = - 4.1 m/s, vyi = 4.1 m/s,
a = - 9.80 m/s2
-4.1 m/s – 4.1 m/s = 0.84 s
-9.8 m/s2
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Step 3: Evaluate the Answer
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
Are the units correct?
Dimensional analysis verifies that the units are
correct.
Do the signs make sense?
All of the signs should be positive.
Are the magnitudes realistic?
0.84 s is fast, but an initial velocity of 4.5 m/s makes
this time reasonable.
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
The steps covered were:
Step 1: Analyze and Sketch the Problem
Establish a coordinate system with the initial position of
the ball at the origin.
Show the positions of the ball at the beginning, at the
maximum height, and at the end of the flight.
Draw a motion diagram showing v, a, and Fnet.
SECTION
6.1
Projectile Motion
The Flight of a Ball (cont.)
The steps covered were:
Step 2: Solve for the Unknown
Find the y-component of vi.
Find an expression for time.
Solve for the maximum height.
Solve for the time to return to the launching height.
Step 3: Evaluate the Answer
SECTION
6.1
Projectile Motion
Forces from Air
• So far, air resistance has been ignored in the analysis
of projectile motion.
• Forces from the air can significantly change the motion
of an object.
– Ex. Flying a kite, parachute for a skydiver.
SECTION
6.1
Projectile Motion
Forces from Air (cont.)
• Picture water flowing from a hose:
– If wind is blowing in the same direction as the
water’s initial movement, the horizontal distance
the water travels increases.
– If wind is blowing in the opposite direction as the
water’s initial movement, the horizontal distance
the water travels decreases.
• Even if the air is not moving, it can have a significant
effect on some moving objects.
– Ex. Paper falling to the ground.
SECTION
6.1
Section Check
A boy standing on a balcony drops one ball and throws
another with an initial horizontal velocity of 3 m/s. Which
of the following statements about the horizontal and
vertical motions of the balls is correct? (Neglect air
resistance.)
A.
The balls fall with a constant vertical velocity and a constant
horizontal acceleration.
B.
The balls fall with a constant vertical velocity as well as a constant
horizontal velocity.
C.
The balls fall with a constant vertical acceleration and a constant
horizontal velocity.
D.
The balls fall with a constant vertical acceleration and an increasing
horizontal velocity.
SECTION
6.1
Section Check
Answer
Reason: The vertical and horizontal motions of a
projectile are independent. The only force
acting on the two balls is the force of
gravity. Because it acts in the vertical
direction, the balls accelerates in the
vertical direction. The horizontal velocity
remains constant throughout the flight of
the balls.
SECTION
6.1
Section Check
Which of the following conditions is met when a
projectile reaches its maximum height?
A. The vertical component of the velocity is zero.
B. The vertical component of the velocity is maximum.
C. The horizontal component of the velocity is maximum.
D. The acceleration in the vertical direction is zero.
SECTION
6.1
Section Check
Answer
Reason: The maximum height is the height at which
the object stops its upward motion and
starts falling down, i.e. when the vertical
component of the velocity becomes zero.
SECTION
6.1
Section Check
Suppose you toss a ball up and catch it while
riding in a bus. Why does the ball fall in your
hands rather than falling at the place where you
tossed it?
SECTION
6.1
Section Check
Answer
Trajectory depends on the frame of reference.
For an observer on the ground, when the bus is
moving, your hand is also moving with the same
velocity as the bus, i.e. the bus, your hand, and the
ball will have the same horizontal velocity. Therefore,
the ball will follow a trajectory and fall back into your
hands.
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
Click image to view movie.
SECTION
6.2
Circular Motion
Centripetal Acceleration
• The angle between position vectors r1 and r2 is
the same as that between velocity vectors v1 and
v 2.
• Thus, ∆r/r = ∆v/v. The equation does not change
if both sides are divided by ∆t.
SECTION
6.2
Circular Motion
Centripetal Acceleration (cont.)
• However, v = ∆r/∆t and a = ∆v/∆t
• Substituting v = ∆r/∆t in the left-hand side and
a = ∆v/∆t in the right-hand side gives the following
equation:
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.
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.
• When an Olympic hammer thrower swings the
hammer, the force is the tension in the chain
attached to the massive ball.
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
Centripetal Acceleration (cont.)
• Rather than labeling this axis x or y, call it c, for
centripetal acceleration. The other axis is in the
direction of the velocity, tangent to the circle. It is
labeled tang for tangential.
• Centripetal force is just another name for the net
force in the centripetal direction. It is the sum of all
the real forces, those for which you can identify
agents that act along the centripetal axis.
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 relationship between the magnitude
of centripetal acceleration (ac) and an object’s
speed (v)?
A.
B.
C.
D.
SECTION
6.2
Section Check
Answer
Reason: From the equation 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 the 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
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
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.
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.)
• You can use the
equations in the figure to
solve problems for relative
motion in two dimensions.
• Velocity of a reference
frame moving relative to
the ground is
.
• Velocity of an
object in the moving frame
is
.
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
End of Custom Shows