Transcript centripetal force is the

Objective - Lesson 1: Motion
Characteristics for Circular Motion
1.
2.
3.
4.
Speed and Velocity
Acceleration
The Centripetal Force Requirement
Mathematics of Circular Motion
Uniform circular motion
• Uniform circular motion is the motion of an
object in a circle with a ___________ or
uniform speed. The velocity is changing
because the direction of motion is
______________.
Speed: _____________
Direction of motion:
_____________ of the path
Velocity:
•magnitude: ___________
•Direction: _____________
Speed and velocity
• Calculation of the Average Speed
vavg = _________
• The distance of one complete cycle around the
perimeter of a circle is known as the ____________.
Circumference = ____ (R is the radius of the circle)
• The time (T) to make one cycle around the circle is
called one _________.
• The average speed of an object in uniform circular
motion is:
vavg = ____________
vavg =
2·π·R
T
• The average speed and
the Radius of the circle are
_____________________.
• The average speed and
the Period of the circle are
_____________________.
The Direction of the Velocity Vector
The best word that can be used to
describe the direction of the
velocity vector is the word
____________________.
The direction of the velocity vector
at any instant is in the direction of a
tangent line drawn to the circle at
the object's location.
• To summarize, an object moving in uniform
circular motion is moving around the perimeter
of the circle with a __________speed. While the
speed of the object is constant, its velocity is
changing. Velocity, being a vector, has a
constant magnitude but a changing direction.
The direction is always directed _________ to
the circle and as the object turns the circle, the
tangent line is always pointing in a new
direction.
2·π·R
vavg =
T
The average speed is directly proportional to the
____________ and inversely proportional to the
_____________.
Check Your Understanding
• A tube is been placed upon the table and shaped into a
three-quarters circle. A golf ball is pushed into the tube at
one end at high speed. The ball rolls through the tube
and exits at the opposite end. Describe the path of the
golf ball as it exits the tube.
example
• A vehicle travels at a constant
speed of 6.0 meters per
second around a horizontal
circular curve with a radius of
24 meters. The mass of the
vehicle is 4.4 × 103 kilograms.
An icy patch is located at P on
the curve. On the icy patch of
pavement, the frictional force
of the vehicle is zero. Which
arrow best represents the
direction of the vehicle's
velocity when it reaches icy
patch P?
a
b
c
d
Acceleration
• An object moving in uniform circular motion is moving in a circle with a
uniform or constant speed. The velocity vector is constant in
magnitude but changing in direction.
• Since the velocity is changing. The object is __________________.
where vi represents the initial velocity and vf represents the
final velocity after some time of t
Direction of the Acceleration Vector
• The velocity change is directed towards point C - the
________ of the circle.
• The acceleration of the object is dependent upon this velocity
change and is in the same direction as this ____________.
The acceleration is directed towards point C as well - the
___________________________.
example
• The initial and final speed of a ball at two different points in
time is shown below. The direction of the ball is indicated by
the arrow. For each case, indicate if there is an acceleration.
Explain why or why not. Indicate the direction of the
acceleration.
a.
b.
c.
d.
example
• Explain why an object moving in a circle at
constant speed can be said to experience
an acceleration.
example
An object is moving in a clockwise direction around a
circle at constant speed.
1. Which vector below represents the direction of the velocity
vector when the object is located at point B on the circle?
2. Which vector below represents the direction of the acceleration
vector when the object is located at point C on the circle?
3. Which vector below represents the direction of the velocity
vector when the object is located at point C on the circle?
4. Which vector below represents the direction of the acceleration
vector when the object is located at point A on the circle?
The Centripetal Force Requirement
• According to Newton's second law of motion, an
object which experiences an acceleration
requires a __________________.
• The direction of the net force is in the same
direction as the ______________. So for an
object moving in a circle, there must be an
inward force acting upon it in order to cause its
inward acceleration. This is sometimes referred
to as the ________________ force
requirement.
• The word centripetal means _______________.
For object's moving in circular motion, there is a
net force acting towards the center which causes
the object to seek the center.
Centrifugal force is a fictitious force
• centrifugal (center fleeing) force
– A ‘fictitious’ or ‘inertial’ force that is
experienced from INSIDE a circular motion
system
• centripetal (center seeking) force
– A true force that pushes or pulls an object
toward the center of a circular path
The Centripetal Force is
Net Force
• Any object moving in a circle (or along a circular
path) experiences a _______________ force.
This is the centripetal force requirement.
• The word centripetal is merely an adjective used
to describe the direction of the force. We are not
introducing a new type of force but rather
describing the direction of the ________ force
acting upon the object that moves in the circle.
examples of centripetal force
As a car makes a
turn, the force of
_____________
acting upon the
turned wheels of
the car provides
centripetal force
required for
circular motion.
As a bucket of water
is tied to a string
and spun in a circle,
the _____________
force acting upon
the bucket provides
the centripetal force
required for circular
motion.
As the moon
orbits the Earth,
the force of
____________
acting upon the
moon provides
the centripetal
force required for
circular motion.
• To summarize, an object in uniform circular
motion experiences an __________ net force.
This inward force is sometimes referred to as a
_______________ force, where centripetal
describes its direction. Without this centripetal
force, an object could never alter its direction.
The fact that the centripetal force is directed
___________________ to the tangential velocity
means that the force can alter the direction of
the object's velocity vector without altering its
magnitude.
Check your understanding
• An object is moving in a clockwise direction around a
circle at constant speed
1. Which vector below represents the direction of the force vector
when the object is located at point A on the circle?
2. Which vector below represents the direction of the force vector
when the object is located at point C on the circle?
3. Which vector below represents the direction of the velocity vector
when the object is located at point B on the circle?
4. Which vector below represents the direction of the velocity vector
when the object is located at point C on the circle?
5. Which vector below represents the direction of the acceleration
vector when the object is located at point B on the circle?
Mathematics of Circular Motion
vavg =
a=
v2
R
2·π·R
T
a=
4π2R
T2
Relationship between quantities
a=
v2
R
This equation shows for a constant mass and radius,
both Fnet and a is directly proportional to the v2.
F ~ _____
a ~ ______
If the speed of the object is doubled, the net force
required for that object's circular motion and its
acceleration are ____________. And if the speed of the
object is halved (decreased by a factor of 2), the net
force required and its acceleration are decreased by a
factor of ________.
example
•
1.
2.
3.
4.
A car going around a curve is acted upon
by a centripetal force, F. If the speed of
the car were twice as great, the
centripetal force necessary to keep it
moving in the same path would be
F
2F
F/2
4F
Centripetal force and mass of the object
This equation shows for a constant speed and radius,
the Fnet is _________________________ to the mass.
If the mass of the object is doubled, the net force required
for that object's circular motion is ____________. And if the
mass of the object is halved (decreased by a factor of 2),
the net force required is decreased by a factor of ______.
Centripetal acceleration and mass of the object
a=
v2
R
Centripetal acceleration is
________________ by the mass of
the object
example
• Anna Litical is practicing a centripetal force
demonstration at home. She fills a bucket with water, ties
it to a strong rope, and spins it in a circle. Anna spins the
bucket when it is half-full of water and when it is quarterfull of water. In which case is more force required to spin
the bucket in a circle? Explain using an equation.
example
•
1.
2.
3.
4.
The diagram shows a 5.0-kilogram cart
traveling clockwise in a horizontal circle of
radius 2.0 meters at a constant speed of 4.0
meters per second. If the mass of the cart was
doubled, the magnitude of the centripetal
acceleration of the cart would be
doubled
halved
unchanged
quadrupled
Centripetal Force, acceleration and the radius
a=
v2
R
This equation shows for a constant speed and mass,
the Fnet and acceleration a is ______________________
to the radius
If the radius of the object is doubled, the net force
required for that object's circular motion and its
acceleration are both ____________. And if the radius
of the object is halved (decreased by a factor of 2), the
net force required and its acceleration are both
increased by a factor of _______________.
example
•
1.
2.
3.
4.
Two masses, A and B, move in circular paths as
shown in the diagram. The centripetal acceleration of
mass A, compared to that of mass B, is
the same
twice as great
one-half as great
four times as great
Equations as a Recipe for ProblemSolving
• A 900-kg car moving at 10 m/s takes a turn around a
circle with a radius of 25.0 m. Determine the acceleration
and the net force acting upon the car.
example
• A 95-kg halfback makes a turn on the football field. The
halfback sweeps out a path which is a portion of a circle with
a radius of 12-meters. The halfback makes a quarter of a
turn around the circle in 2.1 seconds. Determine the speed,
acceleration and net force acting upon the halfback.
example
• Determine the centripetal force acting upon a 40-kg child
who makes 10 revolutions around the Cliffhanger in 29.3
seconds. The radius of the barrel is 2.90 meters.
Lesson 2: Applications of
Circular Motion
1. Newton's Second law - Revisited
2. Amusement Park Physics
Applications of Circular Motion
• Newton's Second Law - Revisited
Where Fnet is the sum (the resultant) of all forces acting on
the object.
Newton's second law was used in combination of circular
motion equations to analyze a variety of physical situations.
Note: centripetal
force!
force is the _____
Steps in solving problems involving
forces
1. Drawing Free-Body Diagrams
2. Determining the Net Force from
Knowledge of Individual Force Values
3. Determining Acceleration from
Knowledge of Individual Force Values
Or Determining Individual Force Values from
Knowledge of the Acceleration
example
• A 945-kg car makes a 180-degree turn with a speed of 10.0 m/s.
The radius of the circle through which the car is turning is 25.0 m.
Determine the force of friction and the coefficient of friction acting
upon the car.
example
• The coefficient of friction acting upon a 945-kg car is 0.850. The
car is making a 180-degree turn around a curve with a radius of
35.0 m. Determine the maximum speed with which the car can
make the turn.
• A 1.50-kg bucket of water is tied by a rope and whirled in a circle with
a radius of 1.00 m. At the top of the circular loop, the speed of the
bucket is 4.00 m/s. Determine the acceleration, the net force and the
individual force values when the bucket is at the top of the circular
loop.
m = 1.5 kg
a = ________ m/s/s
Fnet = _________ N
• A 1.50-kg bucket of water is tied by a rope and whirled in a
circle with a radius of 1.00 m. At the bottom of the circular
loop, the speed of the bucket is 6.00 m/s. Determine the
acceleration, the net force and the individual force values
when the bucket is at the bottom of the circular loop.
m = 1.5 kg
a = ________ m/s/s
Fnet = _________ N
Roller Coasters and Amusement
Park Physics
In a roller coaster, the centripetal force is provided by the
combination of ____________________________.
example
• Anna Litical is riding on The Demon at Great America. Anna experiences a
downwards acceleration of 15.6 m/s2 at the top of the loop and an upwards
acceleration of 26.3 m/s2 at the bottom of the loop. Use Newton's second
law to determine the normal force acting upon Anna's 864 kg roller coaster
car.
example
• Anna Litical is riding on The American Eagle at Great America. Anna is
moving at 18.9 m/s over the top of a hill which has a radius of curvature of
12.7 m. Use Newton's second law to determine the magnitude of the applied
force of the track pulling down upon Anna's 621 kg roller coaster car.
Lesson 3: Universal Gravitation
1. Gravity is More than a Name
2. The Apple, the Moon, and the Inverse
Square Law
3. Newton's Law of Universal Gravitation
4. Cavendish and the Value of G
5. The Value of g
Gravity is More Than a Name
• We know that gravity is a force and we represent it by
the symbol Fgrav. It causes an acceleration of all
objects around it. The acceleration is referred as the
acceleration of gravity. On and near Earth's surface,
the value for the acceleration of gravity is
approximately 9.81 m/s/s. It is the same acceleration
value for all objects, regardless of their mass (and
assuming that the only significant force is gravity).
• but
– How and by whom was gravity discovered?
– What is the cause of this force of gravity?
– What variables affect the actual value of the force of
gravity?
– Is the force of gravity that attracts my body to the
Earth related to the force of gravity between the
planets and the Sun?
The Apple, the Moon, and the Inverse
Square Law
• In the early 1600's, German mathematician and
astronomer Johannes ________________ developed
three laws to describe the motion of planets about the sun.
However, there was no accepted explanation for why such
paths existed.
• Newton was troubled by the lack of explanation for the
planet's orbits. Newton knew that for the motion of the
moon in a circular path required that there be an inward
component of __________. However, the nature of such a
force - its cause and its origin - bothered Newton for some
time.
And according to legend, a
breakthrough came at age 24 in an
apple orchard in England.
Clearly, it was Newton's ability to
relate the cause for heavenly
motion (the orbit of the moon
about the earth) to the cause for
Earthly motion (the falling of an
apple to the Earth) that led him to
his notion of
__________________________
Newton's reasoning
Suppose a cannonball is fired horizontally
from a very high mountain in a region
devoid of air resistance. In the presence
of gravity, the cannonball would
___________________
Now suppose that the cannonball is fired
horizontally again with a greater speed. In
this case, the cannonball would
__________________________________.
Now suppose that there is a speed at which
the cannonball could be fired such that the
trajectory of the falling cannonball matched
the curvature of the earth, then the
cannonball would ____________________
___________________________________
And then at even greater launch speeds, a
cannonball would once more orbit the earth,
but in an ____________ path, like the planets
The motion of the cannonball orbiting to the earth under the
influence of gravity is similar to the motion of the moon
orbiting the Earth. And if the orbiting moon can be
compared to the falling cannonball, it can even be
compared to a falling apple. The same force that causes
objects on Earth to fall to the earth also causes objects in
the heavens to move along their circular and elliptical
paths.
• It was known at the time, that the force of gravity causes
earthbound objects (such as falling apples) to accelerate
towards the earth at a rate of 9.81 m/s2. And it was also
known that the moon accelerated towards the earth at a rate
of 0.00272 m/s2.
• If the same force that causes the acceleration of the apple to
the earth also causes the acceleration of the moon towards
the earth, then there must be a plausible explanation for why
the acceleration of the moon is so much smaller than the
acceleration of the apple. What is it about the force of gravity
that causes the more distant moon to accelerate at a rate of
acceleration that is approximately 1/3600-th the acceleration
of the apple?
• Newton knew that the force of gravity must somehow be
“____________" by distance.
• The riddle is solved by a comparison of the _____________ from
the apple to the center of the earth with the __________ from the
moon to the center of the earth. The moon in its orbit about the
earth is approximately _____________ further from the earth's
center than the apple is. The mathematical relationship becomes
clear. The force of gravity between the earth and any object is
inversely proportional to the square of the distance that separates
that object from the earth's center. The moon, being 60 times
further away than the apple, experiences a force of gravity that is
1/(60)2 times that of the apple. The force of gravity follows an
_____________________.
Inverse square law.
• The relationship between the force of gravity (Fgrav)
between the earth and any other object and the distance
that separates their centers (d) can be expressed by the
following relationship
The force of gravity is inversely related to the square of
the distance. This mathematical relationship is
sometimes referred to as an inverse square law.
Relationships in the equation
• The inverse square law suggests that the force of
gravity acting between any two objects is
___________________________ to the ___________
of the separation __________________ between the
object's centers.
• If the separation distance is increased by a factor of 2,
then the force of gravity is decreased by a factor of four
(______). And if the separation distance is increased by
a factor of 3, then the force of gravity is decreased by a
factor of nine (_______).
Fg
r
Check Your Understanding
1 . Suppose that two objects attract each other with a gravitational force of
16 units. If the distance between the two objects is doubled, what is the
new force of attraction between the two objects?
2. Suppose that two objects attract each other with a gravitational force of
16 units. If the distance between the two objects is tripled, then what is
the new force of attraction between the two objects?
3. Suppose that two objects attract each other with a gravitational force of
16 units. If the distance between the two objects is reduced in half, then
what is the new force of attraction between the two objects?
4. Suppose that two objects attract each other with a gravitational force of
16 units. If the distance between the two objects is reduced by a factor
of 5, then what is the new force of attraction between the two objects?
example
•
A.
B.
C.
D.
An astronaut weighs 8.00 × 102 newtons on
the surface of Earth. What is the weight of
the astronaut 6.37 × 106 meters above the
surface of Earth?
0.00 N
2.00 × 102 N
1.60 × 103 N
3.20 × 103 N
Newton's Law of Universal
Gravitation
• Consider Newton's famous equation Fnet = m • a
• Newton knew that the force that caused the apple's acceleration
(gravity) must be dependent upon the mass of the __________.
And since the force acting to cause the apple's downward
acceleration also causes the earth's upward acceleration (Newton's
third law), that force must also depend upon the mass of the
________.
• So for Newton, the force of gravity acting between the earth and any
other object is directly proportional to the mass of the _______,
directly proportional to the mass of the ________, and inversely
proportional to the _______________________ that separates the
centers of the earth and the object.
• Newton's law of universal gravitation is about the _______________
of gravity. _________ objects attract each other with a force of
gravitational attraction.
Examples
• What happens to Fg in the following cases?
-Double one mass
-Double both masses
-Triple one mass
-Triple both masses
-Halve one mass
-Halve one mass and double the other
Examples
• What happens to Fg in the following cases?
-Double the distance
-Triple the distance
-Halve the distance
-Quarter the distance
Examples
• What happens to Fg in the following cases?
-Double one mass and double the distance
-Double both masses and double the distance
-Halve one mass and halve the distance
Triple one mass and quarter the distance
Newton's Law of Universal
Gravitation
m1m2
Fg ~ 2
r
• Another means of representing the
proportionalities is to express the relationships
in the form of an equation using a constant of
proportionality.
Gm1m2
Fg 
2
r
G represent Gravitational
Constant
G = ____________________
Example
• Determine the force of gravitational attraction between
the earth (m = 5.98 x 1024 kg) and a 70-kg physics
student if the student is standing at sea level, a distance
of 6.38 x 106 m from earth's center.
The Universality of Gravity
Mass of
Object 1
(kg)
Mass of Object 2
(kg)
Separation
Distance
(m)
Student
70 kg
Earth
5.98 x1024 kg
6.60 x 106 m
(low-height orbit)
Student
70 kg
Physics Student
70 kg
1m
Student
70 kg
Physics Book
1 kg
1m
Student
70 kg
Jupiter
1.901 x 1027 kg
6.98 x 107 m
(on surface)
Force of
Gravity
(N)
Cavendish and the Value of G
• The value of G was not experimentally
determined until nearly a century later
(1798) by Lord Henry Cavendish using
a torsion balance.
Lord Henry Cavendish - English
chemist and physicist
The Value of g
Fgrav = m∙g
• We can use the two equations above to derive an
equation for the value of g.
The acceleration of gravity is dependent upon the mass of
the earth (approx. 5.98x1024 kg) and the distance (d) that
an object is from the center of the earth.
The acceleration of gravity is location dependent.
Note that g is inversely proportional to the distance squared
– inverse square law.
The table below shows the value of g at various locations
from Earth's center.
Location
Distance from
Earth's center (m)
Earth's surface
6.38 x 106 m
1000 km above
surface
5000 km above
surface
7.38 x 106 m
10000 km
above surface
1.64 x 107 m
50000 km
above surface
5.64 x 107 m
1.14 x 107 m
Value of g
m/s2
9.8
Using this equation, the following acceleration of gravity
values can be calculated for the various planets.
Planet
Radius (m)
Mass (kg)
Mars
3.38 x 106
6.42 x 1023
Jupiter
6.98 x 107
1.901 x 1027
Neptune
2.27 x 107
1.03 x 1026
Pluto
1.15 x 106
1.2 x 1022
g (m/s2)
Example - Calculating the mass of the
Earth
• Knowing G, and the radius of the Earth, RE =
6.37 x 106 m, we can now actually calculate
the mass of the Earth.
Gravity is a field force
• Gravitational field – a
________________ where an
object would experience a
gravitational force. Every mass is
surrounded by a gravitational
field.
• As the distance from the Earth
increases, the strength of
gravitational
___________________.
• As the distance from the Earth
increases, the arrows are further
apart and the length of arrows
are _________, indicating the
strength of the gravitational force
_______________.
The gravitational field around Earth
Gravitational field strength is a
vector quantity.
Its direction is directed
_____________________ of
Earth, or normal to Earth’s
surface.
Its magnitude at a point equals
the force per unit mass at that
point.
•The concentration of the field lines increases as the
distance from Earth decreases.
example
• The weight of an object was determined at five different
distances from the center of Earth. The results are shown in the
table below. Position A represents results for the object at the
surface of Earth. What is the approximate mass of the object?
Lesson 4: Satellite Motion
– Circular Motion Principles for Satellites
– Mathematics of Satellite Motion
– Weightlessness in Orbit
Circular Motion Principles for Satellites
• A satellite is any object that is orbiting the earth, sun or
other massive body. Satellites can be categorized as
____________ satellites or _____________ satellites.
• The ___________, the _______________ and comets
are examples of natural satellites.
• ________________ launched from earth for purposes of
communication, scientific research, weather forecasting,
intelligence, etc. are man-made satellites.
• Every satellite's motion is governed by the same physics
principles and described by the same mathematical
equations.
Velocity, Acceleration and Force Vectors
• The motion of an orbiting satellite can be described by
the same motion characteristics as any object in circular
motion.
– The ______________ of the satellite would be
directed tangent to the circle at every point along its
path.
– The _______________ of the satellite would be
directed towards the center of the circle - towards the
central body that it is orbiting.
– And this acceleration is caused by a ____________
that is directed inwards in the same direction as the
acceleration. This centripetal force is supplied by
_________________ - the force that universally acts
at a distance between any two objects that have
mass.
Mathematics of Satellite Motion
• If the satellite moves in circular motion, then the net
centripetal force acting upon this orbiting satellite is
given by the relationship
Fnet = ____________________
• This net centripetal force is the result of the
gravitational force that attracts the satellite towards
the central body and can be represented as
Fgrav = ____________________
• Since Fgrav = Fnet,
_________________________________
v2 = _______________
v = _______________
where
G is 6.673 x 10-11 N•m2/kg2,
Mcentral is the mass of the central body about
which the satellite orbits, and
R is the radius of orbit for the satellite.
The speed of satellite is determined by its location ___ and
mass of the central body __________.
Check your understanding
• A satellite is orbiting the earth. Which of the
following variables will affect the speed of the
satellite?
a. mass of the satellite
b. height above the earth's surface
c. mass of the earth
Weightlessness in Orbit
• Astronauts who are orbiting the Earth often
experience sensations of weightlessness. These
sensations experienced by orbiting astronauts
are the __________ sensations experienced by
anyone who has been temporarily suspended
above the seat on an amusement park ride.
• Not only are the sensations the same (for
astronauts and roller coaster riders), but the
__________ of those sensations of
weightlessness are also the ________.
Unfortunately however, many people have
difficulty understanding the causes of
weightlessness.
Test your preconceived notions
about weightlessness:
•
a.
b.
c.
d.
Astronauts on the orbiting space station are
weightless because...
there is no gravity in space and they do not
weigh anything.
space is a vacuum and there is no gravity in a
vacuum.
space is a vacuum and there is no air
resistance in a vacuum.
the astronauts are far from Earth's surface at a
location where gravitation has a minimal affect.
Contact versus Non-Contact Forces
Contact versus Non-Contact Forces
• As you sit in a chair, you experience two forces – ________
• The normal force and results from the _______________
between the chair and you. You can feel this force because of
the contact you have with the chair.
• The force of gravity acting upon your body is a field force, which
is the result of your center of mass and the Earth's center of
mass exerting a mutual pull on each other; this force would even
exist if you were not in contact with the Earth.
• The force of gravity ___________________. Forces that result
from contact ____________. And in the case of sitting in your
chair, you can feel the chair force; and it is this force that
provides you with a sensation of weight. Without the contact
force (the normal force), there is no means of feeling the noncontact force (the force of gravity).
Scale Readings and Weight
Now consider Otis L. Evaderz who
is conducting one of his famous
elevator experiments. He stands on
a bathroom scale and rides an
elevator up and down. As he is
accelerating upward and
downward, the scale reading is
__________________ than when
he is at rest and traveling at
constant speed.
Fnet = m*a
Fnet = 0 N
Fnorm equals
Fgrav
Fnorm = ____N
Fnet = m*a
Fnet = 400 N, up
Fnet = m*a
Fnet = 400 N, down
Fnet = m*a
Fnet = 784 N, down
Fnorm > Fgrav by Fnorm < Fgrav by Fnorm < Fgrav by
400 N
400 N
784 N
Fnorm = ____ N Fnorm = _____ N Fnorm = __ N
Weightlessness in Orbit
• Earth-orbiting astronauts are weightless for the same
reasons that riders of a free-falling amusement park ride
or a free-falling elevator are weightless. They are
weightless because there is ______________________
force pushing or pulling upon their body.
• In each case, gravity is the _____________ acting upon
their body. Being an action-at-a-distance force, it cannot
be felt and therefore would not provide any sensation of
their weight. But for certain, the orbiting astronauts weigh
something; that is, there is a force of gravity acting upon
their body.
• In fact, if it were not for the force of gravity, the astronauts
would not be orbiting in circular motion. It is the force of
gravity that supplies the ____________________
requirement to allow the __________________
acceleration that is characteristic of circular motion.
• The astronauts and their surroundings are falling towards
the Earth under the sole influence of _______________.
1. Otis stands on a bathroom scale and reads the scale while ascending
and descending the John Hancock building. Otis' mass is 80 kg.. Use a
free-body diagram and Newton's second law of motion to solve the
following problems.
a. What is the scale reading when Otis accelerates upward at 0.40 m/s2?
b. What is the scale reading when Otis is traveling upward at a constant
velocity of at 2.0 m/s?
c. As Otis approaches the top of the building, the elevator slows down at a
rate of 0.40 m/s2. Be cautious of the direction of the acceleration. What
does the scale read?
d. Otis stops at the top floor and then accelerates downward at a rate of
0.40 m/s2. What does the scale read?
e. As Otis approaches the ground floor, the elevator slows down (an
upward acceleration) at a rate of 0.40 m/s2. Be cautious of the direction
of the acceleration. What does the scale read?
• Use the results of your calculations above to explain why Otis fells less
weighty when accelerating downward on the elevator and why he feels
heavy when accelerating upward on the elevator.