Forces
Unit 2: Forces
Chapter 6: Systems in Motion
 6.1
Motion in Two Dimension
 6.2
Circular Motion
 6.3
Centripetal Force, Gravitation, and
Satellites
 6.4
Center of Mass
6.1 Investigation: Launch Angle and Range
Key Question:
Which launch angle will give a
marble the best range?
Objectives:

Use the Marble Launcher to find the launch angle that
produces the maximum range for a projectile.
Motion in two dimensions
 Real
objects do not move in straight lines alone; their
motion includes turns and curves.
 To
describe a curve you need at least two
dimensions (x and y).
Displacement
 Distance
is scalar, but
displacement is a vector.
 A displacement
vector
shows a change in
position.
Displacement
 A displacement
vector’s
direction is often given
using words.
 Directional
words include
left, right, up, down, and
compass directions.
Solving displacement problems
 Displacement
vectors can be added just like force
vectors.
 To
add displacements graphically, draw them to scale
with each subsequent vector drawn at the end of the
previous vector.
 The
resultant vector represents the displacement
for the entire trip.
Adding vectors
A mouse walks 5 meters north and 12 meters west. Use
a scaled drawing and a protractor to find the mouse’s
displacement. Use the Pythagorean theorem to check
your work.
1.
2.
3.
4.
Looking for: … the displacement.
Given: … distances and direction (5 m, N) and (12 m, W)
Relationships: Pythagorean theorem a2 + b2 = c2
Solution: Make a drawing with a scale of 1 cm = 2 meters.
Measure the angle from the x axis.
Pythagorean theorem: (5)2 + (12)2 = c2
The mouse walks 13 meters at 157°.
169 = c2
13 = c
Velocity vectors
 Velocity
is speed with direction, so velocity is a
vector.
 As
objects move in curved paths, their velocity
vectors change because the direction of motion
changes.
 The
symbol v is used to represent a velocity
vector.
Velocity vectors
 Suppose
 At
a ball is launched at 5 m/s at an angle of 37°
the moment after launch, the velocity vector for the
ball written as a magnitude-angle pair is v = (5 m/s,
37°).
Velocity vectors
 In
x-y components, the
same velocity vector is
written as v = (4, 3) m/s.
 Both
representations tell
you exactly how fast and in
what direction the ball is
moving at that moment.
Using velocity vectors
A train moves at a speed of 100 km/h heading east. What is
its velocity vector in x-y form?
1.
Looking for: … the velocity vector.
2.
Given: … speed (100 km/h) and direction (east).
3.
Relationships: x-velocity is east and y-velocity is
north.
4.
Solution: v = (100,0) km/h
Projectile motion
 Any
object moving through
air and affected only by the
force of gravity is called a
projectile.
 Flying
objects such as
airplanes and birds are not
projectiles, because they are
affected by forces generated
from their own power.
Projectile motion
 The

path a projectile follows is called its trajectory.
The trajectory of a projectile is a special type of
arch- or bowl-shaped curve called a parabola.
Trajectory and range
 The
range of a projectile is the horizontal distance it
travels in the air before touching the ground.
 A projectile’s
range depends on the speed and angle
at which it is launched.
Two dimensional motion
 Projectile
motion is two-dimensional because there is
both horizontal and vertical motion.
 Both
speed and direction change as a projectile
moves through the air.
A ball rolling off a table
 The
horizontal and vertical components of a
projectile’s velocity are independent of each other.
Horizontal velocity
 The
ball’s horizontal motion looks exactly like the its
motion if it was it rolling along the ground at 5 m/s.
Vertical velocity
 The
vertical (y) velocity
increases due to the
acceleration of gravity.
NOTE: These equations are suitable only for situations where the projectile
starts with zero vertical velocity, such as a ball rolling off a table.
Range of a Projectile
 The
range, or horizontal distance, traveled by a
projectile depends on the launch speed and the
launch angle.
Range of a Projectile
 The
range of a projectile is calculated from the
horizontal velocity and the time of flight.
 The
air time and height are greatest when a ball is
hit at an angle of 90°, but air time and height are
zero when a ball is hit at an angle of 0°.
Projectile motion
A stunt driver steers a car off a cliff at a speed of 20.0 m/s. The
car lands in a lake below 2.00 s later. Find the horizontal distance
the car travels and the height of the cliff.
1.
Looking for: … vertical and horizontal distances.
2.
Given: … the time (2.00 s) and initial horizontal speed (20.0
m/s).
3.
Relationships: Use : dx= vxt dy = 4.9t2
4.
Solution: dx= (20 m/s)(2 s) = 40 m.
dy = (4.9 m/s2)(2 s)2 = (4.9 m/s2)(4 s2) = 19.6 m.
Unit 2: Forces
Chapter 6: Systems in Motion
 6.1
Motion in Two Dimension
 6.2
Circular Motion
 6.3
Centripetal Force, Gravitation, and
Satellites
 6.4
Center of Mass
6.2 Investigation: Launch Speed and Range
Key Question:
How does launch speed affect
the range of a projectile?
Objectives:

Use the Marble Launcher to explore if a relationship exists
between the launch speed and the range of a projectile.
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 merrygo-round because he is
external to the merry-goround's axis.
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 (degrees/s or
radians/s)
Angular speed
 There
are 360 degrees in a full rotation, so one
rotation per minute is the same angular speed as
360 degrees per minute
Calculating angular speed
A merry-go-round makes 10 rotations in 2
minutes. What is its angular speed in rpm?
1.
Looking for: … the angular speed in rotations per minute.
2.
Given: … number of rotations (10) and the time (2 min.)
3.
Relationships: Use: angular speed = rotation
time
4.
Solution: angular speed = 10 rotations = 5 rpm
2 minutes
The relationship between linear and
angular speed
 Each point on a rotating
object has the same angular
speed.
 The
linear speed of each
child is not the same
because they travel different
distances.
The relationship between linear and
angular speed
 The
distance traveled during one revolution equals
the circumference of the circle.
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.
Calculating linear speed
The blades on a ceiling fan spin at 60 rpm. The
fan has a radius of 0.5 m. Calculate the linear
speed of a point at the outer edge of a blade in
m/s.
1.
Looking for: … the linear speed in m/s.
2.
Given: … angular speed (60 rpm) and the radius (0.5 m)
3.
4.
Relationships: Use: v = 2  r
t
Solution: The blades spin at 60 rotations per minute, so they make 60
rotations in 60 seconds. Therefore it takes 1 second to make one
rotation. v = 2 (3.14) (0.5 m) = 3.14 m/s
(1 s)
Linear, rotational and rolling motion

Rolling is a combination of
linear motion and rotational
motion.
— Holding a bicycle wheel up
in the air and moving it to
the right is linear motion.
— If you lift a bicycle’s front
wheel off the ground and
make it spin, the spinning
wheel is rotational motion.
Linear distance equals circumference
 The
distance the bicycle moves depends on the
wheel’s size and angular speed.
 When
the wheel makes one full rotation, the bicycle
goes forward one circumference of the wheel.
Unit 2: Forces
Chapter 6: Systems in Motion
 6.1
Motion in Two Dimension
 6.2
Circular Motion
 6.3
Centripetal Force, Gravitation, and
Satellites
 6.4
Center of Mass
6.3 Investigation: Levers and Rotational
Equilibrium
Key Question:
How do levers work?
Objectives:
Explain the meaning of torque and describe its relationship
to how levers work.
 Use different combinations of weights to balance a lever.


Apply an understanding of rotational equilibrium to
determine an unknown mass.
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.
Centripetal force and direction
 Whether
a force makes an
object accelerate by
changing its speed or by
changing its direction or
both depends on the
direction of the force.
Centripetal force and direction

Imagine tying a ball to the end
of a string an twirling it in a
circle over your head.

The string exerts the
centripetal force on the ball to
move it in a circle.

The direction of the centripetal
force changes as the object
moves around you.
Centripetal force and inertia
 If
you give a ball on a string an initial velocity to
the left at point A, it will try to keep moving straight
to the left.
 But
the centripetal force pulls the ball to the side.
Centripetal force and inertia
 A short
time later, the ball is at point
B and its velocity is 90 degrees from
what it was.
 But
now the centripetal force pulls to
the right.
 The
ball’s inertia makes it move
straight, but the centripetal force
always pulls it towards the center.
Centripetal force and inertia
 Notice
that the velocity is always perpendicular to the
string and therefore to the centripetal force.
 The
centripetal force and velocity are perpendicular
for any object moving in a circle.
Newton’s 2nd law and circular motion
 An
object moving in a circle at a constant speed
accelerates because its direction changes.
 How
quickly an object changes direction depends on
its speed and the radius of the circle.
 Centripetal
acceleration increases with speed and
decreases as the radius gets larger.
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 whose direction and velocity changes.
Newton’s 2nd law and circular motion
 Newton’s
second law
relates force, mass, and
acceleration.
 The
strength of the
centripetal force needed to
move an object in a circle
depends on its mass,
speed, and the radius of the
circle.
Newton’s 2nd law and circular motion
 Newton’s
second law
relates force, mass, and
acceleration.
 The
strength of the
centripetal force needed to
move an object in a circle
depends on its mass,
speed, and the radius of
the circle.
1. Centripetal force is directly
proportional to the mass. A 2kg object needs twice the
force to have the same
circular motion as a 1-kg
object.
Newton’s 2nd law and circular motion
2. Centripetal force is inversely
proportional to the radius of its
circle.
The smaller the circle’s radius,
the greater the force. An object
moving in a 1 m circle needs
twice the force it does when it
moves in a 2 m circle at the
same speed.
Centrifugal force?
 Have
you ever noticed that when a
car makes a sharp turn, you are
pushed toward the outside edge of
the car?
 This
apparent outward force is
called centrifugal force.
 While
it feels like there is a force
acting on you, centrifugal force is
not a true force.
Inertia and circular motion

Suppose a box is in the center of the
bed as a truck travels along a straight
road.

The box and the truck are both moving
in a straight line.

If the truck suddenly turns to the left, the
box tries to keep moving in that same
straight line.

While it seems like the box is being
thrown to the right side of the truck, the
truck is actually turning under the box.
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.
Gravitational force
 The
force of gravity
between Earth and the
Sun keeps Earth in orbit.
Gravitational force

You notice the force of gravity
between you and Earth because the
planet’s mass is huge.

Gravitational forces tend to be
important only when one of the
objects has an extremely large mass,
such as a moon, star, or planet.

If you tell a person on the north pole
and one on the south pole to point
down, they will be pointing in
opposite directions
The gravitational force
 The
force of gravity
between two objects is
proportional to the mass
of each object.
 The
distance between
objects, measured from
center to center, is also
important when
calculating gravitational
force.
Law of universal gravitation
 Newton’s
law of universal
gravitation gives the
relationship between
gravitational force, mass, and
distance.
 The
gravitational constant (G)
is the same everywhere in the
universe (6.67 × 10–11 N·m2/kg2)
Law of universal gravitation
 The
gravitational force of
Earth on the Moon has
the same strength as the
gravitational force of the
Moon on Earth.
Calculating
Use the following information to calculate the force of gravity
between Earth and the Moon: mass of Earth: 5.97 × 1024 m; mass
of the Moon: 7.34 × 1022 kg; distance between Earth’s and the
Moon’s centers: 3.84 × 108 m.
1.
Looking for: … the force of gravity between Earth and the Moon.
2.
Given: … Earth’s (5.97 × 1024 kg) and the Moon’s masses (7.34 × 1022
kg) and the distance between their centers (3.84 × 108 m).
Relationships: Use: F = m1m2
r2
3.
4.
Solution: F = (5.97 × 1024 kg) (7.34 × 1022 kg) = 1.99 X 1020 N
(3.84 × 108 m)2
Satellites
 A satellite
is an object that
circles around another object
with gravity providing the
centripetal force.
Earth, its moon, and
the other planets are
examples of natural
satellites.
 Artificial
satellites that orbit,
or revolve around, Earth
include the Hubble Space
Telescope.
Orbital motion
 An
orbit can be a circle or an
oval shape called an ellipse.
 The
planets move in nearly
circular orbits.
 Comets
travel in elliptical
orbits around the Sun.
Unit 2: Forces
Chapter 6: Systems in Motion
 6.1
Motion in Two Dimension
 6.2
Circular Motion
 6.3
Centripetal Force, Gravitation, and
Satellites
 6.4
Center of Mass
Center of Mass

There are three different axes about which
an object will naturally spin.
 The point at which the three axes intersect is
called the center of mass.
Finding the center of mass

If an object is irregularly shaped, the center of mass can be
found by spinning the object and finding the intersection of
the three spin axes.

There is not always material at an object’s center of mass.
Finding the center of
gravity

Closely related to the center of mass
is the center of gravity.

The center of gravity of an
irregularly shaped object can be
found by suspending it from two or
more points.

For very tall objects, such as
skyscrapers, the acceleration due to
gravity may be slightly different at
points throughout the object.
Center of mass and stability
 Objects
balance because the torque caused by the
force of the object’s weight is equal on each side.
Center of mass and stability

For an object to remain
upright, its center of gravity
must be above its area of
support.

The area of support includes
the entire region surrounded
by the actual supports.

An object will topple over if its
center of mass is not above
its area of support.
Center of mass and stability
 An
object will topple over
if its center of mass is not
above its area of support.
 If
a vector from the center
of mass to the center of
Earth passes through the
area of support, the object
will not topple.
Sailboat Racing: Vectors to Victory

Racing the 1,022 kilometers from
Newport, Rhode Island to
Bermuda in a sailboat is one of
the great adventures in sailing.

With the Gulf Stream current,
giant waves, and unpredictable
weather, staying on course is
difficult

Before the boat ever starts the
race, the planning begins.