DYNAMICS
UNIT 1
SCALARS & VECTORS
Scalar- has magnitude only !
eg. Speed, distance
Vector- has magnitude and direction.
eg. Velocity, Acceleration
Distance - total length of the path travelled by an
object in motion
Position- is the distance and direction of an
object from a particular reference point
Displacement – the change in position of an
object


d  d final  dinitial
When an object changes its position more than
once, total displacement is calculated by adding
the displacements.


 d T   d1   d 2
CONTINUED…
Adding/Subtracting vectors
UNIFORM/NON-UNIFORM MOTION
Uniform – motion of an object at a constant speed in
a straight line
Non-uniform – motion in which the objects speed
changes or the object does not travel in a straight
line.
SPEED/VELOCITY/ACCELERATION
Average Speed:
Average Velocity:
Acceleration:
d
vav 
t


d
vav 
t


v
aav 
t
Information on Linear Motion
Graphs
Position-Time Graph
Slope- represents Velocity
Velocity-Time Graph
Slope – represents Acceleration
Area – represents Displacement
Acceleration-Time Graph
Area- represents change in velocity
RELATIONSHIPS AMONG LINEAR
MOTION GRAPHS

d t
SLOPE
SLOPE

a t

v t
AREA
AREA
KINEMATIC EQUATIONS
PROJECTILE MOTION
WHAT IS PROJECTILE
MOTION?
INSTRUCTIONAL
OBJECTIVES:
• Students will be able to:
• Define Projectile Motion
• Distinguish between the different types of projectile motion
• Apply the concept to a toy car and measure its velocity
WHAT IS A PROJECTILE?
Projectile -Any object which projected by some
means and continues to move due to its own inertia
(mass).
PROJECTILE MOTION
• Two-dimensional motion of an object
• Vertical
• Horizontal
PROJECTILES MOVE IN TWO
DIMENSIONS
Since a projectile
moves in 2dimensions, it
therefore has 2
components just like
a resultant vector.
• Horizontal and
Vertical
TYPES OF PROJECTILE MOTION
• Horizontal
• Motion of a ball rolling freely along
a level surface
• Horizontal velocity is ALWAYS
constant
• Vertical
• Motion of a freely falling object
• Force due to gravity
• Vertical component of velocity
changes with time
• Parabolic
• Path traced by an object
accelerating only in the vertical
direction while moving at constant
horizontal velocity
HORIZONTAL “VELOCITY”
COMPONENT
• NEVER changes, covers equal
displacements in equal time periods. This
means the initial horizontal velocity equals
the final horizontal velocity
In other words, the horizontal
velocity is CONSTANT. BUT WHY?
Gravity DOES NOT work
horizontally to increase or
decrease the velocity.
VERTICAL “VELOCITY” COMPONENT
• Changes (due to gravity), does NOT cover equal
displacements in equal time periods.
Both the MAGNITUDE and DIRECTION change. As
the projectile moves up the MAGNITUDE
DECREASES and its direction is UPWARD. As it
moves down the MAGNITUDE INCREASES and the
direction is DOWNWARD.
COMBINING THE COMPONENTS
Together, these
components
produce what is
called a trajectory or
path. This path is
parabolic in nature.
Component Magnitude
Direction
Horizontal
Constant
Constant
Vertical
Changes
Changes
EXAMPLES OF PROJECTILE MOTION
• Launching a Cannon ball
HORIZONTALLY LAUNCHED
PROJECTILES
Projectiles which have NO upward trajectory and NO
initial VERTICAL velocity.
voy  0 m / s
vox  vx  constant
HORIZONTALLY LAUNCHED
PROJECTILES
To analyze a projectile in 2 dimensions we need 2
equations. One for the “x” direction and one for
the “y” direction. And for this we use kinematic
#2.
2
1
x  voxt  at
2
x  voxt
Remember, the velocity is
CONSTANT horizontally, so
that means the
acceleration is ZERO!
y  1 gt 2
2
Remember that since the
projectile is launched
horizontally, the INITIAL
VERTICAL VELOCITY is
equal to ZERO.
HORIZONTALLY LAUNCHED
PROJECTILES
Example: A plane traveling
with a horizontal velocity of
100 m/s is 500 m above the
ground. At some point the
pilot decides to drop some
supplies to designated
target below. (a) How long
is the drop in the air? (b)
How far away from point
where it was launched will it
land?
y  1 gt 2  500  1 (9.8)t 2
2
2
102.04  t 2  t  10.1 seconds
What do I
know?
What I want to
know?
vox=100 m/s
t=?
x=?
y = 500 m
voy= 0 m/s
g = -9.8 m/s/s
x  voxt  (100)(10.1) 
1010 m
VERTICALLY LAUNCHED PROJECTILES
NO Vertical Velocity at the top of the trajectory.
Vertical
Velocity
decreases
on the way
upward
Vertical Velocity
increases on the
way down,
Horizontal
Velocity is
constant
Component Magnitude
Direction
Horizontal
Vertical
Constant
Changes
Constant
Decreases up, 0
@ top, Increases
down
VERTICALLY LAUNCHED PROJECTILES
Since the projectile was launched at a angle, the
velocity MUST be broken into components!!!
vox  vo cos 
vo

vox
voy
voy  vo sin 
EQUATIONS
• X- Component
x f  xi  v xi t
• Y- Component
1 2
gt
2
 2 g y
y f  y i  v yi t 
v yf  v yi
2
• Vectors
2
v yf  v yi  gt
v xi  vi cos( )
v yi  vi sin(  )
Note: g=
9.8 m/s^2
VERTICALLY LAUNCHED PROJECTILES
There are several things
you must consider when
doing these types of
projectiles besides using
components. If it begins
and ends at ground
level, the “y”
displacement is ZERO: y
=0
VERTICALLY LAUNCHED PROJECTILES
You will still use kinematic #2, but YOU MUST
use COMPONENTS in the equation.
vo

vox
voy
x  voxt
y  voyt  1 gt 2
2
vox  vo cos 
voy  vo sin 
EXAMPLE
A place kicker kicks a football with a velocity of 20.0
m/s and at an angle of 53 degrees.
(a) How long is the ball in the air?
(b) How far away does it land?
(c) How high does it travel?
vox  vo cos 
vox  20 cos 53  12.04 m / s
  53
voy  vo sin 
voy  20sin 53  15.97 m / s
EXAMPLE
A place kicker kicks a
football with a
velocity of 20.0 m/s
and at an angle of 53
degrees.
(a) How long is the ball
in the air?
What I know
vox=12.04 m/s
voy=15.97 m/s
y=0
g = - 9.8
m/s/s
y  voy t  1 gt 2  0  (15.97)t  4.9t 2
2
2
15.97t  4.9t  15.97  4.9t
t  3.26 s
What I want
to know
t=?
x=?
ymax=?
EXAMPLE
A place kicker kicks a
football with a
velocity of 20.0 m/s
and at an angle of 53
degrees.
(b) How far away does
it land?
What I know
vox=12.04 m/s
voy=15.97 m/s
What I want
to know
t = 3.26 s
x=?
y=0
g = - 9.8
m/s/s
ymax=?
x  voxt  (12.04)(3.26) 
39.24 m
EXAMPLE
What I know
A place kicker kicks a
football with a velocity
of 20.0 m/s and at an
angle of 53 degrees.
(c) How high does it
travel?
CUT YOUR TIME IN HALF!
vox=12.04 m/s
voy=15.97 m/s
y=0
g = - 9.8
m/s/s
What I want
to know
t = 3.26 s
x = 39.24 m
ymax=?
y  voy t  1 gt 2
2
y  (15.97)(1.63)  4.9(1.63) 2
y  13.01 m
FACTORS AFFECTING PROJECTILE MOTION
• What two factors would affect projectile motion?
• Angle
• Initial velocity
• Visual
Initial Velocity
Angle
EXAMPLE
An object is fired from the ground at 100 meters per
second at an angle of 30 degrees with the
horizontal
 Calculate the horizontal and vertical components of the
initial velocity
 After 2.0 seconds, how far has the object traveled in the
horizontal direction?
 How high is the object at this point?
SOLUTION
• Part a
 s cos 30   87 m s
 v sin   100 m sin 30   50 m
s
s
vix  vi cos   100 m
• Part b
viy
0
i
x
vix 
t
• Part c
0

x  v x t  87 m
y  viy t 
s
2.0s   174m
  



1
1
2
g t 2  50 m 2.0s   9.8 m 2 2.0s 
s
s
2
2
NEWTON’S LAWS
1) The law of Inertia
- an object at rest or in uniform motion will remain
at rest or in uniform motion unless acted on by an
external force.
2) F = ma
3) For every action force on an object (B) due to
another object (A), there is a reaction force, equal
in magnitude but opposite in direction
FORCE
FBD’s
Fnet  ma
F f  FN
Fg  mg
MASS
Inertial Mass
-measure of how strongly the body is accelerated
(by A) by a given force.
Gravitational Mass
-measure of how strongly the body is affected by the
force of Gravity
FRAMES OF REFERENCE
Inertial frame of reference
-Has a constant velocity, meaning moving at a
constant speed in a straight line, or it is standing still
Non-inertial frame of reference
-Does not have a constant velocity, it is
accelerating.
RELATIVE MOTION
• The motion (or way of moving) of an object viewed
by an observer
Relative Velocity- the velocity of an object relative to
a specific frame of reference
GENERAL RELATIONSHIP



v AC  v AB  vBC
A relative to
C
A relative to
B
B relative to
C
Note: the outside subscripts on the right side of the
equation (A &C) are in the same order as the
subscripts on the left side of the equation and the
inside subscripts on the right side of the equation
are the same (B)
TYPES OF RELATIVE MOTION
PROBLEMS
1) Relative Motion in 1D
2) Relative Motion in 2D with perpendicular vectors
3) Relative Motion in 2D non perpendicular vectors
QUICK PRACTICE
1) A group of teenagers on a ferry walk on the deck
with a velocity of 1.1 m/s relative to the deck. The
ship is moving forward with a velocity of 2.8 m/s
relative to the water.
a) Determine the velocity of the teenagers relative to
the water when they are walking to the
bow(front).
b) Determine the velocity of the teenagers relative to
the water when they are walking to the stern
2) A plane is travelling with a velocity relative to the
air of 3.5 x102 km/h [N35°W] as it passes over
Hamilton. The wind velocity is
62 km/h[S].
a) Determine the velocity of the plane relative to the
ground.
b) Determine the displacement of the plane after 1.2
h.
COMBINING
DYNAMICS AND KINEMATICS
Recall:
Kinematics – the motion of an object with
disregard to the cause
Dynamics – The cause of the motion
FORCES & FBD’S
Common Forces

• Gravity ( F )
g
• Normal ( )
FN
• Tension ( )
• Applied ( FT)

• Friction ( F)
a

Ff
Units: Newton's (N)
1N =
1kg  m / s
2
NEWTON’S LAWS OF MOTION
Newton’s First Law: (Law of Inertia)
If the external net force on
an object is zero, the object
will remain at rest or continue
to move at a constant velocity.
Inertia – a measure of an object’s resistance
to change in velocity
Mass – a measure of the amount of matter in an
object
Newton’s Second Law:


Fnet  ma
Newton’s Third Law:
For every action force, there exists a
simultaneous reaction force that
is equal in magnitude but
opposite in direction
QUICK PRACTICE
1)At an instant when a soccer ball is slightly off the
ground, a player kicks it, exerting a force of 25 N at
40.0°above the horizontal. The force of gravity
acting on the ball is 4.2 N[down]. Determine the net
force.
2) Two children pull a sled across the ice. One child
pulls with a force of 15 N [N 35°E], and the other
pulls with a force of 25 N[N 54°W]. Ignore friction,
find the net force.
FORCE OF FRICTION
Force of Friction – acts in the opposite direction as
the applied force (opposes motion)
F f  FN
Recall: Types of Friction
1) Kinetic Friction
- coefficient of kinetic friction ( k )
2) Static Friction
-coefficient of static friction( s )
QUICK PRACTICE
1) You are pulling a 39 kg box on a level floor by a
rope attached to the box. The rope makes an
angle of 21° with the horizontal. The coefficient of
kinetic friction between the box and the floor is 0.23.
Calculate the magnitude of the tension in the rope
needed to keep the box moving at a constant
velocity.
2D PROBLEMS USING
NEWTON’S 2ND LAW
1) A mass of 1.2x 102 kg with a force of
1.5 x 102 N [N] and a force of 2.2 x 102 N[W] acting
on it. Determine the acceleration of the mass.
Assume no other forces act on the object other
than the ones given.
2) Two ropes are used to lift a 1.5x102 kg beam with a
force of gravity of 1.47 x 103 N[down] acting on it.
One rope exerts a force of tension of 1.8 x 103 N[up
30.0°left] on the beam, and the other rope exerts a
force of tension
1.8 x 103 N[up 30.0°right] on the beam. Calculate the
acceleration of the beam.
SOLVING NEWTON’S 3RD LAW
PROBLEMS
1) A swimmer with a mass of 56kg pushes horizontally
against the pool wall towards the east for 0.75
cm/s. Neglecting friction, determine the
magnitude of
a)The (constant) acceleration
b) The force exerted by the swimmer on the wall
c) The force exerted by the wall on the swimmer
d) The displacement of the swimmer from the wall
after 1.50s.
2) A projectile launcher fires a projectile horizontally
from a platform, which rests on a flat, icy, frictionless
surface. Just after the projectile is fired and while it is
moving through the launcher, the projectile has an
acceleration of 25 m/s2. At the same time, the
launcher has an acceleration of 0.25 m/s2. The mass
of the projectile is 0.20 kg. Calculate the mass of the
launcher.
ATWOOD MACHINES
6
kg
4
kg
• Used to determine
the acceleration in
a two body pulley
system
• Could be used to
help determine the
acceleration due to
gravity on a
different planet
• Resolve the forces
for each mass
• Choose the
direction of overall
acceleration of
each object as
positive
Mass 1 (6 kg)
• Down is positive
• 𝑚1 𝑔 − 𝑇 = 𝑚1 𝑎
•
Mass 2 (4 kg)
• Up is positive
• 𝑇 − 𝑚2 𝑔 = 𝑚2 𝑎
T is the tension in the rope and is the same in both equation
Solve for T in one equation
Substitute into the
• T = 𝑚2 𝑔 + 𝑚2 𝑎
formula
• 𝑚1 𝑔 − 𝑚2 𝑔 − 𝑚2 𝑎 = 𝑚1 𝑎
• 𝑎=
𝑔 𝑚1 −𝑚2
𝑚1 +𝑚2
• 𝑎=
9.81 6−4
6+4
• a= 1.962m/s2
Horizontal Pulley
Hyperphysics
INCLINED PLANES
• Label the direction of N and mg.
N
θ
mg
INCLINED PLANES
• Mark the direction of acceleration a.
N
a
θ
mg
INCLINED PLANES
• Choose the coordinate system with x in the
same or opposite direction of acceleration
and y perpendicular to x.
y
N
x
a
θ
mg
INCLINED PLANES
• Now some trigonometry
y
N
x
a
θ
90- θ
θ
mg
INCLINED PLANES
• Replace the force of gravity with its components.
y
N
x
a
θ
θ
mg
INCLINED PLANES
• Use Newton’s second law for both the x and y directions
Fx  max  ma
y
N
x
a
 mg sin   ma
Fy  may  0
θ
θ
mg
 N  mg cos   0
The force and acceleration in the x-direction have a negative
sign because they point in the negative x-direction.
INCLINED PLANES
• Why is the component of mg along the x-axis –mgsinθ
• Why is the component of mg along the y-axis –mgcosθ
y
N
x
a
θ
θ
mg
INCLINED PLANES
• Why is the component of mg along the x-axis: –mgsinθ
• Why is the component of mg along the y-axis: –mgcosθ
INCLINED PLANES
• Why is the component of mg along the x-axis: –mgsinθ
• Why is the component of mg along the y-axis: –mgcosθ
INCLINED PLANES
• Why is the component of mg along the x-axis: –mgsinθ
• Why is the component of mg along the y-axis: –mgcosθ
sinθ =
cosθ
=
opposite
hypotenus
e
adjacent
hypotenus
e
QUICK PRACTICE
1) A child on a toboggan slides down a hill with an
acceleration of magnitude 1.9 m/s2. Friction is
negligible. Determine the angle between the hill
and the horizontal.
2) A sled takes off from the top of the hill inclined at
6.0° to the horizontal. The sled’s initial speed is
12m/s. The coefficient of kinetic friction between
the sled and the snow is 0.14. Determine how far
the sled will slide before coming to rest.
INCLINED PULLEY
Hyperphysics
UNIFORM CIRCULAR MOTION
Uniform Circular Motion – the motion of an object
with a constant speed along a circular path of
constant radius
Centripetal Acceleration- ( ) – the instantaneous
 toward the center of a
acceleration that is directed
ac
circular path
Note
• the instantaneous velocity, is always tangential. v
• the velocity is continually changing as the direction
of motion is always changing.
• because the velocity is changing, every particle on a
rotating rigid body is accelerating.
• In uniform circular motion (that is with constant
angular velocity), the acceleration is always towards
the center.
ac = centripetal
acceleration
aT = tangential
acceleration
vT = tangential velocity
Uniform Circular Motion:
- Constant speed
- Constant angular
velocity
Non-Uniform Circular Motion:
- Changing speed
- Changing angular velocity
EQUATIONS FOR CENTRIPETAL
ACCELERATION
1)
 v
ac 
r
2)
3)
ac  4 rf
2
4 r
ac  2
T
2
2
2
PERIOD & FREQUENCY
Period – (T)- the time required for a rotating, revolving
or vibrating object to complete one cycle (units: s)
Frequency – (f)- the number of rotations, revolutions
or vibrations of an object per unit of time; the
inverse of period (units: Hz)
QUICK PRACTICE
1) At a distance the of 25 km from the eye (center) of
a hurricane, the wind moves at nearly 50.0 m/s.
Assume that the wind moves in a circular path.
Calculate the magnitude of the centripetal
acceleration of the particles in the wind at this
distance.
QUICK PRACTICE
2) The planet Venus moves in a nearly circular orbit
around the Sun. The average radius of its orbit is 1.08
x 1011 m. The centripetal acceleration of Venus has
a magnitude
of 1.12 x10-2m/s2. Calculate Venus’s period of
revolution around the Sun.
a) In seconds
b) In Earth days
CENTRIPETAL FORCE
Centripetal Force – (Fc) – the net force that
causes centripetal acceleration
mv
Fc 
r
2
QUICK PRACTICE
1) A curved road with a radius of 450 m in the
horizontal plane is banked so that the cars can
safely navigate the curve. Calculate the banking
angle for the road that will allow a car travelling at
97km/h to make it safely around the curve when
the road is covered with black ice.
ROTATING FRAMES OF REFERENCE
Centrifugal Force – A non existent force which is
actually the absence of a centripetal force
Centrifuge – a rapidly rotating device used to
separate substances and simulate the effects of
gravity
Coriolis Force – a fictitious force that acts
perpendicular to the velocity of an object in a
rotating frame of reference
visualizing centrigual/coriolis forces
Artificial Gravity – a situation in which the value of
gravity has been changed artificially to more
closely match Earth’s gravity
Artificial Gravity
Explanation
QUICK PRACTICE
1) A spacecraft travelling to Mars has an interior
diameter of 324m. The craft rotates around its axis
at the rate required to give astronauts along the
interior wall an apparent weight equal in
magnitude to their weight on Earth.
a) Calculate the speed of the astronauts relative to
the center of the spacecraft
b) Determine the period of rotation of the
spacecraft.