86
UNIT 1 DYNAMICS
U N I T 1
DYNAMICS
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 initialfinal dd
When an object changes its position more than once, total displacement is calculated by adding the displacements.
Td 21 dd
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:
tdvav
tdvav
tvaav
Position-Time GraphSlope- represents Velocity
Velocity-Time GraphSlope – represents AccelerationArea – represents Displacement
Acceleration-Time GraphArea- represents change in velocity
Information on Linear Motion Graphs
RELATIONSHIPS AMONG LINEAR MOTION GRAPHS
td
tv ta SLOPE SLOPE
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 DIMENSIONSSince a projectile
moves in 2-dimensions, 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 COMPONENTSTogether, 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 PROJECTILESProjectiles which have NO upward trajectory and NO
initial VERTICAL velocity.
0 /oyv m s
constantox xv v
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.
212oxx v t at
oxx v t
Remember, the velocity is CONSTANT horizontally, so that means the acceleration is ZERO!
212y gt
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?
What do I know?
What I want to know?
vox=100 m/s t = ?y = 500 m x = ?voy= 0 m/s
g = -9.8 m/s/s
2 2
2
1 1500 ( 9.8)2 2102.04
y gt t
t t
10.1 seconds(100)(10.1)oxx v t 1010 m
VERTICALLY LAUNCHED PROJECTILES
Component Magnitude DirectionHorizontal Constant ConstantVertical Decreases up, 0
@ top, Increases down
Changes
Horizontal Velocity is constant
Vertical Velocity decreases on the way upward
Vertical Velocity increases on the way down,
NO Vertical Velocity at the top of the trajectory.
VERTICALLY LAUNCHED PROJECTILESSince the projectile was launched at a angle, the
velocity MUST be broken into components!!!
cossin
ox o
oy o
v vv v
vo
vox
voy
EQUATIONS
• X- Component
• Y- Component
• Vectors
tvxx xiif
gtvv
ygvv
gttvyy
yiyf
yiyf
yiif
221
22
2
)sin()cos(
iyi
ixi
vvvv
Note: g= 9.8 m/s^2
VERTICALLY LAUNCHED PROJECTILESThere 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.
cossin
ox o
oy o
v vv v
vo
vox
voy
oxx v t 212oyy v t gt
EXAMPLEA 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?
cos20cos53 12.04 /
sin
20sin 53 15.97 /
ox o
ox
oy o
oy
v vv m sv v
v m s
v o=20
.0 m/s
53
EXAMPLEA 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 What I want to know
vox=12.04 m/s t = ?voy=15.97 m/s x = ?y = 0 ymax=?g = - 9.8 m/s/s
2 2
2
1 0 (15.97) 4.9215.97 4.9 15.97 4.9
oyy v t gt t t
t t tt
3.26 s
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 What I want to know
vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = ?y = 0 ymax=?g = - 9.8 m/s/s
(12.04)(3.26)oxx v t 39.24 m
EXAMPLEA 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!
What I know What I want to know
vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = 39.24 my = 0 ymax=?g = - 9.8 m/s/s
2
2
12
(15.97)(1.63) 4.9(1.63)
oyy v t gt
yy
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 horizontalCalculate the horizontal and vertical components of the
initial velocityAfter 2.0 seconds, how far has the object traveled in the
horizontal direction?How high is the object at this point?
SOLUTION
• Part a
• Part b
• Part c mssmtvx
txv
x
ix
1740.287
s
ms
mvvs
ms
mvv
iiy
iix
5030sin100sin
8730cos100cos
0
0
22
2 0.28.9210.250
21 s
smss
mtgtvy iy
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 = ma3) 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
Nf FF
mgFg
maFnet
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 REFERENCEInertial 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
BCABAC vvv
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)
A relative to C
A relative to B
B relative to C
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 PRACTICE1) 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 ( )• Normal ( )• Tension ( )• Applied ( )• Friction ( )
Units: Newton's (N)1N =
gF
NF
TF
aF
fF
2/1 smkg
NEWTON’S LAWS OF MOTION
Newton’s First Law: (Law of Inertia)If the external net force on an object is zero, the objectwill 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:
Newton’s Third Law:For every action force, there exists asimultaneous reaction force thatis equal in magnitude but opposite in direction
amFnet
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 FRICTIONForce of Friction – acts in the opposite direction
as the applied force (opposes motion)
Recall: Types of Friction1) Kinetic Friction
- coefficient of kinetic friction ( )2) Static Friction
-coefficient of static friction( )
k
s
Nf FF
QUICK PRACTICE1) 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 tension1.8 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 wallc) The force exerted by the wall on the swimmerd) 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
• 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 planet6
kg4 kg
• Resolve the forces for each mass
• Choose the direction of overall acceleration of each object as positive
Mass 1 (6 kg)• Down is positive
• Substitute into the formula
Mass 2 (4 kg)• Up is positive
• T =
• a= 1.962m/s2
T is the tension in the rope and is the same in both equationSolve for T in one equation
INCLINED PLANES• Label the direction of N and mg.
mgθ
N
INCLINED PLANES• Mark the direction of acceleration a.
mgθ
N
a
INCLINED PLANES• Choose the coordinate system with x in the
same or opposite direction of acceleration and y perpendicular to x.
mgθ
N
a
x
y
INCLINED PLANES• Now some trigonometry
mgθ
N
a
x
y
90- θθ
INCLINED PLANES• Replace the force of gravity with its components.
mgmg sin
θ
θθ
mg cosθ
N
a
x
y
INCLINED PLANES• Use Newton’s second law for both the x and y directions
mamaF xx
mamg sin
mgmg sin
θ
θθ
mg cosθ
N
a
x
y
0 yy maF
0cos mgN
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
mgmg sin
θ
θθ
mg cosθ
N
a
x
y
INCLINED PLANES• Why is the component of mg along the x-axis: –mgsinθ• Why is the component of mg along the y-axis: –mgcosθ
mg
mg sinθ
θθ
mg cosθ
Na
x
y
INCLINED PLANES• Why is the component of mg along the x-axis: –mgsinθ• Why is the component of mg along the y-axis: –mgcosθ
mg
mg sinθ
θ
mg cosθ
Na
x
y
θ
INCLINED PLANES• Why is the component of mg along the x-axis: –mgsinθ• Why is the component of mg along the y-axis: –mgcosθ
mg
mg sinθ
θ
mg cosθ
Na
x
y θsinθ =
cosθ =
oppositehypotenuseadjacenthypotenuse
QUICK PRACTICE1) 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.
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 acceleration that is directed toward the center of a circular pathca
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 accelerationaT = 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) 2)
3)
rvac
2
2
24Trac
224 rfac
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 secondsb) In Earth days
CENTRIPETAL FORCE
Centripetal Force – (Fc) – the net force that causes centripetal acceleration
rmvFc
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 REFERENCECentrifugal 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.
