Ch1 Kinematics and Dynamics

8/12/2019 Ch1 Kinematics and Dynamics

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KINEMATICS AND DYNAMICSNORAIHAN BINTI SALLEH HUDIN

DEPARTMENT OF SCIENCE AND MATHEMATICS

CENTRE FOR DIPLOMA STUDIES, UTHM


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LEARNING OBJECTIVES

The objectives of this chapter are to impart

students:

With the basic knowledge in kinematics anddynamics.

With the concept of kinematics and dynamics in

technical engineering courses.


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LEARNING OUTCOMES

After completing this chapter, students should be able to: Differentiate between linear and non-linear motion and

their examples in everyday life. Able to perform analysis and calculation on problems

in linear motion with constant acceleration. Identify types of forces exists.

Differentiate the three Newtons Law of motion. Compute the magnitude and direction of forces using

free-body diagram (FBD) method. Manipulate the forces concept in solving dynamics

problems.


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INTRODUCTION

Everyday life deals a lot with motion, linear andnon-linear.

Kinematics is the study of positions andmotionsof objects in space as the function oftime.

Dynamics is the study of the effects of forcesonthe motion of an object with mass.


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CONTENT:

4.1 Reference Frame and Coordinate System

4.2 Displacement, Velocity and Acceleration

4.3 Linear Motion with Constant Acceleration4.4 Linear Motion in 2-Dimension: The Projectile Motion.

4.5 The Concepts of Force

4.6 Newtons Laws

4.7 The Concept of Mass and Weight4.8 Linear Momentum

4.9 Applications of Newtons Law of Motion


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4.1 REFERENCE FRAME &

COORDINATE SYSTEM What we see is usually different from what otherpeople see on the same event.

A stationary person observes differently than a moving

observer. We can avoid confusion if we can carefully specify thelocation and motion of each observer.

We can say that the different observer is in differentreference frame.

A reference frame is the local surrounding such as theground, a room or a moving train to which we refer theposition and motion of objects.

Within such a frame, the coordinate system permits

the measurement of the position.


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4.2 DISPLACEMENT, VELOCITY &

ACCELERATION Distanceis a scalar quantity that refers to how

long is the path taken for an object to move fromits initial position to final position.

Displacementis a vector quantity that refers tohow far is the final position of an object from itsinitial position.


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Speedis a scalar quantity that refers to how fastan object is moving.

Velocity is a vector quantity that refers to the rate

at which an object changes its position (taking intoaccount its speed and direction).

Average speed =Distance

Time taken

Average velocity = DisplacementTime taken


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Acceleration is a vector quantity that refers to

the rate at which an object changes its velocity. An object is accelerating if it is changing itsvelocity.

Constant acceleration is when an object is

changing its velocity at a constant rate. Graphical interpretation for linear motion:

Displacement-Time graph

Velocity-Time graph

Acceleration = Final velocity Initial velocityTime taken


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Displacement-Time Graph

Gradient of the graphrepresents thevelocityof the object.


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Velocity-Time Graph

Gradient of the graphrepresents the

acceleration of theobject.

Area under graphrepresents the distancecovered by the object.


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4.3 LINEAR MOTION WITHCONSTANT ACCELERATION This is the case study of a motion of an object along a

straight line under constant acceleration.

Physical quantities related to these problems are: Distance, s

Initial velocity, u

Final velocity, v

Acceleration, a


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Equations that are used to solve problems involving

linear motion with constant accelerations are:

DISTANCE VELOCITY

1

( )2s u v t

21

2

s ut at

v u at

2 2 2v u as


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Example:

A car travelling on a straight road increases its speed from 20km/h to 100 km/h in 45 s. What is the distance covered bythe car?

Solution:

u = 20 km/h

v= 100 km/h

t= 45 s =0.0125 h

1

( )2

1(20 / 100 / ) (0.0125 )

2

0.75

s u v t

s km h km h h

s km


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Example:

A drag racer starting from rest accelerates in a straight line at aconstant rate of 5.5 ms-2for 6 s.

a) What is the racers velocity at the end of this time?

b) If a parachute deployed at this time causes the racer to slowdown uniformly as 2.4 ms-2, how long will the racer take to

come to a stop?

Solution:

a= 5.5 ms-2

t= 6 s

u= 0 ms-1

2

1

0 (5.5 6 )

33

v ua

t

v u at

ms s

ms

a)


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b) a= -2.4 ms-2

u= 33 ms-1

v= 0 ms-1

1

2

(0 33)

2.4

13.75

v ua

t

v ut

a

ms

ms

s


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Example:

1. Two riders on buggies sit 10 m apart on long, straighttrack, facing opposite direction. Starting at the same time,both riders accelerate at a constant rate of 2 ms-2. How farapart will the buggies be after 3 s? (Answer: 28 m)

2. A train at rest starts to move from one station and stop at

another station after 18 minutes. In the first 2 minutes, itmoves with constant acceleration of 0.2 ms-2. After that itsspeed is constant until it is stopped by a constantdeceleration for 1 minute.

a) Sketch a velocity-time graph for the motion.b) Determine the distance between the two station.

(Answer: 23.76 km)


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FREE FALL MOTION

A motion of object moving downward solely under theinfluence of gravity.

Gravitational acceleration, g= 9.8ms-2 (downward)

For upwardmotion:

Gravitational acceleration, gis negative

For downward motion:

Gravitational acceleration, gis positive

At maximum height, the velocity equals to 0 ms-1.

The velocity of an object is the same at the same height

during upward and downward motion.

Time taken for an object to travel from ground to maximumheight is equal to the time taken for the object to travel tofrom maximum height to ground.


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Example:

1. A rock is thrown with the speed of 49 ms-1 vertically upward from the edge of a

cliff 117.6 m high. Calculate:a) The maximum height reached by the rock from the cliff. (Answer: 122.5 m)

b) The time taken to reach maximum height.(Answer: 5 s)

c) The total time taken for the rock to fall to the ground from maximum height.(Answer: 7 s)

d) The rocks velocity just before it strikes the ground. (Answer: 68.6 ms-1

)

2. A stone of thrown from the ground vertically upward with a speed of 30 ms-1.Determine the time interval between two instances when the stone is thrown atthe same height of 25 m above the ground. Take g= 10 ms-2. (Answer: 4 s)

3. An object is thrown vertically upward at the speed of 20 ms-1 from point P 25 mabove the ground. Sketch:

a) The displacement- graph time.

b) the velocity-time graph.


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4.4 LINEAR MOTION IN 2D:THE PROJECTILE MOTION Can be analyzed by resolving the motion into x-

and y-component.

Projectile motion has constant velocity along x-component, which means ax= 0 ms

-2.

The motion is solely under the influence of

gravitational acceleration, ay= 9.81 ms-2

.


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ux= u cos

vx= v cos

uy= u sin vy= v sin

hmax

v y= 0

xmax

ax= 0

ay= 9.8 ms-2

GRAPHICAL MOTION OF PROJECTILE MOTION


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Example:

Suppose a golf ball is hot off the tee with initial velocity of 30 ms-1at an

angle of 35 to the horizontal.a) What is the maximum height, hmaxreached by the ball?

b) What is the range,xmax?

Solution:

ux= (30 cos 35) ms-1= 24.57 ms-1

uy= (30 sin 35) ms-1= 17.21 ms-1

a) At maximum height, vy= 0 ms-1.

2 2

2 22

(0) (17.21) 2( 9.8)

0 296.18 19.6

296.1815.11

19.6

v u ass

s

s m

b)

0 17.211.76

9.81

y yy

y y

y

v uat

v ut s

a

Total flight time = 2t = 2(1.76 s)= 3.52 s

2

1

1; 0

2

(24.57 )(3.52 )

86.5

x x x x

x

s u t a t a

s ms s

m


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Example:

A ball is projected from top of a 25 m building and is thrown with initial

horizontal velocity of 8.25 ms-1.a) How long is the flight time of the ball before striking the ground?

b) How far from the wall of the building does the ball land?

Solution:

hmax= 25 m

ux= 8.25 ms-1

ay= 9.8 ms-2

a) 2 1

2

1

; 02

125 ( 98)

2

2.26

y y y ys u t a t u ms

t

t s

b) 2 2

1

1 ; 02

(8.25 )(2.26 )

18.65

x x x x

x

x

s u t a t a ms

s ms s

s m


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Example:

1. A cannon position on the roof top of a 9 m building in height is

directed horizontally. If a shell is fired at a speed of 150 ms-1

:a) How far will the shell be from the wall of the building when it hits theground? (Answer: sx= 203.3 m)

b) What is the speed of the shell before it strikes the ground?

(Answer: v= 150.6 ms-1)

2. An object is thrown upward at a speed of 30 ms-1at an angle 60.Determine:

a) The time taken by the object to reach maximum height.

(Answer: t= 2.6s)

b) The maximum height reached. (Answer: 33.8 m)

c) The maximum range, sx. (Answer: sx = 77.9 m)d) The time taken for the object to reach the height of 30 m.

(Answer: t= 1.73 s or 3.47 s)

e) The velocity of the object at height 30m. (Answer: v= 17.3 ms-1)

Use g= 10 ms-2.


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4.5 THE CONCEPT OF FORCE

A force may be thought of as any influence whichtends to change the motion of an object.

Practically, force can be defined as a push or a pull. The SI unit for force is Newton, N.

Force is a vector quantity, which means it hasmagnitude and direction.

Another definition of force is by stating Newtonssecond law, which states that force is the result ofthe multiplication of mass and acceleration.


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4.6 NEWTONS LAWS

Newtons laws of motion are three physical lawswhich provide relationships between forces

acting on a body and the motion of the body. They were first compiled by Sir Isaac Newton.

The laws form the basis of classical mechanics.

The three Newtons laws are related to inertia,force and friction.


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NEWTONS FIRST LAW: INERTIA

States that: An object at rest will remain at rest andan object in motion tends to stay in motion with thesame speed and in the same direction unless actedupon by an unbalanced force.

There are two parts in this statement: The behavior of a stationary object.

The behavior of moving object.

In other words, they can be described by saying that

objects tend to keep on doing what they are doing(unless acted upon by an unbalanced force).


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INERTIA

Definition: the tendency of objects to resist changesin their state of motion.

Passengers in a car tend to slide forward when thebrake is applied and the car slows down, especially

during emergency brake. The passengers areexperiencing inertia.

Inertia is a quantity which is solely depend onmass.

The more mass of an object, the more inertia it hasthe more tendencies it has to resist changes in itsstate of motion.


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NEWTONS SECOND LAW: FORCE

This law pertains to the behavior of objects for whichall existing forces are notbalanced.

States that: the acceleration, ais directlyproportional to the net force,Fand is inversely

proportional to the mass, m. The direction of acceleration is the same with the

direction of the net force.

In mathematical form, this law can be representedby the equation:

F = ma


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Example:

A car, which weighs 1000 kg, is out of gas. A man pushes the

car which managed to move at 0.05 ms-2to the nearest gasstation. Calculate the net force applied by the man in order tomake the car move.

Solution:m= 1000 kg

a = 0.05 ms-2

2

2

(1000 )(0.05 )

50

50

F makg ms

kgms

N


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Example:

A 10 N force acts on a wooden box of mass 4 kg, which is placed

on a smooth horizontal surface. Determined the accelerationof the object.

Solution:

m= 4 kg

F= 10 N

4 kg 10 N

2

2

10 (4 )( )

2.5

F ma

kgms kg a

a ms


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NEWTONS THIRD LAW: FRICTION /NORMAL FORCE

Also known as action-reaction principle. States that: for every action there is an equal but

opposite reaction. In another words, when an objectAexerts a force

on objectB, objectBwill exert a force with the samemagnitude but in opposite direction to objectA.

FABFBA

A

B

FAB= - FBA


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Example: When you sit on a chair, your body exertsa downward force on the chair. At the same time,the chair also exert an upward force with the samemagnitude to your body.

These two forces are called action-reaction forces.

500 N

500 N


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TYPES OF FORCES Normal force:

the reactionforce exerted from the surface by which the body is lying on.

Acts perpendicular to the surface of contact

The magnitude is equal to the magnitude of the action force.

Example:

A book is at rest on a horizontal table. The mass of the book is 0.8 kg. What

is the normal force,N acting on the book? (Answer: 7.84 N)

N

W

N

W


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Net force:

The vector sum of all the forces acting on an object.

Also known as resultant force.

When more than one force acts on an object, the subsequentmotion of the object is determined by calculating the net forceacting on the object.

Example:

In a game of tug of war, Team A pulls the rope with a force of 50 Nwhile Team B pulls with 75 N. Calculate the net force acting onthe rope. (Answer: 25 N towards Team B)

F1 F2F3

Fnet= -F1+ F2+ F3


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Equilibrium condition:

The condition when the net force acting on a body is equals to zero.

Conveys the idea that the forces are in balance; there is as muchforce upward as there is downward, as much to the left as to theright.

Any object with constant velocity, whether at rest or moving instraight line is said to be in equilibrium.

F1

F3

F2

F4

F1= F2F3= F4


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Friction:

Defined as the surface force that opposes motion.

Is the contact force parallel to the contact surface. The two types of friction:

Static friction,fs: the minimum force required to get an objectmoving from rest.fs= sN

Kinetic friction,fk: occurs when two objects are moving relative to

each other and rub together.fk= kN = coefficient of friction,N= magnitude of normal force

Example:Calculate the force needed to keep a book sliding at a constant speed, ifthe mass of the book is 2 kg. Given k=0.75. (Answer: 14.7 N)


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FREE BODY DIAGRAM (FBD)

Is a pictorial representation often used to show allcontact and non-contact forces acting on an object.

It is a simplified sketch of a single object with forcevectors drawn to represent every force acting on it.

Begins by drawing the object in question. Identify all forces acting on the object.

Convert each forces into the x- and y-component.

Example: a block of wood rests on a table:N

W


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4.7 THE CONCEPT OF MASS &

WEIGHT Mass:

the quantitative measure of inertia.

Is a scalar quantity. The SI unit is kilograms (kg).

Weight:

Is a vector quantity.

Is a form of force, acting downward, as a result of gravity.

The SI unit is Newton (N).

Defined as the mass times the acceleration of gravity.

W = mg


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4.8 LINEAR MOMENTUM Momentum,pis defined as the product of mass and velocity of an object.

The SI unit is Newton second (Ns).

Principle of Conservation of Momentum:

States that the total momentum in a closed system of objects (which hasno interaction with external agents) is constant.

In a system make out of objects that react (collide or explode), the totalmomentum is constant if no external force is acted upon the system.

A common problem in physics that requires the use of this principle is thecollision of two objects. In this case, the sum of the momentum beforecollision must equals to the sum of momentum after collision.

p = mv


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Total momentum before collision = total momentum after collision

p1= p2m1v1= m2v2

where:

p1and p2is the momentum before collision

p1and p2is the momentum after collision

u1and u2is the vector velocity before collision

v1and v2is the vector velocity after collision

m1and m2is the mass of each object respectively

There are two basic kinds of collisions: Elasticcollision: conserve kinetic energy as well as momentum

before and after collision

Inelasticcollision: does not conserve kinetic energy, but total

momentum before and after collision is conserved.


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Example:

A trolley of mass 2 kg moves to the right at a speed of 5 ms-1. It collides with

a table and rebounds to the left at a speed of 3 ms-1

. Determine:a) The momentum of the trolley before and after the collision.

b) The change in the momentum of the trolley after the collision.

Solution:

m = 2 kgu = 5 ms-1

v = 3 ms-1

1 1 1

1

1

2 2 2

1

1

(2 )(5 )

10

(2 )( 3 )

6

p m v

kg ms

kgms

p m v

kg ms

kgms

a) b)2 1

1

1

( 6 10)

16

p p p

kgms

kgms


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4.9 APPLICATIONS OF NEWTONS

LAW OF MOTION The three main application of Newtons laws of

motion are to solve problems involving:

Body on a horizontal plane Body on an inclined plane

Object on an ideal cord with frictionless pulley


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BODY ON A HORIZONTAL PLANEExample:

Sharifah drags a suitcase with a rope along the floor ofan airport terminal. The rope makes a 40 angle with thefloor. The suitcase has the mass of 30 kg and Sharifahpulls the rope with a force of 70 N.

a) What is the magnitude of the normal force acting onthe suitcase due to the floor?

b) If the coefficient of the kinetic friction between thesuitcase and the floor is 0.13, find the frictional forceacting on the suitcase.

c) What is the magnitude of the net force along the

horizontal?


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Solution:

30 kg

N

W

40fk

F = 70 N

Resolving F into x- and y-component:a)

70cos 40 53.6

70sin 40 45.0

x

y

F N

F N

For y-component:

1

0

(30 )(9.8 ) 45

249

y

y

N F W

N W F

kg ms N

N

b)

(0.13)(249 )

32.4

k kf N

N

N

c)

(53.6 33.4)

20.2

net x k F F f

N

N


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BODY ON AN INCLINED PLANE

Example:Suppose that a crate of pineapple with mass 20 kgslides down a ramp inclined at 30 with respect tothe horizontal. If the coefficient of kinetic friction

between the crate and the ramp is 0.3, what is themagnitude of Fnet?

N

W

fk

Fnet

30

Wx

Wy


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Solution:

Resolving W into x- and y-component:

196cos60 98

196sin 60 169.7

x

y

W N

W N

1

(20 )(9.8 ) 196W mg kg ms N

For y-component: 169.7yN W N

For x-component:

(0.3)(169.7 ) 50.9k kf N N N

(50.1 98)

47.9

net x k

net k x

F W f

F f W

N

N

Fnetis equal to 47.9 N towards negativex-direction.


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OBJECT ON AN IDEAL CORD WITH FRICTIONLESS PULLEY

Ideal cord has zero mass and does not stretch.

Principle: Each object is pulled upwards by the tension, T in the string. The

tension is the same at both end of the string.

The objects are pulled downwards by their weight, mg.

The objects have the same acceleration, a. The direction ofacceleration points downward for object moving downward, and itpoints upward for object moving upward.

aa

W1 W2

T

T

1 1 1

2 2 2

F T W m a

F T W m a


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Example:

Two plastic boxes of, 4 kg and 6 kg each, are connected by a light stringwhich passes over a smooth pulley. Determine

a) The acceleration of each box.

b) The tension in the string.

Solution:

aa

W1W2

T

T

1 1 1

1 1

2 2

2 2

4 4 (1)

2

6 6 (2)

F T W m a

T m g m a

T g a

F W T m a

m g T m a

g T a

a)

b) 4 4

4(9.8) 4(1.96)

47.04

T g a

T

T N

2

2

(1) (2) :

2 10

2(9.8 ) 10

1.96

g a

ms a

a ms


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Example:

The diagram below shows two objects connected by a light string over a

frictionless pulley. The object A has a mass of 25 kg resting on a woodentable with frictional force of 49 N. The object B has a mass of 15 kg. Thesystem is released and object B falls down pulling object A along.

a) Draw a free-body diagram for the forces acting on object A and B.

b) Calculate the acceleration on the system.

c) Calculate the tension on the string.

B

A


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a)

Solution: N

Wa

fk

Wa

Fa

Fb

T

T

B

A

b)

49 25 (1)

(15)(9.8) 15

147 15 (2)

A k A

B B B

F T f m a

T a

F W T m a

T a

T a

1

(1) (2) :

147 49 40

2.45

a

a ms

b)

49 25(2.45)

110.25

T

T N

Ch1 Kinematics and Dynamics

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