Chap 2 linear kinematics

UNIVERSITI PUTRA MALAYSIAASP 0501 Introduction to Mechanics © PUSAT ASASI SAINS PERTANIAN04/11/23UNIVERSITI PUTRA MALAYSIAASP 0501 Introduction to Mechanics © PUSAT ASASI SAINS PERTANIAN

CHAPTER 2: Linear Kinematics

2.3

2.2

2.1

2.4

2.5

UNIVERSITI PUTRA MALAYSIAASP 0501 Introduction to Mechanics © PUSAT ASASI SAINS PERTANIAN

KINEMATICS OF LINEAR MOTIONKINEMATICS OF LINEAR MOTION

• define and calculate displacement, velocity and acceleration.

• use linear kinematic equations for uniform acceleration

• solve free fall problems• solve projectile motion problems

LEARNING OUTCOMES

After completing this chapter you must be able to:


Classical MechanicsClassical Mechanics• Describes the relationship between the motion of

objects in our everyday world and the forces acting on them

• Conditions when Classical Mechanics does not apply– very tiny objects (< atomic sizes)– objects moving near the speed of light– Quantum and relativistic mechanics

• Kinematics or dynamics: – Kinematic describes the motion– Dynamics analyzes the cause of motion – force

2.1 Introduction


Sir Isaac NewtonSir Isaac Newton• 1642 – 1727• Formulated basic

concepts and laws of mechanics

• Universal Gravitation• Calculus• Light and optics

2.1 Introduction


KinematicsKinematics

• The branch of physics involving the motion of an object and the relationship between that motion and other physics concepts

• Kinematics is a part of dynamics– description of motion– Not concerned with the cause of the motion

2.1 Introduction


Types of MotionTypes of Motion

• Translational motion: change of position– Linear motion:

• One dimensional, the x- or y-axis– Curvilinear:

• Two dimensional, x-y axis• Parabolic path

– Circular motion• Circular path

2.1 Introduction


Types of MotionTypes of Motion

• Repeated Motions: – Rotation: change of orientation– Rolling: change of position and orientation– Simple harmonic motion– Damped harmonic motion– Wave motion

2.1 Introduction


Position vectorPosition vector• Defined in terms of a frame of reference

– Defines a point relative to a reference point.

Reference point

Reference point

An object that moves, means there is change in position.

2.2 Kinematics of Linear Motion


DisplacementDisplacement

• The displacement is the quantity that determines final position from the initial position of a moving object.

• Defined as the change in position

• f stands for final and i stands for initial

– The SI units are meters (m)

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

f ix x x

Initial position

Final position




• In linear motion, we may represent linear displacement as

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&f ix x x


Initial position Final position

xi xf



• It is a vector quantity.• The magnitude of displacement is the shortest

distance between the final position from the initial position.

• The direction of displacement is from the initial position to the final position



Distance Distance

• The distance travelled is the length measured along the actual path taken.

• Scalar quantity.



• The displacement of an object is not the same as the distance it travels– Example: Throw a ball straight up and then

catch it at the same point you released it• The distance is twice the height• The displacement is zero


Displacement versus distanceDisplacement versus distance


• The magnitude of displacement of an object is the same as the distance

only if it travels in a straight line.


Displacement versus distanceDisplacement versus distance


Average SpeedAverage Speed– the total distance traveled divided by the total

time elapsed

Average speed =Total distance moved

Total time taken

Average speed totally ignores any variations in the object’s actual motion during the trip

Speed is a scalar quantitySI units are m s-1



Average VelocityAverage Velocity

• Rate of change of position (rate of displacement)

• Average velocity – Ratio of displacement to time taken.

• A vector quantity.

f i

f i

x xxv

t t t



Average VelocityAverage Velocity

• A vector quantity.• Direction will be the same as the

direction of the displacement • The SI units of velocity are m s-1


Initial position

Final position


Average speed vs. Average velocityAverage speed vs. Average velocity

• Consider two cars taking the same time interval but on different paths.

• Cars on both paths have the same average velocity since they had the same displacement in the same time interval

• The car on the blue path will have a greater average speed since the distance it traveled is larger



Instantaneous VelocityInstantaneous Velocity• The velocity of an object at particular instant.• The limit of the average velocity as the time

interval becomes infinitesimally short, or as the time interval approaches zero

• The instantaneous velocity indicates what is happening at every point of time of the motion.

lim

0t

xv

t



Instantaneous VelocityInstantaneous Velocity

• The instantaneous velocity varies from point to point. • In magnitude or direction or both.



Instantaneous VelocityInstantaneous Velocity

• The instantaneous velocity normally written as:– velocity at point …– velocity at time …..

• The magnitude of the instantaneous velocity is commonly known as the speed at that particular position (time).



Uniform VelocityUniform Velocity

• Uniform velocity is constant velocity

• The instantaneous velocities are always the same – All the instantaneous velocities will also equal the average

velocity

constant

t

xv



AccelerationAcceleration

• The average acceleration is defined as the rate at which the velocity changes

• The instantaneous acceleration is the limit of the average acceleration as Δt approaches zero

av t

v

arr

dt

vda

rr



AccelerationAcceleration

• Rate of change of velocity• Changing velocity (non-uniform) means an

acceleration is present• Acceleration is the rate of change of the

velocity

• Units are m s-²

fi

fi

v vva

t t t



Ways an Object Might AccelerateWays an Object Might Accelerate

• The magnitude of the velocity (the speed) can change

• The direction of the velocity can change– Even though the magnitude is constant

• Both the magnitude and the direction can change



Average AccelerationAverage Acceleration

• Vector quantity• When the sign of the velocity and the

acceleration are the same (either positive or negative), then the speed is increasing

• When the sign of the velocity and the acceleration are in the opposite directions, the speed is decreasing



Instantaneous AccelerationInstantaneous Acceleration

• The instantaneous acceleration is the actual acceleration at any particular time or position.

• When the instantaneous accelerations are always the same, the acceleration will be uniform– The instantaneous accelerations will all be equal

to the average acceleration



Graphical representation of VelocityGraphical representation of Velocity

• Velocity can be determined from a position-time graph

• Average velocity equals the slope of the line joining the initial and final positions

• An object moving with a constant velocity will have a graph that is a straight line



Uniform velocity graphUniform velocity graph

• The straight line indicates constant velocity

• The slope of the line is the value of the average velocity

• A positive gradient means a positive velocity

• A negative gradient means a negative velocity



Non uniform velocityNon uniform velocity

• The graph is a curve.• The average velocity is the

slope of the blue line joining the starting and end points.

• The instantaneous velocity is the tangent at each point.



Non-uniform VelocityNon-uniform VelocityPositive direction, Increasing magnitude

Positive direction, decreasing magnitude

Negative direction, deccreasing magnitude

negative direction, Increasing magnitude

time time

time time



Graphical representation of AccelerationGraphical representation of Acceleration

• Average acceleration is the slope of the line connecting the initial and final velocities on a velocity-time graph

• Instantaneous acceleration is the slope of the tangent to the curve of the velocity-time graph



Acceleration: velocity – time graphAcceleration: velocity – time graph

dt

dva



Displacement, Velocity and AccelerationDisplacement, Velocity and Acceleration

x

xt

xv

t

va

Change of position

Change of position

Time interval

Change of velocity

Time interval



Problem solving strategyProblem solving strategy1. Read and understand the question. 2. Visualise the situation. In some cases you may

need to split the motion into sections.3. Draw the vector diagram to represent the motion.

It is advisable to draw a separate diagrams for the displacement vectors and velocity vectors.

4. If you need to determine the resultant vector, use the problem strategy for vector additions.

5. Extract the relevant data from the question. 6. Use the relevant equation to find the required

quantity. 7. Solve the equation to calculate your answer. 8. State your final answer. Do not forget the unit.



Uniformly accelerated linear motionUniformly accelerated linear motion• Acceleration is constant, a

ti tf

xi xfAssume : initial time t i = 0 then time interval t = t f – t i = t

Assume : initial position x i = 0 then displacement x = x f – x i = s

t, s, a

2.3 Uniformly Accelerated Linear Motion



t

u vAssume: Initial velocity u Final velocity v

acceleration a = Change in velocity

Time interval=

v - u

t

2.3 Uniformly accelerated linear motion



u v

t, s, a

a =v - u

tv = u + a t

average velocity < v > = v + u

2

①

=st

s = ½ (u + v) t ②




u v

t, s, a

average velocity < v > = v + u

2

average velocity < v > = displacementtime interval

= st

st =

v + u

2=

(u + at) + u2

s = u t + ½ a t2 ③



v = u + a t …① s = u t + ½ a t2 …③

Uniformly accelerated linear motionUniformly accelerated linear motion

v - ua

v - ua

s = u + ½ a 2

v2 = u2 + 2 a s

u v

t, s, a

④



Kinematic equationsKinematic equations

equation u v s t a

v = u + at

s = ½ (u + v) t

s = ut + ½ at2

v2 = u2 + 2 a s

u v

t, s, a



Problem solving strategyProblem solving strategy2.3 Uniformly accelerated linear motion

1. Read and understand the question. 2. Visualise the situation. In some cases you

may need to split the motion into sections.3. Draw a straight line to represent the linear

motion. Indicate the initial and final positions.

4. In your diagram, indicate the initial velocity, final velocity, acceleration, time interval and displacement, for each section of the motion.

u v

t, s, a


Problem solving strategyProblem solving strategy2.3 Uniformly accelerated linear motion

5. Determine the known and unknown values.6. Use the relevant kinematic equation to determine

the unknown values. Take care of the signs of each values.

7. State your final answer. Do not forget the unit.


DefinitionDefinition

• All objects moving under the influence of gravity only are said to be in free fall

• All objects falling near the earth’s surface fall with a constant acceleration

• The acceleration is called the acceleration due to gravity, and indicated by g

2.4 Free Fall Motion


Acceleration due to GravityAcceleration due to Gravity

• Symbolized by g• g = 9.81 m s-2 (constant)• g is always directed downward

– toward the center of the earth

• Ignoring air resistance and assuming g doesn’t vary with altitude over short vertical distances, free fall is constantly accelerated motion



Kinematic equations for free fall motionKinematic equations for free fall motion

equation u v h t g

v = u + (-g) t

h = ½ (u + v) t

h = ut + ½ (-g) t2

v2 = u2 + 2 (-g) h u

v

- g

h

t

Reference direction: positive upwards



Object thrown vertically upwardObject thrown vertically upward• Initial velocity is not zero• Let up be positive• Use the kinematic equations

v ( + ve)

a (- ve)

t ( +ve)

h ( +ve)

u ( + ve)

v = u + (g ) t

h = u t + ½ (g) t 2

v2 = u2 + 2 (g) h

g



Free Fall – an object droppedFree Fall – an object dropped

• Let up be positive• Use the kinematic equations

u (-ve)

a = - g

t ( +ve)

h ( -ve)

V ( -ve)

(- v) = (- u) + (- g) t

(-h) = (- u) t + ½ (-g) t 2

(- v2) = (-u)2 + 2 (-g) (-h)



Object thrown vertically upwardObject thrown vertically upward

• Initial velocity is upward.• a = g = 9.80 ms-2 is always

downward everywhere in the motion

• The instantaneous velocity at the maximum height is zero

v = 0



Object thrown vertically upward and caught at the same position.

Symmetrical Free FallSymmetrical Free Fall

Initial velocity = final velocity but in opposite direction

Time to reach highest position = time to drop to initial position

u v

tup tdown



Non-symmetrical Free FallNon-symmetrical Free Fall• Object thrown vertically

upward and falls to a lower level.– Upward and downward

portions



Problem solving strategyProblem solving strategy2.4 Free Fall Motion

1. Read and understand the question. 2. Visualise the situation. In some cases you may need to

split the motion into upward and downward sections.3. Draw a vertical straight line to represent the free fall

motion. Indicate the initial and final levels. 4. In your diagram, indicate the initial velocity, final

velocity, acceleration, time interval and displacement, for each section of the motion.

5. Determine the known and unknown values.6. Use the relevant kinematic equation to determine the

unknown values. Take care of the signs of each values.7. State your final answer. Do not forget the unit.


Projectile MotionProjectile Motion• An object may move in both

the x and y directions simultaneously– It moves in two

dimensions• The form of two dimensional

motion we will deal with is called projectile motion

2.5 Projectile Motion


Projectile Motion - accelerationProjectile Motion - acceleration• y-direction

– free fall problem, acceleration ay = g (downward)

– Uniformly accelerated linear motion, so the motion equations all hold

• x-direction– Uniform motion – ax = 0



Projectile Motion – initial velocityProjectile Motion – initial velocity• The initial velocity can be resolve into its x-

(horizontal) and y- (vertical) components–

v

V cos

V sin



Projectile Motion – velocityProjectile Motion – velocity• The velocity x- and y-directions of motion are completely independent of each

other• The velocity can be resolve into its horizontal (x-) and vertical (y-) components.

Velocity components



Velocity of the ProjectileVelocity of the Projectile

• The velocity of the projectile at any point of its motion is the vector sum of its x and y components at that point



Velocity of the ProjectileVelocity of the Projectile• The velocity of the projectile at any point of its

motion is the vector sum of its x and y components at that point.

• The magnitude and direction of the velocity can be calculated using:

2 2 1t a n yx y

x

vv v v a n d

v



Projectile Motion – displacementProjectile Motion – displacement► The displacement in x- and y-directions of motion are completely independent of The displacement in x- and y-directions of motion are completely independent of

each othereach other► The displacement can be resolve into its horizontal (x-) and vertical

(y-) components.

Displacement components



Projectile MotionProjectile Motion

Horizontal component

Vertical component ( + upward)

acceleration ax = 0 ay = - g

initial velocity vox = vo cos voy = vo sin

final velocity (at max height)

vx = vox vy = 0

time interval t t

displacement x y horizontal

vo



Horizontal component Vertical component ( + upward)



final velocity(at max height)

vx=vox vy = 0

time interval t t

displacement x y


For vertical motion (+ upward):② s = u t + ½ a t2

For horizontal motion:

② s = u t + ½ a t2

cos

)0(2

1)cos( 2

o

o

v

xt

ttvx

2

cos)(

2

1

cos)sin(

ooo v

xg

v

xvy

22

2

cos2tan

ov

gxxy parabolic equation

y = ax2 + bx + c




►The shape of the path is a parabolaThe shape of the path is a parabola




horizontal

vo

At max height , vy = 0, using Eqn ①

v = u + at 0 = vo sin (- g) T that is,

g

vT o sin

H






vx=vox vy = 0

time interval T

displacement H




At max height, using Eqn ② s = u t + ½ (-g) t2

g

vT o sin

horizontal

vo

g

v

g

vg

g

vvH

o

oo

2

sin

sin)(

2

1sinsin

22

2

0






vx=vox vy = 0

time interval T

displacement H

H



Projectile MotionProjectile MotionHorizontal component





vx = vox = vo cos vy = 0

time interval 2 T T

displacement R Hhorizontal

vo

At max horizontal displacement , using Eqn ② s = u t + ½ (-g) t2

g

vT o sin

g

v

g

v

g

v

g

vvR

oo

ooo

)2sin(sincos2

sin2)0(

2

1sin2cos

22

2

R



Height and Range of the projectileHeight and Range of the projectile

• Maximum height: (y-axis)

• Range: (x-axis)

g

vH o

2

sin22

g

vR o 2sin2


horizontal

vo

R

H


Range and height of projectile motionRange and height of projectile motion

• Complementary values of the initial angle result in the same range– The heights will be

different• The maximum range

occurs at a projection angle of 45o



Some Variations of Projectile MotionSome Variations of Projectile Motion• An object may be fired

horizontally• The initial velocity is all

in the x-direction vo = vx and vy = 0

• All the general rules of projectile motion apply



Projectile MotionProjectile MotionHorizontal component Vertical component

( + down)

acceleration ax = 0 ay = + g = 9.81 ms-2

initial velocity vox = 40.0 ms-1 voy = 0

final velocity vx = ? vy = ?

time interval T = ? T = ?

displacement R = ? H = 100 m

Find (a) the range R (b) time to reach the ground (c) velocity when touching the ground.



Non-Symmetrical Projectile MotionNon-Symmetrical Projectile Motion• Follow the general

rules for projectile motion

• Break the y-direction into parts– up and down– symmetrical back to

initial height and then the rest of the height



Non-Symmetrical Projectile MotionNon-Symmetrical Projectile Motion


Vertical component ( + up)

acceleration ax = 0 ay = g = - 9.81

initial velocity vox = 20 cos 30 voy = 20 sin 30

final velocity vx = ? vy = 0

time interval t1 + t2 = ? t1 = ?

displacement X = ? h1 = ?



Problem solving strategyProblem solving strategy2.5 Projectile Motion

1. Read and understand the question. 2. Visualise the situation. 3. Draw the path of the projectile motion. 4. Resolved your vector quantities into vertical and

horizontal components5. For each component, determine the known and

unknown values.6. Use the relevant kinematic equation to determine the

unknown values. Note: Horizontal component, a = 0

Vertical component, a = g = 9.81 ms-2 downward7. State your final answer. Do not forget the unit.

Education

Chap 2 linear kinematics