week 9 kinematics

7/23/2019 Week 9 Kinematics

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T e stu y o o jects

in motion under theaction of forces

Determine trajectorieso sa e es

(aerospace) Highways design:

geometry and speed

limit2


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Links with other sub ects and rofessions

Maths, Physics

Statics Dynamics

Mechanics of Solids Structural DynamicsFluid Dynamics

Structural Analysis,

Water Engineering

3Design of structures


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Kinematics and Kinetics

Kinematics: Branch of dynamics which

reference to the forces that cause the motionconcerne w e geome r c aspec s o mo on

-- Geometry of Motion

Kinetics: Study of the relations between forces

-- Force, Mass and Acceleration-- Work and Energy

-- Im l n M m n m

concerned with the forces causing the motion


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Introduce the fundamental of dynamics

curvilinear motion

ne cs o par c es, orce, mass an acce era on Work and energy

Impulse and momentum

5


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Rectilinear motion

Position and displacement Velocity

Acceleration

Rectilinear motion with

o

Variable acceleration


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Rectilinear Motion

he sim lest motions are those in a strai ht line.

Consider the sim lest form with a constant velocit .

The next simplest form is constant acceleration.


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Speed and Velocity

S eed has onl ma nitude.

Velocit has a ma nitude and direction.

Speed is a scalar, Velocity is a vector.

We will deal with velocity.


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Position and Displacement

Consider a particle moving in a straight line.

time is the displacement (s).

Change in position means the final position minusinitial position.


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Displacement and Velocity

Displacement in time tis s. Displacement in time

.

displacement is s.

sss =

herefore velocit s

v

=t

,

Instantaneous velocity

sdtv &

==


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Displacement ,Velocity , Acceleration

Similarly acceleration is the rate of change of velocity

vdva &== 2)( sd

ds

d ssda &&==

2

dt 2dtdt

a == dt

If the acceleration is uniform, then

t

a

= atuv +=

s=average velocity time

2)2()2( atuttts +===


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atuv +=

Squaring both sides

)(22 222222 atutautauatuv ++=++=

asuv 222 +=Note: the following statement is true only for

.

vu +

2average


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sva &&&==

dvdsdvdvds

vdtdsdt

a ===

dvsddv===

2

&

dsdtdt

sssva &&&& ===


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For a constant acceleration

adv

v ==& t Tv

=

dtadv [ ] [ ]Tvu tav 0=

= auv

aTuv +=


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Relation between s, u, aand T

aTuv +=

atuv += atus

sv +=== &

TS T

+= tatus

00

atuts

0

0

2

+=

22a 2a

2

2


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Relation between s, u, vand a

a

v

v =

Sv v2

=

dsavdv sa 02 =

uv=

22 22 =

22


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aSuv = asuv =

asuv 222 +=

Caution: one of the frequent mistakes made by

for problems of variable acceleration.

The above equations derived are only valid for.


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Rectilinear Kinematics: Erratic Motion

Given the s-tGraph, Construct the v-tGraph


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Given the v-tGraph, Construct the a-tGraph


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Given the v-sGraph, Construct the a-sGraph


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EXAMPLE 1

75m/s. When it is 40m from

, .

Determine

.B

the rocket (Answer: 327m)

2.its speed just before it hits

the ground(Answer: -80.1m/s).


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EXAMPLE

Solution:

.

coordinate at ground level withpositive upward.

.

Knowns: 1) vA = +75m/s when t= 0. s = sB when

v = 0 at max ht.

2) aC= -9.81m/s2 (negative since it actopposite sense to positive velocity or

positive displacement)

Solve for s (s0=0 in the selected coordinate system)

A 0 aC=-9.81m/s2 v=75m/s 40

B t = ? aC=-9.81m/s v=0 s=?


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EXAMPLE


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Variable Acceleration

Acceleration is the function of time:

=

for velocity v and displacement s. Displacement s is

.

s=

r u r .


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Variable Acceleration Example 2

Acceleration a = 4t-30, Determine v and sas function

o time t. Initia y at t= , s0= -5 man v0= m s.

tv

304 === tdtav& dttdvv )30400 =

t

ttvv 2

302 =2 2

0 0


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Variable Acceleration Example 2

2ds&

ts

2

dt

s 00

=

ttttss 32 2153

+=

3

32

0

2153 tttss ++=

32 2

3s =


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Position, displacement, velocity and acceleration

Motion of a projectile

Polar components

Circular motion


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Plane curvilinear motion

This describes the motion of a particle along a curved

.

Circular motion and Projectiles

Consider the continuous motion of a particle along a

p ane curve.


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General Curvilinear Motion

Curvilinear motion occurs when the particle moves

a ong a curve pa

. ,from a fixed point O, is designated by theposition


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Position. Suppose P is in a motion relative to the

c osen re erence rame, so e r s a unc on o

time t.We can express this

mo on y

trr=


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Displacement. Suppose during a small time

interval (t, t+t) the particle moves a distance s

along the curve to a new position P, defined byr =r(t+t). The displacementrrepresents the

change in the particles position.

)()( trttrr +=


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Velocity. The average

defined as

tavg

=

r

v

The instantaneous

from this equation by

,

d&

r

dt==


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Direction of v is tangent

o e curve

speed, which may be

magnitude of the

length of the straight line

ds

dt


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Acceleration. If the particle has a velocity v at time = + = +

The increment of velocity isv

=v(t +

t) -

v(t).

The average acceleration

tavg

=a

The instantaneous

acceleration

dd&&& ====

2rv

dtdt 2


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Curvilinear Motion: Rectangular Components

Position. Position vector is defined by

r=xi + yj + zk

, ,

The magnitude ofr is always

222positive and defined as

The direction of r isspecified by the

components of the

unit vector ur= r/r


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Velocity.

kvjviv

dt

zyx

rrr++==v

zvyvxv zyx &&& ===w

magnitude defined as the

222

vvvv ++=and a direction that is specified by the

v

always tangent to the path.


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Acceleration.

kajaia

dt

dzyx

rrr++==

v

a

yva

xva

yy

xx

&&&

&&&

==

==where

zva zz &&& ==

positive value of

zyx aaaa ++=

velocity, a will notbe tangent to the path.


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Projectile motion

An important application of two-dimensional


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Motion of a Projectile

Free-flight motion studied in terms of rectangular

components since projectiles acceleration

alwa s act verticall Consider projectile launched at (x0, y0)

-

Air resistance neglected

Only force acting on the projectile is its weight,

resulting in constant downwards accelerationay= g= - 9.81 m/s

2


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Motion of a Projectile: Horizontal motion

At t= 0:

x= 0 and vx = v0cos 0.

e acce era on n ex-

direction is zero i.e.

0== dva xx

Therefore vx is constant and remains equal to itsinitial value:dx

00

dt

x


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Motion of a Projectile: Horizontal motion

Integrating dtvdtvdx x 00 cos==

tx vx0

000

cos=

tvx 00 cos=,

velocity of the projectile in thex-direction

as functions of time w/o considering the

ro ectiles motion in the -direction.


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Motion of a Projectile: Vertical motion

At t= 0;

y 0 0.

The acceleration in the y-direction is ay= - g

gdta

y

y ==

By integrating, v tywe o a n:

d

v ysin 000

gtv

dt

vy == 00s n


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Motion of a Projectile: Vertical motion

Integrating vy yields:

00 )sin( dtgtvdy

y t=

200

1)sin( gttvy =

horizontal motion.

Eliminating time t from


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Eliminating time t from

( ) 2000021)sin(;cos gttvytvx ==

Parabolic trajectory of the projectile:

022

0

0cos2

tan xv

xy

=


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e muzz e ve oc y

of a long-range rifle

a s u= m s.

Determine the two

ang es o e eva on

which will permite pro ec e o

the mountain target

. w : . ,26.1o)

Example


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Example

Example


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Example

Curvilinear Motion: Polar Coordinates


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Curvilinear Motion: Polar Coordinates

Polar Coordinates

using

1. radial coordinate rwhichextends outward from the

fixed origin O to the

particle and a2. traverse coordinate ,

which is the

counterclockwise angle

e ween a xereference line and the r

/tan 1 x

yxr

=

+=

Curvilinear Motion: C li d i l C di t


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Curvilinear Motion: Cylindrical Coordinates

Vectors in polar coordinates are defined by uisngthe unit vectors u and u extends from P

ur : along increasing r,

when is held fixed u : along increasing

when ris held fixed

Position

):asor written(rr

urrr rr

== urNote that urand u

change directions as P moves

are perpendicular to each other

Rotatin Unit Vector


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Rotatin Unit Vector

Unit vector e rotating at an angular velocity

dt

d=&

Rotation angle during t : = (t + t) - (t)

The time derivative of eis defined b

( ) ( )tttd + eee

tdt t 0

Curvilinear Motion: C li d i l C di t


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Velocity

derivative of r (using product rule)

rrr rr

dt

ruu

urv &&& +===

To evaluate note thaturchanges its directionr

u&V

w.r.t. time although its

magnitude = 1 (rotatingVr

V

of a unit vector)

&& r



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Instantaneous velocity v

&

vv rr += uuv

&

rvr=

Radical componentvris a measure of the rateof increase or decrease in the length of the

radial coordinate

Transverse componentv is the rate of motion

radius r



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Since vrand v are,

magnitude of the velocity

positive value of

( ) ( )22 && rrv += Direction of v is tangent



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Acceleration

uuva &&& rr

dt r+== )(

uuuuu &&&&&&&&&& rrrrr rr ++++=

&&& . . . ruu =



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Instant acceleration has two componets

&

ruauaa rr +=

&&&&

&& rrar =

The term is called the an ular22 / dtd =&&

acceleration since it measures the change made

in the an ular velocit durin an instant of time Use unit rad/s2



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Since arand a are alwaysperpendicular, the magnitude of

the acceleration is simply thepositive value of

2&&&& 2 &&& rrrra ++=

from the vector addition

Acceleration is not

Circular Motion


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Circular Motion

For circular motion, radius ris constant for all .

Coordinate System. Polar coordinates

. ,

00 === rrrr &&&

Circular Motion


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C cu a o o

Velocity and Acceleration. )00constant( === rrr &&&

&

&rvr == 0

&&&&

rv =22

v

r 2

&&&&&

r=

& rrra =+= 2

Circular Motion


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For circular motion with a constant angular=&&&

0==rvr &

2

=

rv

&

&

2

=

v

r

=

=r

Acceleration is towards the centre of the circle

EXAMPLE


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Due to the rotation of the

across the slotted path, a

shape of a cardioids, r=

in radians. If the ballsvelocit is v = 1.2m/s and its

acceleration is 9m/s2 at

instant

= 180

determinethe angular velocity and

an ular acceleration of the

fork.

EXAMPLE


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EXAMPLE


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Motion of a Projectile


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j

0at0If tyx

====

2tan xg

x =

00sin gtvv y = 00 cos2v

002

)sin( gttvy =

Position: rurr rr =

VVelocity:

&

rrr

rv

uvuvv

r

rr

=

+=

r

&rv =

&

rruauaa rr +=

Acceleration:

&&

&&

&&

rra

rra r

2+=

=

urr rr

= rrr

uvuvv +=


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uurr

&

r

& = &rvr=r&

r&

r&r

&

r

&

r

&

r

&

r

===

&rv =r =

rrr( ) ( )&& rrv +=

urururururva rr

&&&&&&

r

&&r&&

r&&

r

&&r

&&

r

&r

++++== rruauaa rr +=

ururururur rrr

&&

&&r

&&&

rrr&

r&

r&& )(

2

=

+++=

&&&&

&&& rrar2=

r =

( )

22 2 &&&&&&& rrrra ++=

week 9 kinematics

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