Kinematics of a particle

34

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CHAPTER.12. Kinematics of a particle. Mechanics of rigid body. mechanics of rigid body. Galilei. Statics. Dynamics. Statics. Dynamics. Newton. Euler. Lagrange. Equilibrium. Σ F = 0. Kinematics. Kinetics. v=ds/dt a=dv/dt. Σ F = m a. 12.1 Introduction. - PowerPoint PPT Presentation

Transcript of Kinematics of a particle

Page 1: Kinematics of a particle
Page 2: Kinematics of a particle

Mechanics of rigid body

Statics Dynamics

Equilibrium

Galilei

Newton

Lagrange

Euler

Kinematics Kinetics

v=ds/dta=dv/dt

Σ F = 0

Σ F = m a

mechanics of rigid body

Statics Dynamics

12.1 Introduction

Page 3: Kinematics of a particle

1.1. KinematicsKinematics Analysis of the geometric aspects of motion. Analysis of the geometric aspects of motion.

2.2. ParticleParticle A particle has a mass but negligible size and shape. A particle has a mass but negligible size and shape.

3.3. Rectilinear KinematicsRectilinear Kinematics Kinematics of objects moving along straight path and Kinematics of objects moving along straight path and characterized by objects position, velocity and characterized by objects position, velocity and acceleration.acceleration.

4.4. PositionPosition(1) position vector r(1) position vector r A vector used to specify the location of particle P at any A vector used to specify the location of particle P at any instant from origin O.instant from origin O.

12.2 Rectilinear Kinematics: Continuous motion

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(2) position coordinate , S An algebraic scalar used to represent the position coordinate of particle P

rS

scalar vector

5. DisplacementChange in position of a particle , vector

(1) Displacement

rrr ' or SSS '

from O to P.

P’

P

r’

r

s

r

s

s’

o

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(2) DistanceTotal length of path traversed by the particle.A positive scalar.

6. Velocity(1) Average velocity

t

rVavg

(2) Instantaneous Velocity

dt

rd

t

rV lim

0t

dt

ds

t

Slimv

0t

)(

speed = magnitude of velocity = | v |Average speed = Total distance/elapsed time = t

ST

(3) Speed

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7. Acceleration

t

Vaavg

(2) (Instantaneous) acceleration

t

vd

t

vlima

0t

2

2

)(dt

Sd

dt

ds

dt

d

dt

dVa

(1) Average acceleration

)(

8. Relation involving a , s and vv=ds/dt, dt=ds/va=dv/dt, dt=dv/a

so, ds/v=dv/a vdv=ads

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dt

dvac dtadv c dtadv c

atvv o

9. Constants acceleration a = ac

t

0c

v

vtav

0

dt

dsv

t

c

s

sdttavds

0 0 )(0

dsavdv c dsavdvs

s

c

v

v 00

s

sc

v

v

2

0

0

sav2

1 )ss(a2vv 0c

20

2

0

t

2

1t

So

SS tav

2

c0

tavs

2

c00 2

1tS

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10. Analysis Procedure

(2) Kinematic Equations A. Know the relationship between any two of the

four variables a, v, a and t. B. Use the kinematic equations to determine the

unknown varaibles

(1)Coordinate System

A. Establish a position coordinate s along the path.

B. Specify the fixed origin and positive direction of the coordinate.

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12.3 Rectilinear Kinematics : Erratic Motion

t t t

S V a

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GivenGiven methodmethod Kinematics egnKinematics egn FindFind

S-t graphS-t graph Measure slopeMeasure slope V=ds/dtV=ds/dt V-t graphV-t graph

V-t graphV-t graph Measure slopeMeasure slope A=dv/dtA=dv/dt a-t grapha-t graph

A-t graphA-t graph Area integrationArea integration v-t graphv-t graph

v-t graphv-t graph Area integrationArea integration s-t graphs-t graph

A-s graphA-s graph Area integrationArea integration v-s graphv-s graph

v-s graphv-s graph Measure slopeMeasure slope A=v(dv/ds)A=v(dv/ds) a-s grapha-s graph

tadtvvv

00

tvdtsss

00

2

12

0 )2(0

vadsvs

s

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1. Curvilinear motion1. Curvilinear motion

The particle moves along a curved path.The particle moves along a curved path.

Vector analysis will be used to formulate the

particle’s position, velocity and acceleration.

p

12-4 General Curvilinear Motion

Page 12: Kinematics of a particle

s

= change in position of particle form p to p’

(t)r r

r

r'rr

o

p

r (t)

s

2. Position2. Position

3. Displacement3. Displacement

o

p

p’s

r

r ’

r

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t

rVavg

td

rd

t

rV

t

lim0

= “tangent” to the curve at Pt .p= “tangent” to the path of motion

4. Velocity

(1) average velocity 平均

(2) Instantaneous velocity (2) Instantaneous velocity 瞬時瞬時

o

p

p’v

r

r ’

r

(3) Speedvv

dt

ds

t

s

t

rlimlim

0t0t

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t

v

t

vva avg

dt

rd

dt

rd

dt

d

dt

vd

t

va

2

0tlim

(2) Instantaneous acceleration

which is not tangent to the curve of motion, but

tangent to the hodograph.

Hodographv

'vv

5. Acceleration

(1) Average acceleration(1) Average acceleration

= time rate of change of velocity vectors

Hodogragh is a curve of the locus of points for the arrowhead of velocity vector.

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kzjyixr

xyz : fixed rectangular coordinate system

12-5 Curvilinear Motion : Rectangular components

x

y

z

θ

r

pathp

s

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1. Position vector

rur

kzyx

zj

zyx

yi

zyx

xzyx

ktzjtyitxtr

222222222

222

)()()()(

Here

222 zyxrr = magnitude of r

r

ru r

= unit vector = direction of r

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2. Velocity

v

zyx

uv

kvjviv

kdt

dzj

dt

dyi

dt

dxdt

kdzk

dt

dz

dt

jdyj

dt

dy

dt

idxi

dt

dx

kzjyixdt

ddt

)t(rd)t(v

0 0 0

222zyx vvvv

v

vu v

tangent to the path

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3. Acceleration

a

au

aaaa

a

zyx

222

a

zyx

2

2

2

2

2

2

zyx

zvx

ua

kajaia

kdt

zdj

dt

ydi

dt

xd

kdt

dVj

dt

dVi

dt

dV

kVjViVdt

ddt

Vda

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: initial velocity

: Constant downward acceleration

: velocity at any instant

0V

a

V

12.6 Motion of a projectile

y

x

v0

v0

a= - g j

v(v0)x

(v0)y

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Position Vector (x,y components)

= x + y

initial position

= xo + yo

Velocity Vector

= = x + y = Vx + Vydt

vd

r

i

i

j

0r

j

V

i

j

j

i

Acceleration Vector

a = = + = -g = ax + ay

V0 = (Vx)o + (Vy)o (known)

dt

vd

idt

dVx j

dt

dVy j

i

j

i

j

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1. Horizontal motion, 1. Horizontal motion, aaxx=0=0

0dt

dvx

)v(dt

dxv xx

One independent eqn

20x0x

20x

2x )v()xx(a2)v(v

Vx = (Vx)0 + axt = (Vx)0

X = X0 + (Vx)0t

Same as 1st Eq.

X = X0 + (Vx)0t

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gt)v(ta)v(v 0yy0yy

two independent eqns

200 2

1)( tatvyy yy

)yy(g2)v(v 02

0y2

y

20y0 gt

2

1t)v(yy

2. Vertical motion, aayy=-g constant=-g constant

Can be derived from above two Eqs.

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Path of motion of a particle is known.

1. Planar motion

12-7 Curvilinear Motion:Normal and Tangential components.

opath

o’n

ρ

un

uttp

Here: t (tangent axis ): axis tangent to the curve at P and positive in the direction of increasing S; ut: unit vectorn (normal axis ): axis perpendicular to t axis and directed from P

toward to the center of curvature o’; un: unit vectoro’ = center of curvature= radius of curvaturep = origin of coordinate system tn

s

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(1) Path Function

(known) sts

(2) Velocity

dt

dsv tt utsuvv

(3) Acceleration

dt

udsusus

dt

d

dt

dva t

tt

o’ρ

un

ut

pds

ut’

un

dut

p

ut’

ut

ttt uduu '

dduud tt nt uud

//

tttt uduudud

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dsd

nnt uu

dt

d

dt

ud

s

nt u

s

dt

ud

nnttn

2

t uauaus

usa

at: Change in magnitude of velocity

an: Change in direction of velocity

vdvdsat

n

2

tn

2

t uv

uds

dvvu

vuva

2

2

2/32

dxyd

dxdy1

If the path in y = f ( x )

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1. Polar coordinates

u

r

pr ru

Reference line

r: radial coordinate , ru

: transverse coordinate , )..( wccu

(2) Position

rurr

(3) Velocity

rrr urururdt

d

dt

rdrv

)(

12-8 Curvilinear Motion: Cylindrical Components

(1) coordinates (r,)

o

Page 27: Kinematics of a particle

1

1

duu

'ru

ruru

uuuu

uuu

rr

rr

'

utt

u r

utt

ulimlim

0t

r

0t

uu r

ururv r

uvuv rr

r

)r()r(v 22

rate of change of the length of the radial coordinate.

angular velocity (rad/s)

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(4) Acceleration

vdt

vda

)uvuv( rr

rr u)u(u

uu

uu'u

uvuvuvuv rrrr

ruu

rr urururura )(

uaua

urrurr

rr

r

)2()( 2

22 )a()a(aa

r0t0t

u)t

(t

ulimlim

angular acceleration

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2. Cylindrical coordinates

Position vector

zrp uzurr

Velocity

zrzrpp uzururuz)ur(rv

Acceleration

zrpp uzururva )(

zr uzurrurr )2()( 2

z

x

yr

pr

3D

ru

uzu

v

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12.9 Absolute Dependent Motion Analysis of Two Particles

1. Absolute Dependent Motion The motion of one particle depends on the corresponding motion of another particle when they are interconnected by inextensible cords which are wrapped around pulleys.

A BA B

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(1)position-coordinate equation

A. Specify the location of particles using position coordinates having their origin located at a fixed point or datum line.

B. Relate coordinates to the total length of card lT

(2)Time Derivatives

Take time derivatives of the position-coordinate equation to yield the required velocity and acceleration equations.

2.Analysis procedure

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3. Example

A B

ASBS

DatumDatum

(1)position-coordinate equation

BAT SDCSl

(2) Time Derivatives

BA SS 0 BA VV BAT SDCSl

BA VV

BA aa

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12.10 Relative-Motion Analysis of Two Particles

1. Translating frames of referenceA frame of reference whose axes do not rotate and are only permitted to translate relative to the fixed frame.

xyz:fixed framex’y’z’:translating frame moving with particle ArA、 rB : absolute positions of particle A & BrB/A : relative position of B with respect to A

x

o y

z

x’

y’

z’

B

A

rB/A

rB

rA

Page 34: Kinematics of a particle

2. position vector

B

A

O

rA

rB

rB/A rA rB rB/A

3. velocity Vector

ABAB

ABAB

VVV

rrdt

d

dt

rd

/

/ )(

VB/A : relative velocity observed from the translating frame.

4. acceleration vector

A/BAB

A/BAB

aaa

)VV(dt

d)V(

dt

d

aB/A:acceleration of B as seen by an observer located at A and translating with x’y’z’ frame.