1 plane kinematics of rigid bodies rigid body it has dimensions. (particle doesn’t have it). ...

1

Plane Kinematics of Rigid Bodies

Rigid Body

It has dimensions. (particle doesn’t have it). distance between 2 points in the body remains unchanged. assumption validity? i.e. there is no real rigid body.

Kinematics

Plane Motion

study of body motion without reference to force.

Definition: All parts of the body move in parallel planes.

2

Plane Motion Definition:

All parts (points) of the body move in parallel planes.

Movement of one cutting face (corresponding to its motion plane) describes movement of the whole body.

Treat the body as thin slab object.

C.G.

Motion Plane

Any slab object (cutting face) is okay,But we usually use the plane where the object’s C.G.is in.

All corresponding points in other motion plane have the same velocity and acceleration

4

Rigid-Body Plane Motion

(Pure) Translation

(Pure) Rotation

General Motion

I

II

III

5

Type of Plane motion

(Pure) Translation Definition: A line between two points in the body remains

parallel through out the motion.

Rectilinear

Translation

Curvilinear

TranslationC.G.

Motion of one point can be used to describe motion of the whole body.

Treat it as a particle

C.G.

Types of Plane Motions (cont.)

Rotation axis

motion plane

All , move along circular path with center at

Not having the same velocity and acceleration (depends on circle radius r)

Use techniques for particle (circular) motion (n-t, r-)

Fixed-axis (Pure) Rotation:

Rotation Axis

Motion Plane

Important Property of Fixed-Axis Rotation

7

Types of Plane Motions (cont.)

General Plane motion: need new techniques in this chapter

8

Angle between any two lines on “rigid” object does not change during the time

5/2 Rigid Body’s Rotation

any Reference

+

a l

ine

“Rotating” concept is basically

a concept on rotation of “line”.

Define angular position (+/-)

for rigid body

( )

( ) angular acceleration

angular velocity

1

b

0

, ,

Any lines on a rigid body in its plane of motion have the same angular displacement, velocity and acceleration

same , ?w a a concept on (whole) rigid body.

any

0d

dt

9

(a) Angular Motion Relations

Sign convention of all variables must be consistent!

Similar to rectilinear motion

d d

d

dt

, , relation any Reference

+

any lin

e

Define angular position (+/-)

( )

( ) angular acceleration

angular velocity

( )or d d s

v

a

10

Angular Motion Relations

o t

22 2 ( )o o 2

0 0

1

2t t

Observed the similarity with the linear motion

2) Graphical meanings

d d

d

dt

d

dt

s

v

a

1) Integrals Calculation

: const

( , , ) ( ), ( ) ?t t t 2 220 0

1: ( )

2Area

11

A flywheel rotating freely at 1800 rev/min clockwise is subjected to a variable

counterclockwise torque which is first applied at time t = 0. The torque

produces a counterclockwise angular acceleration = 4t rad/ss, Determine the

total number of revolutions, clockwise plus counterclockwise, turned by the

flywheel during the first 14 seconds of torque application.

+60 0

t

d dt

21800

60 60 rad/s

o

d

dt

d

dt

d d

260 2t

809.6128.85 rev

2N

220 60 2t 2 9.71 st

2

0 0

( 60 2 )t

d t dt

3. 2 2

260 1216.96 rad

3C W t t

314

2| 60 (14) 14 809.6 rad

3t

4 4.91| |CCW t t

| 1216.96 | | 410.36 |

1627.32 rad

3260

3t t

w = - 60p

a=4tCCW:+

w= + 203.50

a=4t

w = 0

a=4t

14

2

|

60 2(14)

203.50

t

32{ 60 (14) (14) } ( 1219.96)

3410.36 rad

1627.32 = 258.996 rev

2

.C W

CCW

13

Fixed-Axis (Pure) Rotation (scalar notation)

whole body (rigid body) pointsRigid body

,

w a

O

Can we find?

P Pv aOPr

2r v

r r

rv

2

n

va

ta v

Point P

n-t coord:

r

P

rw

rarww

Fixed-Axis (Pure) Rotation Only !

OPr

14

The pinion A of the hoist motor drives gear B, which is attached to the the hoisting drum. The load L is lifted from its rest position and acquires an upward velocity of 2 m/s in a vertical rise of 0.8 m with constant acceleration. As the load passes this position, compute

(a) the acceleration of point C on the cable in contact with the drum (b) the angular velocity and angular acceleration of the pinion A.

2 2

2

2

2 2(0.8)

2.5 m/s

LL

va

s

( ) =2C t Lv v( ) C t La a

=2Lv

2 2

n

2( ) = = 10

0.4C

Cc

va

r

2 2 10 2.5 10.31 Ca

( )6.25C t

Cc

a

r

5CC

c

v

r

B B Bv r

( )B t B Ba r

18.75B BA

A

r

r

15B BA

A

r

r

, , , , ,t t nr v a a

,w a

Av

( )A ta

B Br

B Br

CCW CCW

and L has

the same

C

v ?a

15

If it do really roll without slipping,It should have some “motion constraint”.[ fixed relation between w and v ]

(Next session: we will find out this)

n

n

tt

share thesame na

Non-slipping

(= )tv v = 0

floor- fixed

ta = 0

share thesame ta

(= )tv v

naWhy?

Non-slipping

Without gear teeth, the ball is not guaranteed to roll without slipping.

Two possible motion: - roll without slipping - roll with slipping.

In case of No gear-teeth

18

Fixed-Axis (Pure) Rotation (vector notation)

The body rotating about the O axis

v r From the last section, the magnitude of

the velocity is

Represent “angular velocity” w as Vector Direction: “Right-hand rule”

v r

The cross product can be used to establish the direction:

( )r

r

We consider only Plane motion of rigid body

//

| |

Think in 3D Vector

19

( )d

rdt

//

Fixed Axis Rotation (vector notation) Differentiating the velocity with respect to time:

a v ( ) r r

( )na r

v r

ta r

Direction: r

Direction depends on , r r

r

r r

// ( ) // r r v

r We consider

only “Plane motion” of rigid body

( ) : - r direction r

ˆ ˆ ˆ ˆ( ( ))k k i j

(when r )

2 ˆ ˆ ( )i j

2rr

20

Fixed-Axis (Pure) Rotation (vector notation)

v r

a r r

21

The rectangular plate rotates clockwise.If edge BC has a constant angular velocity of 6 rad/s. Determine the vector expression of the velocity and acceleration of point A, using coordinate as given.

( )a r r

v r

ˆ ˆ ˆ ˆ ˆ6 (0.3 0.28 ) 1.8 1.68v k i j j i

ˆ ˆ ˆ ˆ ˆ6 ( 1.8 1.68 )) 0 (0.3 0.28 )a k j i i j

ˆ ˆ10.8 10.08i j

BC

BC =

Rigid body

( ) : - r direction r


(when r )

2 ˆ ˆ ( )i j

ˆ ˆ0.3 0.28r i j

23

Equations Review:

(Pure) Translation Movement of one point describes movement of the whole body.

(Pure) Rotation rotation of one line describes rotation of the whole body. Whole body shares the same “angular” quantities.

dt

d dt

d dd

( )a r r

v r

v

a

sdt

dsv v

dt

dva adsvdv

rdt

rdv

vdt

vda

Reference

+an

y line

General Plane MotionAbsolute motionRelative motion

24

5/3 Absolute Motion (of General Plane Motion)

Use geometric constraint (which define the configuration of the body) to obtain the velocity and acceleration (in general motion)

cosy b

2cos siny b b

Idea: write a (position) constraint equation which always applies regardless of the system’s configuration, then differentiate the equation to get velocity and acceleration.

2 2 2x y b siny b

2 2 0xx yy

yx y

x

for complex constraint method of relative motion may be easier.

?w ?y ?y x ( )?

25

cosy b

2 2 2x y b siny b

2 2 0xx yy

yx y

x

?w ?y ?y x ( )?

Define the displacement and its positive direction.

The variable must be measured from the fixed reference point or line.

Find the equation of constraint motion.

Equation must be true all during the motion.

Differentiate it to find (angular) velocity and acceleration.

26

5/56. Express the angular velocity and angular acceleration of the connecting rod AB in terms of the crank angle for a given constant .

sin sinl r

cos

cosAB

r

l

cos cosl r

0

2

cos

1 ( sin )

r

l rl

2 2cos sin cos sinl l r r 0

2 2sin sin

cosAB

l r

l

2

2 20

22 3/ 2

2

1sin

(1 sin )

rr l

rll

ABAB

0

,AB AB sin sin

l r

b

27

SP5/4 A wheel of radius r rolls on a flat surface without slipping.Determine angular motion of the wheel in terms of the linear motion of its center O. Also determine the acceleration of a point on the rim of the wheel as the points comes into contact with the surface on which the wheel rolls

s r

cosy r r sinx r r

s r

( )Oa v r r

motion constraint

Ov r

Ov w

Point C’s trajectory

Point O:rectilinear motion

O Ov a

28

cosy r r sin sinOy r v

cos (1 cos )Ox r r v sinx r r 2(1 cos ) sinOx a r 2sin cosOy a r

2 : 0y 0x 0x

2y r

velocity=0

Acceleration in the direction of axis x = 0

SP5/4 A wheel of radius r rolls on a flat surface without slipping.Determine angular motion of the wheel in terms of the linear motion of its center O. Also determine the acceleration of a point on the rim of the wheel as the points comes into contact with the surface on which the wheel rolls

When Point D comes to contact the surface,

It also has a velocity (=0), and acc. as above.

D

C’’

O

Not depend on (a t), (w t) are!

s r Ov r Oa r

29

cosy r r sin sinOy r v

cos (1 cos )Ox r r v sinx r r 2(1 cos ) sinOx a r 2sin cosOy a r

2 : 0y 0x 0x

2y r

velocity=0

Acceleration in the direction of axis x = 0

C’’

O

AvAa

No slipping

B Av v B Aa a

0Av 0Aa

0Bv 0Ba

O

AvAa

B Av v ( )B t Aa a

Analogy

Rel. vel. =0Rel acc. = 0

O

O

v r

a r

Non-slipping Condition

C’’

O

tfloor- fixed

31

“no slipping” implies: 1) Contact point { C , C’ } on two body has no relative velocity. 2) Contact point { C, C’ } on two body has same tangential component of acceleration

( ) C t Lv v

( ) C t La a

32

Each cables do not slip. Load-supporting pulleys are rigid body.

1 2 3 4 1 1 2 2, , , , , : knownr r r r

3r 4r

LFind: , , v ,o o La

1 1 1, , => ,( )A A tr v a

2 2 2, , => ,( )B B tr v a

A B

Ads Bds

( )( )

( )B A

AB

v t v t

3 4

B Ao

d v v

dt r r

3 4

( ) ( )B t A to

d a a

dt r r

2 2 2 2 ( )B B tv r a r

1 1 1 1 ( )A A tv r a r

2 2 1 1

3 4

r r

r r

2 2 1 1

3 r

r r

r r

2 2 1 13

3L A

r

r rv v r

r r

2 2 1 1

33

L Ar

r ra a r

r r

O

3r 4r

O and L has same vertical velocity & acceleartion

( )( ) ( )B AAB v v

( )( ) ( ) ( )B t A tAB a a

33

cosx L

( sin )x L

2

20

2 22 2

cot

x v

L xL x

0

2 2

v

L x

0x v0x

2{(cos ) (sin ) }x L

35

- Calculation methods

- Introduction น�ยามการเคล��อนท��ของว�ตถ�เกร�ง

- Absolute motion

Instantaneous Center of Zero Velocity (ICZV)- Relative acceleration

- Rotation ว�ธ�อธ�บายการเคล��อนท��แบบหม�น

- Relative velocity

Motion relative to rotating axes

ใช�ค�านวณ V A

ใช�ค�านวณ V

ใช�ค�านวณ V

ใช�ค�านวณ A

ใช�ค�านวณ V A

V

Observer is at the point

of rigid body where its

velocity = 0

Translating-only observer

Translating and rotating

Observer

usually forsome t(instant)

using its geometric shape at that instant

General Plane Motion of Rigid Body

usually for any t

using constraint equation

36

5/4RelativeVelocity

Since the distant between the two points on a rigid body is constant,

an observer at one point will see the other point move in a circular motion around it!

General Plane Motion = Translation + Rotation

A B A BV V V

Wait! B really sees A moving circularly?

relative velocity concept

“simultaneous”

B sees A has no movement !?!?!?

Different viewpoint

Motion of point (observer) B, detected by O= Motion of plate moving “translationally”

O

37

Applying the relative concept

Observer B is on the plate

AB A

B

Observer B is sitting on the magic carpet.

B see A no moving at all B see A having a velocity perpendicular with its distance.

A

non-rotating observer (attached to B)Rotating observer (attached to B)

BABA VVV

Only this case

Which one?

Rotating framenon-rotating frame

38

/ /A B A Bv r

Relative Velocity (non-rotating observer)

We use: non-rotating observer (frame) attached to B

Line

Observer at B see A moving in a circle around it

Relative world:Absolute world:

Observer B detects: /A BrObserver O detects: Bv

/A B A Bv v v

/Line BA Bv v

AB

same?

Only when non-rotating observer

A Bv v

r

(see next page)

39

Translating observer see same , w a as absolute Observer

1

1

2 1

2

2 1

2

same wThe rotating of rigid body = The rotating of line compared with “fixed” reference axis

O’s reference line

B’s reference line

same a

O

B

B

40

Translating-Rotating observer sees different , w awith absolute Observer

1

1

2 1

2

2 1

different w

O’s reference line

B’s reference line

different a

The rotating B’s “reference line”,observed by absolute observer.

O

B

B

The rotating of rigid body = The rotating of line compared with “fixed” reference axis

2

41

A

B

Any 2 points on the same rigid body

Above equation (2D: “3D-fake”) can be solved when there are at most 2 unknown scalar quantities

Above equations usually contains 5 scalar quantities (not including position vector r)

Identify the known and unknown

Hint on solving problems

Understanding the equation

/ A B A Bv v v

absolute absolute absolute Non-rotating,Moving with B

Non-rotating,Moving with B

Also works with A as the observer / /( )B A A Bv v

always perpendicular to line AB. Its direction can be deduced from

/A Bv

Important key: /A Bv

/ A Br

42

Relative Vector Analysis on General Plane Motion

G

P

Pw

/P Gv r

Pv

Gv

G OP Pv r v

Gv

Pv r

Fixed-Axis (Pure)Rotation

w

GP

w/P Gv r

Pv

Gv

Gv

G P OPv v r

/A B A Bv v v

| |

| |P G

OP

v v

r

w of observer at G = w (of rigid body) of observer at O

44

2 rad sCB ?OA ?AB

/ = + A B A Bv v v

Solved by Vector Analysis

ˆOA k ˆ2k ˆ

AB k

ˆ0.1

ˆ0.75

ˆ ˆ0.175 0.05

A

B

A B

r j

r i

r i j

OA Ar AB A Br

CB Br

Velocity at A is key point to find w

A, B on the same rigid body (bar AB)

45

ˆ ˆ ˆˆ ˆ ˆ ˆ0.1 2 ( 0.075 ) ( 0.175 0.050 )OA ABk j k i k i j

ˆ ˆ ˆ ˆ0.1 0.15 0.175 0.050 OA AB ABi j j i

i

jk +

BAABBCBAOA rrr From

secrad7

6AB secrad

7

3OA

A

B C

O +

3D vector calculation (i,j,k):Sign indicates angular direction(right hand rule)

xy

k

46

2 rad sCB ?OA ?AB

D

M B CBr ??

BAvBv

Solved by Graphical Method

Av

need to find the angular direction from the figure

tan

0.0429A Bv v

/ / cos

0.156A B Bv v

0.15

0.429AA

A

v

r

/ 0.857A BB

BC

v

r

1 100 50tan 15.95

250 75o

CW

CW

A, B on the same rigid body (bar AB)

A B A BV V V

47

middle link

49

Note: relative velocity technique

Direction is simply found:

Non-rotating

BAvBv

Av

1

2

: same in absolute and relative world

/ A B A Bv v r

To know of the rigid body

AB

AvBv

r1

complicated, automatic sign indication

simplest, be careful about sign / 1

A Bv

Bv

Av

graphical solution:

3D (i, j, k) vector:

AB

Av

r

(only) direction

of is enoughBv

/direction of A Bv

51

1 0.250cos 60

0.500o

wAB

C

BCBC vvv /

/B ABC C Bv r

A B A BV V V

D

M OBr ??

20o

/ 0.49A BABC

AB

v

r

0.2B OBv r

20o

60o

70o

/A Bv

Av

50o

sin 60 0.226sin50

BA

vv

/ sin 70 0.245sin50

BA B

vv CCW

0.2Bv

/ /

10.1225

2C B A Bv v 60o

0.175cv 37.316o

C ABC BCv r

Graphical solution

AB ABCr

0.8

52

ˆ ˆcos 20 sin 20o oA Av v i j

jikvBˆ2.0ˆ25.0ˆ8.0

/ˆ ˆ ˆ0.5 cos sinA B ABCv k i j

vA

vB

1 0.250cos 60

0.500o

0.226 /Av m s

0.491ABC

BCBC vvv /

/B CB C Bv r

ˆ ˆ0.1746 cos127.5 sin127.5o oCv i j

CW

A B A BV V V

D

M OBr ??

20o

Vector solution

Direction?

54

Pick points A and B as coincident points, one on each link

(the points may be imaginary).

AB Another usage of the

relative velocity equation

For constrained sliding contact between two links in a mechanism.

D

Av

Bv

Relative Velocity (Part 2)

The observer on B no longer see A moving around it in a circle.

BAv

2 points need not be in the same rigid bodyB onscrew

A on OD

parerell

/A B A Bv v v

some reason later!

OD bv

OD

c.c.w.

56

1 0.175sintan

0.4 0.175cos

19.402o

q fB O

P0.175

0.175cosq0.4-0.175cosq

ˆ ˆ3(0.263) sin cosQv i j

ˆ ˆsin cosP Pv v i j

/ /ˆ ˆcos sinP Q P Qv v i j

BCBC vvv /

Ans

/0 BC C Br

ji oo ˆ300sinˆ300cos21.1

ˆ ˆ0.262 0.745i j

ˆ ˆ0.5 0.866Pv i j

/ˆ ˆ0.766 0.6428P Qv i j

/P Q P Qv v v

12.348 rad/sPBC

BP

v

r

/ 1.06 m/sP Qv

CW

30o

40o

Qv

Pv

/P Qv

on OAQ

M

/ P Q P Qv v v

D 3(0.263) ??

0.4 0.175cos0.263

cosPO

ˆ ˆ ˆ12.348 0.35 cos sink i j

2.161 m/sPv

Qv

Pv

/P Qv

58

/

ˆ ˆcos sin

ˆ ˆ 2.8 0.75

ˆ ˆ sin cos

Q

Q P

v i j

i j

v i j

OPOPOPOP rivvv //ˆ5.1

ˆˆ ˆ ˆ1.5 15 0.1 sin cosi k i j ji ˆ75.0ˆ8.2

/Q P Q Pv v v

vP

vQ/P

y

vQ

C

O

100

200

x

q

b

1 0.1sintan 23.79

0.2 0.1coso

2.26 /Qv m s2.26

18.23 rad/s0.150sin

sin

QC

QC

v

r

o OPv

15 CWOOP

v

OP

Non-slipping condition

CCW

Plus

| |O OPv OP

,o OP Pv v

Q on slot C

? ?

DO Dv v r

0

V=1.5 q=30

w=?

vO/

: not // OP

( // OP )

P

P O

Note v

v

ˆ2.8

ˆ0.75

i

j

61

5/6RelativeAcceleration

Since the distant between the two points on a rigid body is constant,

an observer at one point will see the other point move in a circular motion around it!

General Plane Motion = Translation + Rotation

A B A Ba a a

Wait! B really sees A moving circularly?

relative acceleration concept

“simultaneous”

B sees A has no movement !?!?!?

Different viewpoint

Motion of point (observer) B, detected by O= Motion of plate moving “translationally”

O

62

/ /

/

( )

A B A B

A B

a r

r

Relative Acceleration (non-rotating observer)

We use: non-rotating observer (frame) attached to B

Line

Observer at B see A moving in a circle around it

Relative world:Absolute world:

Observer B detects: /A Br

Observer O detects: Bv

/A B A Ba a a

/Line Bsame?

Only when non-rotating observer

/Line B

Line

(see the proof at relative velocity part)

63

/ A B A Ba a a

/ / ( ) A B A Br r

Understanding Equations

Above equation (2D: “3D-fake”) can be solved when there are at most 2 unknown scalar quantities

Above equations usually contains 6 scalar quantities (not including position vector r)

, : the same both in absolute and relative (translation-only) world

Identify the known and unknown

Hint on solving problems

Non-rotating

64

/ A B A Ba a a

( )B CB CB B CB Ba r r ˆ ˆ ˆ2 (2 ( 75 ))k k i 2ˆ300 seci mm

Given : 2 sCB rad (constant) ?OA ?AB Find :

CB Br ( )CB CB CBr

OA Ar / AB A Br

( )A OA OA A OA Aa r r 2900ˆ ˆ100 sec

49OA i j mm

/A B A Bv v v

ˆOA k ˆ2k

OA Ar AB A Br

CB Br += 6

rad/s7AB

3 rad/s

7OA

( )A B AB AB A B AB A Ba r r

2 ˆ(2) ( 75 )i 6 6ˆ ˆ ˆˆ ˆ ˆ ˆ( ( 175 50 )) ( 175 50 )7 7 ABk k i j k i j

2900 1800ˆ ˆ ˆ ˆ (50 175 ) sec7 49 ABi j i j mm

ˆ kAB AB ˆ kOA OA

/( )AB AB A Br ( )OA OA Ar

ˆAB k

65

900 1800175

49 49 AB

20.105 secAB rad ANS

24.338 secOA rad ANS

900ˆ ˆ100 49OA i j i300 j

49

1800i

7

900 ˆ ˆ (50 175 )AB i j

ˆ: direction k

Given : 2 sCB rad (constant)

?OA ?AB (at this instant)Find :

900100 300 50

7OA AB

/ A B A Ba a a

CB Br ( )CB CB CBr

OA Ar / AB A Br

ˆ kAB AB ˆ kOA OA

/( )AB AB A Br ( )OA OA Ar

66

/ A B A Ba a a ?OA ?AB (at this instant)Find :

CB Br ( )CB CB CBr

OA Ar / AB A Br

( )OA OA Ar /( )AB AB A Br

ˆ kAB AB ˆ kOA OA

/ = + A B A Bv v v

ˆOA k ˆ2k ˆ

AB k

OA Ar AB A Br

CB Br

Find velocity first,Before acceleration


2 ˆ ˆ ( )i j

( )r

69

5/124 The center O of the disk has the velocity and acceleration shown in the figure. If the disk rolls without slipping on the horizontal surface, determine the velocity of A and the acceleration of B for the instant represented.

s r ov r oa r Non-slipping condition

I.C.Z.V

7.5 rad/sOv

r

212.5 rad/sOa

r

a w OAOOAOA rvvvv //

jiki oo ˆ45sinˆ45cos4.0ˆ5.7ˆ3

smji oo /ˆ6.32sinˆ6.32cos85.9

tOBnOBOB aaaa //

/ /( )O B O B Oa r r

ikikki ˆ2.0ˆ5.12ˆ2.0ˆ5.7ˆ5.7ˆ5

jii ˆ5.2ˆ25.11ˆ5

2/ˆ3.171sinˆ3.171cos44.16 smji oo

You can calculate using point O and D.

You can calculate using point O and and D.D

?Da

71

vAwOA

vA

vE

O

420 CCW

0.2E

OAOE

v

r

kADˆ5.12

BDBDD rv /

jkBDˆ25.0ˆ

kBDˆ5.7

wBD

B

vD

4 m/s (constant)

2.5A OA OAv r

/ D A D Av v v

/D Av r

ˆ ˆ ˆ = 0.2 0.15AD k i j

ˆ2.5Av j

72

/ /( )A OA OA A O OA A Oa r r

ˆ ˆ ˆ ˆ ˆ20 (20 0.125 ) 0 0.125 50k k i i i

( CW )

(constant)

( ) 0E ta

/ / /( )D A AD AD D A AD D Aa r r

/ /( )BD BD D B BD D Br r

0.25 0.15 18.75BD AD

14.06 23.44 0.200 AD

O

wOA=20

246.9 rad/sAD 246.9 rad/sBD

/ˆ ˆ ˆˆ ˆ12.5 (12.5 ( 0.2 0.15 )) AD D Ak k i j k r

ˆ ˆ ˆˆ ˆ7.5 (7.5 0.25 ) .250BDk k j k j

ˆ ˆ0.25 14.06BDi j

aOA

/ D A D Aa a a

00

0.2E t

OEOE

a

r

75

General Plane Motion (non-rotating observer)

/ A B A Ba a a

Cross-Vector Approach Graphical Approach

I.C.Z.V

New techniqueany point: A, B (on same rigid body moving in GPM)

/ A B A Bv v v

/A Br

/

/ A B

A B

r

r

B is special point : I.C.Z.V

76

Note: relative velocity technique

Direction is simply found:

Non-rotating

BAvBv

Av

1

2

: same in absolute and relative world

/ A B A Bv v r

To know of the rigid body

AB

AvBv

r1

complicated, automatic sign indication

simplest, be careful about sign / 1

A Bv

Bv

Av

graphical solution:

3D (i, j, k) vector:

AB

Av

r

(only) direction

of is enoughBv

/direction of A Bv

78

* v

r

Checkpoint: circular motion

v r

Fixed-point rotation(Rotation)

from rotation point

Don’t know it isfixed-point rotationor not (General Motion)

valid method?Yes! but show your reason!

I.C.Z.V concept

?

79

/

A

A Z

v

r

5/5 Instantaneous Center of Zero Velocity

Z is called I.C.Z.V (the point where its velocity at that instant is zero)

If 0

(at this instant)zv

Extension theory using relative velocity.

General Plane motion

A BAv /A Z A Zv v v

w of observer at C = w of rigid body (in Absolute Observer’s perception)

each point on the body can be though of as rotating around point Z.

/A Zr

Z

/ A A Cv v

/B Z B Zv v v

/B Cr

/ B B Cv v

direction

of Bv

/ ( )B Zr

w

Bv

P

- can find w easily by geometry

- can find velocity and its direction of any points easily by geometry

For calculating v and w only

A

B

Av

Bv

Finding an I.C.Z.V.

2

/

( ) DD n

D Z

va

r

General Plane motion

z

A

B

Av

Bv

z

A

B

Av

Bv

A

B

Av

Bv

z

AB

Av

Bv

z

not a rigid body

/ (where Z = I.C.Z.V)A A Zv r

/ /D D Z D Zv v r

z

/

A

A C

v

r

az usually 0 (Even vz = 0)

Da

2 2/

// /

( ) D C DD Z t

D C D C

v va

r r

I.C.Z.V for calculating instantaneous velocity only w of observer at C

= w of rigid body (in your perception)

AB

Av

Bv

I.C.Z.V at Inf.

?

“instantaneous” Translational motion

D

0v v

r

/( )D t D Za r

/Z D Za a

82

w

1500 2

/

OB OBr

mm s

Arm OB of the linkage has a clockwise angular velocity of 10 rad/s in the position shown where = 45°. Determine the velocity of A, the velocity of D, and the angular velocity of link AB for the instant shown.

Thus, we can locate the instantaneous center of velocity, which is point C

BABA VVV

D

M B CBr ??

Direction of (absolute) velocity of two point in the same rigid body

solved byrelative velocityI.C.Z.V

350

45o

350 2

30 = rad/s

7350 2Bv

w of what? OA, AB, BO

30( ) (350) mm/s

7A CBV CA 30

( ) (381) mm/s7D CBv CD

381

You have to findDirection yourself

84

sin075.00.065

0.913.86 rad/s

0.065B

ABCB

v

r

13.86 0.135 1.873 m/sC ABC Cv r

20.135

cosB

C

rr

vBqx

y

Vector Diagram

vC

rC

a

rB

1 0.075costan 16.1

2o

Br

ICZV

vA

CCW

q

60o

q

0.9 m/s

85

0 (fixed)R ?s P A Non-slipping motion

Sun Gear: Fixed-Axis (Pure) rotation

Planet Gear: general Plane motion

sR

sR

ICZV of sun

ICZV of P1

ICZV of P2

1 = 2

sP

R

r

CW

2 = 2

sP

R

r

CW

2sR

ICZV of A

2 = ( )

s

A

R

R r

CCW

86

60 ? to make 2A Fv v

( ) 4 m/s

cos30 3D y

D o

vv

40 rad/s

2(0.1)cos30 3D

AD o

v

80.2 m/s

3A CDv

2Dv

2 m/sFv

ICZV

Av

ICZVBv

( ) 2D yv Dv

30

100 mm

87

I.C.Z.V

s r

ov r oa r

Absolute motion

Non-slipping condition

I.C.Z.V

Ov r

OO

dva

dt

I.C.Z.V

t ta v r

Ov r

2

*n

va

r

rectilinear

*r R r

t

ncircle

5/140

rO

O

dva

dt

2 2r

R r

VD0vH2

vH3

rrv OAH 822

rrv OAH 1643

rrrvC 8)8(11

2 16 242C

rv r r

33

168 rad/s

2Hv r

d r

vH2

= 8r

vC1

=8r

VD2

II

vC1VD1

I

vC2

vH3

VD3

IIIICZV

vC2

ICZVd

CCW

4 rad/s

8 rad/s

12

816

/ 2 / 2Cv r

r r

2

3

C

H

v r d

v d

24

16

r r d

r d

2d r

CW

Practice Before Rotating Observer

91

92

0Bv

AAB

v

L

ICZV of AB

0Aa

5/134 The sliding collar moves up and down the shaft, causing an oscillation of crank OB. If the velocity of A is not changing as it passes the null position where AB is horizontal and OB is vertical, determine the angular acceleration of OB in that position.

0

/ A B A Ba a a

( )OB OB OBr /( )AB AB A Br

0OB

CCW

0AB 2A

OB

Lv

r

2

ˆ : 0AOB

vi r L

L

ˆ : 0ABj L

CW

OB Br / AB A Br

ˆ ˆ ˆ ˆˆ ˆ ˆ0 ( ) A AOB AB

v vk rj k k li k Li

L L

AAB

v

L

ICZV of AB

0Aa

0OB

CCW

93


0

/ A B A Ba a a

OB Br

( )OB OB OBr

0AB

/ AB A Br

/( )AB AB A Br

2/( )A B ABr

/( )A B ABr

( )OB OBr

=0

22 2

/( )( )

A

A B AB AOB

OB

vLr LvL

r r r

CW

2/( )A B ABr

( )OB OBr

95

120 rev/min

(constant)OB

DFind a

DAB DABa

Av

2 (120)0.05 0.2

60Bv

ICZV of DAB is at infinite

0DAB 0.2A Bv v

2( ) 0.32A n A ACa v

14.48o

( )A t A ACa r

/ /( )A B t A B DABa r

2( ) 0.8Ba

22

0

(0.8 0.32)2.479

0.2cos14.48DAB

/ /( )A B DAB DAB A B DAB A Ba a r r

/ /( )A AC AC A C AC A Ca r r

3D vector solution

Graphical solution

0

CW

0.21.6

0.125AC

CW

22(0.2 )

0.80.05Ba

/D B DAB D Ba a r

2 2 ˆˆ ˆ0.8 2.479 0.3( sin cos )i k i j

2ˆ ˆ0.081 0.186i j

0.2A Bv v ˆˆ ˆ ˆ ˆ0.2 ( 0.2sin 0.2cos )A DABv j j k i j

/A B DAB A Bv v r

0DAB

97

middle link

AAB

v

L

ICZV of AB

0Aa

0OB

CCW

100


0

/ A B A Ba a a

OB Br

( )OB OB OBr

0AB

/ AB A Br

/( )AB AB A Br

2/( )A B ABr

/( )A B ABr

( )OB OBr

=0

22 2

/( )( )

A

A B AB AOB

OB

vLr LvL

r r r

CW

2/( )A B ABr

( )OB OBr

2(3)(2 )

sin

102

/A B A Bv v v

0AB ICZV at inf. AB translational.

6 m/sA Bv v 6

2 rad/s3

ACA

v

CA

/A B A Ba a a

B AB AB BA AB BAa r r

AC AC CA AC CA B AB AB BA AB BAr r a r r

2

2(3)(2 )

CA CAr

2Ba

(5)AB

(3)AC23(2 )(tan ) 2 7

3 3AB

CW

21 (3)(2 )3

5 sinAB

CW

7ˆ ˆ ˆˆ ˆ =2k 2k 3 33

A AC AC CA AC CAa r r

j k j

ˆ ˆ-7 12Aa i j

AC AB

AC CD

2(3)(2 )

2(3)(2 ) tan 2

? ?AB Aa

103

: /C A AB AC rel C Av v r v

3 rad/s

0AB

AB

3 0.75: /rel C Av (3)(0.75) tan 3.897

Cv (3)(0.75)

cos 4.5

90.5

CCD

v CCW

: / CD rel C Av

: / : /2C A AB AB AC AB AC AB rel C A rel C Aa a r r v a

20.75 3

2(3)(3.897)

CD CD DC CD DCr r

20.5 9 40.5

(0.5)CD DC CDr

: /rel C Aa

2(3)(3.897)

sin

2

2(3)(3.897)40.5

2(3)(3.897)sin0.75 3 cos

cos sin

128.24881

2(3)(3.897)40.5 tan

sin

1 2(3)(3.897)40.5 tan 233.824

0.5 sinCD

: / CD rel C Aa

CCW

? ? AC AC

1 plane kinematics of rigid bodies rigid body it has dimensions. (particle doesn’t have it). ...

Documents

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plane motion definition

general plane motion

rectilinear motion

acceleration slide

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real rigid body

chapter slide