1

Acceleration analysis (Chapter 4)

• Objective: Compute accelerations (linear and angular) of all components of a mechanism

2

Outline• Definition of acceleration(4.1, 4.2)• Acceleration analysis using relative acceleration

equations for points on same link (4.3)– Acceleration on points on same link– Graphical acceleration analysis– Algebraic acceleration analysis

• General approach for acceleration analysis (4.5)– Coriolis acceleration– Application– Rolling acceleration

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• Definition of acceleration (4.1, 4.2) – Angular = = rate of change in angular velocity– Linear = A = rate of change in linear velocity(Note: a vector will be denoted by either a bold

character or using an arrow above the character)

4

Acceleration of link in pure rotation (4.3)

A

PAt

PA

AnPA

APA

,

Magnitude of tangential component = p, magnitude of normal component = p 2

Length of link: p

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Acceleration of link, general case

A

PAt

PA

AnPA

APA

,

Length of link: p

AA AA

APA

AP

AP=AA+APA

AnPA At

PA

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Graphical acceleration analysis

2

3

4

1

AtA

AnA

B

A

AtBA

AtB

Clockwise acceleration of crank

Four-bar linkage example (example 4.1)

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• Problem definition: given the positions of the links, their angular velocities and the acceleration of the input link (link 2), find the linear accelerations of A and B and the angular accelerations of links 2 and 3. Solution: – Find velocity of A– Solve graphically equation:

– Find the angular accelerations of links 3 and 4

nBA

tBAA

nB

tB

BAAB

AAAAA

AAA

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Graphical solution of equationAB=AA+ABA

2

3

4

1

AtA

AnA

B

A

AtB

AA

AnBA-An

B

AtBA

AtBA

AtB

Steps:•Draw AA, An

BA, -AtBA

•Draw line normal to link 3 starting from tip of –An

B

•Draw line normal to link 4 starting from origin of AA

•Find intersection and draw AtB and At

BA.

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• Guidelines– Start from the link for which you have most information– Find the accelerations of its points– Continue with the next link, formulate and solve equation:

acceleration of one end = acceleration of other end + acceleration difference

– We always know the normal components of the acceleration of a point if we know the angular velocity of the link on which it lies

– We always know the direction of the tangential components of the acceleration

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Algebraic acceleration analysis (4.10)

2

3

4

1

BA

ab c

Given: dimensions, positions, and velocities of links and angular acceleration of crank, find angular accelerations of coupler and rocker and linear accelerations of nodes A and B

1

R2

R3 R4

R1

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0 nB

tB

tBA

nBA

tA

nA AAAAAA

0)( 42

32

22 44

433

322

2 jcecejbebejaeae jjjjjj

0 1432 RRRR

Loop equation

Differentiate twice:

This equation means:

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Solution

424

222

4243

232

2222

3

4

4243

232

2222

3

4

4

3

44

22

sinsinsincos

coscos

coscoscossin

sinsin:where

jcece

jaeae

cbaaF

bEcD

cbaaC

bBcA

BDAEBFCEBDAEAFCD

jjB

jj

A

AA

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General approach for acceleration analysis (4.5)

• Acceleration of P = Acceleration of P’ + Acceleration of P seen from observer moving with rod+Coriolis acceleration of P’

P, P’ (colocated points at some instant), P on slider, P’ on bar

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Coriolis accelerationWhenever a point is moving on a path and the

path is rotating, there is an extra component of the acceleration due to coupling between the motion of the point on the path and the

rotation of the path. This component is called Coriolis acceleration.

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Coriolis acceleration

VPslip

P

OAP’

t

APcoriolis

AP’n

APslip

AP

APslip: acceleration of P as seen by observer moving with rod

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Coriolis acceleration• Coriolis acceleration=2Vslip

• Coriolis acceleration is normal to the radius, OP, and it points towards the left of an observer moving with the slider if rotation is counterclockwise. If the rotation is clockwise it points to the right.

• To find the acceleration of a point, P, moving on a rotating path: Consider a point, P’, that is fixed on the path and coincides with P at a particular instant. Find the acceleration of P’, and add the slip acceleration of P and the Coriolis acceleration of P.

• AP=acceleration of P’+acceleration of P seen from observer moving with rod+Coriolis acceleration=AP’+AP

slip+APCoriolis

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Application: crank-slider mechanism

B2, B3

O2 Link 3, b2

Link 2, a 2, 2

B2 on link 2 B3 on link 3These points coincide at the instant when the mechanism is shown.When 2=0, a=d-b

d

3, 3, 3

Unknown quantities marked in blue

.

normal to crank

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General approach for kinematic analysis

• Represent links with vectors. Use complex numbers. Write loop equation.

• Solve equation for position analysis• Differentiate loop equation once and solve

it for velocity analysis• Differentiate loop equation again and solve

it for acceleration analysis

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Position analysis

2222

2 sincos dbda

)sin(sin 21

3 ba

Make sure you consider the correct quadrant for 3

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Velocity analysis

B2 on crank, B3, on slider

O2rocker

crankVB2 ┴

crank

VB3B2

// crankVB3 ┴

rocker

.

VB3= VB2+ VB3B2

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Velocity analysis

)cos(1

322

3

ba

)cos()sin(

2323

2

aa

a is the relative velocity of B3 w.r.t. B2

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Acceleration analysis

)cos(sincos

3233

CBa

)cos(sincos

3222

3

bBC

3232

222222 coscossinsin2 baaaB

Where:

3232

222222 sinsincoscos2 baaaC

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Relation between accelerations of B2 (on crank) and B3 (on slider)

B2, B3

AB2

AB3Coriolis

┴ crank

AB3slip

// crank AB3

SlipB

CoriolisBBB 3323 AAAA

.

Crank

Rocker

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Rolling acceleration (4.7)

R

r

(absolute)

PC

O

C

O

(absolute)

First assume that angular acceleration, , is zero

No slip condition: VP=0

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Find accelerations of C and P -(R-r)/r (Negative sign means that CCW

rotation around center of big circle, O, results in CW rotation of disk around its own center)

• VC= (R-r) (Normal to radius OC)• An

C=VC2/(R-r) (directed toward the center O)

• AnP=VC

2/(R-r)+ VC2 /r (also directed toward the

center O)• Tangential components of acceleration of C and P

are zero

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Summary of results

AC, length VC2/(R-r)

P

C VC, length (R-r)

r

VP=0

R

AP, length VC2/(R-r)+ VC

2 /r

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Inverse curvature

(R+r)/r• VC=(R+r) (normal to OC)

• AnC=VC

2/(R+r) (directed toward the center O)

• AnP=VC

2/r - VC2 /(R+r) (directed away from the

center O)• Tangential components of acceleration are zero

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Inverse curvature: Summary of results

AC, length VC2/(R+r)

P

AP, length VC2 /r -VC

2/(R+r)

CVC, length (R+r)

r

VP=0

R

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Now consider nonzero angular acceleration, 0

• The results for zero angular acceleration are still correct, but

• ACt=r (normal to OC)

• APt is still zero

• These results are valid for both types of curvature