1

4. Acceleration Analysis of Linkages

4.1. Acceleration Analysis by Vector Mathematics

• The acceleration of a point P moving in the x-y-z system relative to the X-Y-Z system, is obtained by differentiating the velocity equation

• Differentiating equation (1) yields the acceleration Equation is

)2(RRVVV op

)1(RVVV op

2

• Each term on the right hand side of the acceleration equation is evaluated as follows:

• Substituting for

• Note that

)4(

)3(

kzjyixkzjyix

kzjyixdt

dV

aV oo

kji ,,

)6(

)5()()()(

akzjyix

let

kzjyixkzjyixV

)7(

)()()(

V

kzjyixkzjyix

3

The acceleration component can be written as:

• The last term on the right hand side of the acceleration equation is

• Substituting the corresponding values in the acceleration equation

• The different acceleration components are: is Coriolis’ component of acceleration, sense normal to ;

is acceleration of the origin of x-y-z system relative to X-Y-Z system;

is acceleration of P relative to x-y-z system;

V

)8(VaV

)9()(

)(

RV

RVR

)10()(2 RVRaaa op

a

ao

V

V

2

4

: is the tangential acceleration of a point fixed on the x-y-z system coincident with P as the system rotates about O;

: is the normal component of acceleration of the point coincident with P

Where: : is angular velocity of x-y-z system related to X-Y-Z system

: is velocity of P relative to x-y-z system; and : is position vector of P.R

V

R

R

5

4.2. Acceleration Analysis Using Equations of Relative Motion

4.2.1 Acceleration of points on a common link• Consider a link AB rotating with an angular velocity and

angular acceleration α as shown.• The relative acceleration equation is:

Where acceleration of point A acceleration of point B

Where the acceleration term has two components:

along the link from A to B

in the direction perpendicular to AB

)11(/BABA aaa

B

A

a

a

BAa /

ABtBA

ABnBA

Ra

Ra

/

/

6

4.2.2 Acceleration analysis of a block sliding on a rotating link

• Block A slides on the rotating link O2B as shown.• At time t angular position of link 2 is and at t+dt, +d.• The acceleration of the block is found by considering the radial

and tangential components of the change in velocity of the block.

• The rotating coordinate system r - is attached to link O2B. • The radial component of the velocity is V at time t, and V+dV at

time t+dt.

7

• The components of dV in the radial and tangential directions are:

(dV)r in the radial direction, Vd in the tangential (transverse) direction.

• Similarly the change in the transverse component of the velocity in time dt is dV as shown.

• The components of dV are:-rd in the radial direction, and dr + rd in the transverse direction, neglecting

higher order differentials• Thus the total change of velocity is

dV - rd in the radial direction, andVd + dr + rd in the tangential direction.

the radial component of acceleration of block A is

)12(

1

2

ra

rddVdt

ar

8

or, in vector notation,

The tangential component of the acceleration is

Vectorially, the tangential acceleration is written as

the acceleration of the block sliding on the rotating link is given by:

is the sliding velocity of the block along link O2B, and

is the sliding acceleration of the block along the link O2B.

13raar

)14(2

1

rV

rddrVddt

a

152 rVa

162 aVrraA

a

Vwhere

9

• If a point A2 fixed on link O2B is coincident with A for the position shown, the acceleration equation can be written as:

• The term is the relative acceleration b/n two moving points,

• If the link were a curved link:

• In general the relative velocity equation is

)17(2/2 AAAA aaa

2/ AAa

)18(22/ aVa AA

)19(22/ tnAA aaVa

onaccelerati ofcomponent coriolis’ theis2,

2

2

Vand

ra

rawhere

tA

nA

)20(222 tntAnAA aaVaaa

10

4.3. Relative Acceleration of Coincident Points at the Point of Contact of Rolling

Elements.• Link 2 and 3 roll on each other at the point of contact P.

• Considering P2 on link 2 and P3 on link 3, for pure rolling contact,

VP2=VP3

• The relative acceleration aP2/P3 or aP3/P2 has two components• Normal :- toward center of rotation • Tangential:- along the tangent.

11

• For no sliding

• From which it can be deduced that

• and

)22(

..

)21(0

32

2/33/2

tPtP

tPPtPP

aa

ei

aa

)23(3

2

2

3

r

r

)24(323/2 nPnPnPP aaa

12

4.4. Acceleration Analysis by Complex Numbers

• Consider the mechanism shown below.• Replace the elements of the mechanism by position vector

such that their sum is zero yields the acceleration equation on two successive differentiations with respect to time.

• Differentiating the vector sum with respect to time, and introducing the complex notation yield the equation

)25(0RRR 421

)26(044244422 iii ereireir

13

• Again differentiating equation (26) yield the acceleration equation

• Separating the real and imaginary parts of the equation (27) yield

• Solving equations (28) simultaneously, the required accelerations are obtained to be:

)27(02 44442 24444444

222 iiiii ereirerierer

)28(0sincoscos2sinsin

0cossinsin2coscos

4244444444442

222

4244444444442

222

rrrrr

rrrrr

)30(

sin2

cos)cos(

)29()cos(

44

224222

4

24222

2444

r

r

rrr