16 Planar Kinematics of Rigid Body

7/23/2019 16 Planar Kinematics of Rigid Body

1/47

CHAPTER 16

3.1 Kinematics of Rigid Body - 2D


2/47

Planar Kinematics of Rigid Body 16.1 Planar Rigid-Body Motion

16.2 Translation

16.3 Rotation about a fixed axis 16.4 Relative-Motion Analysis: Velocity

16.5 Instantaneous Center of Zero Velocity

16.6 Relative-Motion Analysis: Acceleration


3/47

16.1 Rigid Body Motion There are three types of rigid body planar motion:

1) TranslationThis type of motion occurs if every line segment onthe body remains parallel to it original direction during the motion.

Rectilinear translationoccurs when the paths of motion for any twoparticles of the body are along equidistant straight lines.

Curvilinear translationoccurs when the paths of motion are alongcurves lines which are equidistant.


4/47

2) Rotation about a fixed axis

When a rigid body rotates about a fixed axis,all the particles of the body, except thosewhich lie on the axis of rotation, move along

circular paths.

Rigid Body Motion

3) General plane motion

When a body is subjected to general planemotion, it undergoes a combination of

translation and rotation. The translationoccurs within a reference plane, and therotation occurs about an axis perpendicularto the reference plane.


5/47


6/47

16.2 TRANSLATION

Velocity

ABAB /rrr

vB=vA+ drB/A/dt

The term drB/A/dt= 0, since the magnitude of rB/Aisconstant by definition of a rigid body.

AB vv AB aa

Position

Acceleration


7/47

16.3 Rotation About a Fixed AxisAngular Motion

Consider the body shown and the angular motion of aradial line rlocated with the shaded plane and directedfrom point Oon the axis of rotation to pointP.

Angular Position

The angular positionof ris defined by the angle

Angular Displacement

The change in the angular position, d, is called the

angular displacement.Angular Velocity

The time rate of change in the angular positionis called the angular velocity. Since d

occurs during an instant of time dt, then,dt

d


8/47

Angular Acceleration

The angular accelerationmeasure the time

rate of change of the angular velocity. Themagnitudeof this vector may be written as

( +)

2

2

dt

d

dt

d

The line of action of is the same as that for . However, it senseof direction depends on whether is increasing or decreasing.

dt

dAngular Velocity


9/47

dd ( +)

dt

d

dtd


10/47

Constant Angular Acceleration.

If the angular acceleration of the body is constant, = c

)(2

2

1

020

2

200

0

c

c

c

tt

t

( +)

( +)

( +)

Constant Angular Acceleration


11/47

Position.

The position ofPis defined by the positionvector r, which extends from OtoP.

Velocity.

The velocity ofPhas a magnitude whichcan be found from its polar coordinatecomponents and

Since ris constant, the radial components

and so

rvr rv

0rvr 0 rv

rv


12/47

Acceleration.

The acceleration ofPcan be expressed in terms ofits normal and tangential components

ra

ra

n

t

2

The tangential component of accelerationrepresents the

time rate of change in the velocitys magnitude.If speed ofPincreases => atsame direction asv

If speed ofPdecreases => atopposite direction asv

If speed ofPconstant => at

is zero


13/47

Therefore,

nt aaa

Since atand anare perpendicular to one another, if neededthe magnitude of acceleration can be determined from thePythagorean theorem

22tn aaa


14/47

Example 16.1A cord is wrapped around a wheel which is initially at rest. If aforce is applied to the cord and gives it an acceleration a = (4t)m/s2, where t is in seconds, determine as a function of time (a)the angular velocity of the wheel, and (b) the angular position ofthe line OPin radians.


15/47

Hence the angular acceleration of the wheel is

2/20

)2.0()4()(

sradt

t

ratP

+

= d/dtIntegrating, with the initial condition that = 0, t = 0,

+

sradt

dttd

tdt

d

t

/10

20

)20(

2

00


16/47

Using this result, the angular position of OP can be found

from = d/dt, since this equation relates , , and t.

Integrating, with the initial condition = 0 at t= 0,

radt

dttd

tdt

d

t

3

0

2

0

2

33.3

10

)10(

+


17/47

Example 16.2

The motor is used to turn a wheeland attached blower containedwith the housing. If the pulley Aconnected to the motor begins

rotating from rest with an angularacceleration of A = 2 rad/s

2,determine the magnitudes of thevelocity and acceleration of pointP

on the wheel, after the wheelBhasturned one revolution.


18/47

Angular Motion.

Converting revolution to radians, radB 283.6

Since the belt does not slip, an equivalent length of belt smust be unraveled from both the pulley and wheel at alltimes. Thus,

rad

rrs

A

ABBAA

57.12)4.0(283.6)15.0(;

Since Ais constant, the angular velocity of pulley A is therefore

sradA

A

c

/09.7

)057.12)(2(20

)(22

0202

+


19/47

The belt has the same speed and tangential component ofacceleration as it passes over the pulley and wheel. Thus,

2/750.0

/659.2

sradrra

sradrrv

BBBAAt

BBBAA

Motion of P. As shown in the diagram, we have,

222

22

2

/84.2)827.2()3.0(

/827.2)(

/3.0)(

/06.1

sma

smra

smra

smrv

P

BBnP

BBtP

BBP

Thus,


20/47

The origin of the x,ycoordinate system will be attached to theselected basepointA, which generally has a knownmotion.

The axes of this coordinate system do not rotate with the body;rather they will only be allowed to translate with respect to the

fixed frame.

16.4 RelativeMotion Analysis: Velocity

Position.

The position vector rAspecifies the

location of the base pointA, and therelative-position vector rB/AlocatespointBwith respect to pointA.

By vector addition, theposition ofBisABAB /rrr


21/47

Displacement.

During an instant of time dt, point A and B undergodisplacements drAand drB.

If we consider the general plane motion by its component partsthen the entire body first translates by an amount drAso thatA,the base point, moves to its final position and pointBtoB.

The body is then rotated aboutAby an amount dso thatBundergoes

a relative displacement drB/Aand thus moves to its final positionB.


22/47

Due to the rotation aboutA, drB/A= rB/Ad, and thedisplacement ofBis

ABAB ddd /rrr Velocity due to general motion

ABAB /vvv ABAB /rvv


23/47

Methods :-Vector Method

Velocity Diagram


24/47

Vector Cross Product

Vector Method

k = always be the

rotation axis.


25/47

Example 16.8

The collar Cis moving downward with a velocity of2 m/s. Determine the angular velocities of CBandABat this instant.


26/47

Velocity Equation.

Link CB(general plane motion):

smv

srad

v

v

v

B

CB

CB

CBB

CBCBB

CBB

CBCBCB

/2

/102.020

2.0

2.02.02

)2.02.0(2

/

ijji

jikji

rvv

Example 16.8


27/47

Link AB(rotation about a fixed axis):

sradAB

AB

AB

BABB

/10

2.02

)2.0(2

jki

rv

Same results can be obtained using Scalar Analysis.

Example 16.8


28/47

16.5 Instantaneous Center of Zero Velocity

The velocity of any point Blocated on a rigid body can beobtained in a very direct way if one choose the base point Ato be a point that haszero velocityat the instant considered.

SincevA= 0, thereforevB= x rB/A.

Point A is called the instantaneous center of zerovelocity(IC)and it lies on the instantaneous axis of zerovelocity.

This axis is always perpendicular to the plane of motionand the intersection of the axis with this plane defines thelocation of the IC.


29/47

Since point A is coincident with the IC, then

vB= x rB/Aand so pointBmoves momentarily about theICin a

circular path.The magnitudeofvBis vB= rB/IC.

Due to the circular motion, the direction of vB must always beperpendicularto rB/IC

Consider the wheel as shown, if it rolls without slipping, thenthe point of contact with the ground haszero velocity


30/47

i) Given the velocity vA of a point A on

the body, and the angular velocity ofthe body. TheICis located along the linedrawn perpendicular tovAatA, such thatthe distance from A to the IC is rA/IC =vA/. Note that the IC lies up to the right

of A since vA must cause a clockwiseangular velocity about theIC.

Location of the IC.

ii) Given the line of action of twononparallel velocities vA and vB.

Construct at points A and B linesegments that are perpendicular to vAandvB. Extending these perpendicular totheir point of intersection as shownlocates theIC at the instant considered.


31/47

iii) Given the magnitude and direction of two parallelvelocities vA and vB. Here the location of the IC isdetermined by proportional triangles.


32/47

Example 16.11

Block D moves with a speed of 3 m/s. Determinethe angular velocities of links BD and AB, at theinstant shown.


33/47

From the geometry,

mr

mr

ICD

ICB

566.045cos

4.0

4.045tan4.0

/

/

Since the magnitude ofvDis known, theangular velocity of linkBDis

sradr

v

ICD

DBD /30.5

566.0

3

/

The velocity of B is therefore

smrv ICBBDB /12.2)4.0(30.5)( / 45

From the figure, the angular velocity of AB is

sradr

v

AB

BAB /30.5

4.0

12.2

/


34/47

Example 16.12

The cylinder rolls without slipping between the two movingplates E and D. Determine the angular velocity of thecylinder and the velocity of its center Cat the instant shown.


35/47

Solution

Since no slipping occurs, the contact points A and B on thecylinder have the same velocities as the plate E and D,respectively.

Assuming this point to be a distancexformB,

we have

)25.0(25.0);25.0(

4.0;

xxv

xxv

A

B

Dividing one equation into the other eliminates and yields

mx

xx

154.065.0

1.0

25.0)25.0(4.0


36/47

Hence the angular velocity of the cylinder is

sradx

vB /60.2154.0

4.0

The velocity of point Cis therefore

sm

rv ICCC

/0750.0

)125.0154.0(6.2/


37/47

16.6 Relative-Motion Analysis: Acceleration

An equation that relates the accelerations of two points on arigid body subjected to general plane motion,

dt

d

dt

d

dt

d ABAB /vvv

The terms dvB/dt = aBand dvA/dt = aAare measured from aset offixedx, yaxesand represent the absolute accelerationsof pointsBandA.

aB/A can be expressed in terms of its tangential and

normal components of motion.aB

= aA+ aB/A

nABtABAB )()( // aaaa


38/47

The relative-acceleration components represent

the effect of circular motionobserved fromtranslating axes having their origin at the basepoint A, and can be expressed as

(aB/A)t= x rB/A

(aB/A)n= -2rB/A

ABABAB /2

/ rraa


39/47

Vector Cross Product

Vector Method

k = always be the

rotation axis.


40/47

Example 16.17

The collar is moving downward with an acceleration of 1m/s2. At the instant shown, it has a speed of 2 m/s whichgives links CB and AB an angular velocity AB = CB = 10rad/s. Determine the angular accelerations of CBandABatthis instant.


41/47

Link BC(general plane motion):

jiijjjiji

jikjji

rraa

20202.02.01202.0)2.02.0()10(

)2.02.0()(1202.0

2

/2

/

CBCBAB

CBAB

CBCBCBCBCB

Thus,

202.0120

202.02.0

CB

CBAB

22

2

/95/95

/5

sradsrad

srad

AB

CB

Solving,


42/47

Example 16.18

The crankshaft AB of an engine turnswith a clockwise angular acceleration of20 rad/s2. Determine the acceleration ofthe piston at this instant AB is in the

shown. At this instant AB = 10 rad/sand BC= 2.43 rad/s.


43/47

ac = aB+ aC/B

jaCC

a

2

2

2

/}14.1421.21{

)177.0177.0()10()177.0177.0()20(

sm

BABBABB

ji

jijik

rra

m

mB

}177.0177.0{

}45cos25.045sin25.0{

ji

jir


44/47

=


45/47

Link AB in the mechanism shown has an angular velocity of 8

rad/sec (CCW) constant. At the same instant shown,AB

is verticalandBCis in horizontal position. Determine the

(i) Velocity and acceleration of point C

(ii)Angular velocity and acceleration of link CD


46/47

The motion of the piston rodBcontrols the rotation of linkAB about O

as shown. At this instant shown, the piston rod moves at VB= 2.5 m/sto the left and decreasing at the rate of 15 m/s2. Determine

i) Angular velocity of link OAandAB at this instant, and

ii) Angular acceleration of link OAandAB at this instant


47/47

Date post:	19-Feb-2018
Category:	Documents
Upload:	zackziffi
View:	239 times
Download:	5 times

16 Planar Kinematics of Rigid Body

Documents