kinematics of rigid bodies

1

References:

Vector Mechanics for Engineers: Dynamics, Ferdinand Beer,E. Russell Johnston, David F.Mazurek: McGraw Hill

Engineering Mechanics: Dynamics, Russell C. Hibbeler: Prentice Hall

Kinematics of Rigid Bodies

Lecture Notes: Dr. Ceren BİLGİN GÜNEY

Kinematics of Rigid Bodies

5-2

IntroductionTranslationRotation About a Fixed Axis: VelocityRotation About a Fixed Axis: AccelerationRotation About a Fixed Axis: Representative SlabEquations Defining the Rotation of a Rigid Body About a Fixed AxisGeneral Plane MotionAbsolute and Relative Velocity in Plane MotionInstantaneous Center of Rotation in Plane MotionAbsolute and Relative Acceleration in Plane MotionAnalysis of Plane Motion in Terms of a Parameter

2

Introduction

• Kinematics of rigid bodies: relations between time and the positions, velocities, and accelerations of the particles forming a rigid body.

• Classification of rigid body motions:

- general motion

- motion about a fixed point

- general plane motion

- rotation about a fixed axis

• curvilinear translation

• rectilinear translation

- translation:

5-3

Translation:Translation occurs if every line segment on the body remains parallel to its original direction during the motion. When all points move along straight lines, the motion is called rectilineartranslation. When the paths of motion are curved lines, the motion is called curvilineartranslation.

Translation

5-4

3

Passengers on this amusement ride are subjected to curvilinear translation since the vehicle moves in a circular path but they always remains upright.

5-5

Translation• Consider rigid body in translation:

- direction of any straight line inside the body is constant,

- all particles forming the body move in parallel lines.

• For any two particles in the body,

ABAB rrrrrr +=

ABAB rrr &r&r&r +=• Differentiating with respect to time,

All particles have the same velocity.

AB

AABAB

aa

rrrrrr

&&r

&&r

&&r

&&r

=

=+=• Differentiating with respect to time again,

All particles have the same acceleration.5-6

AB

AABAB

vv

rrrrrr

&r&r&r&r

=

=+=

4

In this case, all the particles of the body, except those on the axis of rotation, move along circular pathsin planes perpendicular to the axis of rotation.

Rotation About a Fixed Axis

5-7

Curvilinear translationall its particles moving alongparallel circles,

Rotationall its particles moving along concentric circles

Rotation should not be confused with certain types of curvilinear translation


5-8

5

Angular velocity, , is obtained by taking the time derivative of angular displacement:

(rad/s) +

The change in angular position, dθ, is called the angular displacement, with units of either radians or revolutions. They are related by

1 revolution = (2π) radians

When a body rotates about a fixed axis, any point P in the body travels along a circular path. The angular position of P is defined by θ.

Similarly, angular accelerationis

= d2θ/dt2 = or = rad/s2 +


ωr

αrdt

dωr

θω

d

dr

ωrαr

dt

dθω =r

5-9

The magnitude of the velocity of P is equal to ωr. The velocity’s direction is tangent to the circular path of P.

In the vectorformulation, the magnitude and direction of can be determined from the cross productof and . Here is a vector from any point on the axis of rotation to P.

rvrrr ×= ω

vr

ωr rr

rr

Rotation About a Fixed Axis: Velocity

5-10

Let’s derive the velocity equation in the next slide

6

Rotation About a Fixed Axis: Velocity• Consider rotation of

rigid body about a fixed axis AA’

Velocity vector of the particle P is tangent to the path with magnitude

dtrdvrr =

dtdsv =

( ) ( )

( ) φθθφ

θφθ

sinsinlim

sin

0&r

tr

dt

dsv

rBPs

t=

∆∆==

∆=∆=∆

→∆ 5-11

rdt

rdv

rrr

r ×== ω

• The same result is obtained if we draw a vector along AA’ kk

r&

rr θωω ==

locityangular vekk ===r

&rr θωω

vrdt

da

rrrr

r ×+×= ωω

Using the vectorformulation, the acceleration of P can also be defined by differentiating the velocity.

( )rdt

d

dt

vda

rvr

r ×== ω

kkk

celerationangular acdt

d

r&&

r&

r

rr

θωα

αω

===

==

componenton accelerati normal )(

componenton accelerati l tangentia

)(

=××=×

××+×=

r

r

rra

rrr

rr

rrrrrr

ωωα

ωωα

• Acceleration of P is combination of two vectors,

Rotation About a Fixed Axis: Acceleration

5-12

dt

rdr

dt

dr

rrr

×+×= ωω

7

Rotation About a Fixed Axis: Representative Slab

• Consider the motion of a representative slab in a plane perpendicular to the axis of rotation.

• Velocity of any point P of the slab,

ωωω

rv

rkrv

=×=×= rrrrr

• Acceleration of any point P of the slab,

rrk

rrarrr

rrrrrr

2ωα

ωωα

−×=

××+×=

• Resolving the acceleration into tangential and normal components,

22 ωωααrara

rarka

nn

tt

=−==×=

rr

rrr

5-13

Equations Defining the Rotation of a Rigid Body

About a Fixed Axis

• Motion of a rigid body rotating around a fixed axis is often specified by the type of angular acceleration.

ωθθω d

dtdt

d == or• Recall

• Uniform Rotation, α = 0:tωθθ += 0

• Uniformly Accelerated Rotation, α = constant:

( )020

2

221

00

0

2 θθαωω

αωθθ

αωω

−+=

++=

+=

tt

t

5-14

θωωθωα

d

d

dt

d

dt

d ===2

2

8

If the lower roll has a constantangular velocity, the speed of the paper being wound onto it increases as the radius of the roll increases

Equations Defining the Rotation of a Rigid Body

About a Fixed Axis

5-15

• Uniform Rotation, α = 0:

tωθθ += 0

Sample Problem

Load B is connected to a double pulley by one of the two inextensible cables shown. The motion of the pulley is controlled by cable C.

Cable C has a constant acceleration of 225mm/s2 and an initial velocity of 300mm/s, both directed to the right.

Determine (a) the number of revolutions of the pulley in 2 s, (b) the velocity and change in position of the load B after 2 s, and (c) the acceleration of the point D on the rim of the inner pulley at t = 0.

SOLUTION:

• Due to the action of the cable, the tangential velocity and acceleration of D are equal to the velocity and acceleration of C. Calculate the initial angular velocity and acceleration.

• Apply the relations for uniformly accelerated rotation to determine the velocity and angular position of the pulley after 2 s.

• Evaluate the initial tangential and normal acceleration components of D.

5-16

9

rev23.2=N

mm 1750

smm1250

=∆↑=

B

B

y

vr

( ) ( ) ↓=→== 22 smm1200smm 225 nDCtD aaarrr

2smm1220=Da

°= 4.79φ

5-17

As the slider block A moves horizontally to the left with vA, it causes the link CB to rotate counterclockwise. Thus vB is directed tangent to its circular path.

Which link is undergoing general plane motion? Link AB or link BC?

General Plane Motion

The connecting rod AB is subjected to general plane motion5-18

10

General Plane Motion

• General plane motionis neither a translation nor a rotation.

• General plane motion can be considered as the sumof a translation and rotation.

• Displacement of particles A and B to A2 and B2

can be divided into two parts: - translation to A2 and- rotation of about A2 to B2

1B′1B′

5-19

Absolute and Relative Velocity in Plane Motion

• Any plane motion can be replaced by a translation of an arbitrary reference point A and a simultaneous rotation about A.

ABAB vvvrrr +=

ωω rvrkv ABABAB =×= rrr

ABAB rkvvrrrr ×+= ω

5-20

11


• Assuming that the velocity vA of end A is known, wish to determine the velocity vB of end B and the angular velocity ω in terms of vA, l, and θ.

• The direction of vB and vB/A are known. Complete the velocity diagram.

θ

θ

tan

tan

AB

A

B

vv

v

v

=

=

θω

θω

cos

cos

l

v

l

v

v

v

A

A

AB

A

=

==

5-21


• Selecting point B as the reference point and solving for the velocity vA of end Aand the angular velocity ω leads to an equivalent velocity triangle.

• vA/B has the same magnitude but opposite sense of vB/A. The sense of the relative velocity is dependent on the choice of reference point.

• Angular velocity ω of the rod in its rotation about B is the same as its rotation about A. Angular velocity is not dependent on the choice of reference point.

5-22

12

5 - 23

A similar analysis can be used when gears are involved.• the teeth in contact must have the same

absolute velocity.

Most mechanisms consist not of one but of several moving parts

When the various parts of a mechanism are pin-connected, • consider each part as a rigid body, • keep in mind that the points where two

parts are connected must have the same absolute velocity.


Note: when a mechanism contains parts which slide on each other, the relative velocity of the parts in contact must be taken into account.

Sample Problem

The double gear rolls on the stationary lower rack: the velocity of its center is 1.2 m/s.

Determine (a) the angular velocity of the gear, and (b) the velocities of the upper rack R and point D of the gear.

SOLUTION:

• The displacement of the gear center in one revolution is equal to the outer circumference. Relate the translational and angular displacements. Differentiate to relate the translational and angular velocities.

• The velocity for any point P on the gear may be written as

Evaluate the velocities of points B and D.

APAAPAP rkvvvvrrrrrr ×+=+= ω

5-24

13

x

y

( )kkrrr

srad8−== ωω

( )ivRrr

sm2= ( ) ( )sm697.1

sm2.1sm2.1

=+=

D

D

v

jivrrr

5-25

Sample Problem

5-26

Small wheels have been attached to the ends of rod AB and roll freely along the surfaces shown. Knowing that wheel A moves to the left with a constant velocity of 1.5 m/s, determine (a) the angular velocity of the rod,(b) the velocity of end B of the rod.

14

5-27

= +

( )srad 26.2=ABωrsm696.1

sm840.1

B/A

B

==

V

V 60°

70°

Instantaneous Center of Rotation in Plane

Motion• Plane motion of all particles in a slab can always be

replaced by the translation of an arbitrary point A and a rotation about A with an angular velocity that is independent of the choice of A.

• The same translational and rotational velocities at A are obtained by allowing the slab to rotate with the same angular velocity about the point C on a perpendicular to the velocity at A.

• The velocity of all other particles in the slab are the same as originally defined since the angular velocity and translational velocity at A are equivalent.

• As far as the velocities are concerned, the slab seems to rotate about the instantaneous center of rotation C.

5-28

15

5 - 29

If the tires of this car are rolling without sliding the instantaneous center of rotation of a tire is the point of contact between the road and the tire.

The velocity of any point located on a rigid body can be obtained in a very direct way by choosing a base point to be the zero velocity at the instant considered


Motion

The instantaneous center (IC) of zero velocity for this bicycle wheel is at the point in contact with ground. The velocity direction at any point on the rim is perpendicular to the line connecting the point to the IC.

Which point on the wheel has the maximum velocity?

Does a larger wheel mean the bike will go faster for the same rider effort in pedaling than a smaller wheel?


Motion

5-30

16


Motion

• If the velocity at two points A and B are known, the instantaneous center of rotation lies at the intersection of the perpendiculars to the velocity vectors through Aand B .

• If the velocity vectors at A and B are perpendicular to the line AB, the instantaneous center of rotation lies at the intersection of the line ABwith the line joining the extremities of the velocity vectors at A and B.

• If the velocity vectors are parallel and the velocity magnitudes are equal, the instantaneous center of rotation is at infinity and the angular velocity is zero.

5-31

Instantaneous Center of Rotation in Plane Motion

• The instantaneous center of rotation lies at the intersection of the perpendiculars to the velocity vectors through A and B .

θω

cosl

v

AC

v AA == ( ) ( )

θθ

θω

tancos

sin

A

AB

vl

vlBCv

=

==

• The velocities of all particles on the rod are as if they were rotated about C.

• The particle at the center of rotation has zero velocity.

• The particle coinciding with the center of rotation changes with time and the acceleration of the particle at the instantaneous center of rotation is not zero.

• The acceleration of the particles in the slab cannot be determinedas if the slab were simply rotating about C.

• The trace of the locus of the center of rotation on the body is the body centrode and in space is the space centrode.

5-32

17

Sample Problem

The double gear rolls on the stationary lower rack: the velocity of its center is 1.2 m/s.

Determine (a) the angular velocity of the gear, and (b) the velocities of the upper rack R and point D of the gear.

5-33

Solve sample problem, using the method of the instantaneous center ofrotation.

• Evaluate the velocities at B and D based on their rotation about C.

5-34

srad8=ω

sm697.1=DV

sm2=RV

45°

SOLUTION:

• The point C is in contact with the stationary lower rack and, instantaneously, has zero velocity. It must be the location of the instantaneous center of rotation.

• Determine the angular velocity about C based on the given velocity at A.

18

Sample Problem

5-35

Small wheels have been attached to the ends of rod AB and roll freely along the surfaces shown. Knowing that wheel A moves to the left with a constant velocity of 1.5 m/s, determine (a) the angular velocity of the rod,(b) the velocity of end B of the rod.

Solve sample problem, using the method of the instantaneous center ofrotation.

( )srad 26.2ω =AB

sm840.1B =V 60°

Absolute and Relative Acceleration in Plane Motion

• Absolute acceleration of a particle of the slab,

ABAB aaarrr +=

• Relative acceleration associated with rotation about A includes tangential and normal components,

ABar

( )( ) ABnAB

ABtAB

ra

rka

rr

rrr

2ω

α

−=

×= ( )( ) 2ω

α

ra

ra

nAB

tAB

=

=

5-36

19

15 - 37

• Givendetermine

, and AA varr

. and αrrBa

( ) ( )tABnABA

ABAB

aaa

aaarrr

rrr

++=

+=

• Vector result depends on sense of and the relative magnitudes of ( )

nABA aa and Aar

• Must also know angular velocity ω.



+→ x components: θαθω cossin0 2 llaA −+=

↑+ y components: θαθω sincos2 llaB −−=−

• Solve for aB and α.

• Write in terms of the two component equations,ABAB aaarrr +=

5-38

20

Analysis of Plane Motion in Terms of a Parameter

• In some cases, it is advantageous to determine the absolute velocity and acceleration of a mechanism directly.

θsinlxA = θcoslyB =

θωθθ

cos

cos

l

l

xv AA

===

&

&

θωθθ

sin

sin

l

l

yv BB

−=−=

=&

&

θαθω

θθθθ

cossin

cossin2

2

ll

ll

xa AA

+−=

+−=

=&&&

&&

θαθω

θθθθ

sincos

sincos2

2

ll

ll

ya BB

−−=

−−=

=&&&

&&

5-39

Sample Problem

The center of the double gear has a velocity and acceleration to the right of 1.2 m/s and 3 m/s2, respectively. The lower rack is stationary.

Determine (a) the angular acceleration of the gear, and (b) the acceleration of points B, C, and D.

SOLUTION:

• The expression of the gear position as a function of θ is differentiated twice to define the relationship between the translational and angular accelerations.

• The acceleration of each point on the gear is obtained by adding the acceleration of the gear center and the relative accelerations with respect to the center. The latter includes normal and tangential acceleration components.

5-40

21

5-41

( ) ( ) 222 sm12.8sm40.6m5 =−= BB ajisarrr

( ) jacrr 2sm60.9=

( ) ( ) 222 sm95.12sm3m6.12 =+= DD ajisarrr

kinematics of rigid bodies

Documents