DC -MECH 242Rigid Body Kinematics I1

2. Rigid Body Kinematics I –

relative motion analysis using

translating axes

(16.1-16.7 –Hibbeler)

Dr. Daniela Constantinescu

danielac@me.uvic.ca

EOW 543

DC -MECH 242Rigid Body Kinematics I2

2.1. Road Map*

Dynamics

Particles

Rigid bodies

Kinematics

Kinetics

I -Relative motion –analysis using a

translating frame of reference

II -Relative motion –analysis using a

rotating frame of reference

Force-mass-acceleration method

Work-energy method

Impulse-momentum method

•Figures numbered 16.* are taken from the course textbook:

Hibbeler, Engineering Mechanics: Dynamics, 11e, Copyrigth2008, Prentice Hall.

DC -MECH 242Rigid Body Kinematics I3

2.1. Road Map (ctd.) Notes:

•Two right-handed rectangular

frames will be used throughout

this section:

-one

fixed frame Oxyz

-shown in

red;

-one

translating frame Ax’y’z’with

axes parallel to the axes of the

fixed frame (x’IIx, y’IIy, z’IIz) and

with origin moving with point A on

the rigid

-shown in green.

•Absolute motion

(positions,

velocities and accelerations) of

points will be

observed from the

fixed reference frame

.

•Relative motion

(positions,

velocities and accelerations) of

points will be

observed from the

translating reference frame

.

DC -MECH 242Rigid Body Kinematics I4

2.2. Objectives:1.To fully describe the planar motion of

rigid bodies.

2.To establish relationships between the

motion of points on the same rigid

body.

3.To establish relationships between the

motion of points on connected rigid

bodies that have non-slipping contacts

to other moving bodies (eg., hinges,

non-slipping cables, meshing teeth,

non-slipping rollers).

DC -MECH 242Rigid Body Kinematics I5

2.3. Basic premise: ChaslesTheorem The motion of a rigid body consists of translation with one point on the body

plus rotation about the same point.

Can be shown using:

(1)

(A & P on the same rigid body, i.e., ).

Velocity formula:

(2)

Acceleration formula:

(3)

Note:

•For a rigid body, the angular velocity and angular

acceleration are the sameat all points on the rigid body.

×=

+=

AP

AP

APAP

r

dt

rd

rrr

/

/

/

r

r

r

r

r

r

ω

ctr

AP

=

/

r

AP

A

APPP

P

rv

dtrd

dt

rd

dt

rd

v

/

/

r

r

r

r

r

r

r

×+=+==

ω

()

APAPA

AP

AP

AP

P

rra

dt

rd

r

dt

d

dt

vd

dt

vd

a

//

/

/

r

rr

r

r

r

r

r

r

r

r

r

r

××+×+=

=×+×+==

ωωα

ω

ω

ω

r

α

r

DC -MECH 242Rigid Body Kinematics I6

2.4. Translation Definition: A rigid body is translating if and

only if any lineof the body

remains parallel to itself

throughout the motion.

Mathematically:

It follows that:

and the velocity and the acceleration

formulae become:

0

≡

ω

r

0≡=

•

ωα

rr

def

Notes:

•All points on a translating rigid body move

along parallel paths (the kinematic constraints

of the rigid body determine whether the paths are rectilinear orcurvilinear).

•A translating rigid body has 3DOF (translation is fully described by the

displacement of one point, ). If translation is planar, the rigid body

has 2DOF.

≡

≡

AP

AP

aa

vv

rr

r

r

)(trr

AA

r

r

=

DC -MECH 242Rigid Body Kinematics I7

2.5. Rotation about a fixed axis Definition: A rigid body is rotating about a fixed

axis if and only iffone lineof the body

remains fixed throughout the motion.

Mathematically:

If A is chosen on the fixed axis of rotation, the

velocity and acceleration formulae become:

Notes:

•All points on a rigid body rotating about a fixed axis move along circles

centered on the axis of rotation.

•A rigid body rotating about a fixed axis has 1DOF (rotation about a fixed axis

is fully described by a single kinematic parameter, the angular displacement

about the axis of rotation ).

)(t

θ

θ

=

APnAPtAPAPP

APAPP

aarra

vrv

/,/,/

2

/

//

rrrr

r

r

r

r

r

r

+=−×=

=×=

ωα

ω

kkkk

dt

d

kk

))))

r

)

)

r

•••

•

====

==

θωα

ω

α

θωω

.

.

DC -MECH 242Rigid Body Kinematics I8

2.6. Planar motion Definition: A rigid body is in plane

motion if and only if a plane

of the body remains in a

fixed planethroughout the

motion.

Chasles’Theorem for planar motion

The planar motion of a rigid body

consists of translation with one

pointon the body plus rotation

about an axis through the same

pointand perpendicular on the

plane of the motion at this point.

Notes:

•For a rigid body in planar motion, the angular velocity and the angular

acceleration have constantdirection (perpendicular on the plane of

motion):

APAP

aavv

r

r

r

r

r

r

,,,⊥

αω

ω

r

α

r

DC -MECH 242Rigid Body Kinematics I9

•If P and A are both in the plane of motion (i.e., ) then:

and the velocity and the acceleration formulae become:

•Graphically, plane motion can be represented as the superposition of a

translation with point A and a rotation about an axis perpendicular on the

plane of motion at A as follows*:

(

)

APAP

rr

/

2

/

r

r

r

r

ωωω

−=××

APnAPtAAPAPAP

APAAPAP

aaarraa

vvrvv

/,/,/

2

/

//

rrrrr

r

rr

r

r

r

r

r

r

++=−×+=

+=×+=

ωα

ω

AP

r

/

r

r

⊥

ω

* Red vectors represent the absolute motion of points (observed from the fixed frame), while green

vectors represent the relative motion of points (observed from atranslating frame with origin at A).

DC -MECH 242Rigid Body Kinematics I10

•At each moment in time, there exists anaxis at reston a rigid body in plane

motion. This axis is called the instantaneous axis of rotationand is

perpendicular on the plane of motion. The point on the instantaneous axis of

rotation in the plane of motion is called instantaneous center of zero

velocity (ICZV).

•A rigid body in planar motion has 3DOF (i.e., planar motion is fully

described by the displacement of one point plus the angular

displacement of the rigid body about an axis perpendicular on the plane of

motion ).

Example mechanism: a system of rigid bodies connected to each other, in

translation, in rotation about a fixed axis, and in general plane motion (i.e.,

translation plus rotation).

)(trr

AA

r

r

=

)(t

θ

θ

=

DC -MECH 242Rigid Body Kinematics I11

2.6.1. Plane motion of a rigid body -properties of the ICZV

1.The ICZV is on a line passing through a point A on the rigid body and

perpendicular on the velocity of point A.

The ICZV can be found if the direction of the velocity of two points on

the rigid body is known.

DC -MECH 242Rigid Body Kinematics I12

2.The ICZV is not necessarilyon

the rigid body.

3.The ICZV is an instantaneouscentre of rotation,

i.e., it has zero velocity, but its acceleration is

always different than zero.

4.If two points on a rigid body:

1.have the same velocity ;

or

2.have parallel velocities that are not

perpendicular on the line between the two

points ,

then the ICZV of the body is at infinityand the body

is instantaneously translating.

≠

=

0

0

ICZVICZV

a

v

r

r

0

=

⇒

ω

r

AB

vv

r

r

=

ABvv

AB

⊥

r

r

DC -MECH 242Rigid Body Kinematics I13

Note the major difference between an instantaneous axis of rotation

(instantaneous centre of zero velocity) and a fixed axis of rotation (fixed

centre of zero velocity):

The instantaneous centre of zero velocity has zero velocityand

acceleration different than zero, i.e., it is at rest instantaneously (it will

start moving at the next moment in time).

The fixed centre of zero velocity has zero velocityand zero

acceleration, i.e., it is at rest throughout the motion (it will remain at rest

at next moment in time).

≠

=

0

0

ICZVICZV

a

v

r

r

≡

≡

0

0

A

A

a

v

r

r

, A on fixed axis of rotation.

DC -MECH 242Rigid Body Kinematics I14

2.7. Problem solving technique

1.Number the rigid bodies, starting from the rigid body with knownmotion.

2.Determine the connections (kinematic constraints) between the rigid

bodies, and between the rigid bodies and the ground.

3.Determine the motion of each rigid body: translation, or rotation about a

fixed axis, or plane motion.

4.Choose a coordinate frame in which to represent all vectors.

5.Analyze of motion of the mechanism using the following steps:

1.Start from the rigid body with known motion.

2.Use to determine:

1.the velocity of point P connecting the body with known motion ( )

to the next body in the mechanism; and

2.the angular velocity of the next body in the mechanism.

3.Repeat step 2 until the angular velocities of all rigid bodies in the

mechanism have been determined.

4.Start from the rigid body with know motion.

5.Use to determine:

1.the acceleration of point P connecting the body with known motion

( ) to the next body in the mechanism; and

2.the angular acceleration of the next body in the mechanism.

6.Repeat step 5 until the angular accelerations of all rigid bodies in the

mechanism have been determined.

APAPAP

rvv

/

r

r

r

r

×

+

=

ω

AP

ω

)(

//APAPAPAPAPAP

rraa

r

r

r

r

r

r

r

×

×

+

×

+

=

ω

ω

α

APAP

αω

r

r

,

DC -MECH 242Rigid Body Kinematics I15

2.7. Problem solving technique (ctd)

Notes:

•The velocity analysis should typically be completed before starting the

acceleration analysis.

•In Equations

and

A and P are points on the same rigid body chosen such that:

1.the motion of A is known, i.e., and are known.

2.the path of P is known locally. This means that:

1.during the velocity analysis, the direction of is known

(tangent to the path);

2.during the acceleration analysis:

1.the direction of its tangential acceleration is known

(tangent to the path);

2.its normal acceleration is known: it has direction

from P to the centre of curvature of its path, and it has

magnitude

.

ρ

/

2

,PPn

va=

A

v

r

A

a

r

P

v

r

Pt

a

,

r

Pn

a

,

r

APAPAP

rvv

/

r

r

r

r

×

+

=

ω

)(

//APAPAPAPAPAP

rraa

r

r

r

r

r

r

r

×

×

+

×

+

=

ω

ω

α

DC -MECH 242Rigid Body Kinematics I16

Example

1.3 rigid bodies: slider P, link BP, link AB.

2.Body connections (kinematic constraints):

1.Slider P: connected to the ground.

2.Link BP: pinned to link AB at B, pinned to the slider at P.

3.Link AB: pinned to link BP at B , pinned to the ground at A.

3.Body motion:

1.Slider P: rectilinear translation.

2.Link BP: plane motion (does not have a grounded pin and does notremain parallel to a

fixed line throughout the motion), i.e., translation plus rotation.

3.Link AB: rotation about the fixed axis through A and perpendicular on the plane of the

diagram (the motion plane).

4.Coordinate frame for vector representation: Origin at A, x-axis horizontal, y-axis

vertical, z-axis out of the page.

DC -MECH 242Rigid Body Kinematics I17

5.Velocity analysis:

1.P is the point with known motion on the link BP:

2.From P, move to B (path of B is circular, with centre at A). UseChasles’theorem

to relate the motion of the points P and B on the link BP and solve:

3.B is the connection between links AB and BC and its motion has been

determined. Use Chasles’stheorem to relate the motion of the points A and B on

the link AB and solve to compute the angular velocity of the link AB:

Link AB is instantaneouslyat rest.

P

v

r

0)

ˆ

)sin(

ˆ

)(cos(

ˆ

00

/

=⇒+×+=

⇒

×

+

=

ABAB

ABABAB

jiABk

rvv

ωθθω

ω

r

r

r

r

−=

=

⇒

=−

⋅+=

⇒×+=−

⇒×+=

BP

v

v

v

BPvv

jPBkivjiv

rvv

P

BP

B

B

BPPB

BPPB

PBBPPB

ω

θ

ωθ

ωθθ

ω

0

0)cos(

)sin(

ˆˆˆ

)

ˆ

)cos(

ˆ

)(sin(

/

r

r

r

r

DC -MECH 242Rigid Body Kinematics I18

6.Acceleration analysis:

1.P is the point with known motion on the link BP:

2.From P, move to B (path of B is circular, with centre at A). UseChasles’theorem to

relate the motion of the points P and B on the link BP:

The one vector equation above is equivalent to 2 scalar equations (plane motion), and

contains 3 unknowns: the magnitude and the direction of the acceleration of B, and the

magnitude of the angular acceleration of the link BP. Additionalinformation about the

path of P needs to be added to allow the equation above to be solved.

The additional information about the path of B is provided by the link AB. Link AB rotates

about a fixed axis through A. Hence, B moves on a circle with the centre at A.

Mathematically, this information is available in Chasles’theorem applied to the link AB.

P

a

r

jBPiBPaa

jBPjBPkiaa

rraa

BPBPPB

BPBPPB

PBBPPBBPPB

ˆˆ

)(

ˆˆˆˆ

2

2

/

2

/

⋅−⋅−−=

⇒

⋅−×+−=

⇒

−×+=

ωα

ωα

ωα

r

r

r

r

r

r

r

DC -MECH 242Rigid Body Kinematics I19

3.Use Chasles’theorem to relate the motion of the points A and B on the link AB:

Combining the two 2 vector equations (i.e., the 4 scalar equations) above, the

magnitude and the direction of the acceleration of B, and the angular accelerations of

the links BP and AB can be determined. Solving for the angular accelerations of the

links BP and AB:

jABiABa

jiABka

rraa

ABABB

ABB

ABABABABAB

ˆ

)cos(

ˆ

)sin(

)

ˆ

)sin(

ˆ

)(cos(

ˆ

/

2

/

θαθα

θθα

ωα

⋅⋅+⋅⋅−=

⇒+×=

⇒−×+=

r

r

r

r

r

r

r

⋅

⋅

−=

⋅⋅+

−=

⇒

⋅⋅=⋅−

⋅⋅−=⋅−−

)cos(

)tan(

)cos(

)sin(

2

2

2

θ

ω

α

θω

α

θαω

θαα

AB

BP

BP

BPa

ABBP

ABBPa

BP

AB

BPP

BP

ABBP

ABBPP

DC -MECH 242Rigid Body Kinematics I20

2.8. Road up to here:

Dynamics

Particles

Rigid bodies

Kinematics

Kinetics

I. Relative motion –analysis using a

translating frame of reference

II. Relative motion –analysis using

a rotating frame of reference

Force-mass-acceleration method

Work-energy method

Impulse-momentum method

Notes:

We have used two reference systems:

one fixed coordinate frame

Oxyz

one translating coordinate frame

Ax’y’z’

with axes parallel to the axes of

the fixed frame (x’IIx, y’IIy, z’IIz) and with origin attached to point A on the

body whose motion we have been interested in

and the following equations:

(i.e., Chasles’theorem) to establish relationships between:

the motion of points on the same rigid body;

the motion of connected rigid bodies that do not slip with respect to each

other at the point where they connect and/or that are connected with sliders

that slide on the ground or on translating rigid bodies.

A and P in the equations above are points on the same rigid body.

AnPAtPAAPAPAPAPAPAP

APAAPAPAP

aaarraa

vvrvv

////

//

)(

rrrr

rr

r

r

rr

r

r

r

r

r

r

++=××+×+=

+

=

×

+

=

ωωα

ω

DC -MECH 242Rigid Body Kinematics I21

Consider a third coordinate frame, glued to the rigid body,

Ax’’y’’z’’.

An observer sitting in the

fixed frameOxyz

and looking at P sees the

absolute motion of P

(translation of P with A and rotation of P about A):

An observer translating with A (sitting in the

translating frameAx’y’z’

) and

looking at P sees only the

rotation of P about A

:

An observer translating with A and rotating with the body (sitting in the

frame

Ax’’y’’z’’glued to the rigid body

) and looking at P sees

P at a fixed location

(A and P are on the same rigid body):

)(

//

/

APAPAPAPAPAP

APAPAP

rraa

rvv

r

rr

r

r

rr

r

r

r

r

××+×+=

×

+

=

ωωα

ω

)(

/////

//

APAPAPAPAPAnPAtPAP

APAPAP

rraaa

rv

r

rr

r

r

rrr

r

r

r

××+×=+=

×

=

ωωα

ω

AP

r

/

r

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