2. Rigid Body Kinematics I – relative motion analysis...

DC - MECH 242 Rigid Body Kinematics I 1

2. Rigid Body Kinematics I –

relative motion analysis using

translating axes

(16.1-16.7 – Hibbeler)

Dr. Daniela Constantinescu

[email protected]

EOW 543


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, Copyrigth 2008, Prentice Hall.


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.


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).


2.3. Basic premise: Chasles Theorem

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 same at all points on the rigid body.

×=

+=

AP

AP

APAP

rdt

rd

rrr

/

/

/

rrr

rrr

ω

ctrAP

=/

r

APA

APPP

Prv

dt

rd

dt

rd

dt

rdv

/

/


2.4. Translation

Definition: A rigid body is translating if and only if any line of the body remains parallel to itselfthroughout the motion.

Mathematically:

It follows that:

and the velocity and the acceleration formulae become:

0≡ωr

0≡=•

ωαrr 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 or curvilinear).

• 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

vvrr

rr

)(trrAA

rr=


2.5. Rotation about a fixed axis

Definition: A rigid body is rotating about a fixed

axis if and only iff one line of 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 ).)(tθθ =

APnAPtAPAPP

APAPP

aarra

vrv

/,/,/

2

/

//

rrrrrr

rrrr

+=−×=

=×=

ωα

ω

kkkkdt

d

kk

))))r

))r

•••

•

====

==

θωαω

α

θωω.

.


2.6. Planar motion

Definition: A rigid body is in plane motion if and only if a planeof the body remains in a fixed plane throughout the motion.

Chasles’ Theorem for planar motion

The planar motion of a rigid body consists of translation with one point on the body plus rotation about an axis through the same point and 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 constant direction (perpendicular on the plane of

motion): APAPaavvrrrrrr

,,,⊥αω

ωr

αr


• 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

/

rrrrωωω −=××

APnAPtAAPAPAP

APAAPAP

aaarraa

vvrvv

/,/,/

2

/

//

rrrrrrrr

rrrrrr

++=−×+=

+=×+=

ωα

ω

APr

/

rr⊥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 a translating frame with origin at A).


• At each moment in time, there exists an axis at rest on a rigid body in plane

motion. This axis is called the instantaneous axis of rotation and 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).

)(trrAA

rr=

)(tθθ =


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.


2. The ICZV is not necessarily on

the rigid body.

3. The ICZV is an instantaneous centre 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 infinity and the body

is instantaneously translating.

≠

=

0

0

ICZV

ICZV

a

vr

r

0=⇒ωr

ABvvrr

=

ABvvAB

⊥rr


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 velocity and

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 velocity and zero

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

at next moment in time).

≠

=

0

0

ICZV

ICZV

a

vr

r

≡

≡

0

0

A

A

a

vr

r

, A on fixed axis of rotation.


2.7. Problem solving technique

1. Number the rigid bodies, starting from the rigid body with known motion.

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.

APAPAPrvv

/

rrrr×+= ω

APω

)(// APAPAPAPAPAP

rraarrrrrrr

××+×+= ωωα

APAPαωrr

,


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

, PPnva =

Avr

Aar

Pvr

Pta

,

r

Pna

,

r

APAPAPrvv

/

rrrr×+= ω

)(// APAPAPAPAPAP

rraarrrrrrr

××+×+= ωωα


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 not remain 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.


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). Use Chasles’ 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’s theorem 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 instantaneously at rest.

Pvr

0)ˆ)sin(ˆ)(cos(ˆ00

/

=⇒+×+=

⇒×+=

ABAB

ABABAB

jiABk

rvv

ωθθω

ωrrrr

−=

=

⇒

=−

⋅+=

⇒×+=−

⇒×+=

BP

vv

v

BPvv

jPBkivjiv

rvv

P

BP

B

B

BPPB

BPPB

PBBPPB

ωθ

ωθ

ωθθ

ω

0

0)cos(

)sin(

ˆˆˆ)ˆ)cos(ˆ)(sin(

/

rrrr


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). Use Chasles’ 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. Additional information 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.

Par

jBPiBPaa

jBPjBPkiaa

rraa

BPBPPB

BPBPPB

PBBPPBBPPB

ˆˆ)(

ˆˆˆˆ

2

2

/

2

/

⋅−⋅−−=

⇒⋅−×+−=

⇒−×+=

ωα

ωα

ωα

r

r

rrrrr


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

rrrrr

⋅

⋅−=

⋅⋅+−=

⇒

⋅⋅=⋅−

⋅⋅−=⋅−−

)cos(

)tan(

)cos(

)sin(2

2

2

θ

ωα

θωα

θαω

θαα

AB

BP

BP

BPa

ABBP

ABBPa

BP

AB

BPP

BP

ABBP

ABBPP


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

////

//

)(rrrrrrrrrr

rrrrrr

++=××+×+=

+=×+=

ωωα

ω


Consider a third coordinate frame, glued to the rigid body, Ax’’y’’z’’.

An observer sitting in the fixed frame Oxyz

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 frame Ax’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

rvvrrrrrrr

rrrr

××+×+=

×+=

ωωα

ω

)(/////

//

APAPAPAPAPAnPAtPAP

APAPAP

rraaa

rvrrrrrrrr

rrr

××+×=+=

×=

ωωα

ω

APr

/

r

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