Dynamics of Rigid Bodies Merian Book 2005

V. Rouillard 20031 Dynamics Introduction

DYNAMICS OF RIGID BODIES(Notes based on textbook)

Textbook: Engineering Mechanics DYNAMICS: 5th Ed.J.L. Meriam & L.G. Kraige

Plane kinematics of rigid bodies

Plane kinetics of rigid bodies Force, Mass & Acceleration Work & Energy Impulse & momentum

Introductory 3-D dynamics of rigid bodies 3-D kinematics of rigid bodies 3-D kinetics of rigid bodies


Some definitions

KINEMATICS: Study of motion without reference to the forces which cause the motion

KINETICS: deals with the forces acting on a body and the resulting motion


Some definitionsRIGID BODY

Rigid body: a body whose changes in shape are negligible compared to the overall dimensions or positional changes of the body as a whole OR distance between any two points within the body remain constant under the application of forces. As a consequence, the angle between any two lines on the rigid body does not change.

Thus a rigid body can be considered as a continuous system of particles, the distance between any two particles and the angle between any two lines do not change.

Actually all bodies deform under the action of forces (this is considered in the subject of SOLID MECHANICS).

V. Rouillard 20034 Dynamics Plane kinematics of rigid bodies

Plane motion

A rigid body executes plane motion when all parts of the body move in parallel planes. Motion occurs in the same plane as that of the bodys centre of mass Body is treated as 2-D (sheet) Motion confined to pane of sheet This describes a significant number of rigid body motion encountered in engineering


Plane motion


Plane motion Translation

Translation: Body remains parallel no rotation Rectilinear translation: body moves in parallel straight lines


Plane motion Translation Translation: Body remains parallel no rotation

Rectilinear translation: body moves in parallel straight lines

Curvilinear translation: body moves along congruent (identical) curves

During translation the motion of the whole body is completely described by the motion of any one particle in the body. Therefore, the translation of any rigid body can be described by with theories for Particle Kinematics


Plane motion Rotation

Fixed-axis rotation: All particles in body move in circular paths about the same axis of rotation. All lines Perpendicular to the axis of rotation rotate thru the same angle in the same time.

Therefore, the circular motion of a single particle can be used to describe the rotational motion of a rigid body.


Plane motion General plane motion

General plane motion: Combination of translation and rotation.


Because is constant, the derivative of (1) with respect to time gives:(2)

i.e. lines 1 and 2 have the same angular velocity.

Plane motion Rotation1 and 2 are the angular positions of arbitrary line 1 and 2 respectively. is the angle between lines 1 and 2.

(1) 2 1= +

d dt d dt or 2 1 2 10 0/ / = + = +


Differentiating (1) again gives:

(3)d / dt d / dt or = + = +2 2 2 22 1 2 10 0

All lines on a rigid body in its plane of motion have the same angular displacement, same angular velocity and the same angular acceleration.


Relationships of angular motion

The angular velocity and acceleration are the first and second derivative of the angular position respectively:

Angular velocity:

Angular Acceleration:

or

= = = =d dt v ds dt s/ ( / )( )1 = = = =d dt a dv dt va/ ( / )( )2 = = = =d dt a d s dt sb2 2 2 22/ ( / )( )

Eliminating dt from (1) and (2):

or d d vdv adsa= =( ) ( )3 ( )( ) d d sds sdsb= =3


Relationships of angular motionFor constant angular acceleration, (2a) and (3a) can be integrated directly:

rearranging d dt

or td dtt

( ):

( )

2a

40 0

0

=

= = +z zrearranging d d

ord d

( )

( )

( ) ( )

3a

50 0

202

0

202

0

2

2

=

= = = +

z zrearranging d dt and substituting for

or t t

t t

d t dtt

( ):

( )

( )

1

6

00

00 0

2

0 0

2

2

2

=

= = +

= + +z z +


Rotation about a fixed axisAll points, except for those on the axis, move in concentric circles about 0.

v r

a r v r v

a r

n

t

=

= = =

=

2 2 /


Rotation about a fixed axis

Using vector notation (2-D), the angular velocity may be expressed by the vector as shown below (right hand rule).

The vector v is obtained by crossing into r as follows:v r r

r v= =

( , , )

remember because of the RHR gives


Rotation about a fixed axisThe acceleration of point A is obtained by differentiating

v r= which gives:

a v r rr r

v r

= = + = + = +

( )

Summary of vector expressions:

v r

a r

a r

=

=

=

n

t

( )

Sample problems: 5/1, 5/2, 5/3


Absolute motionApproach:

use dimensional (geometrical) relationships between points within the rigid body to calculate the temporal derivatives to obtain velocities and accelerations.

Make sue to be consistent when defining directions (polarity)

If the geometry of the application is easily described, then the absolute motion approach is useful and straightforward.



Relative velocityThe vector representation of relative velocities of two particles A and B is written as:

v v v r r rA A A A= + = +B B B Bremember/ /( : )


Relative velocity due to rotation If we choose two points on the same rigid body, the motion of one point as seen by the other must

be circular as the distance between the two points never changes.

A


Relative velocity due to rotation Relative to point B, the rigid body appears to undergo a fixed axis rotation about B with point A

following a circular arc of angle as shown below. Conversely, if point A was chosen as the reference, point B would have been observed to follow a

circular arc of angle - as shown below. r rB A/ /A B=


Relative velocity due to rotation

Dividing the above by the corresponding time interval t, and applying the limit, we obtain the relative velocity equation:

Also, the magnitude of the relative velocity is:

Using r to represent the vector r A/B,the velocity vector may be written as:

r r rA A= +B B/

A

v v vA A= +B B/

v rBA/ =

v rA/ B =


Relative velocity The relative linear velocity, VA/B, is always perpendicular to the line joining the two points. Always sketch the vector polygon Use two orthogonal sets of scalar equations or one set of vector equations.

Sample problems: 5/7, 5/8, 5/9, 5/10


Instantaneous centre of zero velocity

Previously, the velocity of a point on a RB in plane motion was determined by adding the relative velocity due to rotation about a convenient reference point to the velocity of the reference point.

We now achieve the same result by choosing the reference point such that its velocity is momentarily zero.

In terms of velocity, the body may be assumed to be in pure rotation about an axis normal to the plane of motion, which passes through this reference point.

That axis is the Instantaneous axis of zero velocity The intersection of the axis to the plane of motion is the Instantaneous centre of zero velocity


Locating the instantaneous centre of zero velocityThe existence of the is evident for all conditions:

If there is a point about which the point A has absolute circular motion (rotation) then this is along a line normal to the velocity vector VA

The same applies to point B on the rigid body The instantaneous centre of zero velocity lies at the intersection of these lines


Locating the instantaneous centre of zero velocityIf the magnitude of one of the points, say A, is known then the angular velocity can be easily determined:

= =v r v rA A B B/ /as well as the linear velocity of every point in the body.

Once the instantaneous centre of zero velocity is located, the direction of the instantaneous velocity is easily established as it must be perpendicular to the line joining the centre and the point in question.

= =v r v rA A n n/ /


Locating the instantaneous centre of zero velocityIf the velocities of two arbitrary points in the body are parallel, then the centre of zero velocity is established by direct proportion as shown in (b) and (c) below. It can also be seen that as the magnitude of the parallel velocities converge, the centre is driven toward infinity until the motion of the body transforms to pure translation.

Sample problems: 5/11, 5/12


Relative accelerationJust as before, by differentiating the relative velocity equation wrt time, the relative acceleration equation is obtained:

Relative acceleration due to rotationUsing a similar arrangement as before whereby the relative motion between two points on a rigid body is circular, it follows that the relative acceleration will consist of a normal component (due to the change in direction of the relative velocity vector, VA/B) and a tangential component (due to the change in magnitude of the relative velocity vector, VA/B) . Therefore:

/ /v v v a a aA A A A= + = +B B B Bor

a a a aA A A(= + +B B n B t( ) )/ /


Relative acceleration due to rotationThe magnitudes of the relative acceleration components are:

In vector notation, the acceleration components are:

Both relative acceleration components are a function of the absolute angular velocity and absolute angular acceleration.

( ) /

)

/ /

/ /

A A

A A(

B n B

B t B

v r r

v r

= =

= =

2 2

( ) ( )

)

/

/

a r

a r

A

A(

B n

B t

=

=


Relative acceleration due to rotation

Unlike velocities, the acceleration vectors aA and aB arenot generally tangent to the curvilinear paths of A and B.

a a a a a aA A A A

A A

A A

(

(

= + = + +

= =

= =

B B B B n B t

B n B

B t B

v r r

v r

/ / /

/ /

/ /

( ) )

( ) /

)

2 2


Relative acceleration due to rotation

As for the relative velocity equation, the relative acceleration equation can be solved by three alternative methods:

Scalar algebra and geometry Vector algebra Graphical construction


V. Rouillard 200331 Dynamics Plane kinetics of rigid bodies

Semester 1:DYNAMICS OF RIGID BODIES





Kinetics Kinetics deals with the forces acting on a body and the resulting motion of the body. The definitions of a 2-D, plane rigid body still apply with the mass centre of the body located on

the plane of motion.

All forces will be projected on the plane of motion. Three general approaches are used to solve kinetic problems

The force mass acceleration method (Newtons law) The work energy method The impulse and momentum method


KineticsIn additions to the two force equations of motion required to solve plane (2-D) particle motion, plane rigid bodies require a third equation (a moment equation) to describe the rotation of thebody.

V. Rouillard 200334 Dynamics Plane kinetics of rigid bodies Force, Mass & Acceleration






Force, mass & acceleration general equations of motionFor a general body in three dimensions, the resultant of the external forces acting on the body equates to the product of the body mass and the acceleration of its centre of mass:

The resultant moment of the external forces about the centre of mass, G, equals the temporal rate of change of the angular momentum of the body about its centre of mass:

F = amM HG G=


Force, mass & acceleration general equations of motionThe external forces acting on the body can be replaced by a resultant force and a correspondingmoment acting through the centre of mass


Force, mass & acceleration general equations of motionRecall: the angular momentum of the mass system (body) about its centre of mass, G, is the sumof the moments of the linear momenta of all particles about G:

where i is the position vector relative to G of the partical of mass mi.For the rigid body, the velocity of mi relative to G is:

The magnitude of Hg becomes:

The summation which may also be expressed as

is defined as the mass moment of inertia, IG or , of

the body about the z axis through G.

HG i i ix m Eqn= ( . . )4 8

=

HG i i i im m= = 2 2

2z dm


Force, mass & acceleration general equations of motionWe may now write:

HG I=

where the mass moment of inertia, , is a property of the body and is proportional to its resistanceto change in rotational velocity due to the radial distribution of mass around the z-axis through thecentre of mass, G. This resistance is called rotational inertia.

Demonstration


Force, mass & acceleration general equations of motion

By substituting for HG:

G G I I= = =

Summary of Newtons 2nd law (vectorised):

F a = m G I=


Force, mass & acceleration general equations of motionRadius of gyration, kz, is a measure of the distribution of mass about the axis in question.

I r dm

k I m or I mk

z

z z z z

== =z 2

2/

k represents the distance from the axis at which the mass m of the body would be concentrated to produce the moment of inertia.

Parallel axis theorem: If the moment of inertia of a body about an axis passing through the centre of mass, G, is known, the moment of inertia about any parallel axis can be determined as follows:

where d is the distance between the two axes.

In terms of radius of gyration, k:

I I mdz zG= + 2

k k dG2 2 2= +


Force, mass & acceleration general equations of motionAlternative moment equations.

By applying the familiar principle that the sum of the moments about a point P is equal to thecombined moment about P of their sum:

For a rigid body in plane motion: and the cross product is simply the moment of about P , or: . Therefore:

p GH m= + a maHG I=

ma mad

p I mad= +


Force, mass & acceleration general equations of motionAlternatively, the moment equation can be written as:

p p rel pH m Eqn= + ( ) ( . . )a 413For a rigid body in plane motion, if P is fixed to the body then:

If then P becomes the centre of mass G then

``

p p pI m= + a = 0

p G I= =


Force, mass & acceleration general equations of motion

Interconnected bodies: where kinematic motions are related.

F a a a = m1 1 2 2 + =m m p I I m a d m a d I m ad= + + + = + 1 1 2 2 1 1 1 2 2 2


Force, mass & acceleration Translation

Recall: By definition, translation of a rigid body excludes rotational motion: every line on the body always remains parallel to its original position.

Rectilineal translation: All points move in straight lines Curviliear translation: All points move on identical curved paths.



Consequently, the angular velocity, , and the angular acceleration, , are zero such that:

For rectilinear translation, if the x axis is made

to correspond with the direction of motion,

then the two components of

the force vector become:

F a M= = = m and IG 0

F = a

F = a

x x

y y

m

m

= 0



For curvilinear translation, if we choose the axes to correspond to the n (normal) and t (tangential) cordinates, then the two components of the force vector become:

In both cases,

F = a

F = a

n n

t t

m

m


MG I= = 0


Force, mass & acceleration TranslationVehicle friction example

Slip

After Hill et al. Tyre-road friction estimation


Force, mass & acceleration Fixed axis rotation Recall: For fixed axis rotation all lines of the RB undergo the same angular velocity and the same

angular acceleration , For fixed axis rotation, the aceleration components of the mass centre are best expressed with the n-

t (normal and tangent) coordinates such that

a r and a rn t= = 2


Force, mass & acceleration Fixed axis rotationThe general equations for plane motion may be applied:

F m M IG = a = And the force equation can be divided into its two scalar components:

F r and F rn t = m = m 2


Force, mass & acceleration Fixed axis rotation

For fixed-axis rotation, it is convenient to apply the moment equation to the axis of rotation O:

From the kinetic diagram, the moments of the resultants about O becomes:

Applying the parallel axis theorem gives:

When the axis of rotation corresponds to the centre of mass, G, then

leaving

Mo oI=

Mo tI m a r= + I I m ro = + 2

M I m r m r Io o o= + = ( 2 2)

a therefore F= 0 0 = as the remaining resultant of the applied forces. I



Force, mass & acceleration Fixed axis rotationIf we choose a point Q along the line OG, such that the moments about O are equal:

Point Q is called the Centre of Percussion and is the only point through which the resultant of allforces acting on the body passes. As a concequence, the sum of all moments about Q is alwayszero. Also, an impulsive force acting on the body at Q will minimise the reaction at O. Eg: sweetspot.


M m r r I m r q

I m k

m r m k m r q

r k r q

q r kr

r k k then

q kr

o G

G G

G

G

G

G o

o

substituting for k radius of gyration about G

Applying the parallel axis theorem

G

= + ==

+ =+ =

= +

+=

=

=

( )

( )

:

2

2 2

2 2

2 2

2 2 2

2

Demonstration


Force, mass & acceleration General plane motionRecall: Rigid body dynamics in general plane motion combine translation and rotation.

Again, the general equations for plane motion may be applied:

F m M IG = a =

Sample problems: 6/5, 6/6, 6/7, 6/8


Force, mass & acceleration General plane motionConsiderations when solving plane-motion problems:

Choice of coordinate system: Carefully chose between rectangular (x,y), normal-tangential or polar coordinates such that the acceleration of the centre of mass is easily described.

Choice of moment equation: Careful choice may result in great simplification of the problem: About the centre of mass About a point P whose acceleration is known About the point of rotation whose acceleration is zero

Determine whether the motion of particular elements is constrained or not Determine the number of unknowns and develop the same number of independant equations

V. Rouillard 200354 Dynamics Plane kinetics of rigid bodies: Work & Energy






Work and EnergyIn the previous section, the instantaneous relationship between the forces acting on the body andthe resulting acceleration (linear and angular) was established by applying Newtons second law(F = ma).Velocities and displacement were obtained by integration wrt time.

In this section, we will study the cummulative effect of the forces acting on the body with respect to the displacement of the body.

Integration of force as a function of displacement leads to the solution for Work and Energy.


Work-Energy relationships Forces and Couples

U d= z F rRecall: In general terms the work done by a force F is given by:

U F ds U F dst= = z z cos( )

The effective force is done by the force component actingalong the motion of the particle or body:


Work-Energy relationships Forces and Couples

Work done by couples also need to be evaluated:

When the couple M = Fb acts on a rigid body, the body rotates through an angle d during thetime interval dt and the line AB is displaced to AB.

(Equivalent to translation to AB followed by rotation d about A)When a pure couple is applied, the translational work done by the forces are cancelled and theresulting work done is:

dU F b d M d= =

U M d= z

During a finite rotation, the work done by a couple M

acting in the plane parallel to the plane of rotation is:


Work-Energy relationships Kinetic Energy

In translation (both rectilinear and curvilinear) the kinetic energy of the entire rigid body is simplythe sum of the kinetic energy of all individual particles which make up the rigid body:

T m v v m

or

T m v

i i= =

=

12 2 12 212

2



Correspondingly, during fixed-axis rotation, about O,

the kinetic energy is:

T m r m ri i i i= = 12 2 2 12 2 2 sin ( ) ,ce m r I moment of inertia then

T I

i i o

o

==

2

12

2



For general plane motion, the bodys velocity is a combination of the centre of mass linearvelocity, v and and its angular velocity .The velocity vi of a particle with mass mi is composed of the centre of mass velocity v and therelative velocity i

vi iv



Using the same approach as before, we express the kinetic energy of the rigid body as the sum ofthe kinetic energy of all its constituant particles mi.

We apply the cos rule to resolve the velocity components:

T mv m v vi i i i i= = + + 12 2 12 2 2 2 2( cos ) vi ivFactoring v and we obtain:

T v m m v mi i i i i= + + 12 2 12 2 2 cossin

cos( )

ce

v m v m y myCentre of mass

i i i i = = = 0

Then

T mv I

:

= +12 2 12 2



T mv I= +12 2 12 2 Linear + Angular

This expression reveals the individual contribution of the translational velocity v and the rotationalvelocity to the total kinetic energy of the rigid body.

If the total velocity is expressed as a function of the instantaneous centre of zero velocity, thenthe total kinetic energy is:

T mv IC= +12 2 12 2=0


Work-Energy relationships Potential Energy

In general terms, the total work, U, done on a body consists of:

the kinetic energy, T the gravitational potential energy, Vg and the elastic potential energy (also known as strain energy), Ve

The conservation of energy equation applies to any mechanical system and is written as:

U T V Vg e1 2 = + + When rigid bodies are involved, no deformation occurs and the elastic potential energy termreduces to zero such that:

U T Vg1 2 = +

U accounts for other type of energy such as friction (unrecoverable energy) and spring forces (recoverable in reality partially).


Work-Energy relationships Potential Energy

When solving rigid body problems with the work-energy method, a free-body diagram or an active-force diagram should be used.

Also use diagrams to illustrate the initial and final position of the body(ies) for the given interval ofmotion


Work-Energy relationships Power

Recall: Power is the temporal rate of change of work or the rate at which work is done.

Consider a force F acting on a rigid body in plane motion (2-D) the instantateneous powerdeveloped by that force is written as:

P dUdt

ddt

= = = F r F v



Similarly, when a couple M acts on a body, the power developed by the couple at a given instant isthe temporal rate of change of the work done:

P dUdt

ddt

= = = M M

If the directions of M is the same as that of then thepower is positive and energy is delivered to the body.

Conversely, if the direction of M and are opposite, power is ve and energy is taken out of the body.



When a force F and a couple M act on a body simultaneously, then the total power is:

P M= +F v Power can be obtained from the rate of change of the total mechanical energy of the body. For an infinitesimal displacement 1 2, the power is:

dU dT dV dVg e1 2 = + +The total power of the active forces and couples is obtained by dividing throughout by dt:

P dUdt

dTdt

dVdt

dVdt

g e= = + +1 2

Since the kinetic energy T can be written as a combination of the rectilinear and angularkinetic energies, then:

dTdt

ddt

m I m I= + = + +12 12 2 12v v a v v a e j a f = + = +m I Ma v R v

Where R and M are the resultant of all forces and moment about the mass centre, respectively, acting in the body.

V. Rouillard 200369 Dynamics Plane kinetics of rigid bodies: Impulse & Momentum

Impulse & Momentum

Previous lecture: Focused on work and energy which were obtained by integration the equations of motion with

respect to displacement.

The velocity change was expressed in terms of work done or overall change in mechanical energyof the body.

This lecture: The equations of motion will be integrated wrt time This will generate to the equations for impulse and momentum These are useful for forces acting over short durations - transients and impacts


Impulse & Momentum Linear Impulse & Momentum

Recall, for particles, the basic equation of motion is:

F v v = = =m ddt m G ( ) This is Newtons second law in terms of momentum G.

The effect of the resultant force F on the linear momentumof the particle over a finite period of time can be obtained by integration wrt time:

F v vz = = =t

tdt m m G G G

1

2

1 2 1 2

Change in linear

momentumLinear

impulse



Alternatively:

Wich states that the final linear momentum of the body is equal to the initial linear momentum plus the linearimpulse.

G dt Gt

t

11

2

2+ =z F

As well as for particles, this equation applies to rigidbodies where, in plane motion, it may be useful to express it in scalar form:

G F dt G and G F dt Gx xt

t

x y yt

t

y11

2

2 11

2

2+ = + =z z



The linear impulse momentum equation must take into account all external forces acting on thebody.

It is therefore important to construct a complete free-body diagram when solving impulse & momentum problems

Note: Although work is generated only if the resultant of forces induce motion, impulses occurwhether or not motion (work) is generated by the rigid body.


Impulse & Momentum Angular Impulse & Momentum

As with the linear case, Newtons second law for angular motion can be written as a function ofangular momentum about the bodys centre of mass, HG:

M I I ddt

HG G = = = Integrating wrt time over a finite period we obtain the angular impulse-momentum equation:

M dt I I H H Ht

t

G G Gz = = =1

2

2 1 2 1

Which means that the angular impulse about the centre of mass produced by all forces actingon the body results in a corresponding change in the bodys angular momentum.

Note: This equation is valid only for angular momentum about the centre of mass of the rigidbody.

Hint: Be sure to establish a sign convention (ie: ccw +ve) and assign the direction of allmoments and angular momenta. A complete free-body diagram is essential.


Impulse & Momentum Conservation of Momentum

If there is no external impulse (transient force) acting on the body, momentum is conserved.

Example: when two objects collide, the total (combined) momentum after the collision is equalthe total (combined) momentum before the collision.

Recall for linear momentum:

m mii

n

i ii

n

i =( ) ( )v v1 2Where i represents an object in a group of objects.

As usual, for plane (2D) motion, the equation is applied independently for the suitable mutuallyperpendicular axes: x,y or n,t.

Similarly, for rigid bodies in plane motion, the angular momentum is also conserved:

I Iii

n

i ii

n

i =( ) ( ) 1 2


Impulse & Momentum Impact of rigid bodies

Consider to rigid bodies, A & B, travelling with arbitrary linear andangular velocities vA, vB and A, B respectively.

A

B

P

VA

B

A

VBWe shall assume, reasonably, that, unlike particles, the objects willimpact each other at a point other than their centre of mass.

This will generate moments and result in the spinning of one or bothobjects.

We can expect that the impact will cause the linear and angularvelocities of both objects to change to vA, vB and A, B .Potentially, there will be up to six unknowns as the linear velocities willhave components in the mutually perpendicular x and y directions for plane (2D) motion.



A

B

P

VA

B

A

VB

VBPx

VAPxThe coefficient of restitution, e, is dependent on the properties of thematerials involved.

We also apply the conservation of momentum principle in the linear x and y directions:

x

y

If we carefully chose the orientation of the x and y axes, the coefficient of restitution can be determined from the velocities in the plane perpendicular to the line of impact:

e lative velocity afterlative velocity before

v vv v

BPx APx

BPx APx= =

ReRe

mv mv and mv mvx x y y = =



A

B

P

VA

B

A

VB

VBPx

VAPxThis also applies to any arbitrary (convenient) point, in this case, P.

x

y

rA

rB

Similarly, we can apply the conservation of momentum principle for therotational motion. In general terms, the angular momentum is takenabout the axis of rotation O:

H Ho o =

H HP P = ( ) ( )

( ) ( ) r v r v

r v r vA A A A B B B B

A A A A B B B B

m k I m k I

m k I m k I

+ + + = + + +

Recall: term for angular momentum

of particles

In many cases the problem is simplified when one of the bodies is a particle or is pined about a fixed point.

V. Rouillard 200378 Dynamics 3-D Dynamics of Rigid Bodies





V. Rouillard 200379 Dynamics 3-D Dynamics of Rigid Bodies

Many engineering problem require analysis in 3 dimensions.

This adds considerable complexity to the kinematic & kinetic relationships:

A third vectorial component is introduced to linear quantities Two additional vectorial components are introduced for angular quantities

Vector analysis is essential for solving 3-D problems

V. Rouillard 200380 Dynamics 3-D Kinematics of Rigid Bodies





V. Rouillard 200381 Dynamics 3-D Kinematics of Rigid Bodies: Translation

Recall: For translation motion every line in the body remains parallel to its original position

The motion of any point in the body is described as:

r r rA B A B= + /As rA/B remains costant, its time derivative is zero.

v vA B=

All the points in the body have the same velocity and acceleration:

a aA B=

V. Rouillard 200382 Dynamics 3-D Kinematics of Rigid Bodies: Fixed-Axis Rotation

.

.

Note: For fixed-axis rotation, the direction of the velocity vector, , does not change.Any point A which is not on the axis of rotation has a circular path in a plane normal to the axis of rotation and has a velocity:

v h b h b r= + = + = =

( )0

V. Rouillard 200383 Dynamics 3-D Kinematics of Rigid Bodies: Fixed-Axis Rotation

.

.

The velocity, v, of point A is differentiated wrt time to obtain its acceleration:

a r r= + ( )

a rt b= = a rn b= = ( ) 2

V. Rouillard 200384 Dynamics 3-D Kinematics of Rigid Bodies: Rotation @ Fixed Point

When rotation takes place about a fixed point, the direction of the angular velocityvector changes.

This requires a more general approach to rotation:

Rotation and Proper Vectors

For finite rotations in 3D the parallelogram law of addition does not apply and thevectors may not be treated as proper vectors. Finite rotations


When the rotations in 3D are infinitesimal, the paths (short circular arcs) approach straight lines and they obey the law of vector addition:

.

Infinitesimal rotations


Instantaneous Axis of RotationExample:Imagine a cylindrical rotor made of clear plastic impregnated with numerous black dots.

The cylinder rotates about the horizontal shaft which is, in turn, rotated about a vertical axis.

Both angular velocities are constant.

If the system was photographed at any instant, the exposure would reveal a line (O n) of well-defined black dots (A) which, at that instant, have zero velocity. ie: the velocity components from both rotations will be equal in magnitude but opposite in direction.

All other dots (P) will appear blurred (streaked in circular arcs) indicating that their velocity is not zero.

Angular velocity vector


Body and Space ConesIf a series of photographs were taken, it would reveal that the location of the instantaneous axis of rotation changes both in space and relative to the body.

In this example, with time, the instantaneous axis of rotation follows a circular conical path about the cylinder axis (Body cone).

As the rotational motions progress, the instantaneous axis of rotation also follows a second circular conical path about the vertical axis (Space cone) The body cone appears to roll on the space cone. The angular velocity of the cylinder lies along the shared plane.

Space & body cones


Angular AccelerationRecall: In the case where rotation occurs in a single plane, the angular acceleration, , is a scalar which represents a change in magnitude of the angular velocity.

In 3-D motion, the angular acceleration, , is a vectorwhich represents a change in magnitude and directionof the angular velocity.

The tip of the velocity vector, , follows the space curve, p, and changes both in magnitude and direction,

the angular acceleration, , is a tangent to that curve, p.

.


.

.

r =

=

Angular Acceleration

When the magnitude of remains constant, is perpendicular to .If we let be the angular velocity of the rotation (precession) of the vector as it follows the space cone, then:

=


.

.

If we consider n-n to be the instantaneous axis of rotation, the expressions for thevelocity and acceleration of any point A are given by the same vectorialexpressions as for fixed-axis rotation:

v r= a r r= + ( )

a rt b= = a rn b= = ( ) 2


The only difference between fixed-axis rotation and fixed-point rotation is as follows:

With fixed axis rotation, the angular acceleration = has a single component along thefixed axis due to the change in the magnitude of

With fixed-point rotation the angular acceleration of any point A has two components:1. In the direction of due to change in the magnitude of and2. Perpendicular to due to change in the direction (precesssion) of

Although a point on the rotation axis will have zero velocity, it willnot have zero acceleration due to the change in direction of .

V. Rouillard 200392 Dynamics 3-D Kinematics of Rigid Bodies: General Motion

The principles of relative motion are useful when solvingproblems in 3-D.

Translating Reference Axes:We select a convenient point B as the origin of a translatingreference system x-y-z.

The velocity v and acceleration a of any other point on the body (A in this case) are given by the relative velocity and relative acceleration formulae:

These equations, developed for the 2D case, also apply to 3D situations.

In 3-D, observing from B, the body appears to rotate about B and point A appears to lie on a spherical surface of which B is the centre.

Therefore, the general motion is the translation of B (thus the body) + the 3-D rotation of thebody about point B.

v v v a a aA B A/ B A A/ B= + = +and B


The relative motion terms are identical to the velocity andacceleration expressions developed previously. Therefore thevelocity at A becomes:

Although the reference point B can be chosen arbitrarily, it is ofenthe case that this point coincides with the centre of mass of thebody when solving kinetic problems.

v v v v rA B A/ B B= + = + A B/

And the acceleration at A becomes:

A B A B B A / B A / B( ) + = + = +/ r ra a a a


Rotating Reference Axes:For more general cases, the reference axes need to rotateand translate.

As before, the origin of the axes of reference x-y-z, B, istranslating. In addition, the references axes are also madeto rotate with an absolute angular velocity while the body rotates with an angular velocity .

i i j j k k= = =

v v r vA B rel= + + A B/

a a r r v aA B A B A B rel rel= + + + + / /( ) 2

V. Rouillard 200396 Dynamics 3-D Kinetics of Rigid Bodies: Inertia

Principal Axes.The moments and products of inertia can be used to completely characterise the inertialproperties of a 3D body as follows:

I I II I II I I

xx xy xz

yx yy yz

zx zy zz

F

HGGG

I

KJJJ

This matrix is called the inertia tensor (matrix) and has a unique set of values for a body for eachlocation of the point of origin O.

As the orientation of the axes relative to the body is changed, the values for the moments andproducts of inertia will aslo vary.

For a given point of origin there exists a unique orientation of the x,y,z for which the products of inertiavanish and the inertia matrix reduces to:

II

I

xx

yy

zz

0 00 00 0

F

HGG

I

KJJ

In this case Ixx, Iyy and Izz are referred to as the principal moments of inertia where two of thevalues will be the minimum and maximum moments of inertia for the body.


Moments and Products of InertiaRecall: In planar kinetic analysis of rigid bodies, the moment of inertia, IG, about an axis perpendicular to the plane of motion (rotation) thru the centre of mass needed to be calculated.

In 3-D, it is sometimes necessary to calculate six inertial quantities, called the moments andproducts of inertia, which describe the distribution of mass of a body relative to a given coordinatesystem which has a specific orientation and origin.

x

z

Moment of inertia:For the body shown, the moment of inertia of thedifferential element dm of the body about one of thethree coordinate axes is defined as the product of themass and the square of the shortest distance form theaxis to the element. Eg: about the x axis:

dI r dm y z dmxx x= = +2 2 2e j


Ixx is obtained by integrating over the entire mass of the body :

I r dm y z dmxx xm m

= = +z z2 2 2e jSimilarly:

I r dm x z dmyy ym m

= = +z z2 2 2e j

As expected, moments of inertia are always positive quantities.

I r dm x y dmzz zm m

= = +z z2 2 2e jx

z


Products of inertia:The product of inertia of a differential element dm is defined wrt a pair of ortogonal planes as theproduct of the mass of the element and the perpendicular (shortest) distances from the planes to the element. Eg:, wrt the x-z and y-z planes:

dI xy dmxy =dI dIxy yx=Note:

The products of inertia for every combination of planes isobtained by integrating over the entire mass of the body and are expressed as:

I I xy dm

I I yz dm

I I xz dm

xy yxm

yz zym

xz zxm

= =

= =

= =

zzz x

z


I I xy dm I I yz dm I I xz dmxy yxm

yz zym

xz zxm

= = = = = =z z zProducts of inertia can be negative or zero. If one or both ortogonal planes are planes ofsymetry for the body, the product of inertia with respect to these planes will be zero. In suchcases the elements of mass occur in pairs on each side of the plane of symetry. On one side ofthe plane the PoI will be negative and on the other positive. Example:

Left: the y-z plane is a plane of symetry; for point O, Ixz = Ixy= 0 and Iyz= +ve. Right: the x-z and y-z plane are planes of symetry; for point O, Ixz = Ixy= Iyz= 0

x

y

z

xy

z


Parallel axis and parallel plane theorems.Recall: the // axis theorem is used to transfer the moment of inertia of a body about an axis thru its centre of mass to a parallel axis passing thru another point.

I I m y z

I I m x z

I I m x y

xx x x G G G

yy y y G G G

xx z z G G G

= + += + += + +

( ) ( )

( ) ( )

( ) ( )

2 2

2 2

2 2

If G has the coordinates xG, yG, zG then the // axis relationships used to calculate the moments of inertiaabout x, y and z are given by:

x

z

y

x

z

y


The parallel plane theorem is used to transfer the products of inertia of a body from a set ofthree orthogonal planes passing thru the bodys centre of mass to a corresponding set of threeparallel planes passing thru some other point O.

I I mx y

I I my z

I I mz x

xy x y G G G

yz y z G G G

zx z x G G G

= += += +

( )

( )

( ) x

z

y

x

z

y

Defining the perpendicular distances between theplanes as xG, yG, zG , the parallel plane equations are given by:

V. Rouillard 2003103 Dynamics 3-D Kinetics of Rigid Bodies: Angular Momentum

Recall:The angular momentum of a rigid body about its centre of mass, G, is the sum of the moments about G of the linear momenta of all elements in the body:

H mG i i i= vwhere vi is the absolute velocity of mass element mi.For the 3D rigid body:

v vi i= + where x i is the relative velocity of mass element mi wrt G as seen from the main axes X Y Z.

The angular momentum equation may be written as:

H mG i i i i i= + m v ( )Since mii = mimean = 0 and substituting dm for mi and for i:

HG dm= z ( )


When the 3D rigid body rotates about a fixed point, O, with an angular velocity , the angular momentum about O

H mO i i i= r vwhere vi = x ri .When substituting dm for mi and r (distance b/w centre of mass andthe fixed point O) for ri the angular momentum about O is:

HO dm= z r r( )


Note that both HG and HO take the form:

H dm= z ( )Expressing H, and in terms of x,y and z components gives:

H H H x y z x y zx y z x y z dmi j k i j k i j k i j k+ + = + + + + + +z b g c h b g Expanding the cross product, combining the respective i, j and k components and recognising thatthe integrals represent the moments and products of inertia, we obtain:

H I I I

H I I I

H I I I

x xx x xy y xz z

y yx x yy y yz z

z zx x zy y zz z

= = + = +

H i

j

k

= + + + +

I I I

I I I

I I I

xx x xy y xz z

yx x yy y yz z

zx x zy y zz z

c hc hc h

or

If the orientation of the x,y and z axes were such that they represented the principal axes ofinertia, the products of inertia Ixy = Iyz = Izx = 0 and the three components of the angularmomentum become:

H I H I H Ix xx x y yy y z zz z= = =


Angular momentum transfer principle:The angular momentum of a rigid body can be represented by thesum of two resultant momenta:

The linear momentum vector G thru the centre of mass The angular momentum H about the centre of mass

H H GP G= + r

The vectors G and H are analogous to a force and a couple

V. Rouillard 2003107 Dynamics 3-D Kinetics of Rigid Bodies: Kinetic Energy

Recall:The general expression for kinetic energy:

T m mi i= + 12 2 12 2v Translational + Angular

Where v is the velocity of the centre of mass and i is the position vector of a representative masselement of mass mi wrt the centre of mass.

The translational term may be written in terms of linear momentum G as follows:

12

2 12

12m mv r r v G= =


In the angular term, = Where is the angular velocity of the body. The angular term in the kinetic energy eqn. becomes:

12

2 12m mi i i ( ) ( ) =

Since P x QR = PQ x R then:

= b g b g b giFactoring out:

12

2 12

12m mi i i i G = = b g H

The general expression for kinetic energy becomes:

T G= + 12 12v G H


H i j kG xx x xy y xz z yx x yy y xz z zx x zy y zz zI I I I I I I I I= + + + + c h c h c hSubstituting for

The vectorised form for the kinetic energy becomes:

( ) ( ) = + + + + +2 2 21 12 2 xx x yy y zz z xy x y xz x z yz y zT I I I I I Iv GWhen the axes coincide with the principal axes of inertia, the kinetic energy reduces to:

( ) = + + +2 2 21 12 2 xx x yy y zz zT I I Iv GWhen the body rotates about a fixed point O or if there is a point O which momentarily has zerovelocity then:

= + = 1 1 12 2 2O OT v G H H =0





Introductory 3-D dynamics of rigid bodies 3-D kinematics of rigid bodies 3-D kinetics of rigid bodies (Groscopic Motion)

V. Rouillard 2003111 Dynamics 3-D Kinetics of Rigid Bodies: Gyroscopic Motion

Introduction:Gyroscopic motion occurs whenever the axis about which a body is spinning is rotating about another axis.

The complete description of this phenomenon is very complex

The phenomenon is significantly simplified when the spinning and precession velocities are constant.

Applications: Inertial guidance systems (when gyroscope is mounted in gimbal

rings: The spinning disk is free from external moments and willmaintain its orientation in space.

Gyro-compass: When a suspended mass is introduced, theearths rotation causes precession such that the rotation of spin axis stays parallel to the earths axis of rotation (North South)

Gyro-stabiliser: Large gyro mounted in ships to counteract rollingmotion

Gyroscoping motion is important when designing (large) bearingswhich undergo forced precession.

V. Rouillard 2003114

When a couple M (about the x axis) is applied to the rotor with a spin velocity p, the rotor shaft rotates about the y axis with the direction as shown!

The rotor shaft does not rotate with M as it would if therotor was at rest.

This angular velocity (precession) is small relative to thespin velocity p.The direction of rotation of the spin axis (p) the torque axis (M) and the precession axis () conform with the righthand rule.

z


The rotating vectors of gyroscopic motion can be explained by drawing an analogy with the vectorsused to describe the curvilinear motion of particles.

Fy

Consider a mass particle m with a constant speed vthru the x-y plane.

When a force F is applied perpendicular to its linearmomentum G, it results in a change in momentum dG=d(mv)= mdvdG, hence dv is a vector in the direction of force F which can be written as:

F G F G= = or dt dIn the limit (dt 0), tan(d) = d = Fdt/mv or F=mv(d/dt)In vector notation:

F v= =m ( )F man n


Applying the same approach to a rotating body.

For a high spin velocity p and relatively lowprecession velocity about y, the angularmomentum is:

M H=

H p= Izz

As the precession velicity is low relative to the spin velocity, its angular momentum can beneglected.

The moment M applied perpendicular to the angular momentum vector H causes a change dH (=dIzzp) in the angular momentum vector.As with the case with a particle, the the change in the angular momentum of a gyroscope isin the direction of the moment M.Vectors M, H and dH are analogous to vectors F, G and dGAs the rotation vector p (z axis) is rotated in the direction of M (x axis), it is not suprising to observe the rotor axis to precess about the y axis (due to the x product of the vectors)


During a time interval dt, the angular momentum vector Iphas undergone a directional change d.As before, in the limit, when dt 0, tan(d) = d Thus, in scalar form:

d Mdtp

M d p = =I

or Idtxx

xx

Since d /dt = then:M p= Ixx

As M, and P are mutually perpendicular vectors, the equation for gyroscopic motion can be written in vector forms as:

M p= I


Since the change in angular momentum dH, thus dp is always in thedirection of M, the spin vector p will always tend to re-orientate toward thetorque vector M.

When a rotor is forced to precess, say as in a ship turbine when the shipis changing direction, or a single propeller aeroplane during take-off, a gyroscopic couple M will be generated the magnitude and direction ofwhich will be in accordance with:

Note that M represents the moment resulting from all forces acting on the rotor.

M p= I


So far it has been assumed that the precession velocity was small relative to the spin velocityWe will now study how the magnitude of affects the nomentum relationshipsThe precession velocity will be assumed constant.

Since the rotor precesses about the y axis and it has a moment of inertia about this axis, there existsan additional angular momentum component about the y axis. (This was previously neglected due to


Precession non-perpendicular to spin:When a symetrical top with a large spinning velocity p is supported at O, the spin axis is at an angle with the vertical axis z about which precession occurs.The small angular momentum due to precession is neglected so that the total angular momentum is that associated with spin:

The moment about O is due to the gravitationalacceleration acting on the tops centre of mass:

M g )O = m r sin(Due to precession, during an interval dt, the angularmomentum vector HO undergoes a change in thedirection of MO (toward the x axis)

H p= Izz

M H H= =ddt

or d M dtO O O


The incremental angle of precession around the z axis is:

Substituting for MO = mgrsin() and = d/dt:

= grk p2

Since I = mk2 , the precession velocity isgiven by:

d M dtI p

O = sin( )

mgr I por

mgr I p

sin( ) sin =

=

a f

Which is independent of the top angle .This equation is an approximation based on theassumption that the angular momentum associatedwith precession is negligible compared to thatassociated with spin.


Based on this analysis, the top will precess at a constant angle only if the precssion velocity is:

= grk p2

When this exact condition is not met, precession becomes unsteady and mayoscillate.

As the spin velocity decreases, theamplitude of oscillation of tends to increase.

This oscillation of the spin axis is known as nutation.

Dynamics IntroductionDynamics IntroductionDynamics IntroductionDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid bodiesDynamics Plane kinematics of rigid 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AccelerationDynamics Plane kinetics of rigid bodies Force, Mass & AccelerationDynamics Plane kinetics of rigid bodies Force, Mass & AccelerationDynamics Plane kinetics of rigid bodies Force, Mass & AccelerationDynamics Plane kinetics of rigid bodies Force, Mass & AccelerationDynamics Plane kinetics of rigid bodies Force, Mass & AccelerationDynamics Plane kinetics of rigid bodies Force, Mass & AccelerationDynamics Plane kinetics of rigid bodies Force, Mass & AccelerationDynamics Plane kinetics of rigid bodies Force, Mass & AccelerationDynamics Plane kinetics of rigid bodies Force, Mass & AccelerationDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodies: Work & EnergyDynamics Plane kinetics of rigid bodiesDynamics Plane kinetics of rigid bodies: Impulse & MomentumDynamics Plane kinetics of rigid bodies: Impulse & MomentumDynamics Plane kinetics of rigid bodies: Impulse & MomentumDynamics Plane kinetics of rigid bodies: Impulse & MomentumDynamics Plane kinetics of rigid bodies: Impulse & MomentumDynamics Plane kinetics of rigid bodies: Impulse & MomentumDynamics Plane kinetics of rigid bodies: Impulse & MomentumDynamics Plane kinetics of rigid bodies: Impulse & MomentumDynamics Plane kinetics of rigid bodies: Impulse & MomentumDynamics 3-D Dynamics of Rigid BodiesDynamics 3-D Dynamics of Rigid BodiesDynamics 3-D Kinematics of Rigid BodiesDynamics 3-D Kinematics of Rigid Bodies: TranslationDynamics 3-D Kinematics of Rigid Bodies: Fixed-Axis RotationDynamics 3-D Kinematics of Rigid Bodies: Fixed-Axis RotationDynamics 3-D Kinematics of Rigid Bodies: Rotation @ Fixed PointDynamics 3-D Kinematics of Rigid Bodies: Rotation @ Fixed PointDynamics 3-D Kinematics of Rigid Bodies: Rotation @ Fixed PointDynamics 3-D Kinematics of Rigid Bodies: Rotation @ Fixed PointDynamics 3-D Kinematics of Rigid Bodies: Rotation @ Fixed PointDynamics 3-D Kinematics of Rigid Bodies: Rotation @ Fixed PointDynamics 3-D Kinematics of Rigid Bodies: Rotation @ Fixed PointDynamics 3-D Kinematics of Rigid Bodies: Rotation @ Fixed PointDynamics 3-D Kinematics of Rigid Bodies: General MotionDynamics 3-D Kinematics of Rigid Bodies: General MotionDynamics 3-D Kinematics of Rigid Bodies: General MotionDynamics 3-D Kinematics of Rigid BodiesDynamics 3-D Kinetics of Rigid Bodies: InertiaDynamics 3-D Kinetics of Rigid Bodies: InertiaDynamics 3-D Kinetics of Rigid Bodies: InertiaDynamics 3-D Kinetics of Rigid Bodies: InertiaDynamics 3-D Kinetics of Rigid Bodies: InertiaDynamics 3-D Kinetics of Rigid Bodies: InertiaDynamics 3-D Kinetics of Rigid Bodies: InertiaDynamics 3-D Kinetics of Rigid Bodies: Angular MomentumDynamics 3-D Kinetics of Rigid Bodies: Angular MomentumDynamics 3-D Kinetics of Rigid Bodies: Angular MomentumDynamics 3-D Kinetics of Rigid Bodies: Angular MomentumDynamics 3-D Kinetics of Rigid Bodies: Kinetic EnergyDynamics 3-D Kinetics of Rigid Bodies: Kinetic EnergyDynamics 3-D Kinetics of Rigid Bodies: Kinetic EnergyDynamics 3-D Kinematics of Rigid BodiesDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic MotionDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic MotionDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic MotionDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic MotionDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic MotionDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic MotionDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic MotionDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic MotionDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic MotionDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic MotionDynamics 3-D Kinetics of Rigid Bodies: Gyroscopic Motion

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Dynamics of Rigid Bodies Merian Book 2005