Rigid Body Dynamics

Rigid Body Dynamics K. Craig 1

Rigid Body Dynamics:Kinematics and Kinetics


Topics

Introduction to Dynamics Basic Concepts Problem Solving Procedure Kinematics of a Rigid Body

Essential Example Problem Kinetics of a Rigid Body

Supplement: Rigid Body Plane Kinetics Essential Example Problem


Introduction

Dynamics The branch of mechanics that deals with the motion

of bodies under the action of forces. Newtonian Dynamics

This is the study of the motion of objects that travel with speeds significantly less than the speed of light.

Here we deal with the motion of objects on a macroscopic scale.

Relativistic Dynamics This is the study of motion of objects that travel with speeds

at or near the speed of light. Here we deal with the motion of objects on a microscopic or

submicroscopic scale.


Newtonian Dynamics Kinematics

This is the study of the geometry of motion. It describes the motion of bodies without reference to the forces which either cause the motion or are generated as a result of the motion. It is used to relate position, velocity, acceleration, and time without reference to the cause of the motion.

Kinetics This is the study of the relation existing between

the forces acting on a body, the mass distribution of the body, and the motion of the body. It is used to predict the motion caused by given forces or to determine the forces required to produce a given motion.


Basic Concepts

Space Space is the geometric region occupied by

bodies. Position in space is determined relative to some geometric reference system by means of linear and angular measurements.

The basic frame of reference (perspective from which observations are made) for the laws of Newtonian mechanics is the primary inertial system which is an imaginary set of rectangular axes assumed to have no translation or rotation in space.


Measurements show that the laws of Newtonian mechanics are valid for this reference system as long as any velocities involved are negligible compared with the speed of light (186,000 miles per second). Measurements made with respect to this reference system are said to absolute.

A reference frame attached to the surface of the earth has a somewhat complicated motion in the primary system, and a correction to the basic equations of mechanics must be applied for measurements made relative to the earths reference frame.


In the calculation of rocket- and space-flight trajectories, the absolute motion of the earth becomes an important parameter. For most engineering problems of machines and structures which remain on the earths surface, the corrections are extremely small and may be neglected. For these problems, the laws of Newtonian mechanics may be applied directly for measurements made relative to the earth, and, in a practical sense, such measurements will be referred to as absolute.

Time Time is the measure of the succession of events

and is considered an absolute quantity in Newtonian mechanics.


Mass Mass is the quantitative measure of the inertia or

resistance to change in motion of a body. It is also the property which gives rise to gravitational attraction and acceleration. In Newtonian mechanics, mass is constant.

Newtons Law of Universal Gravitation The force of attraction between two bodies of

mass M and m, respectively, separated by a distance r, is given by:

M mr

re

311

r2 2GMm mF e G 6.673 10

r kg s= =

G


Mass Moment of Inertia The mass moment of inertia of a rigid body is a

constant property of a body and is a measure of the radial distribution of the bodys mass with respect to an axis through some point. It represents the bodys resistance to change in angular motion about the axis through the point.

Force Force is the vector action of one body on another.

There are two types of forces in Newtonian mechanics: Direct contact forces between two bodies. Forces which act at a distance without physical

contact, of which there are only two: gravitational and electromagnetic.


Particle A particle is a body of negligible dimensions. Also,

when the dimensions of a body are irrelevant to the description of its motion or the action of the forces acting on it, the body may be treated as a particle. It can also be defined as a rigid body that does not rotate.

Rigid Body A rigid body is a body whose changes in shape are

negligible compared with the overall dimensions of the body or with the changes in position of the body as a whole.

Coordinate A coordinate is a quantity which specifies position. Any

convenient measure of displacement can be used as a coordinate.


Degrees of Freedom This is the number of independent coordinates

needed to completely describe the motion of a mechanical system. This is a characteristic of the system itself and does not depend upon the set of coordinates chosen.

Constraint A constraint is a limitation to motion. If the number of

coordinates is greater than the number of degrees of freedom, there must be enough equations of constraint to make up the difference.

Generalized Coordinates These are a set of coordinates which describe

general motion and recognize constraint.


A set of coordinates is called independent when all but one of the coordinates are fixed, there still remains a range of values for that one coordinate which corresponds to a range of admissible configurations. A set of coordinates is called complete if their values corresponding to an arbitrary admissible configuration of the system are sufficient to locate all parts of the system. Hence, generalized coordinates are complete and independent.

Newtons Laws of Motion (for a particle) 1st Law: Every particle continues in its state of rest or

of uniform motion in a straight line unless compelled to change that state by forces acting on it. That is, the velocity of a particle can only be changed by the application of a force.


2nd Law: The time rate of change of the linear momentum of a particle is proportional to the resultant force (sum of all forces) acting upon it and occurs in the direction in which the resultant force acts.

3rd Law: To every action there is an equal and opposite reaction, i.e., the mutual forces of two bodies acting upon each other are equal in magnitude, opposite in direction, and collinear.

These laws have been verified by countless physical measurements. The first two laws hold for measurements made in an absolute frame of reference, but are subject to some correction when the motion is measured relative to a reference system having acceleration.


Units SI Units

The primary dimensions are: mass, M, length, L, and time, T.

The units are: mass (kg), length (m), and time (sec).

This is an absolute set of units based on mass, which is invariant.

Force, F, has dimensions of ML/T2 with the unit newton (N).

2kg m1 N 1

s=


US Customary Units The primary dimensions are: force, F, length,

L, and time, T. The units are: force (lb), length (ft), and time

(sec) This is a relative set of units dependent upon

the local force of gravitational attraction. Mass, M, has dimensions FT2/L with the unit

slug.2lb s1 slug 1

ft=


When close to the surface of the earth, g = 9.81 m/s2 in SI units, and g = 32.2 ft/s2 in US Customary units.

Some useful conversions:

Weight = mg = magnitude of the gravitational force acting on mass m near the surface of the earth.

1 ft 0.3048 m1 lb 4.448 N1 slug 14.59 kg

===


Scalar A scalar is any quantity that is expressible as a

real number. Vector

A vector is any quantity that has both magnitude and direction.

Because the study of Newtonian mechanics focuses on the motion of objects in three-dimensional space, we are interested in three-dimensional vectors.

A unit vector has a magnitude of unity.


There are three types of vectors: Free Vector: no specified line of action or point

of application Sliding Vector: specified line of action but no

specified point of application Bound Vector: specified line of action and

specified point of application. A bound vector is unique, i.e., only one vector can have a specified direction, magnitude, line of action, and origin.

Note that vector algebra is valid only for free vectors. Consequently, the result of any algebraic operation on vectors, regardless of the type of vector, results in a free vector.


Matrices An array of numbers arranged in rows and

columns is called a matrix. A m n matrix has m rows and n columns. Our use of matrices will initially be restricted to

coordinate transformations and later to the concept of the inertia matrix.

1

1

1

i i1 0 0 j 0 cos sin j 0 sin cosk k

=

1

1

1

i i j cos j sin k k sin j cos k

== + = +


Notation and Reference Frames A reference frame is a perspective from which

observations are made regarding the motion of a system.

A moving body, such as an automobile or airplane, frequently provides a useful reference frame for our observations of motion. Even when we are not moving, it is often easier to describe the motion of a point by reference to a moving object. This is the case for many common machines, such as linkages.

An engineer needs to be able to correlate observations of position, velocity, and acceleration of points on moving bodies, as well as the angular velocities and angular accelerations of those moving bodies, from both fixed and moving reference frames.


Reference Frames and Notation

Reference Frames:R ground: xyzR1 shaft: x1y1z1R2 disk: x2y2z2

x1

y1x2

y2

O

z1

y

z

y1

O

1

1

1


=

2 1

2 1

2 1

i icos sin 0 j sin cos 0 j 0 0 1k k

=

Pa ????G1

2

R

R

????????

GG

Meaningless!

1

1 2

RR

R R

GGR PaG

Proper Notation


1 1

1 1

2 2

2 2

i unit vector in x direction

j unit vector in y direction

i unit vector in x direction

j unit vector in y direction

====

2 1 1

2 1 1

i cos i sin j j sin i cos j

= + = +

1 2 2

1 2 2

i cos i sin j j sin i cos j

= = +

2 1

2 1

i icos sin sin cosj j

= 1 2

1 2

i icos sin sin cosj j

=


Procedure for theSolution of Engineering Problems

GIVEN State briefly and concisely (in your own words) the information given.

FIND State the information that you have to find. DIAGRAM A drawing showing all quantities involved

should be included. BASIC LAWS Give appropriate mathematical

formulation of the basic laws that you consider necessary to solve the problem.

ASSUMPTIONS List the simplifying assumptions that you feel are appropriate in the problem.


ANALYSIS Carry through the analysis to the point where it is appropriate to substitute numerical values.

NUMBERS Substitute numerical values (using a consistent set of units) to obtain a numerical answer. The significant figures in the answer should be consistent with the given data.

CHECK Check the answer and the assumptions made in the solution to make sure they are reasonable. Check the units, if appropriate.

LABEL Label the answer (e.g., underline it or enclose it in a box).


Kinematics of a Rigid Body

Angular Velocity of a Rigid Body Differentiation of a Vector in Two Reference

Frames Addition Theorem for Angular Velocities Angular Acceleration of a Rigid Body Reference Frame Transformations Velocity and Acceleration of a Point Coriolis Acceleration and Centripetal Acceleration Essential Example Problem


Angular Velocity of a Rigid Body R is the ground reference frame

with coordinate axes xyz fixed in R R1 reference frame is a rigid body

moving in reference frame R with coordinate axes x1y1z1 fixed in R1

is any vector fixed in R1 form a right-handed set of

mutually perpendicular unit vectors fixed in R1

Angular velocity is the time rate of change of orientation of the body. It is not in general equal to the derivative of any single vector.

y

z O xR

x1

y1

z1

R1A

GG

1 1 1 i j k

1

RRRd

dt = G GG

Defining equation for


Simple Angular Velocity of a Rigid Body When a rigid body R1 moves in a reference frame

R in such a way that there exists throughout some time interval a unit vector whose orientation in both R1 and R is independent of the time, then rigid body R1 is said to have simple angular velocity in R throughout this time interval.

For example: 1 1R RR R

1

k kangular speed of R in R

= = =G

Here R1 has simple angular velocity in R (1) and R2 has simple angular velocity in R1

(2). R2 does not have simple angular velocity in R.


Differentiation of a Vector in Two Reference Frames If R and R1 are any two reference frames, the first

time derivatives of any vector (not fixed in either R or R1) in R and in R1 are related to each other as follows:

y

z O xR

x1

y1

z1

R1A

G

11

RRRRd d ( )

dt dt = + G G GG


Addition Theorem for Angular Velocities Consider multiple reference frames: R1, R2, , RN The following relation applies, whether the angular

velocities are simple or not:

There exists at any one instant only one Also

This addition theorem is very powerful as it allows one to develop an expression for a complicated angular velocity by using intermediate reference frames, real or fictitious, that have simple-angular-velocity relations between each of them.

N N 1 N1 1 2R R RR R RR R = + + + G G G G"NRRG

N NR RR R = G G


2 1 1 2R R R RR R = + G G G

simple angular velocity

NOT simple angular velocity


Angular Acceleration The angular

acceleration of reference frame R1 in reference frame R is given by:

There is no addition theorem for angular accelerations.

When R1 has simple angular velocity in R, e.g.,

y

z O xR

x1

y1

z1

R1A

11 11

R RR RR RRR d d

dt dt = =G GG

1 1

1 1

R RR R

R RR R

k

k

= =

= = =

GG

1 k k=


Reference Frame Transformations

R1 x1y1z1R2 x2y2z2

1 2 1 2 1 2

1 2

R R R R R R1 2

R R

k k = = = G

1 1 2 2x 1 y 1 x 2 y 2 V V i V j V i V j= + = +G

1R dVdt

GWhat Is ?

Define:

2 1

2 1

2 1

i cos sin 0 i j sin cos 0 j k 0 0 1 k

=

1 2

1 2

1 2

i cos sin 0 i j sin cos 0 j k 0 0 1 k

=


1 1

1 1 1 1

R R

x 1 y 1 x 1 y 1 V i V j VdV d

dt dti V j+ + = =

G

2

1 1

2

2

1 1

1

22

1

1

1

1

x 2 y2 2

x 2 y2 2

x 2 y2 2 1

x y

R

2

x y 2

1 x 1 y 1

R

RR R

V i V j

V i V j

V i V j ( k V)

d V cos V sin i

dV ddt dt

d (

dtd V sin V cos jdt

k (V i V j )

V)dt

+

++ +

= + + + +

= = +

+

=

G

GGG

1 1

1

i and j are fixed in R

One Approach

Another Approach

Are the two approachesequivalent?

2 2

2

i and j are fixed in R


1 1 1 1

1 1 1 1

1 1

1 1

1 1

x x y y 2

x x y y 2

x 1 y 1

x 2 2 y 2 2

x 2 2 y 2

V cos V sin V sin V cos i

V sin V cos V cos V sin j

V j V i

V cos i sin j V sin i cos j

V sin i cos j V cos i

= + + + + +

= + + + +

1 1

1 1 1 1 1 1

1 1

2

x 1 y 1

x 1 y 1 x 1 y 1 x 1 y 1

x 1 y 1

sin j

V j V i

V i V j V j V i V j V i V i V j

+

= + + + = +

Same Result !


Velocity and Acceleration of a Point The solution of nearly every

problem in dynamics requires the formulation of expressions for velocities and accelerations of points of a system under consideration. y

z O

P

xR

x1

y1

z1

R1A

Reference FramesR - Ground xyzR1 - Body x1y1z1

( )1

1

1

1

1 1

R P

R RR RR

R R

RR A

AP

R P

P R A

P

2 v

a r

a

a

r

= ++

++

G GG

G G

GGG

G G

( )1 1R RR P R A R AP Pv v r v= + +G GG G G

Relative AccelerationCentripetal AccelerationTangential Acceleration

Coriolis Acceleration


DerivationOP OA AP

RR P OP

R ROA AP

R1R A AP R R1 AP

R A R1 P R R1 AP

r r rdv (r )

dtd d(r ) (r )

dt dtdv (r ) ( r )

dtv v ( r )

= +=

= +

= + + = + +

G G G

GG

G G

G G GGG GG G

y

z O

P

xR

x1

y1

z1

R1 A

OArG OPrG

APrG

( )1 1R RR P R A R AP Pv v r v= + +G GG G G


( )1 11 1 1 1

R RR P R A R R AP

R R R RR AP P R P

a a r

r a 2 v

= + + + +

G G G G GG G G G G

( )1 1R R R RR P R P R A R AP Pd da ( v ) v r vdt dt = = + + G G GG G GR

R A R Ad ( v ) adt

= GG

( )1 1 11 1 1

R RR R RR AP R AP R AP

R R RR AP R R1 P R AP

d d r ( r ) ( r )dt dt

( r ) [ v ( r )]

= + = + +

G G G G G GG G G G GG

1 1 1 1 1 1 1

R R1R R R R R R RP P R P P R Pd d( v ) ( v ) ( v ) a ( v )

dt dt= + = + G G GG G G G


Anatomy of Coriolis and Centripetal Acceleration Situation: A turntable, with its center pivot O fixed to

ground, is rotating clockwise at a constant angular rate. An ant is at point P on the turntable walking at a constant speed v, relative to the turntable, towards some food at point 2.

What is the absolute acceleration of the ant? R PaG


1 1

1 1 1 1

R RR P R O R R OP

R R R RR OP P R P

R R1 R R1 R R11

21

1

1

1 1 1

a a ( r )

r a 2 v

k ( k rj ) 2 k

2 vr

j

ij

v

= + + + +

=

+ = +

G G G G GG G G G G

= Centripetal Acceleration + Coriolis Acceleration


The approximate acceleration of the ant with respect to the R reference frame is the difference between its velocity at points 2 and 1 divided by t. Then we take the limit as t 0. The result is:

Acceleration in the y direction:

R PaG

[ ] [ ]radial2 2 2

2radialradial t 0

v 2 4 v vcos (r r)sin v

v r t v( t) v

va lim rt

= + = + =

= =

cos 1sin

Centripetal Acceleration due to term 4 v has no effect on aradial depends on ants position


Acceleration in the x direction:

[ ] [ ][ ]

tangential

tangentialtangential t 0

v 1 3 r vsin (r r)cos r

v t r v t rv

a lim v v 2 vt

= + = + + = + +

= = + = cos 1sin

CoriolisAcceleration

independent of ants position effect of on v (term 1 ) is always

exactly the same as the effect of v on (term 3 ).

effect of changing the orientation of v is exactly the same as the effect of v carrying r to a different radius, changing its magnitude.


Rigid Body 3D Kinematics Example


R

R1 R2O = 30

r = 0.06 m

Rigid Body Kinematics Essential Example

Given:

Find:

Reference Frames:R ground: xyzR1 shaft: x1y1z1R2 disk: x2y2z2

x1

y1x2

y2

O

z1

y

z

y1

O

1

1 2

RR

R R1

5i constant4k constant

= = = =GG

R PaG

1

1

1


=

( ) ( )OP 1 1 r r cos i r sin j= + G


( )2 2 22 2 2

R R RR P R O R R OP R OP

R R RP R P

a a r r

a 2 v

= + + + +

G G G G G G GG G G

2

2

R O

R P

R P

a 0a 0v 0

===

GGG

Point O at end of rotating shaft is fixed in R

Point P is fixed in R2 (disk)

( )

( )( )( )

2 1 1 2

22

1

R R R RR R1

RR R RRR

1

RRR1

1

1 1 1

5i 4k

d d 5i 4kdt dt

dk 0 4 4 kdt

4 5i k 20j

20 jcos k sin

= + = + = = +

= + = = = = +

G G GGGG G


After Substitution and Simplification:

( ) ( ) ( )R P 1 1 1 a 16rcos i 41rsin j 40r cos k= + + GAlternate Solution:

( )1 1 11 1 1

R R RR P R O R R OP R OP

R R RP R P

a a r r

a 2 v

= + + + +

G G G G G G GG G G

1

11

R O

RR

RR RRR

a 05i constantd 0dt

= = =

= =

GG

GG( ) ( )OP 1 1 r r cos i rsin j= + G


( )1 1 1 2 1 2 1 2R R R R R R R RP O OP OPa a r r = + + G G G G G G G(P is fixed in R2)

( )

1

1 2

1 1 2 11 2

1 1 1 2

1

R O

R R1

R R R RR R

1

R R R RP O OP

R O

a 0 4k

d d 4k 0dt dt

v v r

v 0

= =

= = = = + =

GG

GGG GG G

G ( ) ( )OP 1 1 r r cos i rsin j= + GAfter Substitution and Simplification:

( ) ( ) ( )R P 1 1 1 a 16rcos i 41rsin j 40r cos k= + + GSame Result


Kinetics of a Rigid Body Rigid Body Degrees of Freedom Linear Momentum Angular Momentum

Mass Moments of Inertia & Parallel Axis Theorem Principal Axes and Planes of Symmetry Translation Theorem for Angular Momentum

Equations of Motion Eulers Equations

Kinetic Energy and Work-Energy Principle Impulse-Momentum Principle Essential Example Problem


Rigid Body Degrees of Freedom If a system of particles becomes a continuum and

the measured distances between points in the system remains constant, the system is said to be a rigid body.

The same laws of motion that influence a system of particles must also govern the motion of a rigid body. The difference is that with a continuum present, the summation of physical quantities for discrete particles now becomes an integration over the whole volume.

An unconstrained rigid body has 6 degrees of freedom (3 translational and 3 rotational) and 6 equations of motion are needed to specify its motion.


Linear Momentum of a Rigid BodyC mass centerO reference point in body Bxyz body-fixed axesXYZ ground axes

R O R B

B B

R O R B

B B

R O R B

R O R B R C

L vdm v ( r) dm

v dm r dm

m v ( mr )

m v ( r ) m v

= = + = + = +

= + =

G G GG G

G GG

GGGGGG G B B

1m dm r r dmm

= = G GR CL m v=G G

L linear momentum of rigid body=G

total mass

center of masslocation


Angular Momentum of a Rigid BodyC mass centerO reference point in body Bxyz body-fixed axesXYZ ground axes

OHG Angular Momentum of B

about point O

OB

R O R B

B

R O R B R B

B B B

H (r v)dm

r v ( r) dm

v r dm r ( r)dm r ( r)dm

=

= + = + =

G G G

G G GG

G G G G G G GG

R O

B

v 0 if point O is fixed in R

r dm 0 if point O coincides with C

==

GG


R BO

B

R B

B

H r ( r)dm

H ( )dm

=

=

G G G G

G G G GPoint O is fixed in R

Point C is the mass center of B

O x y z

R B R B R B R Bx y z

H H i H j H k i j k

= + + = + +

GG

y

Y

Z

O

XGround R

xz

C

Rigid Body B

dm

rGrG

GHere we assume either point O is fixed in R or coincident with point C.

O

R B

H

GG

Independent of the orientation of the xyz body-fixed axes, but their components are

not.


We can show by integration that:R B

x x xy xz xR B

y yx y yz yR B

z zx zy z z

H I I IH I I IH I I I

= 2 2

xB

2 2y

B

2 2z

B

I (y z )dm

I (x z )dm

I (x y )dm

= +

= +

= +

xy yxB

xz zxB

yz zyB

I (xy)dm I

I (xz)dm I

I (yz)dm I

= =

= =

= =

MassMoments

ofInertia

MassProducts

ofInertia

Inertia Matrix

Note: the elements of the inertia matrix are for a particular point and a particular orientation of

the xyz body-fixed axes.


Parallel Axis Theorem of Inertia Matrix There is an inertia matrix associated with every

point of a rigid body. Consider two parallel coordinate systems fixed to

a rigid body: x1y1z1 and x2y2z2. Let point 1 coincide with the mass center C.

[ ] [ ]2 2

2 22 C

2 2

2 1

2 1

2 1

b c ab acI I m ab c a bc

ac bc a b

x x ay y bz z c

+ = + + + = += += +


Principal Axes It is often convenient to deal with rigid-body

dynamics problems using the coordinate system fixed in the body for which all products of inertia are zero simultaneously, i.e., the inertia matrix is diagonal.

The 3 mutually perpendicular axes are called principal axes.

The 3 mass moments of inertia are called principal moments of inertia.

The 3 planes formed by the principal axes are called principal planes.


Plane of Symmetry Many rigid bodies have a plane of symmetry. For example, if the xy plane is a plane of

symmetry, then for every mass element with coordinates (x, y, z) there exists a mass element with coordinates (x, y, -z).

Hence

Translation Theorem for Angular Momentum The angular momentum of a body B about any

point P(on or off the body, fixed or moving) can be expressed as:

yz xzI I 0= =

PCPH (r L) H= +

GG GG


P

C

Rigid Body B

dm

rG

PCrG

GDerivationP

B

R C R B

B

PC R C R B

B

H (r v)dm

r v ( ) dm

(r ) v ( ) dm

=

= + = + +

G G G

G G GG

G G G GG

Y

Z XGround R

PC R C R C

B B

PC R B R B

B B

(r v )dm ( v )dm

r ( ) dm ( ) dm

= + +

+

G GG G

G G G G G G

PC R C R C

B

PC R B R B

B B

r m v v dm

r dm ( ) dm

= + +

G GG G

G G G G G G

PCPH (r L) H= +

GG GG

0

0


Equations of Motion The six scalar equations of motion for a rigid body are

given by the two vector equations:

is the resultant of all external forces acting on the body.

is the resultant moment of external forces and couples about the mass center C (fixed point O).

R R R C

R R

O O

d d vF L mdt dtd dM H M H

dt dtor

= =

= =

GG G

G G G G

FG

OM(M )G G


Lets express these equations in terms of the body-fixed xyz coordinate system.

x y z

x y z

O O O O

H H i H j H k H H i H j H k

= + += + +

GG

R C R C R C R B R B R Bx y z x y z

R C R C R B R C R Bx z y y z

R C R C R B R C R By x z z x

R R C B R CR B R

R C R C R B R C R Bz

R C

y x x y

C

v i v

d v d v ( v )d

j v k) ( i j k)

( v v v )i( v v v ) j

t dt

( v v v

( v

)k

+ + + + + = + + + + +

= + =

G

G

G G

G

R B R B R B R Bx y z

R C R C R C R Cx y z

i j k v v i v j v k

= + + = + +

GG


x y z

R B R Bx z y y z

R B R By x z z x

R

R B

B R Bz y x x y

R B

R B H i H j H k)(H H

dH dH ( H)dt

H )i(H H H ) j(H H H )k

dt( ( H)+ + +

= + + + + +

= + =

G

G GGGG

R Bx x xy xz x

R By yx y yz y

R Bz zx zy z z


=

The inertia matrix is constant with respect to

time since it is expressed in the body-

fixed coordinate system. So we can write:

R Bx x xy xz x

R By yx y yz y

R Bz zx zy z z


=


The velocity terms refer to: The angular velocity terms refer to:The xyz axes are body-fixed axes.

R CvGR BG

Six Scalar Equations of Motionx x z y y z

y y x z z x

z z y x x y

F m v v v

F m v v v

F m v v v

= + = + = +

x x x xy y x z xz z x y

2 2z y y z yz y z

y y y xy x y z yz z x y

2 2x z x z xz z x

z z z xz x y z yz y x z

2 2y x x y xy x y

M I I ( ) I ( )

(I I ) I ( )

M I I ( ) I ( )

(I I ) I ( )M I I ( ) I ( )

(I I ) I ( )

= + + + + +

= + + + + +

= + + + + +

The moments and inertia

terms are with respect to

axes fixed in the body with

origin at C, the mass center.


If we assume that the xyz body-fixed axes are principal axes (1, 2, and 3), then all the products of inertia are zero,

and the mass moments of inertia are identified as:

x 1 y 2 z 3I I I I I I= = =The three rotational equations then are:

1 1 3 2 2 3 1

2 2 1 3 1 3 2

3 3 2 1 1 2 3

I (I I ) MI (I I ) MI (I I ) M

+ = + = + =

Eulers Equations

Note: If only and are nonzero in the general equations, then:

z z2

x xz z yz z

2y yz z xz z

z z z

M I I

M I I

M I

= = + =

For these to be zero, the xy plane must be a

plane of symmetry:

xz yzI I 0= =


Kinetic Energy

x

C

Rigid Body B

dm

y

zY

Z XGround R

T = Kinetic Energy

R B = G G

[ ]

2

B

B

translation rotation

1 1dT v dm (v v)dm2 21 (v ) (v ) dm2

1T (v ) (v ) dm2

1 1m(v v) ( ) ( ) dm2 2T T

= =

= + + = + +

= + = +

G GiG GG G G GiG GG G G GiG G G G G Gi i

G

vG is the velocity of particle dm with respect to R

B

dm 0 = GNote:


translation1T m(v v)2

= G Gi

[ ][ ]

rotationB

B

2 2 2x x y y z z

xy x y yz y z zx z x

1T ( ) ( ) dm21 ( ( )) dm21 H21 (I I I )2

I I I

=

=

=

= + + + + +

G G G Gi

G G G GiGG i

[ ] [ ][ ] [ ]

T

T

1T m v v2

1 I2

=

+

vectoridentity


[ ] [ ] [ ]T O1T I2= If the body has a fixed point O in inertial space and the origin of the xyz coordinate system is

at this point, then the total kinetic energy T is entirely due to rotational motion about the

fixed point.


Work-Energy Equation

FG

MG

The resultant of all external forces acting on the rigid bodyThe resultant moment of external forces and couples acting on the rigid body about the center of mass C

1 2U The work done by all external forces and couples in time interval from t1 to t22 2

1 1

t t

1 2 t t

1 2 2 1

U (F v)dt (M )dt

U T T

= + = GG G Gi i


Impulse-Momentum Principle Integration of the force equation with respect to

time yields the theorem that the linear impulse of a rigid body is equal to the change in linear momentum.

Similarly, integration of the moment equation with respect to time yields the theorem that the angular impulse of a rigid body is equal to the change in angular momentum.

[ ]21

R R C t

2 1t

d vF m Fdt m v(t ) v(t )dt

= = GG G G G

2

1

R t

2 1t

dM H Mdt H(t ) H(t )dt

= = G G G G G


Supplement:Rigid Body Plane Kinetics


Rigid Body 3D Kinetics Example


Rigid Body Dynamics:Kinematics and KineticsTopicsIntroductionSlide Number 4Basic ConceptsSlide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Kinematics of a Rigid BodySlide Number 26Slide Number 27Slide Number 28Slide Number 29Slide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Slide Number 40Slide Number 41Rigid Body 3D Kinematics ExampleSlide Number 43Slide Number 44Slide Number 45Slide Number 46Kinetics of a Rigid BodySlide Number 48Slide Number 49Slide Number 50Slide Number 51Slide Number 52Slide Number 53Slide Number 54Slide Number 55Slide Number 56Slide Number 57Slide Number 58Slide Number 59Slide Number 60Slide Number 61Slide Number 62Slide Number 63Slide Number 64Slide Number 65Slide Number 66Supplement:Rigid Body Plane KineticsSlide Number 68Slide Number 69Slide Number 70Slide Number 71Slide Number 72Slide Number 73Slide Number 74Slide Number 75Slide Number 76Slide Number 77Rigid Body 3D Kinetics ExampleSlide Number 79Slide Number 80Slide Number 81Slide Number 82Slide Number 83Slide Number 84Slide Number 85Slide Number 86

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Rigid Body Dynamics