Rigid Body Dynamics: Kinematics and Kinetics Body Dynamics K. Craig 4 • Newtonian Dynamics – Kinematics • This is the study of the geometry of motion. It describes the motion

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 earth’s 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 earth’s 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.

• Newton’s 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−= − = ×

⋅


• 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 body’s mass with respect to an axis through some point. It represents the body’s 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.

• Newton’s 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

ˆ î i1 0 0ˆ ˆj 0 cos sin jˆ ˆ0 sin cosk k

⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= α α⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− α α⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

1

1

1

ˆ î iˆ ˆ ˆj cos j sin kˆ ˆ ˆk sin j cos k

=

= α + α

= − α + α


• 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

φ x1

y1

x2

y2

O

z1

y

z

y1

O

α

1

1

1


⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= α α⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− α α⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

2 1

2 1

2 1

ˆ î icos sin 0ˆ ˆj sin cos 0 jˆ ˆ0 0 1k k

⎡ ⎤ ⎡ ⎤φ φ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= − φ φ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

Pa ????1

2

R

R

????????

ω

ω

Meaningless!

1

1 2

RR

R R

ω

ωR Pa

Proper Notation


φ

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

ˆ ˆ î cos i sin jˆ ˆ ˆj sin i cos j

= φ + φ

= − φ + φ

1 2 2

1 2 2

ˆ ˆ î cos i sin jˆ ˆ ˆj sin i cos j

= φ − φ

= φ + φ

2 1

2 1

ˆ î icos sinˆ ˆsin cosj j

⎡ ⎤ ⎡ ⎤φ φ⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥− φ φ⎣ ⎦⎣ ⎦ ⎣ ⎦

1 2

1 2

ˆ î icos sinˆ ˆsin cosj j

⎡ ⎤ ⎡ ⎤φ − φ⎡ ⎤=⎢ ⎥ ⎢ ⎥⎢ ⎥φ φ⎣ ⎦⎣ ⎦ ⎣ ⎦


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

R1

A

ββ

1 1 1ˆ ˆ î j k

1

RRRd

dtβ = ω × β

Defining equation for ω


• 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

ˆ ˆk k

angular speed of R in R

ω = ω = θ

θ =

Here R1 has simple angular velocity in R (ω1) and R2 has simple angular velocity in R1

(ω2). 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 β (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

R1

A

β

11

RRRRd d ( )

dt dtβ β= + ω × β


• 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 −ω = ω + ω + + ωNRRω

N NR RR Rω = − ω


2 1 1 2R R R RR Rω = ω + ω

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

R1

A

11 11

R RR RR RRR d d

dt dtω ω

α = =

1 1

1 1

R RR R

R RR R

k

k

ω = ω

α = α

ω = θ α = ω = θ

1ˆ ˆk k=


• Reference Frame Transformations

φ

R1 → x1y1z1R2 → x2y2z2

1 2 1 2 1 2

1 2

R R R R R R1 2

R R

ˆ ˆk kω = ω = ω

ω = φ

1 1 2 2x 1 y 1 x 2 y 2ˆ ˆ ˆ ˆV V i V j V i V j= + = +

1R dVdt

What Is ?

Define:

2 1

2 1

2 1

ˆ î cos sin 0 iˆ ˆj sin cos 0 jˆ ˆk 0 0 1 k

⎡ ⎤ ⎡ ⎤φ φ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= − φ φ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

1 2

1 2

1 2

ˆ î cos sin 0 iˆ ˆj sin cos 0 jˆ ˆk 0 0 1 k

⎡ ⎤ ⎡ ⎤φ − φ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= φ φ⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦


1 1

1 1 1 1

R R

x 1 y 1 x 1 y 1ˆ ˆ ˆ ˆV i V j VdV d

dt dti V j+ +⎡ ⎤= =⎣ ⎦

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

ˆ ˆV i V j

ˆ ˆV i V j

ˆ ˆ ˆV i V j ( k V)

d ˆV cos V sin i

dV ddt dt

d (

dtd ˆV sin V cos jdt

ˆ ˆ ˆk (V i V j )

V)dt

⎡ +

+

+ + φ ×

⎡ ⎤= φ + φ +⎣ ⎦

⎡ ⎤− φ + φ +

⎤= ⎣ ⎦

⎡ ⎤= + ω ×⎣ ⎦

⎣ ⎦

⎡ ⎤φ × +⎣ ⎦

=

1 1

1

ˆ î and j are fixed in R

One Approach

Another Approach

Are the two approachesequivalent?

2 2

2

ˆ î 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

ˆV cos V sin V sin V cos i

ˆV sin V cos V cos V sin j

ˆ ˆV j V i

ˆ ˆ ˆ ˆV cos i sin j V sin i cos j

ˆ ˆ ˆV sin i cos j V cos i

⎡ ⎤= φ − φ φ+ φ+ φ φ +⎣ ⎦⎡ ⎤− φ − φ φ+ φ− φ φ +⎣ ⎦⎡ ⎤φ − φ⎣ ⎦

⎡ ⎤ ⎡ ⎤= φ − φ + φ + φ +⎣ ⎦ ⎣ ⎦⎡ ⎤φ − φ − φ + φ φ⎣ ⎦

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

ˆsin j

ˆ ˆV j V i

ˆ ˆ ˆ ˆ ˆ ˆV i V j V j V i V j V iˆ ˆV i V j

⎡ ⎤− φ +⎣ ⎦⎡ ⎤φ − φ⎣ ⎦

= + − φ + φ +φ − φ

= + 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

R1

A

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⎡ ⎤α

⎡ ⎤ω ×⎣

⎡ ⎤ω × ω

×

×⎣= ⎦+

+

+⎦+ ⎣

⎦

( )1 1R RR P R A R AP Pv v r v= + ω × +

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 )

= +

=

= +

= + + ω ×

= + + ω ×

y

z O

P

xR

x1

y1

z1

R1A

OAr OPr

APr

( )1 1R RR P R A R AP Pv v r v= + ω × +


( )1 1

1 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

⎡ ⎤= + ω × ω ×⎣ ⎦⎡ ⎤ ⎡ ⎤+ α × + + ω ×⎣ ⎦ ⎣ ⎦

( )1 1

R RR RR P R P R A R AP Pd da ( v ) v r v

dt dt⎡ ⎤= = + ω × +⎣ ⎦

RR A R Ad ( v ) a

dt=

( )1 1 1

1 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 )]

ω × = α × + ω ×

⎡ ⎤= α × + ω × + ω ×⎣ ⎦

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= + ω × = + ω ×


• 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 Pa


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

ˆ ˆ ˆ ˆ ˆk ( k rj ) 2 k

ˆˆ 2 vr

j

ij

v

⎡ ⎤= + ω × ω ×⎣ ⎦⎡ ⎤ ⎡ ⎤+ α × + + ω ×⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤= − ω × − ω

− ω ω

× + − ω ×⎣ ⎦ ⎣ ⎦= += 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 Pa

[ ] [ ]radial

2 2 2

2radialradial t 0

v 2 4 v vcos (r r)sin v

v r t v( t) v

va lim rtΔ →

Δ = + − = θ−ω + Δ θ −

⎡ ⎤= −ω Δ −ω Δ −⎣ ⎦Δ

= = − ωΔ

cos 1sin

θ ≈θ ≈ θ

Centripetal Acceleration• due to term 4• v has no effect on aradial• depends on ant’s 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 vtΔ →

Δ = + − ω = θ+ω + Δ θ − ω

= ωΔ +ω +ω Δ − ω

Δ= = ω +ω = ω

Δ

cos 1sin

θ ≈θ ≈ θ

CoriolisAcceleration

• independent of ant’s position• effect of ω on v (term 1 ) is always

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

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


Rigid Body 3D Kinematics Example


R

R1 R2

O θ = 30º

r = 0.06 m

Rigid Body Kinematics Essential Example

Given:

Find:

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

φ x1

y1

x2

y2

O

z1

y

z

y1

O

α

1

1 2

RR

R R1

ˆ5i constantˆ4k constant

ω = =

ω = =

R Pa

1

1

1


⎡ ⎤ ⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥= α α⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− α α⎢ ⎥⎣ ⎦⎣ ⎦ ⎣ ⎦

( ) ( )OP1 1ˆ ˆr r cos i r sin j= θ + θ


( )2 2 2

2 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

⎡ ⎤ ⎡ ⎤= + ω × ω × + α ×⎣ ⎦⎣ ⎦⎡ ⎤+ + ω ×⎣ ⎦

2

2

R O

R P

R P

a 0a 0v 0

=

=

=

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

ˆ ˆ 5i 4k

d d ˆ ˆ5i 4kdt dt

dk ˆ 0 4 4 kdt

ˆ ˆ ˆ 4 5i k 20j

ˆ ˆ 20 jcos k sin

ω = ω + ω = +

ω ⎡ ⎤α = = +⎣ ⎦

= + = ω ×

= × = −

= − α + α


After Substitution and Simplification:

( ) ( ) ( )R P1 1 1ˆ ˆ â 16rcos i 41rsin j 40r cos k= − θ + − θ + θ

Alternate Solution:

( )1 1 1

1 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

⎡ ⎤ ⎡ ⎤= + ω × ω × + α ×⎣ ⎦⎣ ⎦⎡ ⎤+ + ω ×⎣ ⎦

1

11

R O

RR

RR RRR

a 0ˆ5i constantd 0dt

=

ω = =

ωα = =

( ) ( )OP1 1ˆ ˆr r cos i rsin j= θ + θ


( )1 1 1 2 1 2 1 2R R R R R R R RP O OP OPa a r r⎡ ⎤ ⎡ ⎤= + ω × ω × + α ×⎣ ⎦⎣ ⎦(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 0ˆ 4k

d d ˆ4k 0dt dt

v v r

v 0

=

ω =

ω ⎡ ⎤α = = =⎣ ⎦

= + ω ×

= ( ) ( )OP1 1ˆ ˆr r cos i rsin j= θ + θ

After Substitution and Simplification:

( ) ( ) ( )R P1 1 1ˆ ˆ â 16rcos i 41rsin j 40r cos k= − θ + − θ + θ

Same 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– Euler’s 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

⎡ ⎤= = + ω ×⎣ ⎦

= + ω ×

= + ω ×

⎡ ⎤= + ω × =⎣ ⎦

∫ ∫

∫ ∫

B B

1m dm r r dmm

= =∫ ∫

R CL m v=

L linear momentum of rigid body=

total mass

center of masslocation


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

OH Angular Momentum of Babout 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

= ×

⎡ ⎤= × + ω ×⎣ ⎦

= − × + × ω × = × ω ×

∫

∫

∫ ∫ ∫R O

B

v 0 if point O is fixed in R

r dm 0 if point O coincides with C

=

=∫


R BO

B

R B

B

H r ( r)dm

H ( )dm

= × ω ×

= ρ× ω ×ρ

∫

∫

Point 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

ˆ ˆ ˆH H i H j H kˆ ˆ î j k

= + +

ω = ω + ω + ω y

Y

Z

O

XGround R

xz

C

Rigid Body B

dm

rr

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

O

R B

H

ω

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

⎡ ⎤ω⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥ = ω⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ω⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2 2x

B

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= × +


P

C

Rigid Body B

dm

r

PCr

ρ

DerivationPB

R C R B

B

PC R C R B

B

H (r v)dm

r v ( ) dm

(r ) v ( ) dm

= ×

⎡ ⎤= × + ω ×ρ⎣ ⎦

⎡ ⎤= +ρ × + ω ×ρ⎣ ⎦

∫

∫

∫

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

= × + ρ× +

⎡ ⎤ ⎡ ⎤× ω ×ρ + ρ× ω ×ρ⎣ ⎦ ⎣ ⎦

∫ ∫

∫ ∫PC R C R C

B

PC R B R B

B B

r m v v dm

r dm ( ) dm

⎡ ⎤⎡ ⎤= × − × ρ +⎢ ⎥⎣ ⎦

⎣ ⎦⎡ ⎤

⎡ ⎤× ω × ρ + ρ× ω ×ρ⎢ ⎥ ⎣ ⎦⎣ ⎦

∫

∫ ∫PC

PH (r L) H= × +

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

∑ = =

∑ = ∑ =

F

OM(M )


• Let’s express these equations in terms of the body-fixed xyz coordinate system.

x y z

x y z

O O O O

ˆ ˆ ˆH H i H j H kˆ ˆ ˆH H i H j H k

= + +

= + +

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

ˆ ˆ ˆ ˆ ˆ ˆv i v

d v d v ( v )d

j v k) ( i j k)

ˆ( v v v )iˆ( v v v ) jˆ

t dt

( v v v

( v

)k

⎡ ⎤+ + + ω + ω + ω ×⎣ ⎦= + ω − ω

+ + ω − ω

+ +

= + ω ×

=

ω − ω

R B R B R B R Bx y z

R C R C R C R Cx y z

ˆ ˆ î j kˆ ˆ ˆv v i v j v k

ω = ω + ω + ω

= + +


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 Bˆ ˆ ˆH i H j H k)ˆ(H H

dH dH ( H)dt

H )iˆ(H H H ) jˆ(H H H )k

dt( ( H)+ + +

= + ω − ω

+ + ω − ω

+ + ω

= + ω ×

=

−

ω ×

ω

R Bx x xy xz x

R By yx y yz y

R Bz zx zy z z


⎡ ⎤⎡ ⎤ ω⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥ = ω⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ω⎣ ⎦ ⎣ ⎦ ⎣ ⎦

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


⎡ ⎤⎡ ⎤ ω⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥ = ω⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ ω⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦


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

R CvR Bω

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

⎡ ⎤= + ω − ω⎣ ⎦⎡ ⎤= + ω − ω⎣ ⎦⎡ ⎤= + ω − ω⎣ ⎦

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 ( )

= ω + ω −ω ω + ω +ω ω +

− ω ω + ω −ω

= ω + ω +ω ω + ω −ω ω +

− ω ω + ω −ω= ω + ω −ω ω + ω +ω ω +

− ω ω + ω −ω

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

ω + − ω ω =

ω + − ω ω =

ω + − ω ω =Euler’s Equations

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

zω zω

2x xz z yz z

2y yz z xz z

z z z

M I I

M I I

M I

= ω − ω

= ω + ω

= ω

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 Bω = ω

[ ]

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

= =

⎡ ⎤= +ω×ρ +ω×ρ⎣ ⎦

⎡ ⎤= +ω×ρ +ω×ρ⎣ ⎦

= + ω×ρ ω×ρ

= +

∫

∫

i

i

i

i i

ρ

v is the velocity of particle dm with respect to R

B

dm 0ρ =∫Note:


translation1T m(v v)2

= i

[ ]

[ ]

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

= ω×ρ ω×ρ

= ω ρ× ω×ρ

= ω

= ω + ω + ω

+ ω ω + ω ω + ω ω

∫

∫

i

i

i

[ ] [ ]

[ ] [ ]

T

T

1T m v v2

1 I2

=

⎡ ⎤+ ω ω⎣ ⎦

vectoridentity


[ ] [ ] [ ]T

O

1T I2

= ω ω

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

F

M

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 t2

2 2

1 1

t t

1 2 t t

1 2 2 1

U (F v)dt (M )dt

U T T

→

→

= + ω

= −

∫ ∫i 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.

[ ]2

1

R R C t

2 1t

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

∑ = ⇒ ∑ = −∫

2

1

R t

2 1t

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

⎡ ⎤∑ = ⇒ ∑ = −⎣ ⎦∫


Supplement:Rigid Body Plane Kinetics












Rigid Body 3D Kinetics Example









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