Chapter 21: Three Dimensional Kinetics of a Rigid Body

Chapter 21: Three Dimensional Kinetics

of a Rigid Body

©2007 Pearson Education South Asia Pte Ltd

Chapter Objectives

• To introduce the methods for finding the moments of

inertia and products of inertia of a body about various

axes.

• To show how to apply the principles of work and

energy and linear and angular momentum to a rigid

body having three-dimensional motion.

• To develop and apply the equations of motion in three

dimensions.

• To study the motion of a gyroscope and torque-free

motion.


• Moments and Products of Inertia

• Angular Momentum

• Kinetic Energy

• Equations of Motion

• Gyroscopic Motion

• Torque-Free Motion

Chapter Outline


21.1 Moments and Products of Inertia

Moment of Inertia

• Consider the rigid body

• The moment of inertia for a differential element dm of the body about any one of the three coordinate axes is defined as the product of the mass of the element and the square of the

shortest distance from the axis

to the element

• From the figure,

)( 22 zyrx


21.1 Moments and Products of

Inertia

Moment of Inertia

• For mass moment of inertia of dm about the

x axis,

• For each of the axes,

mmzzz

mmyyy

mmxxx

xxx

dmyxdmrI

dmzxdmrI

dmzydmrI

dmzydmrdI

)(

)(

)(

)(

222

222

222

222



Inertia

Moment of Inertia

• Moment of inertia is always a positive

quantity

• Hence, summation of the product of mass

dm is always positive



Inertia

Product of Inertia

• Defined with respect to a set of orthogonal

planes as the product of the mass of the

element and the perpendicular (or shortest)

distances from the plane to the element

• For product of inertia dIxy for element dm,

dIxy = xy dm

• Note dIyx = dIxy



Inertia

Product of Inertia

• For each combination of planes,

mdxzII

mdyzII

mdxyII

mzxxz

mzyyz

myxxy



Inertia

Product of Inertia

• Unlike the moment of inertia, which is always positive, the product of inertia may be positive, negative or zero

• Result depends on the sign of the two defining coordinates, which vary independently from one another

• If either one or both of the orthogonal planes are planes of symmetry for the mass, the product of inertia with respect to these planes will be zero



Inertia

Product of Inertia

• In such cases, the elements of mass will occur in pairs on each side of the plane of symmetry

• On one side of the plane, the product of inertia will be positive while on the other side, the product of inertia will be negative, the sum therefore yielding zero

• Consider the y-z plane of

symmetry, Ixy = Ixz = 0



Inertia

Product of Inertia

• Calculation of Iyz will yield a positive result

since all elements of mass are located using

only positive y and z coordinates

• Consider the x-z and y-z being planes of

symmetry, Ixy = Iyz = Ixz = 0



Inertia

Parallel Axis and Parallel Plane Theorems

• If G has coordinates xG, yG, zG defined from the z, y, z axes, the parallel axis equations used to calculate the moments of inertia about the x, y, z axes are

• Products of inertia of a composite body are computed in the same manners as the body’s moments of inertia

)()(

)()(

)()(

22

''

22

''

22

''

GGGzzzz

GGGyyyy

GGGxxxx

yxmII

zxmII

zymII



Inertia

Parallel Axis and Parallel Plane Theorems

• Parallel axis theorem is used to transfer products of inertia of the body from a set of three orthogonal planes passing through the body’s mass center to a corresponding set of parallel planes passing through some other points O

• For parallel axis equations,

GGGzxxz

GGGzyyz

GGGyxxy

zmxII

zmyII

ymxII

)(

)(

)(

''

''

''



Inertia

Inertia Tensor

• Inertial properties of a body are completely characterized

by nine terms, six of which are independent of one

another

• For an inertia tensor,

• An inertia tensor has a unique set of values for a body

when it is computed for each location of the origin O and

orientation of the coordinates origin

zzzyzx

xzyyyx

xzxyxx

III

III

III



Inertia

Inertia Tensor

• For point O, specify a unique axes inclination for which the products of inertia for the body are zero when computed with respect to these axes

• For a diagonalized inertia tensor,

• For principal moments of inertia for the body

zzzyyyxxx

z

y

x

IIIIII

I

I

I

,,

00

00

00



Inertia

Inertia Tensor

• Of these three principal moments of inertia,

one will be a maximum and one will be a

minimum

• If the coordinates axes are orientated such

that two of the three orthogonal planes

containing the axes are planes of symmetry

for the body, then all the products of inertia

for the body are zero with respect to the

coordinate planes, hence the coordinate

axes are principle axes of inertia



Inertia

Moment of Inertia About an Arbitrary Axis

• Consider the body where nine elements of the inertia tensor have been computed for the x, y, z axes having an origin at O

• Determine the moment of inertia of the body

about the Oa axis, for which the

direction is defined by vector ua

• IOa = ∫b2 dm where b is the

perpendicular distance from

dm to Oa



Inertia


• Position of dm is located using r, b = rsinθ, which

represents the magnitude of the cross-product ua

x r

• For moment of inertia,

• Provided

kzjyixr

kujuiuu

dmrxurxudmrxuI

zyxa

am

am

aOa

2



Inertia


• Hence,

• For moment of inertia,

zxxzzyyzyyxxyzzzyyyxxx

mzx

mzy

myx

mz

my

mx

yxxzm

zyOa

yxxzzya

uuIuuIuuIuIuIuI

dmxzuudmyzuudmxyuu

dmyxudmzxudmzyu

dmxuyuzuxuyuzuI

kxuyujzuxuiyuzurxu

222

222

])()()[(

)()()(

222

222222222

222



Inertia


• If the inertia tensor is specified for the x, y, z axes, the moment of inertia of the body about the inclined Oa axis can be found

• Direction cosines ux, uy and uz can be determined

• Direction angles α, β and γ made between the positive Oa axis and the positive x, y, z axes can be determined from the direction cosines ux, uy and uz, respectively



Inertia

Example 21.1

Determine the moment of inertia of the bent rod

about the Aa axis. The mass of each of the three

segments are shown in the figure.



Inertia

Solution

• For moment of inertia of a slender rod, I = 1/12

ml2

• For each segment of the rod,

2

2

2

2

2

2

.200.0

.160.0

.160.0

.400.0

.453.0

.480.0

mkgI

mkgI

mkgI

mkgI

mkgI

mkgI

xz

yz

xy

zz

yy

xx

View Free Body Diagram

ppt-qns/21.1.ppt



Inertia

Solution

• For unit vector of the Aa axis,

• Thus,

• Hence,

2

222

222

.169.0

222

408.0,816.0,408.0

408.0816.0408.0

2.04.02.0

2.04.02.0

mkg

uuIuuIuuIuIuIuII

uuu

kji

kji

r

ru

zxxzzyyzyxxyzzzyyyxxxAa

zyx

D

DAa


21.2 Angular Momentum

• Consider the rigid body having a mass m and center of mass at G

• X, Y, Z coordinate system represent an inertial frame of reference and its axes are fixed or translating with a constant velocity

• Angular momentum as

measured from this

reference will be

computed relative to

the arbitrary point A



• Position vectors rA and ρA are drawn from the

coordinates to point A and from A to the ith particle

of the body

• If the particle’s mass is mi, for angular momentum

about point A,

(HA)i = ρA x mivi

where vi represent the particle’s velocity measured

from the X, Y, Z coordinate system

• If the body has an angular velocity ω,

vi = vA + ω x ρA



• Hence,

(HA)i = ρA x mi(vA + ω x ρA )

= (ρAmi) x vA + ρA x (ω x ρA)mi

• Summing all the particles of the body,

HA = (∫m ρA dm) x vA + ∫m ρA x (ω x ρA) dm



Fixed Point O

• If A becomes fixed point O in the body,

vA = 0

HO = ∫m ρO x (ω x ρO) dm



Center of Mass G

• If A is located at the center of mass G,

∫m ρA dm = 0

HG = ∫m ρG x (ω x ρG) dm


Arbitrary Point A

• In general, A may be some point other than O or G

HA = ρG/A x mvG + HG

• Angular momentum consists of two parts – the moment of the linear momentum

mvG of the body about point A

added (vectorially) the angular

momentum HG




Rectangular Components of H

• Choosing a second set of x, y, z axes having an arbitrary

orientation relative to the X, Y, Z axes,

H = ∫m ρ x (ω x ρ) dm

• Expressing in terms of x, y, z components,

Hxi + Hyj + Hzk = ∫m (xi + yj + zk )

x [(ωxi + ωyj + ωzk ) x (xi + yj + zk )]dm

• Expanding and equating the i, j and k components,

Hx = Ixxωx - Ixyωy - Ixzωz

Hy = Iyyωy - Iyxωx - Iyzωz

Hz = Izzωz - Izxωx - Izxωx



Rectangular Components of H

• If the x, y, z coordinate axes are oriented such as they become the principal axes of inertia for the body at that point

• If these axes are used, for products of inertia,

Ixy = Iyz = Izx = 0

• If the principal moments of inertia about the x, y, z axes are represented as Ix = Ixx, Iy = Iyy, Iz = Izz, for components of angular momentum,

Hx = Ixωx, Hy = Iyωy, Hz = Izωz



Principle of Impulse and Momentum

• For principle of impulse and momentum,

• In 3D, each vector term can be represented by 3

scalar components and 6 scalar equations

• 3 equations relate the linear impulse and

momentum in the x, y, z directions and the other

3 equations relate the body’s angular impulse

and momentum about the x, y, z axes

21

21

)()(

)()(

2

1

2

1

O

t

tOO

G

t

tG

HdtMH

vmdtFvm


21.3 Kinetic Energy

• Consider the rigid body which has a mass m and center of mass at G

• For kinetic energy of the ith particle of the body having a mass mi and velocity vi measured relative to the inertial X, Y, Z frame of reference,

• Provided the velocity of an

arbitrary point A of the body

is known,

AAi

iiiiii

xvv

vvmvmT

)(2

1

2

1 2


21.3 Kinetic Energy

• For kinetic energy of the particle,

• The kinetic energy for the entire body is obtained by

summing the kinetic energies of the body

• The last term on the right can be re-written as

cxbacbxa

dmxxdmxvmvvT

mxxmxvmvv

xvxvmT

mAA

mAAiAA

iAAiAAiAA

AAAAii

)()(2

1)()(

2

1

)()(2

1)()(

2

1

)()(2

1


21.3 Kinetic Energy

• Hence,

mAA

mAAiAA

dmx

dmxvmvvT

)(2

1

)()(2

1


21.3 Kinetic Energy

Fixed Point O

• If A is a fixed point O in the body,

222

2

1

2

1

2

1

2

1

0

zzyyxx

O

A

IIIT

HT

v


21.3 Kinetic Energy

Center of Mass G

• If A is located at the center of mass G of the

body,

• Kinetic energy consists of the translational kinetic

energy of the mass center and the body’s

rotational kinetic energy

2222

2

2

1

2

1

2

1

2

1

2

1

2

1

0

zzyyxxG

GG

A

IIImvT

HmvT

dm


21.3 Kinetic Energy

Principle of Work and Energy

• Used to solve problems involving force,

velocity and displacement

• Only one scalar equation can be written for

each body

T1 + ∑U1-2 = T2


21.4 Equations of Motion

Equations of Translation Motion

• Defined in terms of acceleration of the body’s mass center, which is measured from an inertial X, Y, Z reference

• For equations of translation motion in vector form,

• For scalar equations of translation motion,

• For sum of all external forces acting on the body,

kFjFiFF

amFamFamF

amF

zyx

zGzyGyxGx

G

)(,)(,)(



Equations of Rotational Motion

which states that sum of moments about a fixed point O of all the external forces acting on a system of particles (contained in a rigid body) is equal to the time rate of change of the total angular momentum of the body about point O

• When moments of external forces acting on the particles are summed about the system’s mass center G, one again obtain summation ∑MG to the angular momentum HG

OO HM



• Consider a system of particles where X, Y, Z

represents an inertial frame of reference and

the x, y, z axes with origin at G, translate with

respect to this frame

• In general, G is accelerating,

so by definition, the translating

frame is not an inertial reference





• For the angular momentum of the ith particle with respect to this frame,

• Taking the time derivatives,

• By definition,

• Thus, the first term on the right side is zero since the cross-product of equal vectors equals zero.

GiGi

GiiGiGiiGiGi

GiiGiGi

rv

vmxrvmxrH

vmxrH

//

////

//

)(

)(




• Since

• Hence,

• When the results are summed, for the time

change of the total angular momentum of the

body computed relative to point G,

)(

)()(

//

//

//

GiiGiG

GiiGiGi

GiGi

amxrH

amxrH

va




• For relative acceleration for the ith particle,

where ai and aG represent accelerations of the ith particle and point G measured with respect to the inertial frame of reference

• By vector cross-product,

• By definition of the mass center, since position vector r relative to G is zero,

rmrm

axrmamxrH

aaa

iGii

GiGiiGiiGiG

GiGi

)()(

)()(

/

////

/




• Using the equation of motion,

• For the rotational equation of motion for the body,

• If the scalar components of the angular momentum HO or HG are computed about x, y, z axes that are rotating with an angular velocity Ω, which may be different from the body’s angular velocity ω, then the time derivative

GG

iGiG

HM

FxrM

)( /




must be used to account for the rotation of the x, y, z axes as measured from the inertial X, Y, Z axes

• For the time derivative of H,

• There are 3 ways to define the motion of the x, y, z axes

GxyzGG

OxyzOO

HxHM

HxHM

dtHdH

)(

)(

/



x, y, z Axes having motion Ω = 0

• If the body has general motion, the x, y, z axes may be chosen with origin at G, such that the axes only translate to the inertial X, Y, Z frame of reference

• However, the body may have a rotation ω about these axes, and therefore the moments and products of inertia of the body would have to be expressed as functions of time



x, y, z Axes having motion Ω = ω

• The x, y, z axes may be chosen with origin at G,

such that they are fixed in and move with the

body

• The moments and the products of inertia of the

body relative to these axes will be constant

during the motion

GxyzGG

OxyzOO

HxHM

HxHM

)(

)(




• For a rigid body symmetric with respect to the x-y

reference plane, and undergoing general plane

motion,

0//,0

)()()(

)(

)()()(

)(

)()(

)()(

22

22

22

dtddtdII

III

IIIM

III

IIIM

II

IIIIM

yxyxyzxz

xzyyzyxxyzyxzx

yxyyxxzzzz

zyxxyxzzxyxzyz

xzxxzzyyyy

yxyzxzyyz

xzyxyzyzzyyxxxx




• Hence,

• If the x, y and z axes are chosen as principal

axes of inertia, the products of inertia are zero,

Ixx = Iz

• For the Euler equations of motion,

yxyxzzz

xzxzyyy

zyzyxxx

zzzzyx

IIIM

IIIM

IIIM

IMMM

)(

)(

)(

,0




• Since the x, y, z components are rotating at

Ω = ω,

• Since

• Hence,

xyz

xyz

x

x

)(

0

)(



x, y, z Axes having motion Ω ≠ ω

• Choose the x, y, z axes having an angular velocity Ω which is different from the angular velocity ω of the body

• This is particular suitable for the analysis of spinning tops and gyroscopes, which are symmetrical about their spinning axes

• When this is the case, the moments and products of inertia remain constant during the motion



x, y, z Axes having motion Ω ≠ ω

• Any one of these sets of moment equations represents a series of three first order nonlinear differential equations

• These equations are coupled since the angular velocity components are present in all the terms

yxyxyxzzz

xzxzxzyyy

zyzyzyxxx

IIIM

IIIM

IIIM



Procedure for Analysis

FBD

• Draw a FBD of the body at the instant considered and specify the x, y, z coordinate system

• Origin of this reference must be located either at the body’s mass center G or at point O, considered fixed in an inertial reference frame and located either in the body or a massless extension of the body

• Unknown reactive forces can be shown having a positive sense of direction




FBD

• Depending on the nature of the problem, decide which type of rotational motion Ω = 0, Ω = ω, or Ω ≠ ω

• When choosing, one should keep in mind that the moment equations are simplified in such a manner that they represent principal axes of inertia for the body at all times

• Compute the necessary moments and products of inertia for the body relative to the x, y, z axes




Kinematics

• Determine the x, y, z components of the body’s angular velocity and compute the time derivatives of ω

• If Ω = ω then, ώ = (ώ)xyz and we can either find the components of ω along the x, y, z axes when the axes are oriented in a general position and take the time derivatives of these components (ώ)xyz or we can find the time derivatives of ω with respect to the X, Y, Z axes, ώ, and then determine the components of ώx, ώy, ώz




Equations of Motion

• Apply either the two vector equations or the

six scalar component equation appropriate

for the x, y, z coordinate axes chosen for the

problem


21.5 Gyroscopic Motion

• Starting the X, Y, Z and the x, y, z axes in

coincidence, the final position of the top id

determined by

1. Rotate the top about the Z (or z) axis through

an angle Φ (0 ≤ Φ ≤ 2π)

2. Rotate the top about the x axis through an

angle θ (0 ≤ θ ≤ π)

3. Rotate the top about the z axis through an

angle ψ (0 ≤ ψ ≤ 2π) to obtain the final position

• Sequence must be maintained



• Since these finite rotation are not vectors and in

this case, the differential rotation are vectors and

thus, the angular velocity ω of the top can be

expressed in terms of the time derivatives of the

Euler angles

• The angular components are

known as precession,

nutation and spin





• It can be seen that these vectors are not all perpendicular to one another

• However, ω of the top can still be expressed in terms of these three components

• The body (top) is symmetric with respect to the z or spin axis

• If the top is orientated so that at the instant, the spin angle equals 0 and the x, y, z axes follow the motion of the body only in nutation and precession, i.e. Ω = ωp + ωn, the nutation and spin are always directed along the x and z axes respectively



• Hence, for the angular velocity of the body specified only in terms of the Euler’s angle θ,

• Since the motion of the axes is not affected by the spin component,

• The x, y, z axes represent the principal axes of inertia of the body for any spin of the body about these axes

kji

kji

kji

kji

zyx

zyx

)cos()sin(

)cos()sin(



• Hence, the moments of inertia are constant and will be represented by Ixx = Iyy = I and Izz = Iz

• Since Ω ≠ ω,

• Each moment summation applies only at the fixed point O or at the center of mass G of the body

)sincos(

)cos()cos2sin(

)cos(sin)cossin( 2

zx

zy

zx

IM

IIM

IIM



• Since the equations represent a coupled set of nonlinear second order differential equations, in general, a closed-form solution may not be obtained

• Instead, the Euler’s angles may be obtained graphically as functions of time using numerical analysis and computer techniques

• As special case, steady precession, occurs when the nutation angle, precession and spin all remains constant



• The equations reduce to

• Furthering simplifying,

or

• Note that the effects the spin has on the moment about the x axis

)cos(sin

sincossin

0,0

)cos(sincossin

2

2

IIM

IIM

MM

IIM

zzx

zzx

zy

zx


• When θ = 90°, consider the spinning rotor,

equations reduce to

or

• It can be seen that the vectors ∑Mx, Ωy and ωz

all act along their respective positive axes and

therefore are mutually perpendicular


zyzx

zx

IM

IM


• One would expect the rotor to fall under the influence of gravity, however, provided that product of IzΩyωz is correctly chosen to counter moment ∑Mx = WrG of the rotor’s weight about O

• This unusual moment is referred as the gyroscopic effect

• Studying the action of the gyroscope, or known as gyro, a gyro is a rotor that spins at a very high rate about its axis of symmetry

• This rate of spin is greater than its precessional rate of rotation about the vertical axis




• Hence, the angular momentum of the gyro can

be assumed directed among its axis of spin

• Thus, for the gyro rotor, ωz >> Ωy, and the

magnitude of the angular momentum about point

O reduces to the form HO = Izωz

• Since both the magnitude and directions of HO

are constant as observed from the x, y, z, direct

application yields

∑Mx = Ωy x HO


• Ωy always swings HO (or ωz) towards the sense

of ∑Mx

• In effect, the change in direction of the gyro’s

angular momentum, dHO is equivalent to the

angualr impulse caused by the gyro’s weight

about O

dHO = ∑Mx dt

• When a gyro is mounted in gimbal

rings, it become free of external

momentums applied to its base




• In theory, its angular momentum H will never precess but, instead maintain its same fixed orientation along the axis of spin when the base is rotated

• This type of gyroscope is called a free gyro and is useful as a gyrocompass when the spin axis of the gyro is directed north

• In reality, the gimbal mechanism is never completely free of friction, so such a device is useful only for the local navigation of ships and aircraft


21.6 Torque-Free Motion

• When the only external force

acting on a body is caused by

gravitation, the general motion of

the body is referred to as torque-free motion

• In order to describe the

characteristics of this motion, the

distribution of the body’s mass will

be assumed axisymmetric, as

shown in figure, where the z axis

represents an axis of symmetry



• The origin of the x, y, z coordinates is located

at the mass center G, such that Izz = Iz and Ixx = Iyy = I for the body

• Since gravitation is the only external force

present, the summation of moments about

the mass center is zero

• This required the angular momentum of the

body to be constant,

constantGH



• At the instant considered, it will assumed that

the inertial frame of reference is oriented so

that the positive Z axis is directed along HG

and the y axis lies in the plane formed by the

z and Z axes.

• The Euler angle formed between Z and z is

θ, and therefore, with this choice of axes the

angular momentum may be expressed as

kjH cossin GGG HH



• Since

• Equating the respective i, j, k components of

the above two equations yields

• or

kjiH zyxG III

kjz

GG

z

Gz

Gyx

I

H

I

H

I

H

I

H

cossin

cossin0



• Equating the respective i, j, k components of

into

• We obtain

kji

kji

)cos()sin(

zyx

z

Gz

Gyx

I

H

I

H

cossin0

z

GG

I

H

I

H

sincos

sinsin0



• Solving, we get

cos

const

G

z

z

G

HII

II

I

H



• Thus for torque-free motion of an

axisymmetrical body, the angle θ formed

between the angular-momentum vector and

the spin of the body remains constant

• Furthermore, the angular momentum HG,

precession Φ, and spin Ψ for the body

remain constant at all times during the

motion

cosz

z

I

II

. .



• As shown, the body precesses about the Z axis, which is fixed in direction, spinning about z axis.

• These two components of angular motion may be studied by using a simple cone model

• The space cone defining the precession

is fixed from rotating, since the

precession has a fixed direction,

while the body cone rotates around

the space cone’s outer surface

without slipping



• On the basis, an attempt should be made to

imagine the motion

• The interior angle of each cone is chosen such

that the resultant angular velocity of the body is

directed along the line of contact of the two

cones

• This line of contact represents the instantaneous

axis of rotation for the body cone and hence the

angular velocity of both the body cone and the

body must be directed along this line



• Since the spin is a function of the moments

of inertia I and Iz of the body, the cone model

is satisfactory for describing the motion,

provided I > Iz

• Torque-free motion which meets these

requirements is called regular precession

• If I > Iz, the spin is negative and precession is

positive



• This motion is represented by the satellite motion shown below (I < Iz)

• The cone model may again be used to represent; however, to preserve the correct vector addition of spin and precession to obtain the angular velocity ω, the inside surface of the body cone must roll on the outside surface of the (fixed) space cone

• This motion is called

retrograde precession


Chapter Review

Moments and Products of Inertia

• A body has 6 components of inertia for any specified x, y, z axes

• 3 are moments of inertia about each of the axes, Ix, Iy, Iz, and 3 are products of inertia , each defined from two orthogonal planes Ixy, Iyz and Ixz

• If either one or both of the planes are planes of symmetry, then the product of inertia with respect to these planes will be zero

• Moments and products of integrations can be determined by direct integration


Chapter Review


dmxzIIdmyxdmrI

dmyzIIdmzxdmrI

dmxyIIdmzydmrI

mzxxzzzz

mzyyzyyy

myxxyxxx

222

222

222


Chapter Review


• If these quantities are to be determined with

respect to axes or planes that do not pass

through the mass center, parallel axes

theorems must be used

• Provided the 6 components of inertia are

known, the moment of inertia about any axis

may be determined using the transformation

equation

zxxzzyyzyyxxyzzzyyyxxxOa uuIuuIuuIuIuIuII 222222


Chapter Review

Principle Moments of Inertia

• At any point on or off the body, the x, y, z axes can be oriented so that the products of inertia will be zero.

• The resulting moments of inertia are called the principal moments of inertia, one of which will be a maximum and the other a minimum moment of inertia for the body.

z

y

x

I

I

I

00

00

00


Chapter Review


• The angular momentum for a body can be

determined about any arbitrary point A.

• Once the linear and angular momentum for

the body have been formulated, then the

principle of impulse and momentum can be

used to solve problems that involve force,

velocity, and time.


Chapter Review


Mass) of(Center

Point) (Fixed

Point)(Arbitary

/

21

21

2

1

2

1

dmH

dmH

HmvH

HdtMH

vmFdtvm

GGG

OOO

GGAGA

O

t

t

OO

G

t

t

G


Chapter Review


zzzyzyxzxz

zyzyyyxyxy

zxzyxyxxxx

IIIH

IIIH

IIIH

where


Chapter Review

Principle of Work and Energy

• Kinetic energy of a body is usually

determined relative to a fixed point or the

body’s mass center

• Provided the axes are principal axes of

inertia, for a fixed point,

• And relative to the mass center,

222

2

1

2

1

2

1zzyyxx IIIT

2211

2222

2

1

2

1

2

1

2

1

TUT

IIImvT zzyyxxG


Chapter Review

Equations of Motion

• There are 3 scalar equations of translational motion for a rigid body moving in 3D

• There are 3 scalar equations of rotational motion depending upon the location of the x, y, z reference

• Most often, these axes are oriented so that the axes are principal axes of inertia

zGzyGyxGx amFamFamF )(,)(,)(


Chapter Review

Equations of Motion

• If the axes are fixed in and move with the rotation

ω of the body, the equations are referred to as

the Euler’s equations of motion (Ω = ω),

• If the axes have a rotation (Ω ≠ ω), yxyxzzz

xzxzyyy

zyzyxxx

IIIM

IIIM

IIIM

)(

)(

)(

yxyxyxzzz

xzxzxzyyy

zyzyzyxxx

IIIM

IIIM

IIIM


Chapter Review

Gyroscopic Motion

• The angular motion is best described using the

changes in motion of the three Euler angles

• These angular velocity components are the

precession, nutation and the spin

• If spin equals zero and the precession and

nutation are constant, the motion is referred as

steady precession

• For rotational equations of motion,

0,0

)cos(sincossin2

zy

zx

MM

IIM


Chapter Review

Gyroscopic Motion

• It is the spin of the gyro motor that is responsible

for holding the rotor from falling downward, and

instead causing it to precess about a vertical axis

• This phenomenon is called the gyroscopic effect


Chapter Review

Torque-Free Motion

• A body that is subjected to a gravitational force

that will have no moments on it about its mass

center, and so the motion is described as torque-

free motion

• The angular momentum for the body will remain

constant and this causes the body to have both a

spin and a precession


Chapter Review

Torque-Free Motion

• The behavior depends on the size of the moment

of inertia of a symmetric body about the spin axis

Iz versus that about a perpendicular axis I

• If I > Iz, regular precession occur

• If I < Iz, retrograde precession occur

cos

constant

G

Z

Z

G

HH

II

I

H

Documents

Chapter 21: Three Dimensional Kinetics of a Rigid Body