vector mechanics for engineers: chapter...

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Tenth Edition

CHAPTER

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

18 Ferdinand P. Beer

E. Russell Johnston, Jr.

Phillip J. Cornwell

Lecture Notes:

Brian P. SelfCalifornia Polytechnic State University

Kinetics of Rigid Bodies in

Three Dimensions

Kinematics of Rigid

Bodies in Three

Dimensions


Vector Mechanics for Engineers: Dynamics

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Contents

18 - 2

Introduction

Rigid Body Angular Momentum in

Three Dimensions

Principle of Impulse and

Momentum

Kinetic Energy

Sample Problem 18.1

Sample Problem 18.2

Motion of a Rigid Body in Three

Dimensions

Euler’s Equations of Motion and

D’Alembert’s Principle

Motion About a Fixed Point or a Fixed

Axis

Sample Problem 18.3

Motion of a Gyroscope. Eulerian

Angles

Steady Precession of a Gyroscope

Motion of an Axisymmetrical Body

Under No Force



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18 - 3

Three dimensional analyses are needed to determine the forces

and moments on the gimbals of gyroscopes, the rotors of

amusement park rides, and the housings of wind turbines.



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Introduction

18 - 4

GG HM

amF

• The fundamental relations developed for

the plane motion of rigid bodies may also

be applied to the general motion of three

dimensional bodies.

IHG • The relation which was used

to determine the angular momentum of a

rigid slab is not valid for general three

dimensional bodies and motion.

• The current chapter is concerned with

evaluation of the angular momentum and

its rate of change for three dimensional

motion and application to effective

forces, the impulse-momentum and the

work-energy principles.



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Rigid Body Angular Momentum in Three Dimensions

18 - 5

• Angular momentum of a body about its mass center,

n

iiii

n

iiiiG mrrmvrH

11

Δ

• The x component of the angular momentum,

n

iiiiz

n

iiiiy

n

iiiix

n

iiixiziiyixi

n

iiyiiziix

mxzmyxmzy

mzxzxyy

mrzryH

111

22

1

1

ΔΔΔ

Δ

Δ

dmzxdmxydmzyH zyxx 22

zxzyxyxx III

zzyzyxzxz

zyzyyxyxy

IIIH

IIIH



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18 - 6

zzyzyxzxz

zyzyyxyxy

zxzyxyxxx

IIIH

IIIH

IIIH

• Transformation of into is characterized

by the inertia tensor for the body,

GH

zzyzx

yzyyx

xzxyx

III

III

III

• With respect to the principal axes of inertia,

z

y

x

I

I

I

00

00

00

zzzyyyxxx IHIHIH

• The angular momentum of a rigid body

and its angular velocity have the same

direction if, and only if, is directed along a

principal axis of inertia.

GH



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18 - 7

• The momenta of the particles of a rigid body can

be reduced to:

vm

L

momentumlinear

GHG about momentumangular

zzyzyxzxz

zyzyyxyxy

zxzyxyxxx

IIIH

IIIH

IIIH

• The angular momentum about any other given

point O is

GO HvmrH



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18 - 8

• The angular momentum of a body constrained to

rotate about a fixed point may be calculated from

GO HvmrH

zzyzyxzxz

zyzyyxyxy

zxzyxyxxx

IIIH

IIIH

IIIH

• Or, the angular momentum may be computed

directly from the moments and products of inertia

with respect to the Oxyz frame.

n

iiii

n

iiiO

mrr

mvrH

1

1

Δ

Δ



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Principle of Impulse and Momentum

18 - 9

• The principle of impulse and momentum can be applied directly to the

three-dimensional motion of a rigid body,

Syst Momenta1 + Syst Ext Imp1-2 = Syst Momenta2

• The free-body diagram equation is used to develop component and

moment equations.

• For bodies rotating about a fixed point, eliminate the impulse of the

reactions at O by writing equation for moments of momenta and

impulses about O.



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Concept Question

18 - 10

At the instant shown, the

disk shown rotates with an

angular velocity 2 with

respect to arm ABC, which

rotates around the y axis as

shown (1). Determine the

directions of the angular

momentum of the disc

about point A (choose all

that are correct).

+x +y +z

-x -y -z



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Concept Question

18 - 11

A homogeneous disk of

mass m and radius r is

mounted on the vertical

shaft AB as shown

Determine the directions of

the angular momentum of

the disc about the mass

center G (choose all that

are correct).

+x +y +z

-x -y -z



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Kinetic Energy

18 - 12

• Kinetic energy of particles forming rigid body,

)22

2(

Δ

Δ

222

212

21

1

2

212

21

1

2

212

21

xzzxzyyz

yxxyzzyyxx

n

iii

n

iii

II

IIIIvm

mrvm

vmvmT

• If the axes correspond instantaneously with the

principle axes,

)( 222212

21

zzyyxx IIIvmT

• With these results, the principles of work and

energy and conservation of energy may be applied

to the three-dimensional motion of a rigid body.



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Kinetic Energy

18 - 13

• Kinetic energy of a rigid body with a fixed point,

)22

2( 22221

xzzxzyyz

yxxyzzyyxx

II

IIIIT

• If the axes Oxyz correspond instantaneously with

the principle axes Ox’y’z’,

)( 22221

zzyyxx IIIT



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Concept Question

18 - 14

22

1

2zI

At the instant shown, the disk

shown rotates with an angular

velocity 2 with respect to

arm ABC, which rotates

around the y axis as shown

(1). What terms will

contribute to the kinetic

energy of the disk (choose all

that are correct)?

21

1

2yI 2

1

1

2xI

2 21

1

2mL

21

1

2zI



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Sample Problem 18.1

18 - 15

Rectangular plate of mass m that is

suspended from two wires is hit at D in

a direction perpendicular to the plate.

Immediately after the impact,

determine a) the velocity of the mass

center G, and b) the angular velocity of

the plate.

SOLUTION:

• Apply the principle of impulse and

momentum. Since the initial momenta

is zero, the system of impulses must be

equivalent to the final system of

momenta.

• Assume that the supporting cables

remain taut such that the vertical velocity

and the rotation about an axis normal to

the plate is zero.

• Principle of impulse and momentum

yields to two equations for linear

momentum and two equations for

angular momentum.

• Solve for the two horizontal components

of the linear and angular velocity

vectors.



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Sample Problem 18.1

18 - 16

SOLUTION:

• Apply the principle of impulse and momentum. Since the initial momenta is zero,

the system of impulses must be equivalent to the final system of momenta.

• Assume that the supporting cables remain taut such that the vertical velocity and the

rotation about an axis normal to the plate is zero.

kvivv zx

ji yx

Since the x, y, and z axes are principal axes of inertia,

jmaimbjIiIH yxyyxxG

2

1212

121



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Sample Problem 18.1

18 - 17

• Principle of impulse and momentum yields two equations for linear momentum and

two equations for angular momentum.

• Solve for the two horizontal components of the linear and angular velocity vectors.

xmv0

0xv

zvmtF Δ

mtFvz Δ

kmtFv

Δ

x

x

mb

HtbF

2

121

21 Δ

mbtFx Δ6

y

y

ma

HtaF

2

121

21 Δ

matFy Δ6

jbiamab

tF

Δ6



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Sample Problem 18.1

18 - 18

kmtFv

Δ

jbiamab

tF

Δ6

jmaimbH yxG

2

1212

121



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Sample Problem 18.2

18 - 19

A homogeneous disk of mass m is

mounted on an axle OG of negligible

mass. The disk rotates counter-

clockwise at the rate 1 about OG.

Determine: a) the angular velocity of

the disk, b) its angular momentum about

O, c) its kinetic energy, and d) the

vector and couple at G equivalent to the

momenta of the particles of the disk.

SOLUTION:

• The disk rotates about the vertical axis

through O as well as about OG.

Combine the rotation components for

the angular velocity of the disk.

• Compute the angular momentum of the

disk using principle axes of inertia and

noting that O is a fixed point.

• The kinetic energy is computed from the

angular velocity and moments of inertia.

• The vector and couple at G are also

computed from the angular velocity and

moments of inertia.



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Sample Problem 18.2

18 - 20

SOLUTION:

• The disk rotates about the vertical axis through O as well

as about OG. Combine the rotation components for the

angular velocity of the disk.

w =w1i -w

2j

Noting that the velocity at C is zero,

vC

=w ´ rC

= 0

0 = w1i -w

2j( ) ´ Li - rj( )

= Lw2

- rw1( )k

w2

= rw1L

jLri

11



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Sample Problem 18.2

18 - 21

• Compute the angular momentum of the disk using

principle axes of inertia and noting that O is a fixed point.

jLri

11

kIjIiIH zzyyxxO

002

412

12

412

12

21

mrmLIH

LrmrmLIH

mrIH

zzz

yyy

xxx

jLrrLmimrHO

1

2

412

12

21

• The kinetic energy is computed from the angular velocity

and moments of inertia.

2

12

4122

12

21

222

21

LrrLmmr

IIIT zzyyxx

212

22

81 6

L

rmrT



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Sample Problem 18.2

18 - 22

jLri

11

• The vector and couple at G are also computed from the


kmrvm

1

jLrmrimr

kIjIiIH zzyyxxG

2

41

12

21

j

L

rimrHG

21

2

21



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Motion of a Rigid Body in Three Dimensions

18 - 23

GHM

amF

• Angular momentum and its rate of change are

taken with respect to centroidal axes GX’Y’Z’of fixed orientation.

• Convenient to use body fixed axes Gxyz where

moments and products of inertia are not time

dependent.

• Transformation of into is independent of

the system of coordinate axes.

GH

• Define rate of change of change of with

respect to the rotating frame,GH

kHjHiHH zyxGxyzG

Then,

GGxyzGG HHH



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Euler’s Eqs of Motion & D’Alembert’s Principle

18 - 24

GGxyzGG HHM

• With and Gxyz chosen to correspond

to the principal axes of inertia,

yxyxzzz

xzxzyyy

zyzyxxx

IIIM

IIIM

IIIM

Euler’s Equations:

• System of external forces and effective forces

are equivalent for general three dimensional

motion.

• System of external forces are equivalent to

the vector and couple, . and GHam



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Group Problem Solving

18 - 25

Two L-shaped arms, each of mass

2 kg, are welded at the third points

of the 600 mm shaft AB. Knowing

that shaft AB rotates at the constant

rate = 240 rpm, determine (a) the

angular momentum of the body

about B, and b) its kinetic energy.

SOLUTION:

• The part rotates about the axis AB.

Determine the mass moment of inertia

matrix for the part.

• Compute the angular momentum of the

disk using the moments inertia.

• Compute the kinetic energy from the




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18 - 26

2 kg, 200 mm 0.2 mm a

(2 )(240)8 rad/s, 0, 0

60z x y

Given:

Find: HB, T

There is only rotation about the

z-axis, what relationship(s) can

you use?

( )B x xzH I ( )B y yzH I ( )B z zH I

Split the part into four different segments, then determine

Ixz, Iyz, and Iz. What is the mass of each segment?

•Mass of each of the four segments 1 kg2

mm



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18 - 27

Fill in the table below to help

you determine Ixz and Iyz.1

2

34

Ixz Iyz

1

2

3

4

S

( )( )m a a

( 0.5 )( )m a a

( )(2 )m a a

(0.5 )(2 )m a a

(0.5 )( )m a a

(0)( )m a

(0)(2 )m a

(0.5 )(2 )m a a

2

2 22 11 4 2

1

12z zI I m a m a a

21.5m a21.5m a

Determine Iz by using the parallel

axis theorem – do parts 1 and 4 first

Parts 2 and 3 are also

equal to one another

22 3

1

3z zI I m a



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18 - 28

Calculate Hx

(HB

)x

= -Ixzw = -

3

2¢ma2w

= -3

2(1)(0.2)2(8p )

= -1.508 kg×m2/s

(HB

)y

= -Iyzw = -

3

2m¢a2w

= -3

2(1)(0.2)2(8p )

= –1.508 kg.m2 /s

2

2

2

10( )

3

10(1)(0.2) (8 )

3

3.351 kg m /s

B z zH I m a

2 2 2(1.508 kg m /s) (1.508 kg m /s) (3.35 kg m /s)BH i j k

Calculate Hy

Calculate Hz

Total Vector



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18 - 29

2 2 2 21 1 1 1

2 2 2 2x x y y z z xy x y xz x z yz y z zT I I I I I I I

2 2 2 21 10 1 10(1)(0.2) (8 )

2 3 2 3T m a

42.1 JT

Calculate the kinetic energy T



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18 - 30

Retracting the landing gear while the wheels are still spinning

can result in unforeseen moments being applied to the gear.

When turning a motorcycle, you must “steer” in the opposite

direction as you lean into the turn.



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Motion About a Fixed Point or a Fixed Axis

18 - 31

• For a rigid body rotation around a fixed point,

MOå = H

O

= HO( )Oxyz

+W´HO

• For a rigid body rotation around a fixed axis,

x xz y yz z zH I H I H I= - = - =

2

iIjIkIjIiI

kIjIiIk

kIjIiI

HHM

yzxzzyzxz

zyzxz

zyzxz

OOxyzOO

zz

xzyzy

yzxzx

IM

IIM

IIM

2

2

Z or z S axis of rotation

XY S fixed axes

xy S rotate with the body



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Rotation About a Fixed Axis

18 - 32

zz

xzyzy

yzxzx

IM

IIM

IIM

2

2

• For a rigid body rotation around a fixed axis,

• If symmetrical with respect to the xy plane,

zzyx IMMM 00

• If not symmetrical, the sum of external moments

will not be zero, even if = 0,

022 zxzyyzx MIMIM

• A rotating shaft requires both static and

dynamic balancing to avoid excessive

vibration and bearing reactions.

0 0



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Concept Question

18 - 33

The device at the right

rotates about the x axis

with a non-constant

angular velocity. Which

of the following is true

(choose one)?

a) You can use Euler’s equations for the provided x, y, z axes

b) The only non-zero moment will be about the x axis

c) At the instant shown, there will be a non-zero y-axis bearing force

d) The mass moment of inertia Ixx is zero



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Sample Problem 18.3

18 - 34

Rod AB with mass m = 20 kg is

pinned at A to a vertical axle which

rotates with constant angular velocity

= 15 rad/s. The rod position is

maintained by a horizontal wire BC.

Determine the tension in the wire and

the reaction at A.

• Expressing that the system of external

forces is equivalent to the system of

effective forces, write vector expressions

for the sum of moments about A and the

summation of forces.

• Solve for the wire tension and the

reactions at A.

SOLUTION:

• Evaluate the system of effective forces

by reducing them to a vector

attached at G and couple

am

.GH



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Sample Problem 18.3

18 - 35

SOLUTION:

• Evaluate the system of effective forces by reducing them

to a vector attached at G and coupleam

.GH

a = an

= -rw2I = - 12Lcosb( )w2I

= - 112.5m s2( ) I

ma = 20 -112.5( ) I = - 2250 N( ) I

kIjIiIH zzyyxxG

0sincos

0 2

212

21

zyx

zyx mLIImLI

imLHG

cos2

121

HG

= HG( )Gxyz

+w ´HG

= 0+ -w cosb i +w sin b j( ) ´ 112mL2w cosb i( )

= 112mL2w2 sin b cosb k = 649.5N × m( )k



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Sample Problem 18.3

18 - 36

• Expressing that the system of external forces is equivalent

to the system of effective forces, write vector expressions

for the sum of moments about A and the summation of

forces.

effAA MM

1.732J ´ -TI( ) +1I ´ -196.2J( ) = 0.866J ´ -2250I( ) +649K

1.732T -196.2( )K = 1948.5+649.5( )K

1613NT =

effFF

1613 196.2 2250X Y ZA I A J A K I J I+ + - - = -

( ) ( )697 N 196.2 NA I J= - +



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18 - 37

A four-bladed airplane propeller has a

mass of 160 kg and a radius of

gyration of 800 mm. Knowing that the

propeller rotates at 1600 rpm as the

airplane is traveling in a circular path

of 600-m radius at 540 km/h,

determine the magnitude of the couple

exerted by the propeller on its shaft

due to the rotation of the airplane

• Determine the mass moment of inertia for

the propeller

• Calculate the angular momentum of the

propeller about its CG

SOLUTION:

• Determine the overall angular velocity of

the aircraft propeller

• Calculate the time rate of change of the

angular momentum of the propeller about

its CG

• Calculate the moment that must be

applied to the propeller and the resulting

moment that is applied on the shaft



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18 - 38

Establish axes for the rotations of the

aircraft propeller

Determine the x-component of its

angular velocity

540 km/h 150 m/sv

2 rad1600 rpm

60 s

167.55 rad/s

x

150 m/s0.25 rad/s

600 my

v

•

Determine the y-component of its

angular velocity

Determine the mass moment of inertia

2 2 2(160 kg)(0.8 m) 102.4 kg mxI mk



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18 - 39

Angular momentum about G equation:

Determine the couple that is

exerted on the shaft by the

propeller

Determine the moment that must

be applied to the propeller shaft

G x x y yI I H i j

HG

= (HG

)Gxyz

+Ω´HG

= 0+wyj´ (I

xwxi+ I

ywyj)

HG

= -Ixwxwyk = -(102.4 kg ×m2)(167.55 rad/s)(0.25 rad/s)k

HG

= -(4289 N ×m)k=-(4.29 kN × m)k

Time rate of change of angular

momentum about G :

M=HG

= -(4.29 kN × m)k

(4.29 kN m)on shaft M M k



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Concept Question

18 - 40

The airplane propeller has a

constant angular velocity +x.

The plane begins to pitch up at

a constant angular velocity x

y

zThe propeller has zero

angular acceleration

TRUE FALSE

Which direction will the pilot have to steer to counteract the

moment applied to the propeller shaft?

a) +x b) +y c) +z

d) -y e) -z



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18 - 41

Gyroscopes are used in the navigation system of the Hubble

telescope, and can also be used as sensors (such as in the

Segway PT). Modern gyroscopes can also be MEMS (Micro

Electro-Mechanical System) devices, or based on fiber optic

technology.



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Motion of a Gyroscope. Eulerian Angles

18 - 42

• A gyroscope consists of a rotor with its mass center

fixed in space but which can spin freely about its

geometric axis and assume any orientation.

• From a reference position with gimbals and a

reference diameter of the rotor aligned, the

gyroscope may be brought to any orientation

through a succession of three steps:

a) rotation of outer gimbal through j about AA’,

b) rotation of inner gimbal through q about

c) rotation of the rotor through y about CC’.

• j, q, and y are called the Eulerian Angles and

spin of rate

nutation of rate

precession of rate

q



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Motion of a Gyroscope. Eulerian Angles

18 - 43

• The angular velocity of the gyroscope,

kjK

q

kji

jiK

qqq

qq

cossin

cossinwith

• Equation of motion,

OOxyzOO HHM

jK

kIjIiIHO

q

qqq

cossin

M xå = - ¢I j sinq + 2qj cosq( ) + Iq Y +j cosq( )M yå = ¢I q -j 2 sinq cosq( ) + Ij sinq Y +j cosq( )

M z = Id

dtå Y +j cosq( )



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Steady Precession of a Gyroscope

18 - 44

Steady precession,

constant are ,, yq

ki

kIiIH

ki

zO

z

qq

q

q

cossin

sin

sin

jII

HM

z

OO

qq

sincos

Couple is applied about an axis

perpendicular to the precession

and spin axes

When the precession and spin

axis are at a right angle,

jIM O

q

90

Gyroscope will precess about an

axis perpendicular to both the

spin axis and couple axis.



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Motion of an Axisymmetrical Body Under No Force

18 - 45

• Consider motion about its mass center of an

axisymmetrical body under no force but its own

weight, e.g., projectiles, satellites, and space craft.

constant 0 GG HH

• Define the Z axis to be aligned with and z in a

rotating axes system along the axis of symmetry.

The x axis is chosen to lie in the Zz plane.

GH

xGx IHH q sinI

HGx

q

sin

yy IH 0 0y

zGz IHH q cosI

HGz

q

cos

• q = constant and body is in steady precession.

• Note: q

tantan

I

I

z

x



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18 - 46

Two cases of motion of an axisymmetrical body

which under no force which involve no precession:

• Body set to spin about its axis of symmetry,

aligned are and

0

G

xx

H

H

and body keeps spinning about its axis of

symmetry.

• Body is set to spin about its transverse axis,

aligned are and

0

G

zz

H

H

and body keeps spinning about the given

transverse axis.



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18 - 47

The motion of a body about a fixed point (or its mass

center) can be represented by the motion of a body

cone rolling on a space cone. In the case of steady

precession the two cones are circular.

• I < I’. Case of an elongated body. < q and the

vector lies inside the angle ZGz. The space

cone and body cone are tangent externally; the

spin and precession are both counterclockwise

from the positive z axis. The precession is said to

be direct.

• I > I’. Case of a flattened body. > q and the

vector lies outside the angle ZGz. The space

cone is inside the body cone; the spin and

precession have opposite senses. The precession

is said to be retrograde.



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Concept Question

18 - 48

The rotor of the gyroscope shown

to the right rotates with a constant

angular velocity. If you hang a

weight on the gimbal as shown,

what will happen?

a) It will precess around to the right

b) Nothing – it will be in equilibrium

c) It will precess around to the left

d) It will tilt down so the rotor is horizontal

vector mechanics for engineers: chapter...

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