From Riley’s DynamicsChapter 16

Kinetics of Rigid Bodies:Newton’s Laws

(Q) What are the Euler’s Equations of Motion?

Newton’s Law applies only to the motion of a single particle

translation

G G

RR

only translation translation + rotation

GamR

Newton’s 2nd Law Euler’s Equations of Motion

dmf

dmF

Axyz

OXYZ

on force internal:

on force external:

system coordinate fixed-body:}{

system coordinate fixed-space:}{

Euler’s Equations of Motion Rotation of a Rigid Body

moment

∴ Starting Point Moment of F & f about A

Newton’s 2nd Law Substitution yields

What’s this?

then

&let Now,

.0&0 body, rigid Since

2)(

Kinematics :Recall

//

AAdmB

av

avrraa

BrelBrel

BrelBrelABABAB

After integration, we can get the general form of the Euler’s equations of motion.

Very general equation about rotation. Need to unify the coordinate systems to {Axyz}.

(Q) Simplified Version Plane Motion

//

motion plane

Mass center G lies in the xy-plane.

z

z

yxAza

0

r

dm

mmm AA dmrrdmrrdmarM )]}([{)]([)( Recall

Now, kyaxajzaiza

aa

zyx

kji

ar AxAyAxAy

AyAx

A

)()()(

0

After the similar calculation, we have

where

kMjMiMM AzAyAxA

product of inertia

moment of inertia

Using

0

(Note) The 1st 2 equations are required to maintain the plane motion about z-axis, especially for non-symmetrical geometry case.

Motion of Equations sEuler'-Newton)( equations) 3 (Above GamR

If the body is symmetricabout the plane of motion,

section.)next the(See

0 AzAzx II

If (symmetry) + (acceleration of the point A = 0)

If(symmetry) + (A = G)

JT known- wellThe

(Q) More about the Moment of Inertia

For the particle dm

For the entire body

It uses the information about its geometry.

∴ THE SAME MASS BUT DIFFERENT GEOMETRY

DIFFERENT MOMENT OF INERTIA

IF widely distributedTHEN larger moment of inertia

There are various ways of choosing this small mass element for integration.A specific mass element may be easier to use than other elements.

You may treat the rigid body as a system of particles.

0

0

2nd

momentof area

If the density of the body is uniform,

Practical approach

a rigid body summation of several simple shape rigid bodies

composite body

(Q) What is the radius of gyration?gyration [ʤaiəreiʃən] n. U,C 선회 , 회전 , 선전 ( 旋轉 ); 〖동물 〗 ( 고둥 따위의 ) 나선 . ㉺∼ al [-ʃənəl] ―a. 선회의 , 회전의 .

km

m

I : moment of inertia about the axis

(the moment of inertia aboutthe axis) = mk2

=

NO useful physical interpretation!!

Maybe baseball Home Run !!!!!

(Q) What is the Parallel-Axis Theorem for Moments of Inertia?

0

0measurement of the locationof the mass center from themass center

= m

z’

2

22

2'

2

32

1

mR

mRmR

mRII zz

(Q) More about the Product of Inertia

dm

x

yRz

x

y

In 2-D space

dmRdI zz2 (Note)

(Q) What is the effect of symmetry on the product of inertia?

x

y

z

x

y

y

z

x

z

0

m

xy dmxyI

0

m

yz dmyzI

0

m

xz dmxzI

x

yz

x

z z

y

x

y

0

m

xz dmxzI

0

m

xy dmxyI

0

m

yz dmyzI

(Q) What is the Parallel-Axis Theorem for Product of Inertia?

From definition

or

But, mass center from the mass center

0 0

and

Therefore,

AXES. PRINCIPAL

are axes then

,0 If (Note)

xyz

III zxyzxy

(Q) What is the Rotation Transformation of Inertia Properties?

Consider

x’y’

z’

x

y

z

vectorarbitrary an

frame new }{

frame old}{

xyz

zyx

We know that

)2(

)1(

kji

kji

zyx

zyx

We can represent i’, j’, and k’ w.r.t. i, j, and k.

etc. axis,-y to of projection

where

)3(

ir

krjrirk

krjrirj

krjriri

yx

zzyzxz

zyyyxy

zxyxxx

Substitution yields

krrr

jrrr

irrr

krjrir

krjrir

krjrir

kji

kji

zzzzyyzxx

yzzyyyyxx

xzzxyyxxx

zzyzxzz

zyyyxyy

zxyxxxx

zyx

zyx

)(

)(

)(

)(

)(

)(

zzzzyyzxxz

yzzyyyyxxy

xzzxyyxxxx

rrr

rrr

rrr

or

z

y

x

zzzyzx

yzyyyx

xzxyxx

z

y

x

rrr

rrr

rrr

a vector inthe new frame

a vector inthe old frame

[R] rotation transformationmatrix from old to new frame

oldnew

(Example)

x’

y’

x

y

Θ

Θ

z

y

x

zzzyzx

yzyyyx

xzxyxx

z

y

x

rrr

rrr

rrr

1,0,0

0,cos,sin

0,sin,cos

'''

'''

'''

zzzyzx

yzyyyx

xzxyxx

rrr

rrr

rrr

100

0cossin

0sincos

R

1

100

010

001

100

0cossin

0sincos

,

100

0cossin

0sincos

(Note)

RR

IRR

RR

T

T

T

It means that [R] is an orthonormal matrix.

Rotation about z’-axis

y’

z’

y

z

Θ

Θ

z’

x’

z

x

Θ

Θ

Rotation about x’-axis

cossin0

sincos0

001

zzzyzx

yzyyyx

xzxyxx

rrr

rrr

rrr

cos0sin

010

sin0cos

zzzyzx

yzyyyx

xzxyxx

rrr

rrr

rrr

Rotation about y’-axis

Now, the rotational kinetic energy is ATArot HHT

2

1

2

1

matrix inertia

where

momentumangular

zzyzxz

yzyyxy

xzxyxx

A

III

III

III

I

IH

IT Trot 2

1

.2

1 is -axes''' w.r.t.

and2

1 is axes- w.r.t.Let

ITzyxT

ITxyzT

Trottot

Trottot

Since energy is invariant

II TT

Let : knownold frame

Let : unknownnew frame

T

TTT

TT

T

RIRI

IRIR

II

RR

R

have weSo,

becomes

Therefore

'

that know We1

new from old to newold

This term will be derived in the nextchapter.

(Example)a = 240 mm

b = 120 mm

c = 90 mm

x

y

z

m = 60 kg

Claim: [I] = ?

(Idea)

x

y

z

x’

z’

y’

G

x

z

y

IzyxI

IzyxI

zyxI

using w.r.t. Calculate(3)

ˆ using w.r.t. Calculate(2)

ˆˆˆ w.r.t.ˆ Calculate(1)

G

x

z

y

0

m-kg 03285.0)(12

1

m-kg 01125.0)(12

1

m-kg 036.0)(12

1

ˆˆˆˆˆˆ

222ˆˆ

222ˆˆ

222ˆˆ

zxzyyx

zz

yy

xx

III

bamI

cbmI

camI

b

c

a

03285.000

001125.00

00036.0

I

G

x

z

y

b

c

a

x’

y’

z’

Gr

m 06.0

m 12.0

m 045.0

2

2

2

c

a

b

z

y

x

rG

By using the parallel axis theorem,

0162.0

0432.0

0324.0)12.0)(045.0(60

1314.0)(

045.0)(

m-kg 144.0)06.012.0(6036.0)(

ˆˆ''

ˆˆ''

ˆˆ''

22ˆˆ

2ˆˆ''

22ˆˆ

2ˆˆ''

22222ˆˆ

2ˆˆ''

zxmII

zymII

yxmII

yxmImdII

zxmImdII

zymImdII

zxzx

zyzy

yxyx

zzzzzzz

yyyyyyy

xxxxxxx

1314.00432.00162.0

0432.0045.00324.0

0162.00324.0144.0

I

x

y

z

x’

y’

z’

Θ

Θ

ab

c 56.20

240

90tan

a

b

100

09363.03512.0

03512.09363.0

100

0cossin

0sincos

zzzyzx

yzyyyx

xzxyxx

rrr

rrr

rrr

R

zzyzxz

yzyyxy

xzxyxxT

III

III

III

RIRI

1314.00461.00

0461.00359.00081.0

00081.01531.0

Therefore,

Slender rod

Thin rectangular plate

Thin circular plate

Quiz #1

X’

Y’

Z’

{x’y’z’} 좌표 시스템에 대해 표현된 Inertia matrix 를 구하시오 .

(Q) How to analyze the General Plane Motion of NonSymmetric Bodies?

GamR

For Plane Motion

GamR

For Plane Motion

120 mm dia.m = 7.5 kg

30 mm dia.m = 1.2 kgl = 220 mm= 300-120/2-40/2

40 mm dia.8.5 kg

600 rpm ccwincreasing in speedat the rate of60 rpm per second

Bearing A resists any motionin the z-direction.

Claim:5 reactions& T ?

2rad/sec 283.6sec 60

min 1

rev 1

rad 2

sec min

rev60rpm/sec 60

rad/sec 83.62sec 60

min 1

rev 1

rad 2

min

rev600rpm 600

120 mm dia.m = 7.5 kg

30 mm dia.m = 1.2 kgl = 220 mm= 300-120/2-40/2

40 mm dia.8.5 kg

The same result for this sphere sincezG and xG are minus sign.

The same result for this bar sincezG and xG are minus sign.

For the entire system

Or next page

z

x

z’

x’

20

0

6459.012

1

4

1

2556.12

1

22

2

xzGzyGyxG

yGxG

zG

III

mtmRII

mRI

2556.100

06459.00

006459.0

I

z

yzy

xzxyxT

I

II

III

RIRI

R

z

y

x

z

y

x

R

z

y

x

185.10196.0

06459.00

196.007176.0

94.00342.0

010

342.0094.0

20cos020sin

010

20sin020cos

cos0sin

010

sin0cos

Now,

new toold from

oldnew toold fromoldnew

Sym.

(Q) How to analyze the 3-D Motion of a Rigid Body?

GamR

Recall

How?

X

Y

Z

O

x

z

yA

GGr

Ar

dm

r

All vectors are representedw.r.t. the body-fixed {xyz}.

dmf

dmF

Axyz

OXYZ

on force internal:

on force external:

system coordinate fixed-body:}{

system coordinate fixed-space:}{

Euler’s Equations of Motion Rotation of a Rigid Body

moment

∴ Starting Point Moment of F & f about A

Newton’s 2nd Law Substitution yields

What’s this?

then

&let Now,

.0&0 body, rigid Since

2)(

Kinematics :Recall

//

AAdmB

av

avrraa

BrelBrel

BrelBrelABABAB

After integration, we can get the general form of the Euler’s equations of motion.

Very general equation about rotation. Need to unify the coordinate systems to {Axyz}.

If we use the Cartesian coordinate system,

In vector-matrix form,

Gmy

Axxmx Idmzy )( 22

AxyI

AzAyzAxz

AyzAyAxy

AxzAxyAx

A

Az

Ay

Ax

A

G

G

G

G

z

y

x

A

xy

xz

yz

z

y

x

AAGA

III

III

III

I

a

a

a

a

z

y

x

r

IIamrM

,,

where

0

0

0

Or

.0 G,A If

Gr

z

y

x

G

xy

xz

yz

z

y

x

GA IIM

0

0

0

.0 and 0 , symmetric andG A If GzxGyzGxyG IIIr

= 75 rad/sconstant

= 25 rad/sconstant

= 75 rad/sconstant

= 25 rad/sconstant

or moremathematically

= 75 rad/sconstant

= 25 rad/sconstant

0, Therefore

)(

)(

}){(

)(

000

zyzyx

zy

zy

zy

zyy

y

y

zzyy

zy

i

i

jk

jkj

j

j

kkjj

kj

∴ Solvable!

Therefore,