mechanics of rigid bodies

Mechanics of Rigid Bodies

Rigid body

• Rigid body: a system of mass points subject to the holonomic constraints that the distances between all pairs of points remain constant throughout the motion

• If there are N free particles, there are 3N degrees of freedom

• For a rigid body, the number of degrees of freedom is reduced by the constraints expressed in the form:

• How many independent coordinates does a rigid body have?

4.1

ijij cr

The independent coordinates of a rigid body

• Rigid body has to be described by its orientation and location

• Position of the rigid body is determined by the position of any one point of the body, and the orientation is determined by the relative position of all other points of the body relative to that point

4.1


• Position of one point of the body requires the specification of 3 independent coordinates

• The position of a second point lies at a fixed distance from the first point, so it can be specified by 2 independent angular coordinates

4.1

0


• The position of any other third point is determined by only 1 coordinate, since its distance from the first and second points is fixed

• Thus, the total number of independent coordinates necessary do completely describe the position and orientation of a rigid body is 6

4.1

0

Orientation of a rigid body

• The position of a rigid body can be described by three independent coordinates,

• Therefore, the orientation of a rigid body can be described by the remaining three independent coordinates

• There are many ways to define the three orientation coordinates

• One common ways is via the definition of direction cosines

4.1

Direction cosines

• Direction cosines specify the orientation of one Cartesian set of axes relative to another set with common origin

• Orthogonality conditions:

4.1

333231

232221

131211

cosˆcosˆcosˆ'ˆ

cosˆcosˆcosˆ'ˆ

cosˆcosˆcosˆ'ˆ

kjik

kjij

kjii

1'ˆ'ˆ'ˆ'ˆ'ˆ'ˆˆˆˆˆˆˆ

0'ˆ'ˆ'ˆ'ˆ'ˆ'ˆˆˆˆˆˆˆ

kkjjiikkjjii

ikkjjiikkjji

Orthogonality conditions

• Performing similar operations for the remaining 4 pairs we obtain orthogonality conditions in a compact form:

4.1

'ˆ'ˆ ii)cosˆcosˆcosˆ()cosˆcosˆcosˆ( 131211131211 kjikji

'ˆ'ˆ ji

)cosˆcosˆcosˆ()cosˆcosˆcosˆ( 232221131211 kjikji

132

122

112 coscoscos 1

231322122111 coscoscoscoscoscos 0

ikl

lkli

3

1

coscos

Orthogonal transformations

• For an arbitrary vector

• We can find components in the primed set of axes as follows:

• Similarly

4.1

321ˆˆˆ GkGjGiG

3211ˆ'ˆˆ'ˆˆ'ˆ'ˆ' GkiGjiGiiGiG

3131211

2131211

1131211

ˆ)cosˆcosˆcosˆ(

ˆ)cosˆcosˆcosˆ(

ˆ)cosˆcosˆcosˆ(

Gkkji

Gjkji

Gikji

313212111 coscoscos GGG

3332321313

3232221212

coscoscos'

coscoscos'

GGGG

GGGG

Orthogonal transformations

• Therefore, orthogonal transformations are defined as:

• Orthogonal transformations can be expressed as a matrix relationship with a transformation matrix A

• With orthogonality conditions imposed on the transformation matrix A

4.2

ijijj

jiji aGaG cos ;'3

1

AGG '

ikl

lkliaa

3

1

Properties of the transformation matrix

• Introducing a matrix inverse to the transformation matrix

• Let us consider a matrix element

• Orthogonality conditions

4.3

jlk

klkjaa

3

1

1AA 1ki

lliklaa

3

1

3

1

3

1k lliklkj aaa

3

1

3

1k lliklkj aaa

3

1

3

1l kklkjli aaa

3

1ljllia

3

1kkikja

jia

ija

ija

AA~1


• Calculating the determinants

• The case of a negative determinant corresponds to a complete inversion of coordinate axes and is not physical (a.k.a. improper)

4.3

1~ AA 1~ AAAA 1AA ~

AA~

AA~

AA2

A 1 1

1 A


• In a general case, there are 9 non-vanishing elements in the transformation matrix

• In a general case, there are 6 independent equations in the orthogonality conditions

• Therefore, there are 3 independent coordinates that describe the orientation of the rigid body

4.14.2

ikl

lkli

3

1

coscos1'ˆ'ˆ'ˆ'ˆ'ˆ'ˆ

0'ˆ'ˆ'ˆ'ˆ'ˆ'ˆ

kkjjii

ikkjji

333231

232221

131211

aaa

aaa

aaa

A

Example: rotation in a plane

• Let’s consider a 2D rotation of a position vector r

• The z component of the vector is not affected, therefore the transformation matrix should look like

• With the orthogonality conditions

• The total number of independent coordinates is

4 – 3 = 1

4.2

100

0

0

2221

1211

aa

aa

A

jlk

klkjaa

2

1

2,1, lj


• The most natural choice for the independent coordinate would be the angle of rotation, so that

• The transformation matrix

4.2

33

212

211

'

cossin'

sincos'

xx

xxx

xxx

100

0cossin

0sincos

A


• The three orthogonality conditions

• They are rewritten as

4.2

0cossinsincos

1cossin

1sincos22

22

0

1

1

22211211

22221212

21211111

aaaa

aaaa

aaaa


• The 2D transformation matrix

• It describes a CCW rotation of the coordinate axes

• Alternatively, it can describe a CW rotation of the same vector in the unchanged coordinate system

4.2

100

0cossin

0sincos

A

The Euler angles

• In order to describe the motion of rigid bodies in the canonical formulation of mechanics, it is necessary to seek three independent parameters that specify the orientation of a rigid body

• The most common and useful set of such parameters are the Euler angles

• The Euler angles correspond to an orthogonal transformation via three successive rotations performed in a specific sequence

• The Euler transformation matrix is proper

4.4

Leonhard Euler(1707 – 1783)

1A

The Euler angles

• First, we rotate the system around the z’ axis

• Then we rotate the system around the x’’ axis

4.4

'

'

'

100

0cossin

0sincos

z

y

x

''' Dxx

''

''

''

cossin0

sincos0

001

z

y

x

''CxX

The Euler angles

• Finally, we rotate the system around the Z axis

• The complete transformation can be expressed as a product of the successive matrices

4.4

Z

Y

X

100

0cossin

0sincos

BXx

BXx ''BCx 'BCDx 'Ax

'Axx

The Euler angles

• The explicit form of the resultant transformation matrix A is

• The described sequence is known as the x-convention

• Overall, there are 12 different possible conventions in defining the Euler angles

4.4

coscossinsinsin

sincoscoscoscossinsincossincoscossin

sinsinsincoscossincossinsincoscoscos

BCDA

Euler theorem

• Euler theorem: the general displacement of a rigid body with one point fixed is a rotation about some axis

• If the fixed point is taken as the origin, then the displacement of the rigid body involves no translation; only the change in orientation

• If such a rotation could be found, then the axis of rotation would be unaffected by this transformation

• Thus, any vector lying along the axis of rotation must have the same components before and after the orthogonal transformation:

4.6

RARR '

Euler theorem

• This formulation of the Euler theorem is equivalent to an eigenvalue problem

• With one of the eigenvalues

• So we have to show that the orthogonal transformation matrix has at least one eignevalue

• The secular equation of an eigenvalue problem is

• It can be rewritten for the case of

4.6

RAR 1RAR 0)( R1A

0)( R1A

1

1

0 1A 1

0 1A

Euler theorem

• Recall the orthogonality condition:

• n is the dimension of the square matrix

• For 3D case:

• It can be true only if

4.6

1AA ~

A1AAA~~~

A1A1A~~

)( A1A1A~~

)(

A1A1A~~

A1A1A~

1A

A11A~

~A11A A11A )( 1A1A

1A1A n)1(

1A1A 3)1( 1A1A

0 1A ... DEQ

Euler theorem

• For 2D case (rotation in a plane) n = 2:

• Euler theorem does not hold for all orthogonal transformation matrices in 2D: there is no vector in the plane of rotation that is left unaltered – only a point

• To find the direction of the rotation axis one has to solve the system of equations for three components of vector R:

• Removing the constraint, we obtain Chasles’ theorem: the most general displacement of a rigid body is a translation plus a rotation

4.6

0)( R1A

1A1A n)1( 1A1A Michel Chasles(1793–1880)

Infinitesimal rotations

• Let us consider orthogonal transformation matrices of the following form

• Here α is a square matrix with infinitesimal elements

• Such matrices A are called matrices of infinitesimal rotations

• Generally, two rotations do not commute

• Infinitesimal rotations do commute

4.8

RAARAA 1221

α1A

))(( 21 α1α1 2121 αα1α1α1 21 αα1

))(( 12 α1α1 1212 αα1α1α1 12 αα1


• The inverse of the infinitesimal rotation:

• Proof:

• On the other hand:

• Matrices α are antisymmetric

• In 3D we can write:

• Infinitesimal change of a vector:

4.8

α1A 1

Aα1 )( ))(( α1α1 αα1αα11 1

AA~1 α1α1 ~ αα ~

0

0

0

12

13

23

dd

dd

dd

α

rrr 'd rrα1 )( αr

3

1

)(j

jiji rdr


• is a differential vector, not a differential of a vector

• is normal to the rotation plane

4.8

0

0

0

12

13

23

dd

dd

dd

α

3

1

)(j

jiji rdr

12213

31132

23321

)(

)(

)(

drdrdr

drdrdr

drdrdrk

kjjijki drdr

3

1,

)(

)(drrd

)(d

)(d

dnrrd )(

dnd

)(

drdr )sin(


• These matrices are called infinitesimal rotation generators

4.8

0

0

0

12

13

23

dd

dd

dd

α

d

nn

nn

nn

0

0

0

12

13

23

3

1iiind Mα

010

100

000

1M

000

001

010

;

001

000

100

32 MM

dnd

)(

3

1ikijkijji MMMMM

Example: infinitesimal Euler angles

• For infinitesimal Euler angles it can be rewritten as

4.8

coscossinsinsin



A

10

1)(

01

d

ddd

dd

A α1

00

0)(

00

d

ddd

dd

α

0

0

0

12

13

23

dd

dd

dd

α

)(ˆˆ)( ddkdid

Rate of change of a vector

• Dividing by dt

4.9

AGG '

3

1

'j

jiji GaG

3

1kkijkij d

iii ddGdG ΩG'

ωGGG

dt

d

dt

d 'Ωω ddt ωGGG '

3

1

'j

ijjjiji daGdGadG

ijijijijij daa

3

1

)(j

ijjjijij daGdGda

3

1jijjjij GdG

3

1

3

1j kkijkji dGdG

3

1,kjkjijk

i

dG

dG

Example: the Coriolis effect

• Velocity vectors in the rotating and in the “stationary” systems are related as

• For the rate of change of velocity

• Rotating system: acceleration acquiresCoriolis and centrifugal components

4.10

GωGG '

Gaspard-GustaveCoriolis

(1792 - 1843)

rωvv rs

)( 2 rωωvωaa rsr

rωrr rs

srsss vωvv )()( ) (

) (rωvω

dt

rωvdr

r

r

)( 2 )( rωωvωv rrr

constω

Example: the Coriolis effect

• On the other hand

• This is the Lorentz acceleration

• What is the relationship between those two?

ωva rc

2

m

BqvaL

m

BqL

LL va

Kinetic energy of a system of particles

• Kinetic energy of a system of particles

• Introducing a center of mass:

• We can rewrite the coordinates in the center-of-mass coordinate system:

• Kinetic energy can be rewritten:

1.2

i

ii rmT 2)(2

1

ii

iii

m

rmR

Rrr ii

' Rrr ii

'

i

ii rmT 2)(2

1 i

iii RrRrm )'()'(2

1

i

ii

iii

ii RmRrmrm 22 )(2

1 )'()'(

2

1

M

rmi

ii

RMrmi

ii


• On the other hand

• In the center-of-mass coordinate system, the center of mass is at the origin, therefore

1.2

i

ii

iii

ii mRrmRrm 22 )(2

1 ')'(

2

1

MRrmdt

dRrm

iii

iii

22 )(2

1 ')'(

2

1

i

ii

iii

ii RmRrmrmT 22 )(2

1 )'()'(

2

1

RMrmi

ii

'' RMrm

iii

MRrmTi

ii22 )(

2

1 )'(

2

1


• Kinetic energy of the system of particles consists of a kinetic energy about the center of mass plus a kinetic energy obtained if all the mass were concentrated at the center of mass

• This statement can be applied to the case of a rigid body: Kinetic energy of a rigid body consists of a kinetic energy about the center of mass plus a kinetic energy obtained if all the mass were concentrated at the center of mass

• Recall Chasles’ theorem!

1.25.1

MRrmTi

ii22 )(

2

1 )'(

2

1


• Chasles: we can represent motion of a rigid body as a combination of a rotation and translation

• If the potential and/or the generalized external forces are known, the translational motion of center of mass can be dealt with separately, as a motion of a point object

• Let us consider the rotational part or motion

5.1

MRrmTi

ii22 )(

2

1 )'(

2

1

i

iiR rmT 2)'(2

1

Rotational kinetic energy

• Rate of change of a vector

• For a rigid body, in the rotating frame of reference, all the distances between the points of the rigid body are fixed:

• Rotational kinetic energy:

5.3

i

iiR rmT 2)'(2

1 i

iii rrm ''2

1

' )'()'( irisi rωrr

0)'( rir ')'( isi rωr

i

iii rωrωm )'()'(2

1

i

iiiR rωrωmT )'()'(2

1

i j

jijii rωrωm3

1

)'()'(2

1

i j nmnimjmn

lklikjkli rωrωm

3

1

3

1,

3

1,

''2

1

Rotational kinetic energy5.3

i j nmnimjmn

lklikjkliR rωrωmT

3

1

3

1,

3

1,

''2

1

i nmlkj

nilimkjmnjkli rrωωm3

1,,,,

''2

1 knlmnlkm

jjmnjkl

3

1

i nmlk

nilimkknlmnlkmi rrωωm3

1,,, '')(

2

1

i lkllikik

lkliki ωrrωrωm

3

1,

3

1,

22 '')'()(2

1

3

1,

2 ]'')'[(2

1

lklikikli

iilk rrrmωω

]'')'[( 2likikli

iikl rrrmI

3

1,2

1

lklklk ωIω

2

~Iωω

Inertia tensor and moment of inertia

• (3x3) matrix I is called the inertia tensor

• Inertia tensor is a symmetric matrix (only 6 independent elements):

• For a rigid body with a continuous distribution of density, the definition of the inertia tensor is as follows:

• Introducing a notation

• Scalar I is called the moment of inertia

5.3

2

~IωωRT

lkkl II

dVrrrIV

lkklkl ])[( 2

nω ω 2

~IωωRT

2

~ ωω Inn

2

2Iω

Inn~I

]'')'[( 2likikli

iikl rrrmI

Inertia tensor and moment of inertia

• On the other hand:

• Therefore

• The moment of inertia depends upon the position and direction of the axis of rotation

5.3

2

2IωTR

i

iiiR rωrωmT )'()'(2

1 i

iii rnrnmω

)'()'(2

2

i

iii rnrnmI )'()'(

Parallel axis theorem

• For a constrained rigid body, the rotation may occur not around the center of mass, but around some other point 0, fixed at a given moment of time

• Then, the moment of inertia about the axis of rotation is:

5.3

Rrr ii

'

i

iii rnrnmI )()(0

i

iii RrnRrnm ))'(())'((

ii

iii

iii

Rnm

Rnrnmrnm

2

2

)(

)()'(2)'(

CMI

MRnRnrmni

ii2)()()'(2


• Parallel axis theorem: the moment of inertia about a given axis is equal to the moment of inertia about a parallel axis through the center of mass plus the moment of inertia of the body, as if concentrated at the center of mass, with respect to the original axis

5.3

20 )( RnMII CM


• Does the change of axes affect the ω vector?

• Let us consider two systems of coordinates defined with respect to two different points of the rigid body: x’1y’1z’1 and x’2y’2z’2

• Then

• Similarly

5.1

RRR

12

sss RRR )( )()( 12 RωRR rs

11 )( )(

sss RRR )( )()( 21

RωRR rs

22 )( )(

RωRR ss

112 )()(

RωRR ss

221 )()(

0)( 21 Rωω


• Any difference in ω vectors at two arbitrary points must be parallel to the line joining two points

• It is not possible for all the points of the rigid body

• Then, the only possible case:

• The angular velocity vector is the same for all coordinate systems fixed in the body

5.1

0)( 21 Rωω

21 ωω

Example: inertia tensor of a homogeneous cube

• Let us consider a homogeneous cube of mass M and side a

• Let us choose the origin at one of cube’s corners

• Then

5.3

dVrrrIV

lkklkl ])[( 2

a a a

drdrdrrrrI0 0

321

0

112

11 ])[( a a a

drdrdrrr0 0

321

0

23

22 ])()[(

a a

drdrrra0

32

0

23

22 ])()[(

3

2 5a

3

2 2Ma 3322 II

Example: inertia tensor of a homogeneous cube

5.3

dVrrrIV

lkklkl ])[( 2

a a a

drdrdrrrI0 0

321

0

2112 ][

3

2

4

1

4

14

1

3

2

4

14

1

4

1

3

2

2MaI

a a

drdrrra0

21

0

21 ][4

5a

4

2Ma

322331132112 IIIIII

Angular momentum of a rigid body

• Angular momentum of a system of particles is:

• Rate of change of a vector

• For a rigid body, in the rotating frame of reference, all the distances between the points of the rigid body are fixed:

• Angular momentum of rigid body:

5.1

i

iii rrmL )(

irisi rωrr )()(

0)( rir

isi rωr )(

i

iii rωrmL ))((

i lk nminmlmnikjklij rωrmL

3

1,

3

1,

i nmlk

iminiklmnjkl mωrr3

1,,,

Angular momentum of a rigid body

• Rotational kinetic energy:

5.1

i nmlk

iminiklmnjklj mωrrL3

1,,,

i nmk

iminikkmjnknjm mωrr3

1,,

)(

3

1

2 ])[(k

ikijjkii

ik rrrmω

3

1kkjkωI

IωL

2

~Lω

2

~ωL

2

~IωωRT

Free rigid body

• For a free rigid body, the Lagrangian is:

• Recall

• Then

• We separate the Lagrangian into two independent parts and consider the rotational part separately

• Then, the equations of motion for rotation

5.55.6

MRrmTLi

ii22 )(

2

1 )'(

2

1 MRωIωlk

lklk2

3

1,

)(2

1

2

1

ii ddtω

CMlk

lklk TIL

3

1,2

1

i

R

i

R LL

dt

d

0

3

1

kkikI

dt

d 03

1

k

kikωIdt

d

Free rigid body

• Angular momentum of a free rigid body is constant

• In the system of coordinates fixed with the rotating rigid body, the tensor of inertia is a constant – it is often convenient to rewrite the equations of motion in the rotating frame of reference:

5.55.6

03

1,

3

1

llk

kiklk

kik LωωIdt

d

0dt

dLi 0dt

Ld

0

Lω

dt

Ld

dt

Ld

rs

03

1,,

3

1

lkmlmlk

iklk

kik ωωIωI

03

1

k

kikωIdt

d

Principal axes of inertia

• Inertia tensor is a symmetric matrix

• In a general case, such matrices can be diagonalized – we are looking for a system of coordinates fixed to a rigid body, in which the inertia tensor has a form:

• To diagonalize the inertia tensor, we have to find the solutions of a secular equation

5.4

3

2

1

00

00

00

I

I

I

I

0

333231

232221

131211

IIII

IIII

IIII


• Coordinate axes, in which the inertia tensor is diagonal, are called the principal axes of a rigid body; the eigenvalues of the secular equations are the components of the principal moment of inertia

• After diagonalization of the inertia tensor, the equations of motion for rotation of a free rigid body look like

• After diagonalization of the inertia tensor, the rotational kinetic energy a rigid body looks like

5.4

03

1,

kkjkj

ijkii IωωωI

3

1

2

2

1

iiiR ωIT


• To find the directions of the principal axes we have to find the directions for the eigenvectors ω

• When the rotation occurs around one of the principal axes In, there is only one non-zero component ωn

• In this case, the angular momentum has only one component

• In this case, the rotational kinetic energy has only one term

5.4

knkkk ωIL

22

1 23

1

2 nn

iiiinR

ωIωIT

Stability of a free rotational motion

• Let us choose the body axes along the principal axes of a free rotating rigid body

• Let us assume that the rotation axis is slightly off the direction of one of the principal axes (α - small parameter):

• Then, the equations of motion

5.6

332211ˆˆˆ iiiωω

03

1,

kkjkj

ijkii IωωωI

0)(

0)(

0)(

122133

313122

233211

IIωωωI

IIωωωI

IIωωωI

0)(

0)(

0)(

122133

313122

233211

IIωI

IIωI

IIωI

Stability of a free rotational motion5.6

0)(

0)(

0)(

122133

313122

233211

IIωI

IIωI

IIωI

0

0

0

3

12213

2

31312

1

I

IIω

I

IIω

ω

0))(( 312132

21

)3(2)3(2

1

IIIIII

ω

constω

0)3(2)3(2 K ))(( 312132

21 IIIIII

ωK

Stability of a free rotational motion

• The behavior of solutions of this equation depends on the relative values of the principal moments of inertia

• Always stable

• Exponentially unstable

5.6

0)3(2)3(2 K ))(( 312132

21 IIIIII

ωK

0K 2K3121 ; IIII

3121 ; IIII

0)3(22

)3(2

)cos( )3(2)3(2)3(2 tA

0K 2K213 III

312 III

0)3(22

)3(2 teA )3(2)3(2

teA )3(2)3(2

Classification of tops

• Depending on the relative values of the principle values of inertia, rigid body can be classified as follows:

• Asymmetrical top:

• Symmetrical top:

• Spherical top:

• Rotator:

321 III

321 III

321 III

0;0 321 III

Example: principal axes of a uniform cube

• Previously, we have found the inertia tensor for a uniform cube with the origin at one of the corners, and the coordinate axes along the edges:

• The secular equation:

3

2

4

1

4

14

1

3

2

4

14

1

4

1

3

2

2MaI

03

2

483

2

12

11 2242222

I

MaMaaMI

MaI

Ma

0

3

2

44

43

2

4

443

2

222

222

222

IMaMaMa

MaI

MaMa

MaMaI

Ma


• To find the directions of the principal axes we have to find the directions for the eigenvectors

• Let us consider

03

2

483

2

12

11 2242222

I

MaMaaMI

MaI

Ma

12

11 2

1

MaI

6;

12

11 2

3

2

2

MaI

MaI

6

2

3

MaI

33

23

13

3

ω

ω

ω

ω333 1ωIω I


13

2

33

2

23

2

13

2

6443

2ω

Maω

Maω

Maω

Ma

23

2

33

2

23

2

13

2

643

2

4ω

Maω

Maω

Maω

Ma

33

2

33

2

23

2

13

2

63

2

44ω

Maω

Maω

Maω

Ma

12

33

23

33

13 ω

ω

ω

ω

12

33

23

33

13 ω

ω

ω

ω

233

23

33

13 ω

ω

ω

ω

2313 ωω

3313 ωω 332313 ωωω

Free symmetrical top

• For a free symmetrical top:

5.6

03

1,

kkjkj

ijkii IωωωI 0)(

0)(

0)(

122133

313122

233211

IIωωωI

IIωωωI

IIωωωI

321 III

0

0)(

0)(

33

313121

133211

ωI

IIωωωI

IIωωωI

constω

I

IIωωω

I

IIωωω

3

1

31312

1

31321

)(

)(

2

1

31311

)(

I

IIωωω 2

1ω tAω

tAω

sin

cos

2

1

Motion of non-free rigid bodies

• How to tackle rigid bodies that move in the presence of a potential or in an open system with generalized forces (torques)?

• Many Lagrangian problems of such types allow separation of the Lagrangians into two independent parts: the center-of-mass and the rotational

• For the non-Lagrangian (open) systems, we modify the equations of motion via introduction of generalized forces (torques) N:

5.5

ikkjkj

ijkii NIωωωI

3

1,

Heavy symmetrical top with one point fixed

• For this problem, it is convenient to use the Euler angles as a set of independent variables

• Let us express the components of ω as functions of the Euler angles

• The general infinitesimal rotation associated with ω can be considered as consisting of three successive infinitesimal rotations with angular velocities

5.74.9

ωωω ;;

ωωωω


4.9

0

0

A

z

y

x

'''' AxBCDxBCxBXx

0

0

coscossinsinsin



A

cos

sincos

sinsin

z

y

x


4.9

0

0

B

z

y

x

'''' AxBCDxBCxBXx

0

0

0

sin

cos

z

y

x

100

0cossin

0sincos

B


4.9

'''' AxBCDxBCxBXx

0

0

z

y

x

0

0

0

sin

cos

z

y

x

cos

sincos

sinsin

z

y

x


4.9

z

y

x

0

0

0

sin

cos

z

y

x

cos

sincos

sinsin

z

y

x

cos

sinsincos

cossinsin


• The Lagrangian:

• Using the Euler angles

5.7

VTL

RotationnTranslatio TTT 22

)( 233

22

211 II

dVgrV dVrg

RMrmi

ii

MRg

cosgRMV

21 II


5.7

cosgRMV

cos

sinsincos

cossinsin

2

)cos( 23

I

2

)cossinsin()sinsincos( 22

1

I

2

)(

2

22

211

233

IIT

2

sin

2

)cos( 222

1

2

3

II


• The Lagrangian is cyclic in two coordinates

• Thus, we have two conserved generalized momenta

5.7

cos

2

sin

2

)cos( 222

1

2

3 gRMIIL

0;0

LL

aIconstIL

p

bIconstIIL

p

13

12

12

3

)cos(

)sin()coscos(


• The Lagrangian does not contain time explicitly

• Thus, the total energy of the system is conserved

• To solve the problem completely, we need three additional quadratures

• We will look for them, using the conserved quantities

5.7

cos

2

sin

2

)cos( 222

1

2

3 gRMIIL

0t

L

constgRMIIE

cos

2

sin

2

)cos( 222

1

2

3


• One variable only: we can find all the quadratures!

5.7

aII 13 )cos(

bIII 12

12

3 )sin()coscos(

cos313 IaII

bIIIaII 12

12

312

3 )sin()coscos()cos(

ba 2sincos

2sin

cosab )(1 f

cos)(13

1 faI

I )(2 f

cos

2

sin)(

2

))(cos)(( 2221

1

221

3 gRMf

Iff

IE


• We have an equivalent 1D problem with an effective potential!

5.7

cos

2

sin)(

2

))(cos)(( 2221

1

221

3 gRMf

Iff

IE

cos

2

sin)(

2

))(cos)((

2

2211

2213

21 RMg

fIffIIE

cos2

sin)(

2

))(cos)(()(

2211

2213 RMg

fIffIVeff

aII 13 )cos(

cossin2

)cos(

2

)(2

21

3

21 gRM

abI

I

aI

cos

sin2

)cos()('

2

21 gRM

abIVeff


• In the most general case, the integration involves elliptic functions

• Effective potential is a function with a minimum: motion in θ is bound by two values

5.7

)('2

'2

1 effV

IE

1

2 )](''[2

I

VE

dt

d eff

)](''[/2)(

1

effVEI

dt


• When θ is at its minimum, we have a precession

• Otherwise, the top is bobbing: nutation

• The shape of the nutation trajectory depends on the behavior of the time derivative of φ

5.7

2sin

cosab

Charged rigid body in an electromagnetic field

• Let us consider the following vector potential (C – constant vector)

• How is magnetic field related to vector C?

5.9

iiiii

ziyixii Arqrqrrrm

L )()(2

)( 2,

2,

2,

AB

3

1,

)(kj

kjijkii rCrCA rCA

nn AB )(

m

i

imnmi r

A

3

1,

3

1,,, mkji m

kjijknmi r

rC

3

1,,

)(mkj m

kjmjnkmknj r

rC


• Constant magnetic field

5.9

3

1,,

)(mkj m

kjmjnkmknjn r

rCB

3

1

3

1 m m

nm

m m

mn r

rC

r

rC

3

1

3m

mnmn CC nC2 nB

2

BC

2

rBA

i

iiiii

ziyixii rBrqrq

rrrmL

2)(

2

)( 2,

2,

2,


• Let us assume a uniform charge/mass ratio

• Recall rotational kinetic energy

5.9

i

iiiii

ziyixii rBqrrq

rrrmL

2)(

2

)( 2,

2,

2,

)()( acbcba

i

iii

rBqr

2

iiii

i

i mrrm

Bq)(

2

i

iii mrrm

Bq)(

2

L

m

Bq

2

2

~LωRT

2

~)/( LBmq

Radius of gyration

• FYI: radius of gyration is

5.4

M

IR 0

20MRI

2

2IωTR 2

)( 20ωRM

mechanics of rigid bodies

Documents

rigid body position

transformation matrix

body relative

orientation coordinates

rigid body4

matrix inverse

matrix relationship

relative position