Classical Mechanics notes (8 of 10)

7/30/2019 Classical Mechanics notes (8 of 10)

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Chapter 8. Dynamics of Rigid Bodies(Most of the material presented in this chapter is taken from Thornton and Marion, Chap.11.)

8.1 Notes on NotationIn this chapter, unless otherwise stated, the following notation conventions will be used:

1.Einsteins summation convention. Whenever an index appears twice (and onlytwice), then a summation over this index is implied. For example,

2.i i i i ii i

x x x x = (8.1)

2.The index i is reserved for Cartesian coordinates. For example, , for 1,2,3ix i = ,represents either , , orx y z depending on the value of i . Similarly, ip can represent

, , or x y zp p p . This does not mean that any other indices cannot be used for

Cartesian coordinates, but that the index i will only be used for Cartesian

coordinates.

3.When dealing with systems containing multiple particles, the index will be usedto identify quantities associated with a given particle when using Cartesian

coordinates. For example, if we are in the presence of n particles, the position

vector for particle is given by r , and its kinetic energy T by

, ,1 , 1,2, ... , and 1,2,3.2

i iT m x x n i = = = (8.2)

Take note that, according to convention 1 above, there is an implied summation on

the Cartesian velocity components (the index i is used), but not on the masses

since the index appears more than twice. Correspondingly, the total kinetic

energies is written as

( )2 2 2, ,1 1

1 1.

2 2

n n

i iT m x x m x y z = =

= = + + (8.3)

4. The Kronecker tensor is defined by

1, for

0, for jk

j k

j k

==

(8.4)


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position can be specified using one degree of freedom (or coordinate) for it can only

rotate about the axis connecting 1 2andr r . We thus have used up the six degrees of

freedom. It is interesting to note that in the case of a linear rod, any point 3r must lay on

the axis joining 1 2andr r ; hence a linear rod has only five degrees of freedom.

Usually, the six degrees of freedom are divided in two groups: three degrees fortranslation (to specify the position of the centre 1r , and three rotation angles to specify

the orientation of the rigid body (normally taken to be the so-called Euler angles).

8.3 The Inertia TensorLets consider a rigid body composed of n particles of mass , 1, ... ,m n = . If the

body rotates with an angular velocity about some point fixed with respect to the bodycoordinates (this body coordinate system is what we referred to as a noninertial or

rotating coordinate system in Chapter 7), and if this point moves linearly with a

velocity V with respect to a fixed (i.e., inertial) coordinate system, then the velocity of

the th particle is given by (from Chapter 7)

,= + v V r (8.9)

where we omitted the term

,

rotating

0,rd

dt

=

rv (8.10)

since we are dealing with a rigid body. We have also dropped the f subscript, denoting

the fixed coordinate system, as it is understood that all the non-vanishing velocities willbe measured in this system; again, we are dealing with a rigid body.

The total kinetic energy of the body is given by

( )

( ) ( )

2

2

22

1

2

1

2

1 1.

2 2

T T m v

m

m V m m

= =

= +

= + +

V r

V r r

(8.11)

Although this is an equation for the total kinetic energy is perfectly general, considerablesimplification will result if we choose the origin of the body coordinate system to

coincide with the centre of mass. With this choice, the second term on the right hand side

of the last of equations (8.11) can be seen to vanish from


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( ) 0,m m

= =

V r V r

(8.12)

since the centre of mass Rof the body, of mass , is defined such that

0.m

= r (8.13)

The total kinetic energy can then be broken into two components: one for thetranslational kinetic energy and another for the rotational kinetic energy. That is,

rans rot ,T T T= + (8.14)

with

( )

2 2trans

2

rot

1 12 2

1

2

T m V MV

T m

= =

=

r

(8.15)

The expression for otT can be further modified, but to do so we will now resort to tensor

(or index) notation. So, lets consider the following vector equation

( ) ( ) ( )2

, = r r r (8.16)

and rewrite it using the Levi-Civita and the Kronecker tensors

( )( )

( ), , , ,

, ,

, , , , .

ijk j k imn m n ijk imn j k m n

jm kn jn km j k m n

j j k k j j k k

x x x x

x

x x x x

=

=

=

(8.17)

where we have defined the notation so that ,1 ,2 ,3 , ,( , , ) or i ix x x r x = =r . We could do

the same things and perhaps be clearer if we simply use the

identity2 22 2

( ) ( ) = A B A B A Bi

Inserting this result in the equation for otT in equation (8.15) we get

( )2 2 2

rot

1.

2T m = r r

(8.18)


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Alternatively, keeping with the tensor notation we have

( )

rot , , , ,

, , , ,

, , , ,

,

1

2

1

2

1.

2

j j k k i i j j

i j ij k k i i j j

i j ij k k i j

i j

T m x x x x

m x x x x

m x x x x

=

=

=

(8.19)

where we used i j ij = . We now define the components ijI of the so-called inertia

tensor{{{{ }}}}I by

, , , ,ij ij k k i jI m x x x x

= = = = (8.20)

and the rotational kinetic energy becomes

rot

1

2ij i jT I ==== (8.21)

or in vector notation

{{{{ }}}}rot1

2

T I= = = = (8.22)

For our purposes it will be usually sufficient to treat the inertia tensor as a regular 3 3matrix. Indeed, we can explicitly write I{{{{ }}}} using equation (8.20) as

{{{{ }}}}

(((( ))))

(((( ))))

(((( ))))

2 2

,2 ,3 ,1 ,2 ,1 ,3

2 2

,2 ,1 ,1 ,3 ,2 ,3

2 2

,3 ,1 ,3 ,2 ,1 ,2

m x x m x x m x x

m x x m x x m x x

m x x m x x m x x

I

+ + + +

= + = + = + = +

+ + + +

(8.23)

It is easy to see from either equation (8.20) or equation (8.23) that the inertia tensor is

symmetric, that is,

.ij jiI I==== (8.24)


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The diagonal elements 11 22 33, , andI I I are called the moments of inertia about the

1 2 3-, -, and -axesx x x , respectively. The negatives of the off-diagonal elements are the

products of inertia. We note that the inertia tensor is defined with respect to some originand coordinate directions within the fixed body, and would be different if we had chosen

these otherwise: the moment of inertia tensor depends on the location of the axis of

rotation.

Finally, in most cases the rigid body is continuous and not made up of discrete particles

as was assumed so far, but the results are easily generalized by replacing the summation

by a corresponding integral in the expression for the components of the inertia tensor

( ) ( ) 1 2 3,ij ij k k i jV

I x x x x dx dx dx = r (8.25)

where ( ) r is the mass density at the position r , and the integral is to be performed

over the whole volume V of the rigid body.

Example

Calculate the inertia tensor for a homogeneous cube of density , mass , and side

length b . Let the origin (through which is axis of rotation will pass) be at one corner andlet the three adjacent edges lie along the coordinate axes (see Figure 8-1).

Solution.We use equation (8.25) to calculate the components of the inertia tensor. Because of the

symmetry of the problem, it is easy to see that the three moments of inertia

11 22 33, , andI I I are equal and that same holds for all of the products of inertia (off-diagonal elements). So,


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Figure 8-1 A homogeneous cube with sides b with the origin at one corner.

( )

( )

( )

2 2 2 211 1 2 3 1 1 2 3

0 0 0

2 2

2 3 1 2 30 0 0

2 2

3 2 2 3 10 0 0

3 4 42

3 30

5 2

3 3 3

2 2.

3 3

b b b

b b b

b b b

b

I x x x x dx dx dx

x x dx dx dx

dx dx x x dx

b b bb dx bx b

b Mb

= + +

= +

= +

= + = +

= =

(8.26)

And for the negative of the products of inertia

( )

12 1 2 1 2 30 0 0

2 2

5 2

2 2

1 1.

4 4

b b b

I x x dx dx dx

b bb

b Mb

=

=

= =

(8.27)

It should be noted that in this example the origin of the coordinate system is not locatedat the centre of mass of the cube.

8.4 Angular MomentumGoing back to the case of a rigid body composed of a discrete number of particles; we

can calculate the angular momentum with respect to some point O fixed in the body

coordinate system with


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.

= L r p (8.28)

Relative to the body coordinate system the linear momentum of the th particle is

,m m= = p v r (8.29)

where weve used the result from Chapter 7 that

fixed rotating

d d

dt dt

= +

Q QQ (8.30)

and the fact thatrotating

0rd

dt

= =

rv since the particles in a rigid body dont move with

respect to the body frame (that is, the rotating frame). The total angular momentum

becomes

( ).m= L r r

(8.31)

Resorting one more time to tensor notation we can calculate ( ) r r as

( ), , , ,

, ,

, , , , ,

ijk j klm l m kij klm j l m

il jm im jl j l m

j j i j j i

x x x x

x x

x x x x

=

=

=

(8.32)

or alternatively in vector notation

( ) ( )2 .r = r r r r (8.33)

Then, the total angular momentum is given by

( )2m r = L r r

(8.34)

Using the tensor notation the components of the angular momentum are

( )

( )

, , , ,

, , , , ,

i k k i j j i

j ij k k j i

L m x x x x

m x x x x

=

=

(8.35)


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and upon using equation (8.20) for the inertia tensor

i ij jL I = (8.36)

or in tensor notation

{{{{ }}}}L I= = = = (8.37)

Finally, we can insert equation (8.37) for the angular momentum vector into equation

(8.22) for the rotational kinetic energy to obtain

rot

1

2T L= = = = (8.38)

Figure 8-2 A dumbbell connected by masses 1 2andm m at the ends of its shaft. Note

that the angular velocity is not directed along the shaft.

Example

The dumbbell. A dumbbell is connected by two masses 1 2andm m located at distances

1 2andr r from the middle of the shaft, respectively. The shaft makes an angle with a

vertical axis, to which it is attached at its middle (i.e., the middle of the shaft). Calculate

the equation of angular motion if the system is forced to rotate about the vertical axis

with a constant angular velocity (see Figure 8-2).


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Solution. We define the inertial and the body coordinate systems such that their

respective origins are both connected at the point of junction between the vertical axis

and the shaft of the dumbbell. We further define the body coordinate system as having its

3-axisx orientated along the shaft and its 1-axisx perpendicular to the shaft but located in

the plane defined by the axis of rotation and the shaft. The remaining 2 -axisx is

perpendicular to this plane and completes the coordinate system attached to the rigid

body. For the inertial system, we chose the 3 -axisx to be the vertical, and the other two

axes such that the basis vectors can expressed as

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( )

1 1 2 3

2 1 2 3

3 1 3

cos cos sin sin cos

cos sin cos sin sin

sin cos .

t t t

t t t

=

= +

= +

e e e e

e e e e

e e e

(8.39)

From equation (8.20), we can evaluate the components of the inertia tensor. We can in

the first time identify the components that are zero (because 1,1 1,2 2,1 2,2 0x x x x= = = = )

12 21 13 31 23 32 33 0.I I I I I I I= = = = = = = (8.40)

The only two remaining components are

2 2

11 22 1 1,3 2 2,3

2 2

1 1 2 2 .

I I m x m x

m r m r

= = +

= +(8.41)

The components of the angular velocity in the coordinate of the rigid body are

( )

( )

1

2

3

sin

0

cos .

=

=

=

(8.42)

Inserting equations (8.41) and (8.42) in equation (8.36) we find for the 1L component

( )( )2 21 1 11 1 1 1 2 2sin ,i iL I I m r m r = = = + (8.43)

and for the 2 3andL L

2 2 22 2

3 3 33 3

0

0.

i i

i i

L I I

L I I

= = =

= = =(8.44)


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If should be noted from equations (8.42) and (8.43) that the angular velocity and the

angular momentum do not point in the same direction. To calculate the equations of

motion, we express the angular momentum with the inertial coordinates instead of thecoordinates of the rigid body system. From equation (8.43) we can write

( )( ) ( )2 2

1 1 2 2 1 11 1sin sin ,m r m r I = + =L e e (8.45)

but we can transform the basis vector 1e using equation (8.39) (or its inverse)

( ) ( ) ( ) ( ) ( ) ( )11 1 2 3sin cos cos cos sin sin .I t t = + + L e e e (8.46)

We also know that

,d

dt=

LN (8.47)

where N is the torque. Assuming that the angular speed is constant, we find

( ) ( ) ( )

( ) ( ) ( )

2

1 11

2

2 11

3

sin cos sin

sin cos cos

0.

I t

I t

N

=

=

=

(8.48)

An interesting consequence of the fact that the angular momentum and angular velocity

vectors are not aligned with each other is that we need to apply a torque to the dumbbell

to keep it rotating at a constant angular velocity.

8.5 The Principal Axes of InertiaWe now set on finding a set of body axes that will render the inertia tensor diagonal in

form. That is, given equation (8.23) for{{{{ }}}}I , we want to make a change in the body basisvectors (i.e., a change of variables) that will change the form of the inertia tensor to

{{{{ }}}}1

2

3

0 0

0 0 .

0 0

I

I

I

I

====

(8.49)

This would mean that all the products of inertia are zero and will provide a significant

simplification for the expressions of the angular momentum and the kinetic energy, as

measured in the inertial reference frame. That is, these two quantities will be given by

1 1 1 2 2 2 2 3 3, , ,L I L I L I = = == = == = == = = (8.50)


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and

( )2 2 2rot 1 1 2 2 3 31

.2

T I I I = + + (8.51)

The set of axes always exist (though we wont prove this here) that allow this

transformation, and they are called the principal axes of inertia. When the equations for

the components of the angular momentum can be expressed as in equation (8.49), then

, andL are directed along the same axis.

The problem of finding the principal axes is mathematically equivalent to solving a set of

linear equations. More precisely, we have from equation (8.37) that

{{{{ }}}} ,L I= = = = (8.52)

but we are actually looking for a way to reduce this equation to the following form

{ } .I= =L I (8.53)

Mathematically speaking, I, which is called a principal moment of inertia, is an

eigenvalue of the inertia tensor, and , which will give us the corresponding principalaxis of inertia, is an eigenvector. The system of equations (8.53) can be written as

1 1 11 1 12 2 13 3

2 2 21 1 22 2 23 3

3 3 31 1 32 2 33 3 ,

L I I I I

L I I I I

L I I I I

= = + +

= = + +

= = + +

(8.54)

or, after some rearranging

( )

( )

( )

11 1 12 2 13 3

21 1 22 2 23 3

31 1 32 2 33 3

0

0

0.

I I I I

I I I I

I I I I

+ + =

+ + =

+ + =

(8.55)

The mathematical condition necessary for this set of equation to have a nontrivial

solution is that the determinant of the coefficient vanishes

( )( )

( )

11 12 13

21 22 23

31 32 33

0.

I I I I

I I I I

I I I I

=

(8.56)


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The expansion of this determinant leads to the so-called secular or characteristic

equation for the eigenvalues I(i.e., 1 2 3, , andI I I in equation (8.50)); it is a third order

polynomial. Once the characteristic equation has been solved, the principal axes can bedetermined by inserting the eigenvalues back in equation (8.55) and evaluating the ratios

of the angular velocity components ( 1 2 3: : ), therefore, determining the

corresponding eigenvectors.

It is important to realize that in many cases, the rigid body under study will exhibit somesymmetry that will allow one to guess what the principal axes are. For example, a

cylinder will have one of its principal axes directed along the centre axis of the cylinder.

The two remaining axes will be directed at right angle to this axis (and to each other).

Finally, here are a few definitions: a body that has i) 1 2 3I I I= = is called a spherical

top, ii) 1 2 3I I I= is a symmetric top, iii) 1 2 3I I I is an asymmetric top, and

finally, if 1 2 30,I I I= = the body is a rotor.

Example

Find the principal moment of inertia and the principal axes of inertia for the cube ofFigure 8-1.

Solution.In the previous example solved on page 165, we found that the products of inertia did not

equal zero. Obviously, the axes chosen were not the principal axes. To find the principal

moments of inertia, we must solve the characteristic equation

2 1 1

3 4 41 2 1

0,4 3 4

1 1 2

4 4 3

I

I

I

=

(8.57)

where 2b . Before trying to evaluate the determinant from equation (8.57), it isalways good to see if we can bring it to a simpler form with a subtraction of a column or

a row to another column or row. This is permitted since the determinant is not affected by

such operations. In our case, we see that, for example, the third element of the first two

rows is the same; we will therefore subtract the second row to the first. The determinantthen becomes


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11 110

12 12

1 2 10,

4 3 4

1 1 2

4 4 3

I I

I

I

+

=

(8.58)

which implies that

1 1 0

11 1 2 10.

12 4 3 4

1 1 2

4 4 3

I I

I

=

(8.59)

Expanding this equation gives

2 2 2

2 2

11 2 1 1 2 10

12 3 4 4 3 4

11 2 1 2 12 0.

12 3 4 3 4

I I I

I I I

+ =

=

(8.60)

The square-bracketed part of the last expression is a second order polynomial in

( )2 3 I . If we solve for the roots of this polynomial we get

( )

2 22 1 1 1 1

83 2 4 4 4

11 3 .

8

I

= +

=

(8.61)

The last of equation (8.60) can now be factored as

11 11 10,

12 12 6I I I

=

(8.62)

and the three principal moments of inertia are

1 2 3

1 11, .

6 12I I I = = = (8.63)


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To find the direction of the principal axis of inertia, we insert the eigenvalues of equation

(8.63) into equation (8.55). For 1I , we have (after some manipulation)

1 2 3

1 2 3

2 0

2 0,

=

+ = (8.64)

which implies that 1 2 3 = = and the corresponding eigenvector is directed along

1 2 3+ +e e e . For 2 3andI I , because they are equal, the orientation of their correspondingprincipal axes is arbitrary; they need only to lie in a plane perpendicular to the main

principal axis determined for 1I .

8.6 Similarity TransformationsThe diagonalization of the inertia tensor (or any other matrix for that matter) discussed in

the previous section can sometimes be achieved in a different manner. For example, inthe case of the cube discussed in the last example, because of the symmetry of the

problem we could have guessed what the principal axes were. When this can be done, itis then straightforward (if sometimes tedious) to determine the transformation (rotation)

matrix that brings us from the initial set of coordinate axes to the principal axes, and use

it to render the inertia tensor diagonal. In a way, this technique follows a path that is

reversed from what was done in the previous section. That is, instead of, first, renderingthe inertia tensor diagonal and then, determining the principal axes (i.e., the

eigenvectors), we now guess the orientation of the principal axes and then diagonalize the

inertia tensor.

So, lets consider the angular momentum

{{{{ }}}} ,L I= = = = (8.65)

in the initial set of coordinate axes, which we now transform to a new set of axes. We

will have a new equation for the angular momentum as measured in this new system; we

define the angular momentum with an equation similar to equation (8.65)

{{{{ }}}} .L I = = = = (8.66)

If we denote the transformation matrix that links the two coordinate systems by such

that

,L L = = = = (8.67)

and

. = = = = (8.68)


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We have relations similar to equations (8.67) and (8.68) for any corresponding vectors

between the two systems of coordinates. For example, the position vectors are related

through

.r r = = = = (8.69)

Note that L, L, , and are all vectors, while Iand I are tensors (or matrices). Now, is also a matrix, one that embodies any rotation necessary to make the transformationfrom our old coordinates to our new ones. Only rotation? What about translation, if theorigin we choose is not the centre of mass, for example? One usually shouldchoose the

centre of mass as the origin, because it means that the kinetic energy is separable into

translational and rotational parts. However, regardless of the origin we choose, we can

find principal axes of inertia that diagonalize the inertia tensor. So no matter what originwe have, we only need to consider rotations. Scaling of the axes is another possibility one

might consider, but it turns out only the direction of the axes are of interest, not their

units.

One simple example of is presented for clarification purposes. Consider a coordinatechange that rotates thex andy axes by degrees around thez-axis. Then we would havesimply

cos sin 0

sin cos 0

0 0 1

=

Combining equations (8.66) to (8.68) we can write

{{{{ }}}} {{{{ }}}} (((( )))) ,L L I I = = = = = = = = = = = = (8.70)

or

{{{{ }}}}(((( ))))1 .L I = = = = (8.71)

Comparison with equation (8.65) reveals that

{{{{ }}}} {{{{ }}}} 1I I = = = = (8.72)

A transformation of this type is called a similarity transformation. In instances where

we deal with orthogonal transformations (rotations maintain the orthogonality of the

coordinate axes), we have 1 T = = = = , with T the transpose of , and

{{{{ }}}} {{{{ }}}} .TI I = = = = (8.73)


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Finally, if the transformation matrix is such that it takes the initial coordinate axes into

the principal axes, then, the transformed inertia tensor will be diagonal.

Example

Use the results of the preceding example of the cube (equations (8.64) and the followingparagraph) to render its inertia tensor (using equations (8.26) and (8.27)) diagonal.

Solution.From equations (8.26) and (8.27), and the corresponding set of coordinate axes defined inFigure 8-1, we can write the inertia tensor of the cube as

{{{{ }}}}

2 1 1

3 4 4

1 2 1,

4 3 4

1 1 2

4 4 3

I

= = = =

(8.74)

where 2b . If the three initial basis vectors are denoted by 1 2 3, , ande e e , we knowfrom previous results that the main principal axis can be chosen to be

( )1 1 2 31

.3

= + +e e e e (8.75)

and it doesnt matter where the other axes are except that they be perpendicular to the

diagonal. Note that here were cheating a bit. We already know what the principal axesare from our previous result, while earlier we said wed guess their directions from

symmetry arguments. However the choice of the diagonal would not be obvious if we

hadnt already solved the problem (maybe we would have chosen axes parallel to thesides?) so this approach has its pitfalls.

Back to the problem. So what rotation matrix will rotate a cube onto its diagonal? Well,its pretty obvious that no single rotation about one of our original axes will do the trick.

In fact, it takes two rotations in sequence to do it (though we wont work through it here).

The total transformation matrix can be written as

1 1 1

1 3 30 .

2 23

1 12

2 2

=

(8.76)


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and can be verified that 2 1= with the two individual rotations given by

1 2

1 12 10

02 23 31 1

0 , 0 1 0 .2 2

1 20 0 1 0

33

= =

(8.77)

Using equation (8.73) with equations (8.74) and (8.76), we find

{{{{ }}}} {{{{ }}}}'

3 12 1 1 11 1 1 2 23 4 4

3 3 1 2 1 3 10 . 13 2 2 4 3 4 2 2

1 1 21 1 1 0 224 4 32 2

1 11 3 11 1

1 1 1 6 12 2 12 2

3 3 1 11 3 11 10

3 2 2 6 12 2 12 2

1 1 1 22 12 02 2 6 12

TI I= == == == =

= = = =

= = = =

2

110 00 0

62

33 110 0 0 0 ,

3 12 12

33 110 0 0 0

12 12

= = = = = = = =

(8.78)

this is the same result as was obtained in equation (8.63). You may note however that this

was not particularly easy, no easier than solving the characteristic equation, because wehad to come up with the rotation matrix that we needed (here we just pulled it out of the

air, but in practice, wed have to figure it out). Also, if our guess at the correct orientation

of the axes was wrong, wed have to start over. So this technique might get us to theanswer faster, or it might not.


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8.7 Moments Inertia for Different Body Coordinate SystemsWe consider two sets of coordinate axes that are oriented in the same direction, but have

different origins. The -axesix have their origin O located at the centre of mass of the

rigid body, and the -axesiX have their origin Q located somewhere else inside, or

outside, of the body (see Figure 8-3).The elements of the inertia tensor{{{{ }}}}J relative to the -axesiX are (see equation (8.20))

( ), , , , .ij ij k k i jJ m X X X X

= (8.79)

If the vector a connects the origin Q to the centre of mass (and origin) O , then the

general vector R for the position of a point within the rigid body is written as

= +R a r , or using components

, , .i i iX a x = + (8.80)

Figure 8-3 The -axesiX are fixed in the rigid body and have the same orientation as

the -axesix , but its origin Q is not located at the same point O , which is the centre of

mass of the body.

Inserting equation (8.80) into equation (8.79) we get


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( ) ( ) ( ) ( )

( )

( ) ( )

( ) ( )

, , , ,

, , , ,

, , ,

, , ,

2

2 .

ij ij k k k k i i j j

ij k k i j

ij k k k k i j j i i j

ij ij k k i j ij k k i j j i

J m a x a x a x a x

m x x x x

m x a a a a x a x a a

I m a a a a m x a a x a x

= + + + +

=

+ + + +

= + +

(8.81)

The last term on the right hand side of the last of equations (8.81) can be broken down to

( ), , ,

, , ,

2

2

ij k k i j j i

ij k k i j j i

m x a a x a x

a m x a m x a m x

(8.82)

Note that we are not summing over i orj, we are simply looking for the ijth

element so

theres no problem pulling the and as outside the sum. From the definition of thecentre of mass itself, each component of it (as determined in the centre of mass frame)

must be zero so

, 0 for 1,2 or 3im x i

= = (8.83)

and the whole last term of (8.81) goes to zero. We then find the final result that

( )2ij ij ij i jJ I M a a a= + (8.84)

with 2and k km a a a

= = .

We see from equation (8.84) that the inertia tensor components are minimum when

measured relative to the centre of mass.

Example

Find the inertia tensor of a homogeneous cube of side b relative to its centre of mass.

Solution.We previously found that the components of the inertia tensor of such a cube relative toone of its corners are given by

11 12

2 1, ,

3 4J J = = (8.85)

for the moments and the (negative of) the products of inertia, respectively. Since the

centre of mass of the cube is located at 1 2 3 2a a a b= = = relative to a corner, we can use


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equation (8.84) to calculate the components ijI of the inertia tensor relative to the centre

of mass

22

11 11

3

4 4

2 1

3 2

1,

6

bI J M b

=

=

=

(8.86)

and

2

12 12 0.4

bI J M= + = (8.87)

The inertia tensor{{{{ }}}}I is, therefore, seen to be diagonal and proportional to the unit tensor

{{{{ }}}}1 with

{{{{ }}}} {{{{ }}}}21

.6MbI 1==== (8.88)

8.8 The Euler AnglesWe stated in section 8.2 that of the six degrees of freedom of a rigid body, three are

rotational in nature (the other three are for the translation motion of the centre of mass).

In this section, we set to determine the set of angles that can be used to specify therotation of a rigid body.

We know that the transformation from one coordinate system to another can be

represented by a matrix equation such as

.=x x (8.89)

If we identify the inertial (or fixed) system with x and the rigid body coordinate systemwith x , then the rotation matrix describes the relative orientation of the body inrelation to the fixed system. Since there are three rotational degrees of freedom, is

actually a product from three individual rotation matrices; one for each independentangle. Although there are many possible choices for the selection of these angles, we will

use the so-called Euler angles , , and .

The Euler angles are generated in the following series of rotation that takes the fixed x system to the rigid body x system (see Figure 8-4).


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1. The first rotation is counterclockwise through an angle about the 3-axisx . Ittransforms the inertial system into an intermediate set of -axesix . Thetransformation matrix is

( ) ( )( ) ( )

cos sin 0

sin cos 0 ,

0 0 1

=

(8.90)

with 0 2 , and

. =x x (8.91)

Figure 8-4 The Euler angles are used to rotate the fixed x system to the rigid body x system. (a) The first rotation is counterclockwise through an angle about the 3-axisx .

(b) The second rotation is counterclockwise through an angle about the 1-axisx . (c)

The third rotation is counterclockwise through an angle about the 3 -axisx .

2. The second rotation is counterclockwise through an angle about the 1 -axisx (also called the line of nodes). It transforms the inertial system into an intermediate

set of -axesix . The transformation matrix is

( ) ( )( ) ( )

1 0 0

0 cos sin ,

0 sin cos

=

(8.92)

with 0 , and


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181

. =x x (8.93)

3. The third rotation is counterclockwise through an angle about the 3 -axisx . Ittransforms the inertial system into the final set rigid body -axesix . The

transformation matrix is

( ) ( )( ) ( )

cos sin 0

sin cos 0 ,

0 0 1

=

(8.94)

with 0 2 , and

.=x x (8.95)

Combining the three rotations using equations (8.91), (8.93), and (8.95) we find that thecomplete transformation is given by

,=x x (8.96)

and the rotation matrix is

.= (8.97)

Upon calculating this matrix, we find that its components are

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( )( )

11

21

31

12

22

32

13

23

33

cos cos cos sin sin

sin cos cos sin cos

sin sin

cos sin cos cos sin

sin sin cos cos cos

sin cos

sin sin

cos sin

cos .

=

=

=

= +

= +

=

=

==

(8.98)

with 0 2 , 0 , and 0 2 .


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Correspondingly, the rate of change with time (i.e., the angular speed) associated with

each of the three Euler angles are defined as , , and . The vectors associated with

, , and can be written as

3

1

3 3.

=

=

= =

e

e

e e

(8.99)

Taking the projections of the unit bases vectors appearing in equation (8.99) on the rigidbody bases vectors, we find

( ) ( ) ( ) ( ) ( )

( ) ( )

1 2 3

1 2

3

sin sin sin cos cos

cos sin

.

= + +

=

=

e e e

e e

e

(8.100)

Combining the three equations (8.100), we can express the components of the total

angular velocity vector as a function of , , and

( ) ( ) ( )

( ) ( ) ( )

( )

1

2

3

sin sin cos

sin cos sin

cos

= +

=

= +

(8.101)

8.9 Eulers EquationsTo obtain the equations of motion of a rigid body, we can always start with the

fundamental equation (see the dumbbell example earlier)

fixed

,d

dt

=

LN (8.102)

where N is the torque, and designation fixed is used since this equation can only

applied in an inertial frame of reference. We also know from our study of noninertialframes of reference in Chapter 10 that

fixed body

,d d

dt dt

= +

L LL (8.103)

or


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body

.d

dt

+ =

LL N (8.104)

Using tensor notation we can write the components of equation (8.104) as

.i ijk j k iL L N + = (8.105)

Now, if we chose the coordinate axes for the body frame of reference to coincide with the

principal axes of the rigid body, then we have from equations (8.50)

1 1 1 2 2 2 3 3 3, , .L I L I L I = = = (8.106)

Since the principal moments of inertia 1 2 3, , andI I I are constant with time, we can

combine equations (8.105) and (8.106) to get

( )

( )

( )

1 1 2 3 2 3 1

2 2 3 1 3 1 2

3 3 1 2 1 2 3.

I I I N

I I I N

I I I N

=

=

=

(8.107)

Alternatively, we can combine these three equations into one using indices

( ) ( ) 0i j i j ijk k k k k

I I I N = (8.108)

where no summation is implied on the andi j indices. Equations (8.108) are the so-calledEuler equations of motion for a rigid body. We note that our angular momentum vector

composed ofi is the same in either the body frame or the external fixed frame (from Ch7).

We also note that the equations of motion of a rigid body only depend on its shape

though its principal moments of inertiaI1, I2, I3. Therefore any two rigid bodies that havethe same principal moments of inertia (regardless of their shape) will move in the same

way in response to the same torques. The motion of a particular body is sometimes

modeled from its equivalent ellipsoid, a simple body with the same principal axes of

inertia.

Examples

1. The dumbbell. We return to problem of the rotating dumbbell that we solved earlier

Referring to equation (8.45) we found that the angular momentum was given by (using

the rigid body coordinate system)


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( )( )2 21 1 2 2 1sin ,m r m r = +L e (8.109)

with the principal moments of inertia of the system given by

2 2

1 2 1 1 2 2

3 0,

I I m r m r

I

= = +

= (8.110)

and the angular velocity components by

( )

( )

1

2

3

sin

0

cos .

=

=

=

(8.111)

Since the system of axes chosen correspond to the principal axes of the dumbbell, then

we can apply Eulers equations of motion (i.e., equations (8.107)). With the constraintthat 0= , we find

( ) ( ) ( )1

2 2 2

2 1 3 1 1 1 2 2

3

0

sin cos

0,

N

N I m r m r

N

=

= = +

=

(8.112)

or

( ) ( ) ( )2 2 2

1 1 2 2 2sin cos .m r m r = +N e (8.113)

Upon using equations (8.39) we can rewrite the torque in the inertial coordinate system

(that shares a common origin with the body system) as

( ) ( ) ( ) ( ) ( )2 2 21 1 2 2 1 2sin cos sin cos .m r m r t t = + + N e e (8.114)

This is the same result as what was obtained with equations (8.48) without resorting to

Eulers equation.

2. Force-free motion of a symmetric top.

We consider a symmetric top, whose principal moments of inertia are 1 2 3I I I= , whenno forces or torques are acting on it. We allow however that the initial spin of the top isnon-zero and may not be along one of the principal axes. In this case, the Eulers

equations of motion are


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( )

( )1 3 2 3 1 1

3 1 3 1 1 2

3 3

0

0

0,

I I I

I I I

I

=

=

=

(8.115)

where 1 2(as opposed to )I I was used throughout since we have assumed 1 2I I= . Becausethere are no forces involved, the top will be either at rest or in uniform motion withrespect to the inertial frame of reference. We will assume the top to be at rest, and located

at the origin of the inertial coordinate axes. We want to find equations for the time

evolution of the components of the angular velocity vector .

From the third of equations (8.115) we find that 3 0 = , or

3 constant. = (8.116)

The equations for the other two components of the angular velocity yield

1 2

2 1

0

0,

+ =

=

(8.117)

with

3 13

1

.I I

I

=

(8.118)

Equations (8.117) form a system of coupled first order differential equations that can be

solved in a similar fashion as was done for the problem of the Foucault pendulum in the

previous chapter (see page 156). So, we multiply the second equation by i and add it to

the first to get

0,i = (8.119)

with 1 2i = + . The solution to equation (8.119) is of the form

( ) ,i tt Ae = (8.120)

or, alternatively,

( ) ( )

( ) ( )1

2

cos

sin .

t A t

t A t

=

= (8.121)

Since 3 is a constant, we find that the speed of rotation is also a constant. That is,


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2 2 2 2 2 2

1 2 3 3 constant.A = + + = + = (8.122)

However we note that, from equation (8.121), while the magnitude of the angular

momentum is constant, the direction of the angular momentum vector is NOT aconstant, even though we have force free motion.

Because equations (8.121) are that of a circle, we find that the angular velocity vector precesses about the tops axis of symmetry with a constant angular frequency .Precession means that to an observer attached to the rigid body coordinate system, traces a cone around the 3-axisx , called the body cone. Also, since the top is not

subjected to any forces or torques, the angular momentum L is conserved, as is the

energy (which in this case reduces to the kinetic energy of rotation otT ). Therefore,

rot

1constant,

2

T = =L (8.123)

with

fixed

0.d

dt

=

L(8.124)

Equations (8.123) and (8.124) imply that the angular velocity vector makes a constantangle with the (also constant) angular momentum vector L . Therefore, not only does precess about the body 3 -axisx , but it also precesses around the axis that specifies the

direction of L . The angular velocity vector traces a cone around the -axisL , calledthe space cone.

How are these two separate motions possible? Probably the easiest way to see this is toconsider the motion in the fixed inertial frame. This is the frame where L is constant,

and is the same in both frames. We note that the body itself is rotating around bydefinition. So one imagines precessing around L while itself precesses around the

3-axisx of the body (this is the axis with the different principal moment of inertia, from

our initial assumptions). The end result is that the body cone rolls along the space

cones during their mutual precessions.

We can find out the relative orientation of 3, , and the -axisxL by calculating thefollowing double product


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Figure 8-5 The relative orientation of 3, , and the -axisxL that result from the

force-free motion of a symmetric top. We let L to lie along the 3-axisx in the fixed

coordinate system and 1 3I I> . We can imagine the body cone rolling around the spacecone.

( )3 3

3 1 2 2 1

1 1 2 2 2 1 0,

i ijk j k

ij i j

L

L L L

I I

=

= = = =

L e

(8.125)

since 1 2I I= for a symmetric top. We therefore have that 3, , and the -axisxL all lie onthe same plane. An example of this is shown in Figure 8-5 for the case where we let L to

lie along the 3-axisx in the fixed coordinate system and 1 3I I> .

Finally, we can calculate the rate at which the rigid body will precess (about the 3-axisx )in the inertial system. We know from the definition of the Euler angles that the rotation

rate about the 3 -axisx is given by . Furthermore, the second Euler angle is that made

between the 3 3(or ) and thex x L axes, with 0 = since this angle is constant. Then,using the first two of equations (8.101) we have

( )2 2 2 21 2 sin . + = (8.126)

If we define the components of the angular velocity vector as a function of the angle it makes to the 3-axisx , we have

( )

( )

2 2

1 2

3

sin

cos .

+ =

=

(8.127)

Since I1 = I2 we can also express the components of the angular momentum as


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( )

( )

2 2

1 2 1

3 3

sin

cos ,

L L I

L I

+ =

=(8.128)

or, alternatively,

( )

( )

2 2

1 2

3

sin

cos .

L L L

L L

+ =

=(8.129)

Combining equations (8.126) to (8.129) we find that

( )( ) 1

sin.

sin

L

I

= = (8.130)

Example:

Consider the asteroid 4179 Toutatis, with centre of mass axes oriented as in the figure.Thex, y andzaxes correspond to the principal axes of smallest, intermediate and largest

moment of inertia respectively, thus the x-axis is aligned with the long axis of Toutatis

and the z-axis with the shortest (Why does the shortestaxis have the largestmoment ofinertia? Consider equation (8.23) to see why this is). The ratios of the moments of inertia

are

0.31335; 0.94471;yx

z z

II

I I= =

At a given instant, the asteroids angular velocity vector is measured to be 21o

from thex-

axis and in thexy-plane. The asteroid angular velocity vector spins at rate of once every130.156h.

a) Which two axes have the most similar moments of inertia?b) Assuming that the two axes of a) have the same moment of inertia (use theaverage), compute the future motion of the rotation axis of Toutatisassuming it is a symmetric force-free top

Solution:a) Clearly they andzaxes have the most similarIvalues.

b) As a result of a) the x-axis must be the axis of symmetry, so we have to bear this in

mind. In our derivation we assumed 1 2 3I I I= . In this case, thex-axis (which is usuallylabel with a 1) is the one with the different moment of inertia. But it doesnt matter what

the labeling of the axes are, we can use our results from above but instead label things

3 1 2; ;x y zI I I I I I= = = and we also have 1 20.972355y zI I I I . The initialangular velocity is

3 2(cos sin ) (cos sin )y + = +xe e e e

where = angular velocity = 1.34x10-5 radians s-1 and =21o. From our results on theforce-free top, we know that, in the body frame, the rotation axis will precess along thebody cone at an angular rate given by equation (8.118)


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6 13 13

1

0.313351 cos 8.848 10 rad s

0.972355

I I

I

= = =

thus the body cone will precess about the x-axis in 741077 sec or 8.6 days.In inertial space, the space cone will precess at a rate given by equation (8.130) or

( )2

2 2 2 2 2 2

1 1 2 2 3 3

1 1 1

2 2 2

1 2 3

2 2

6 1

1

10.945474 0.945474 0.098188

0.972355

0.945474 0 0.945474 sin 0.098188 cos0.972355

0.121425 0.0855780.972355

0.468 6.27 10 rad s

i iIL I I II I I

= = = + +

= + +

= + +

= +

= =

So the space cone precesses over a period of 11.6 days.

From Dynamics of Orbits close to Asteroid 4179 Toutatis, Scheeres et al, 1998, Icarus,132, 53-79.


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8.10The Tennis Racket TheoremThe so-called tennis racket theorem is concerned with the stability of the rotational

motion of a rigid body about its principal axis. We want to find out if, when a smallperturbation is applied to the body, the motion either returns to its initial state or performs

small oscillations about it (i.e., if it does not do this, then it is unstable).

We consider a general rigid body with principal moments of inertia such that 3 2 1I I I> > .We assume that the body coordinate axes are aligned with its principal axes, and first

consider an initial rotation about the 1-axisx , and then apply small perturbations about the

other two axes such that the angular velocity vector becomes

1 1 2 3 , = + +e e e (8.131)

with 1 1and . Using equations (8.107) we can write the equations of motion

for the system

( )

( )

( )

2 3 1 1

3 1 1 2

1 2 1 3

0

0

0.

I I I

I I I

I I I

=

=

=

(8.132)

If we will only keep terms of no higher than first order in and , the first of equations

(8.132) imply that

1 constant, = (8.133)while the other two can be rewritten as

3 11

2

1 21

3

,

I I

I

I I

I

=

=

(8.134)

where the quantities in parentheses are constant. Taking the time derivative of the first ofequations (8.134) and inserting the second in it we get


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( ) ( )

3 11

2

3 1 1 2 2

1

2 3

.

I I

I

I I I I

I I

=

=

(8.135)

The solution to this second order differential equation is of the following form

( ) 1 1 ,i t i t t Ae Be = + (8.136)

with

( ) ( )1 3 1 21 1

2 3

.I I I I

I I

= (8.137)

Since 1 3 1 2andI I I I< < , 1 is real and the motion resulting from the perturbation is abounded oscillation. Furthermore, upon inserting the result in equations (8.134) we find a

similar result for ( )t . The rotation motion about the 1-axisx is therefore stable.

If we study next rotations about the 2 3and -axisx x , we obtain similar results and identify

the frequency of oscillations, stemming form the perturbations, by permutation of the

indices in equation (8.137). That is,

( ) ( )

( ) ( )

2 1 2 3

2 2

3 1

3 2 3 1

3 3

1 2

.

.

I I I I

I II I I I

I I

=

=

(8.138)

But because 3 2 1I I I> > , we find that 3 is also real, while 2 is imaginary. Just as the

motion about the 1-axisx was found to be stable, so is the motion about the 3-axisx . On

the other hand, because 2 is imaginary a perturbation will increase exponentially with

time when the initial rotation is about the intermediate 2 -axisx . Motion about this axis is

thus unstable. We can readily test this result by spinning a rigid body (like a tennis

racket!) about each of its principal axes.

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Classical Mechanics notes (8 of 10)