2

= (15.49)

i

= Iω, (15.50)

where I is the moment of inertia or inertia tensor, which is represented by

a 3 — 3 symmetric matrix

mi (ri 1 ’ ri rT ),

2

I= (15.51)

i

i

15.7 Rigid body motion 409

written out as

⎛ ⎞

yi + zi ’xi yi ’xi zi

2 2

mi ⎝ ’yi xi x2 + zi ’yi zi ⎠ .

2

I= (15.52)

i

’zi xi ’zi yi x2 + yi2

i

i

Since I is a tensor and L and ω are vectors,11 the relation L = Iω is valid

in any (rotated) coordinate system. For example, this relation is also valid

for the primed quantities in the body-¬xed coordinates:

L = Iω, L =Iω, (15.53)

I = RIRT .

L = RL, ω = Rω, (15.54)

However, we must be careful when we transform a vector v from a rotating

coordinate system: the rotation itself produces an extra term ω — v in the

time derivative of v:

v = RT v + ω — v.

™ ™ (15.55)

It is possible to relate the time derivative of the rotation matrix to the

angular momentum in the body-¬xed system. Since, for an arbitrary vector

v = RT v , the derivative is

dT

R v = RT v + RT v ,

™

™ ™

v= (15.56)

dt

which, comparing with (15.55), means that

RT v = ω — v.

™ (15.57)

Hence

RT v = RT (ω — v ) = RT © v ,

™ (15.58)

where © is a second-rank antisymmetric tensor:

⎛ ⎞

’ωz ωy

0

© = ⎝ ωz ’ωx ⎠ ,

def

(15.59)

0

’ωy ωx 0

and thus

RT = RT © .

™ (15.60)

11 Tensors and vectors are de¬ned by their transformation properties under orthogonal transfor-

mations. In fact, L and ω are both pseudovectors or axial vectors because they change sign

under an orthogonal transformation with determinant ’1., but this distinction with proper

vectors is not relevant in our context.

410 Lagrangian and Hamiltonian mechanics

™

Other ways of expressing R are

™

R = © T R, (15.61)

RT = ©RT ,

™ (15.62)

R = R©T .

™ (15.63)

Recall the equation of motion (15.42) for the angular momentum; in ma-

trix notation in the space-¬xed coordinate system this equation is:

d

™

L = (Iω) = T. (15.64)

dt

Only in the body-¬xed coordinates is I stationary and constant, so we must

refer to those coordinates to avoid time dependent moments of inertia:

dT

R I ω = T, (15.65)

dt

or

RT I ω + RT I ω = T,

™ ™

™

I ω = RT ’ RRT I ω ,

™

Iω = T ’©Iω .

™ (15.66)

™

The latter equation enables us to compute the angular accelerations ω. This

equation becomes particularly simple if the body-¬xed, primed, coordinate

system is chosen such that the moment of inertia matrix I is diagonal. The

equation of motion for the rotation around the principal X -axis then is

Iyy ’ Izz

Tx

ωx =

™ + ωy ωz , (15.67)

Ixx Ixx

and the equations for the y- and z-component follow from cyclic permutation

of x, y, z in this equation. Thus the angular acceleration in the body-¬xed

coordinates can be computed, provided both the angular velocity and the

torque are known in the body-¬xed frame. For the latter we must know

the rotation matrix, which means that the time-dependence of the rotation

matrix must be solved simultaneously, e.g., from (15.61).

The kinetic energy can best be calculated from the angular velocity in the

body-¬xed coordinate system. It is given by

i mi r i = 2 ω · L = 2 ω · (I · ω)

1

™2 1 1

K= 2

1T 3

1T 1 2

= 2 ω Iω = 2 ω I ω = 2 ±=1 I± ω± (principal axes). (15.68)

15.7 Rigid body motion 411

Proof Since r i = ω — r i the kinetic energy is

™

1 1

mi (ω — r i )2 .

mi r 2 =

™i

K=

2 2

i i

With the general vector rule

(a — b) · (c — d) = (a · b)(a · b) ’ (a · d)(b · c), (15.69)

this can be written as