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Fig. 6.5 Axial symmetry of the velocity ﬁeld around the instantaneous axis of motion.

6.4

Rigid bodies: geometry and kinematics

221

The kinematics of rigid systems has far-reaching developments, which we do

not consider here.

We recall the following special kinds of rigid motion.

(a) Plane rigid motions: these are the motions deﬁned by the presence of a plane

in the system superimposed on a ﬁxed plane. They can be characterised by

the condition that

· ω = 0,

ω has constant direction,

(6.12)

and can be studied in a representative plane. The intersection of the latter

with the instantaneous axis of rotation is called the instantaneous centre of

rotation.

(b) Precessions: rigid motions with a ﬁxed point O : v(O)

≡ 0. Clearly v · ω = 0

and the point O, called the pole of the precession, belongs to the instantaneous

axis of rotation for all times.

motions

with

ﬁxed

axis

particular

case

Proposition 6.3). The rotation if said to be uniform if

ω = constant.

6.4

Phase space of precessions

We state a result that we have already discussed.

heorem 6.4 The tangent space to the space of orthogonal 3 × 3 matrices com-

puted at the identity matrix is the space of 3

× 3 skew-symmetric matrices (6.7),

denoted by so(3).

We should add that when v(0) = 0, equation (6.4) reduces to ˙x =

Ω

x, where

Ω

is the matrix given by (6.7). Setting x(0) =

ξ, we can describe the precession by

the equation x = A(t)

ξ, with A(0) = 1 and AA

= 1. It follows that ˙x = ˙

A(t)

and ﬁnally, from (6.4) ˙

ξ =

Ω

ξ, ∀ξ ∈ R

. Hence the matrix A satisﬁes the

Cauchy problem

A =

Ω

A(0) = 1

(6.13)

whose solution, if

Ω

= constant, is given by

A(t) = exp(t

Ω

) =

∞

Ω

(6.14)

It is easy to compute the powers of the matrix

Ω

. The elements of

Ω

are

(

Ω

)

= ω

for i = j, (

Ω

)

−(ω

+ ω

), with i, j, k all distinct,

Ω

−ω

Ω

and

Ω

= (

−1)

−1

2(n−1)

Ω

2n+1

= (

−1)

Ω

, n

≥ 1.

See for instance Fasano et al. (2001).

222

Rigid bodies: geometry and kinematics

6.4

This yields the explicit expression for the series (6.14):

exp(t

Ω

) = 1 +

Ω

− cos ωt) +

Ω

sin ωt.

(6.15)

In the particular case of a rotation around the axis x

the matrix

Ω

reduces to

Ω

3

= ω

⎛

⎝

−1 0

⎞

⎠ ,

Ω

⎛

⎝

−1

−1 0

⎞

⎠ ,

and we ﬁnd the well-known result that

A(t) = ω

⎛

⎝

cos ωt

− sin ωt 0

sin ωt

cos ωt

⎞

⎠ .

Ω

is not constant, equation (6.13) is equivalent to the integral equation

A(t) = 1 +

0

Ω

(t )A(t ) dt ,

(6.16)

which can be solved by iteration, and yields the formula

A(t) = 1 +

∞

n

=1

Ω

)

Ω

) . . .

n−1

Ω

(6.17)

We now give an equivalent, more concise formulation of the latter expression. Note

that the order of multiplication in (6.17) is relevant, as in general

Ω

(t

i

)

Ω

) =

Ω

)

Ω

) (obviously if

Ω

is constant, the series (6.17) reduces to (6.14)). Let

θ(t) =

t < 0,

t > 0.

(6.18)

The general term of the series (6.17) can be written as

0

dt

. . .

θ(t

− t

)

· · · θ(t

−1

− t

)

Ω

)

· · ·

Ω

)

. . .

∈

(n)

θ(t

(1)

− t

(2)

)

· · · θ(t

(n−1)

− t

(n)

)

Ω

(1)

)

· · ·

Ω

(n)

(6.19)

where

P(n) denotes the group of permutations of the indices {1, . . . , n}; it is

clear that the number of elements σ of

P(n) is n!. We then deﬁne the T -product

(time-ordered product) of the matrices

Ω

), . . . ,

Ω

(t

n

) as

T [

Ω

) . . .

Ω

)] =

∈

(n)

θ(t

(1)

− t

(2)

)

· · · θ(t

(n−1)

− t

(n)

)

Ω

(1)

)

· · ·

Ω

(n)

)

(6.20)

6.5

Rigid bodies: geometry and kinematics

223

in such a way that if σ denotes the permutation of

{1, . . . , n} such that t

(1)

(2)

· · · > t

(n−1)

> t

(n)

we have

T [

Ω

) . . .

Ω

)] =

Ω

(1)

)

· · ·

Ω

(n)

Recalling (6.19) and the deﬁnition (6.20) we ﬁnd that the solution (6.17) of

equation (6.13) can be written

Ω

(t) = 1 +

∞

. . .

T [

Ω

)

· · ·

Ω

)].

(6.21)

This solution is also known as the T -exponential (or time-ordered exponential)

Ω

(t) = T

− exp

Ω

(t ) dt

= 1 +

∞

. . .

T [

Ω

)

· · ·

Ω

)].

(6.22)

Again we note that if

Ω

is constant then equation (6.22) becomes (6.14). In the

special case that, for any t , t

∈ [0, t] one has

Ω

(t )

Ω

(t ) =

Ω

(t )

Ω

(t ), equation

(6.22) simpliﬁes to

exp

Ω

(t ) dt

where the order of multiplication of the matrices

Ω

) is not important.

6.5

Relative kinematics

After studying the motion of a frame S with respect to a frame

, we summarise

the main results of relative kinematics. This is concerned with the mutual relations

between kinematic quantities of a point in motion as observed by S (called

relative) and as observed by

(called absolute).

We use the subscript R for relative quantities, the subscript T for quantities

corresponding to the rigid motion of S with respect to

, and no subscript for

absolute quantities. The following relations hold:

v = v

+ v

(6.23)

a = a

+ a

(6.24)

where

ω is the rotational velocity of S with respect to

and the term a

ω × v

is called the Coriolis acceleration.

The former can be found by diﬀerentiating the vector P

− O =

in the

reference frame and using equation (6.5):

d(P

− O)

ω ×

224

Rigid bodies: geometry and kinematics

6.5

Equation (6.23) expresses the relation between the absolute derivative (in

) and

the relative derivative (in S):

ω × W,

(6.25)

where W is any vector, variable in both S and

.

Applying (6.24) to v

we ﬁnd

= a

ω × v

(6.26)

For the derivative of v

= v(0) +

ω × (P − O) we then obtain

= a

ω × v

(6.27)

where the last term has its origin in the fact that the relative motion in general

produces a variation of v

, because the point in question moves in the velocity

ﬁeld of the motion S with respect to

(note that this eﬀect vanishes if the

relative motion is in the direction of

ω, since v

does not vary in this direction).

We can now consider a triple S = (O , x

, x

), in motion with respect to

both S and

(Fig. 6.6), and ﬁnd the relations between the characteristic vectors

of the relative motion and of the absolute motion.

We start by expressing that for every P

∈ S

v

R

(P ) = v

(O ) +

× (P − O ),

(6.28)

where

is the angular velocity of S with respect to S. Let

be the angular

velocity of S with respect to

. Then we have

(P ) = v

(O) +

× (P − O).

(6.29)

From this it follows by adding (6.28) and (6.29) that

v(P ) = v

(O ) + v

(O) +

× (O − O) + (ω

)

× (P − O ),

(6.30)

which contains v

(O) +

× (O − O) = v

(O ).

Hence we ﬁnd the expression for the absolute velocity ﬁeld:

v(P ) = v(O ) +

ω × (P − O ),

(6.31)

with

v(O ) = v

(O ) + v

(O )

(6.32)

6.5

Rigid bodies: geometry and kinematics

225

j

3

j

2

j

Σ

x

2

x

1

x

3

O

Ј

S

Fig. 6.6 Composition of rigid motions.

and

ω = ω

(6.33)

The absolute rigid motion is commonly called compound rigid motion.

Example 6.3 : composition of precessions with the same pole

We immediately ﬁnd that the compound motion is a precession whose angular

velocity is the sum of the angular velocities of the component motions. When

two uniform rotations are composed, the vector ω = ω

+ ω

forms constant

angles with ω

ﬁxed in

(known as the precession axis) and with ω

ﬁxed in

S (spin axis). The resulting precession is called regular.

Clearly

ω can be thought of as the sum of the rotations driven by the variations

of the associated Euler angles:

ω = ˙ϕe

3

+ ˙

θN + ˙

(6.34)

226

Rigid bodies: geometry and kinematics

6.6

«

3

«

2

«

1

e

3

e

1

e

2

N

N

1

u

w

c

Fig. 6.7

where e

is the unit vector ﬁxed in S around which we have the rotation, N is

the unit vector of the node line, around which e

rotates, and

is the ﬁxed

unit vector around which the node lines rotates.

We can ﬁnd the decomposition of

ω in the body frame by appealing to Fig. 6.7,

where the unit vectors N

, N, e

form an orthogonal frame, so that

3

(normal

to N) is in the plane of e

and N

= sin θN

+ cos θe

(6.35)

N = cos ϕe

− sin ϕe

(6.36)

= sin ϕe

+ cos ϕe

(6.37)

It follows that by projecting

ω onto the vectors of the body frame, we ﬁnd

ω = ( ˙θ cos ϕ + ˙ψ sin θ sin ϕ)e

+ (

− ˙θ sin ϕ + ˙ψ sin θ cos ϕ)e

+ ( ˙

ϕ + ˙

ψ cos θ)e

.

(6.38)

6.6

Relative dynamics

As an appendix to the study of relative kinematics we now present some obser-

vations on the relative dynamical picture. What we have just presented allows

us to complete the list of fundamental facts of the mechanics of a point particle

6.6

Rigid bodies: geometry and kinematics

227

(Chapter 2), by answering the following question: how can we write the equation

of motion of a point with respect to a non-inertial observer?

Consider the motion of a point particle (P, m) under the action of a force F

measured by a given inertial observer

, for whom the equation ma = F is valid.

Given a second observer S in motion with respect to the ﬁrst one, using the

notation of (6.24) we can write

R

= F

− ma

(6.39)

where a

is the acceleration as measured by the non-inertial observer and a

and a

are computed with respect to the motion of this observer relative to the

ﬁxed inertial observer.

Ignoring for the moment the force F, the non-inertial observer measures some

apparent forces: the force connected to the acceleration of S relative to

, F

−ma

, and the Coriolis force F

=

−2mω × v

Recall that

(P, t) = a(O) + ˙

ω × (P − O) + ω × [ω × (P − O)].

(6.40)

In particular, in the case that the reference frame of the observer S is in uniform

rotation with respect to that of

, the acceleration a

is reduced to the last

term of (6.40) (centripetal acceleration) and correspondingly, F

is a centrifugal

force.

If the reference frame S translates with respect to

(

ω = 0) we have a

a(O). If S is chosen in such a way that P has zero relative velocity (let S

denote such an observer) then a(O) = a and the apparent force observed by S

acting on P is the so-called inertial force

−ma.

In this case, the equation

0 = F

− ma

(6.41)

is interpreted by S

as the equilibrium equation.

If the point is constrained and φ is the constraint reaction, instead of

equation (6.41) we have

0 = F

− ma + φ,

(6.42)

which again is interpreted by S

as the balance of forces in the equilibrium

position.

Remark 6.3

Despite their name, apparent forces are measurable in practice by a non-inertial

observer. This is true in particular for the inertial force; imagine carrying an

object in your hand and giving it an upwards or downwards acceleration (less

than the acceleration due to gravity, so the object does not leave your hand): in

the two cases one perceives that the force exerted on the hand (corresponding to

−φ) is either increased or decreased. These variations correspond to the inertial

force.

228

Rigid bodies: geometry and kinematics

6.7

The interpretation of (6.42) as an equilibrium equation in S

, with the

introduction of inertial forces, is often referred to as d’Alembert’s principle.

An important property in the study of relative dynamics is that the Galilean

relativity principle, stated in Section 2.2 of Chapter 2 for the case of inertial

observers, is valid for any class of observers whose relative motion is a uniform

translation.

roposition 6.4 The equation of motion (6.39) for a point particle (P, m) is

invariant in the class of observers that move with respect to one another with

zero acceleration.

Proof

If S and S are two observers whose relative motion has zero acceleration, the

sum a

+ a

in S with respect to an inertial observer

is equal to the sum

+ a

in S . Indeed, the relative accelerations a

(with respect to S) and a

(with respect to S ) are equal.

Note also that a

= a

+ 2

ω × v

, where v

is the translation velocity of S

with respect to S. Similarly a

= 2

ω × v

= 2

ω × (v

− v

). Hence the variation

T

− a

= 2

ω × v

is balanced by the variation of a

− a

. Since the observers

S and S cannot distinguish the individual sources of apparent forces, but can

only measure their sum, it follows that it is impossible for them to be aware of

their relative motion solely on the evidence of mechanical observations.

6.7

Ruled surfaces in a rigid motion

efinition 6.3 A ﬁxed ruled surface (respectively body ruled surface) of a given

rigid motion is the locus of the lines which take the role of instantaneous axes of

motion in the ﬁxed (respectively, in the body) reference frame.

A trivial example is given by the purely rolling motion of a cylinder on a

plane; the ﬁxed ruled surface is the plane; the body ruled surface is the cylinder.

These surfaces are important because they can generate a prescribed rigid

motion when the moving ruled surface rolls onto the ﬁxed one (this fact is

exploited in the theory of gears).

heorem 6.5 In a generic rigid motion the body ruled surface rolls onto the

ﬁxed one. Along the contact line, the sliding velocity is equal to the invariant

component of the velocity ﬁeld along the angular velocity.

Proof

We only need to prove that the two ruled surfaces at every instant are tangent

to each other (this fact characterises the rolling motion); indeed, the last claim

of the theorem is evidently true, as the contact line coincides at every instant

with the axis of instantaneous motion.

Recalling equation (6.11), which gives the intersection point P

∗

between the

axis of motion and the plane orthogonal to it and passing through O, we can

6.7

Rigid bodies: geometry and kinematics

229

write the equation for the axis of motion r(t) as

− O = λω + ω × v(0)/ω

,

λ

∈ (−∞, +∞).

(6.43)

The two ruled surfaces are generated by the absolute and relative motion of the

line r(t) which moves with respect to the two reference frames so as to coincide

at every instant with r(t). To understand more clearly the diﬀerent role played

by r(t) and by r(t), consider the case of the cylinder rolling on a ﬁxed plane:

the points of the line r(t), considered in either

or S, have zero velocity, while

the points of the line r(t) move, sweeping the plane (absolute motion) or the

cylinder (relative motion).

We must then study the absolute and relative motion of the point P (λ, t)

∈ r(t),

deﬁned by (6.43): the absolute velocity is tangent to the ﬁxed ruled surface, while

the relative velocity is tangent to the body ruled surface. The velocity of the rigid

motion on the instantaneous axis of motion is v

= v(O)

· ω/|ω|. It follows that

v(P )

− v

(P ) = v

(6.44)

If we ignore the degenerate case when the instantaneous axis of motion is sta-

tionary, we have that neither v(P ) nor v

(P ) are parallel to

ω. Equation (6.44)

then implies that their diﬀerence must be either zero (v(O)

· ω = 0) or parallel

ω, and hence that the three vectors ω, v(P ), v

(P ) lie in the same plane.

Since the plane determined by the axis of motion and by v(P ) is tangent to

the ﬁxed ruled surface, and the plane determined by the axis of motion and by

R

(P ) is tangent to the body surface, this proves that the two planes coincide

and that the two ruled surfaces are tangent.

orollary 6.1 For the motions with ω of constant direction (in particular

for plane rigid motions) the ruled surfaces are cylinders. For plane motions the

intersections of the ruled surfaces with the representative plane are called conjugate

proﬁles.

orollary 6.2 In any precession, the ruled surfaces are cones with their vertex

in the precession pole (Poinsot cones).

The Poinsot cones are circular in the case of regular precession (Example 6.3)

and they both degenerate to a line in the case of rotations.

Example 6.4

We determine the conjugate proﬁles for the motion of a rod with endpoints

sliding along two orthogonal lines (Fig. 6.8).

The instantaneous centre of motion C can be trivially determined by using

Chasles’ theorem (see Problem 2), and considering the vectors normal to the

direction of the velocities of the two extremes A, B of the rod. We deduce that

in every conﬁguration of the pair of reference axes (O, x

, x

) the point C is

characterised as follows:

(a) in the ﬁxed reference frame (

Ω

, ξ

) it is the point at a distance 2 = AB

from the point

Ω

;

230

Rigid bodies: geometry and kinematics

6.8

j

2

j

1

A

x

2

x

1

C

O

B

Fig. 6.8

(b) in the reference frame (O, x

, x

) it is the point at a distance

from the

point O.

Hence the conjugate proﬁles in the ﬁxed and in the body plane are the circle

of centre

Ω

and radius 2 and the circle of centre O and radius , respectively.

The motion of the pole can be generated by laying the rod on the diameter of a

circle of radius

and making it rotate without sliding on a ﬁxed circle of twice

the radius, as in Fig. 6.8

6.8

Problems

1. Compute the direction cosines of a moving reference frame with respect to

a ﬁxed reference frame as a function of the Euler angles.

2. Prove that in a plane rigid motion the instantaneous centre of rotation lies

on the normal to the velocity of each point which is distinct from it (Chasles’

theorem).

3. A disc of radius R moves on a plane, rolling and sliding along a straight

line. Assuming that the velocity of the centre and the sliding velocity are known,

6.9

Rigid bodies: geometry and kinematics

231

ﬁnd the instantaneous centre of motion. Find also the conjugate proﬁles when

the above velocities are constant.

4. A rod AB moves on a plane while being tangent to a given circle of radius

R. The rod has one of its endpoints constrained to slide on a line tangent to

the circle. Prove that the conjugate proﬁles are two parabolas.

5. Compose two uniform rotations around distinct parallel axes (consider both

the case of rotations with the same or opposite direction).

6. Compose two uniform rotations around incident axes (the result is a regular

precession).

7. Compose two uniform rotations around skew lines, proving that the two

ruled surfaces of the compound motion are one-sheeted hyperboloids.

8. A disc of radius r rolls without sliding on a circle of radius R, while being

orthogonal to the circle plane and moving in such a way that its axis intersects

at every instant the normal to the circle plane passing through its centre. Prove

that the motion is a precession and that the Poinsot cones are circular.

9. Determine the generalised potentials of the centrifugal force and of the

Coriolis force.

6.9

Additional solved problems

Problem 1

It is well known that if A, B are two n

× n matrices (n ≥ 2) then e

= e

if and only if [A, B] = AB

− BA = 0. More generally, show that if

[C, A] = [C, B] = 0, where C = [A, B], then

= e

+B+[A,B]/2

(6.45)

Solution

We start by showing that if F is a function analytic in x

∈ R and entire (i.e.

its radius of convergence is +

∞) then

[A, F (B)] = [A, B]

(B).

Indeed it can be seen that if [C, B] = 0, then [C, B

] = 0 and, by induction on

≥ 0, that

[A, B

n

] = [A, B]nB

−1

From this it follows that

[A, F (B)] =

∞

n

=1

[A, B

] = [A, B]

∞

n

=1

−1

= [A, B]

(B),

232

Rigid bodies: geometry and kinematics

6.9

where F (x) =

∞

. Now let G(t) = e

e

tB

; it is immediate to verify that

(t) = (A + e

−tA

)G(t).

Since [B, e

] = [B, A]te

we have

= (A + B + t[A, B])G(t).

Finally, since A + B under our assumptions commutes with [A, B] the latter

diﬀerential equation admits the solution (note that G(0) = 1)

G(t) = exp

(A + B)t + [A, B]

2

2

from which equation (6.45) follows by choosing t = 1.

Problem 2

Let A and B be two n

× n matrices (n ≥ 2) and let t ∈ R. Prove that

= e

(A+B)+(t

2)[A,B]+(t

12)([A,[A,B]]+[B,[B,A]])+

)

(6.46)

Solution

By deﬁnition e

= 1 +

∞

/n!, from which it follows that

= 1 + t(A + B) +

2

2

+ B

+ 2AB)

+ B

+ 3AB

+ 3A

B) +

O(t

(6.47)

exp

t(A + B) +

[A, B] +

([A, [A, B]] + [B, [B, A]]) +

O(t

)

= 1 + t(A + B) +

[A, B] +

([A, [A, B]] + [B, [B, A]])

1

2

(A + B)

[(A + B)[A, B] + [A, B](A + B)]

(A + B)

O(t

(6.48)

and to obtain (6.46) one only needs to identify the coeﬃcients of t

, t

and

. In the case of t

and t

this is obvious, while for t

/2 we ﬁnd

+ B

+ 2AB = (A + B)

+ [A, B] = A

+ AB + BA + B

+ AB

− BA.

6.9

Rigid bodies: geometry and kinematics

233

For t

/12 the computation is more tedious. The coeﬃcient of t

/12 in (6.48) is

[A, [A, B]] + [B, [B, A]] + 3(A + B)[A, B] + 3[A, B](A + B) + 2(A + B)

(6.49)

By iterating the identity BA = AB

− [A, B] we ﬁnd

(A + B)

= (A + B)(A

+ AB + BA + B

) = (A + B)(A

+ 2AB + B

− [A, B])

= A

3

+ B

+ 2A

B + AB

+ BA

+ 2BAB

− (A + B)[A, B]

= A

+ B

+ 3A

B + 3AB

− (2A + B)[A, B] − [A, B](A + 2B).

(6.50)

Comparing (6.49) with the coeﬃcient of t

/6 in (6.47), recalling also (6.50), it

follows that the proof of (6.46) is complete if we can show that

[A, [A, B]] + [B, [B, A]] + 3(A + B)[A, B] + 3[A, B](A + B)

− 2(2A + B)[A, B] − 2[A, B](A + 2B) = 0.

This reduces to

[A, [A, B]] + [B, [B, A]] + (B

− A)[A, B] + [A, B](A − B) = 0

with immediate veriﬁcation.

Problem 3

Consider a plane rigid motion and suppose that the conjugate proﬁles, as well

as the angular velocity ω(t), are known. Determine at every instant t:

(i) the locus of points P for which a(P )

v(P );

(ii) the locus of points P for which a(P )

⊥ v(P ).

Solution

Let C(t) be the instantaneous centre of rotation. This is the point at time t

where the conjugate proﬁles are tangent. By deﬁnition, the velocity v(C) = 0 in

the rigid motion (while in general the acceleration is not zero). Introduce the

point C(t) (moving with respect to both the ﬁxed and the body reference frame),

which at every instant coincides with C(t). In its absolute motion, C(t) travels

over the ﬁxed conjugate proﬁle, while its trajectory in the relative motion is the

body conjugate proﬁle. It is easily seen that C(t) has equal absolute and relative

velocities. Writing the absolute velocity ﬁeld at every instant t in the form

v(P ) = ω

× (P − C)

(6.51)

we obtain, by diﬀerentiating, the acceleration ﬁeld

a(P ) = ˙

× (P − C) + ω × [ω × (P − C) − v],

(6.52)

where v = dC/dt. The condition for a point to belong to the locus (i) is

a(P )

· (P − C) = 0, which reduces to

− C)

+ ω

× v · (P − C) = 0.

(6.53)

234

Rigid bodies: geometry and kinematics

6.9

Setting P = C = C in expression (6.52) we ﬁnd the acceleration of the instantan-

eous centre C, i.e. a(C) =

−ω × v, orthogonal to v. Setting (P − C)/|P − C| = e

and 2R = a(C)/ω

, expression (6.53) can be written

|P − C| = 2R · e,

clarifying the following structure of the locus (i).

The locus of points P for which a(P )

v(P ) at time t is a circle tangent to

the conjugate proﬁles, and of radius R = v/2ω.

The points having such a property ﬁnd themselves at a point of the trajectory

with vanishing curvature. For this reason, the locus (i) is also called the circle

of inﬂection points. Analogously, imposing the condition a(P )

× (P − C) = 0 one

arrives at the equation for the locus (ii):

ω(P

− C)

+ ω[v

· (P − C)] = 0.

If ˙

ω = 0 this locus reduces to the line passing through C and orthogonal to v.

If ˙

ω =

/ 0, setting 2R =

|ω/ ˙ω|v, we conclude that the locus of points P for which

a(P )

⊥ v(P ) at time t, if ˙ω =

/ 0, is a circle through C and orthogonal to the

conjugate proﬁles of radius R =

|ω/ ˙ω|v.

If ˙

ω = 0 it degenerates to a line normal to these proﬁles. The kinematic meaning

of the locus (ii) lies in the fact that the magnitude of the velocity of its points

has at that time zero time derivative, and hence the name stationary circle. We

note ﬁnally that the intersection H of the two circles, diﬀerent from C, has zero

acceleration. Hence the acceleration ﬁeld of the body frame can be written in

the form

a(P ) =

[ω

× (P − H)].

For this reason H is called the pole of accelerations.

To complete the problem, we now ﬁnd the relation between v and ω. Consider

the case that the principal normal vectors to the conjugate proﬁles at the point of

contact have opposite orientation. Let k

, k

be their curvatures. Considering the

osculating circles, it is easy to ﬁnd that if the point C undergoes a displacement

ds = v dt, the angular displacement of the respective normal vectors to the

two curves at the contact point is dϕ

= k

ds, dϕ

= k

ds, in the ﬁxed and

body frames, respectively. Consequently the variation of the angle that a ﬁxed

direction forms with a body direction is ω dt = dϕ

+dϕ

= (k

)v dt, yielding

ω = (k

)v. Hence we conclude that the radius of the circle of inﬂection points

is R =

+ k

), while the radius of the stationariness circle can be written as

R =

1

2

(ω

+ k

))/

| ˙ω|. In the case that the two principal normal vectors have

the same orientation, we substitute k

+ k

with

− k

| (obviously we must

have k

f

=

/ k

7 THE MECHANICS OF RIGID BODIES: DYNAMICS

7.1

Preliminaries: the geometry of masses

In contrast with kinematics, the dynamics of rigid bodies depends on the speciﬁc

distribution of masses. Hence it is necessary to review some results on the

geometry and kinematics of masses; we shall limit ourselves to the essential

facts.

1

Rigid bodies can be treated equally well by a discrete or by a continuum model.

The latter consists of deﬁning a mass density function in the region occupied

by the system. This is due to the fact that for rigid continua it makes sense

to consider forces applied to single points, or rather, to substitute a force ﬁeld

(such as weight) with equivalent systems of forces applied to various points of

the system, or rigidly connected to it. This should be contrasted with the case

of deformable continua, and for which one must deﬁne force densities.

To simplify notation, we consider in this chapter discrete rigid systems,

but the results can be easily extended to continua: it is enough to substi-

tute any expression of the type

f (P

), where the sum extends to

all points (P

, m

), . . . , (P

, m

) of the system, with an integral of the form

R ρ(P )f(P ) dV , where ρ is the density and R is the region occupied by the

body.

We start by recalling the notion of centre of mass:

m(P

− O) =

− O)

(7.1)

(O is an arbitrary point in R

, m =

). The moment of inertia with

respect to a line r is given by

[(P

− O) × e]

(7.2)

∈ r, e is a unit vector of r). The centrifugal moment or product of inertia

with respect to a pair of non-parallel planes π, π with normal vectors n, n is

ππ

−

[(P

− O) · n][(P

− O) · n ]

(7.3)

A more detailed description can be found in Fasano et al. (2001).

Recall that in the gravity ﬁeld the centre of mass coincides with the baricentre.

236

The mechanics of rigid bodies: dynamics

7.2

(O

∈ π ∩ π ). Given a reference frame (O, x

, x

), we denote by I

, I

the moments of inertia with respect to the axes and by I

, i =

/ j, the products

of inertia with respect to the pairs of coordinate planes x

= 0, x

= 0. Note

that the matrix (I

) is symmetric.

It is also possible to deﬁne other quadratic moments, but we will not consider

them here. For each of them Huygens’ theorem holds, which in the cases of

interest here, takes the form

= I

+ m[(P

− O) × e]

(7.4)

ππ

= I

− m[(P

− O) · n][(P

− O) · n ],

(7.5)

where r

is the line parallel to r and passing through the centre of mass P

, and

, π

are planes parallel to π, π , respectively, and passing through P

. The

other symbols in (7.4), (7.5) keep the same meaning as in formulas (7.1)–(7.3).

It is customary to also deﬁne the radius of gyration with respect to a line:

r

= (I

/m)

1/2

(7.6)

and hence equation (7.4) can also be written as

= δ

+ δ

(7.7)

where δ is the distance between P

and r.

Remark 7.1

Equation (7.4) implies that among all lines in a given direction, the one for

which the moment of inertia has a minimum is the one passing through P

7.2

Ellipsoid and principal axes of inertia

The distribution of moments of inertia with respect to the set of lines through

a given point O plays an important role in the dynamics of rigid bodies.

Given a reference frame (O, x

, x

), we now compute the moment of inertia

with respect to the line passing through O with direction cosines α

, α

Applying the deﬁnition (7.2) we ﬁnd

I(α

, α

) =

i,j

(7.8)

Since in equation (7.2) we have that I

> 0 for any unit vector e, the quadratic

form (7.8) is positive deﬁnite (at least excluding the degenerate case of a rigid

system with mass distributed along a straight line passing through O, in which

the matrix is positive semideﬁnite). It follows that the level sets are the ellipsoid

of centre O (ellipsoid of inertia) given by the equation

I

ij

= λ

(7.9)

7.2

The mechanics of rigid bodies: dynamics

237

D

efinition 7.1 The symmetry axes of the ellipsoid of inertia are called the

principal axes of inertia.

If the ellipsoid of inertia is an ellipsoid of rotation around one of the axes, all

straight lines through O and orthogonal to this axis are principal axes of inertia.

A principal reference frame (O, X

, X

) is the triple of the three principal

axes of inertia. We denote by

the moment of inertia with respect to the

principal axis X

. The direction of the principal axes of inertia is determined by

the eigenvectors of the matrix (I

) and the principal moments of inertia

are

the eigenvalues of the same matrix (cf. Proposition 7.8).

roposition 7.1 A necessary and suﬃcient condition for a reference frame to

be the principal frame is that the inertia products with respect to each of the pairs

of coordinate planes are zero.

Proof

If I

= 0 for i =

/ j, setting I

, the level sets of (7.8) are determined by

= λ

(7.10)

and the equation is put in the canonical form characteristic of the triple of the

symmetry axes. Conversely, the quadratic form (7.8) written with respect to the

principal frame is diagonal and the ellipsoid equation is of the form (7.10).

roposition 7.2 A necessary and suﬃcient condition for a straight line to be

a principal axis of inertia relative to O is that the products of inertia relative to

the plane passing through O and orthogonal to the line and to any other plane

through the line are zero.

Proof

Suppose that a straight line is a principal axis of inertia and consider an arbitrary

frame (O, x

, x

), with x

being the line itself. The plane of equation x

= 0

is then a symmetry plane for the ellipsoid. It follows that its equation must be

invariant with respect to the variable change from x

to

−x

, i.e. I

= I

= 0.

Conversely, if in the same frame one has I

= I

= 0, the ellipsoid is

symmetric with respect to the plane x

= 0. Hence x

is a principal axis.

It is easier to determine the principal axis when there are material symmetries.

efinition 7.2 A rigid system has a plane of material orthogonal symmetry

if in addition to the geometric symmetry with respect to the plane, one has the

property that symmetric points have the same mass (or, for continua, the same

density).

This deﬁnition can be extended to the case of material symmetry with respect

to a straight line or to a point.

238

The mechanics of rigid bodies: dynamics

7.2

P

roposition 7.3 Let π be a plane of material orthogonal symmetry. Then for

any O in π, the straight line through O and normal to π is a principal axis of

inertia relative to O.

Proof

The ellipsoid of inertia relative to O is symmetric with respect to π.

Remark 7.2

The latter proposition can be applied to the limiting case of plane systems.

roposition 7.4 Suppose that a system has two distinct (non-parallel) planes

of material symmetry, π and π . Then there are two possible cases:

(a)

the two planes are orthogonal;

(b)

the two planes are not orthogonal.

In the case (a), given O

∈ r = π ∩ π , the principal frame relative to O is given

by r and by the two straight lines through O normal to r lying in the planes π, π .

In the case (b) the ellipsoid of inertia is a surface of revolution with respect to r.

Proof

Case (a) is trivial. In case (b) we note that the two straight lines through O

and normal to π and π are principal axes (Proposition 7.3). It follows that the

ellipsoid has a plane of symmetry containing two non-orthogonal symmetry axes.

This necessarily implies the rotational symmetry with respect to r.

Another useful property for the determination of the principal axes of inertia

is the following.

roposition 7.5 Let (P

, X

(0)

, X

(0)

, X

(0)

) be a principal frame relative to the

centre of mass. The principal axes of inertia with respect to the points of the axes

(0)

, X

(0)

, X

(0)

can be obtained from it by translation (Fig. 7.1).

Proof

We recall Huygens’ theorem (1.5) and use it to verify that the products of inertia

in the translated frame are zero.

Problem 7.1

Determine the principal frame for a regular material homogeneous polygon,

relative to a generic point of the plane of the polygon.

The following remark is in some sense the converse of Proposition 7.5.

roposition 7.6 If a straight line is a principal axis of inertia with respect to

two of its points (distinct), then it must contain the centre of mass.

Proof

Let O and O be the two points, and consider two frames with parallel axes

S = (O, x

, x

), S = (O , x

, x

), where x

is the line referred to in the

statement. Recalling Proposition 7.2, we can write

(i)

= 0,

7.3

The mechanics of rigid bodies: dynamics

239

X

(0)

0

X

3

X

1

X

2

O

Fig. 7.1

and similarly

(i)

− z) =

(i)

− z) = 0,

where z is the third coordinate of O in S.

It follows that

i

=1

(i)

= 0,

and hence that P

lies on the axis x

efinition 7.3 A system such that the central ellipsoid of inertia is a surface

of revolution is called a gyroscope. The axis of revolution of the central ellipsoid

is called the gyroscopic axis.

7.3

Homography of inertia

Let us ﬁx a triple S = (O, x

, x

). Then the product of the symmetric matrix

⎛

⎝

I

⎞

⎠

(7.11)

240

The mechanics of rigid bodies: dynamics

7.3

with the vector

⎛

⎝

⎞

⎠ deﬁnes a map σ : R

→ R

, called the homography of

inertia.

roposition 7.7 The map σ is independent of the choice of the frame S, and

depends only on its origin O.

Proof

We now show that the vector σx is intrinsically deﬁned. Setting

f (x) =

i, j

(7.12)

it is easily veriﬁed that

σx =

∇f.

(7.13)

If x is such that f (x) = λ

, then σx is orthogonal in x to the ellipsoid of inertia.

Since σ is a linear map, we can deduce that in general σx is normal to the plane

tangent to the ellipsoid of inertia at the intersection point with the straight line

through the origin parallel to x (Fig. 7.2).

We also note that for every unit vector e:

· σe = I

(7.14)

where r is the line passing through O with unit vector e. More generally

· σx = I

(7.15)

for any vector (I

is computed with respect to the line through O and parallel

to x). Thus the quadratic form x

· σx is positive deﬁnite (we are considering the

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