Alain J. Brizard
Saint Michael's College
Rigid Body Motion
1
Inertia Tensor
1.1
Discrete Particle Distribution
We begin our description of rigid body motion by considering the case of a rigid discrete
particle distribution in which the inter-particle distances are constant.
The position of each particle ® as measured from a ¯xed laboratory (LAB) frame is
r0® = R + r ® ;
where R is the position of the center of mass (CM) in the LAB and r® is the position of
the particle in the CM frame. The velocity of particle ® in the LAB frame is
v0® = V + ! £ r ® ;
(1)
where ! is the angular velocity vector associated with the rotation of the particle distribution about an axis of rotation which passes through the CM and V is the CM velocity in
the LAB frame.
1
The kinetic energy of particle ® (with mass m® ) in the LAB frame is
´
m® 0 2
m® ³ 2
jv® j =
jVj + 2 V ¢ ! £ r ® + j! £ r® j2 ;
2
2
K® =
and thus the total kinetic energy of the particle distribution is
K =
X
Ã
K® =
®
X
!
m®
®
1
+
2
(
Ã
X
jVj2
+ V £!¢
m® r®
2
®
Ã
!
2
!
X
m® r®2
Ã
X
¡ !! :
®
!
!)
m® r® r®
:
®
P
P
Since the total mass of the particle distribution is M = ® m® and ® m® r® = 0 (by
de¯nition), the total kinetic energy is, therefore, expressed as K = Ktrans + K rot , where
the translational (trans) kinetic energy
M
jVj 2
2
K trans =
(2)
represents the kinetic energy associated with the translational motion of the CM position,
while the rotational (rot) kinetic energy
1
=
2
Krot
(
!
2
Ã
X
!
Ã
m® r2®
¡ !! :
®
X
!)
m® r ® r®
(3)
®
represents the internal kinetic energy associated with the rotational motion of the particle
distribution about its CM point.
We now introduce the inertia tensor of the particle distribution
I =
X
³
´
m® r2® 1 ¡ r® r® ;
(4)
®
where 1 denotes the unit diagonal tensor (i.e., in Cartesian coordinates, 1 = bxbx + byby + bzbz).
The inertia tensor can also be represented as a matrix
0 P
I =
B
B
B
B
B
B
@
¡
¡
®
m® (y®2 + z®2 )
P
P
®
m® (y® x® )
®
m® (z® x® )
¡
P
¡
P
®
m® (x® y®)
®
m® (x2® + z®2 )
P
®
m® (z® y® )
¡
¡
P
P
®
P
®
®
m® (x® z® )
m® (y® z® )
m® (x2® + y®2 )
1
C
C
C
C;
C
C
A
(5)
where the symmetry property of the inertia tensor (I ji = I ij ) is readily apparent. In terms
of the inertia tensor (4), the rotational kinetic energy (3) now becomes
Krot =
1
! ¢ I ¢ !:
2
2
(6)
1.2
Parallel-Axis Theorem
A translation of the origin from which the inertia tensor is calculated leads to a di®erent
inertia tensor. Let Q ® denote the position of particle ® in a new frame of reference (with
its origin located at point P in the Figure below)
and let ½ = r® ¡Q ® is the displacement from point CM to point P . The new inertia tensor
³
X
J =
m® Q2® 1 ¡ Q® Q®
´
®
can be expressed as
J =
X
³
´
X
m® ½2 1 ¡ ½ ½ +
®
½¢
Ã
X
!)
m® r®
(
1 +
Ã
½
®
Since M =
P
®
´
®
(
¡
³
m® r2® 1 ¡ r® r®
m® and
P
®
X
!
m® r ®
®
Ã
+
X
!
m® r®
)
½
:
®
m® r® = 0, we ¯nd
J = M
³
½2 1 ¡ ½ ½
´
+ I:
(7)
Hence, once the inertia tensor I is calculated in the CM frame, it can be calculated anywhere
else. Eq. (7) is known as the Paralle-Axis Theorem.
1.3
Continuous Particle Distribution
For a continuous particle distribution
3
the inertia tensor (4) becomes
I =
Z
³
´
dm r 2 1 ¡ rr ;
(8)
where dm(r) is the in¯nitesimal mass element at point r.
Consider, for example, the case of a cube of mass M and volume b3, with dm =
(M=b3 ) dx dy dz.
The inertia tensor (8) in the LAB frame (with the origin placed at one of its corners)
has the components
J 11 =
J
12
Z b
Z b
³
´
M Zb
2
2
2
dx
dy
dz
¢
y
+
z
= M b2 = J 22 = J 33
b3 0
3
0
0
Z b
Z b
M Zb
1
= ¡ 3
dx
dy
dz ¢ x y = ¡ M b2 = J 23 = J 31
0
0
0
b
4
4
(9)
(10)
and thus the inertia matrix for the cube is
0
J =
¡3
¡3
¡3
8
¡3
¡3
¡3
8
B
B
b2 B
M
12
8
B
B
B
@
1
C
C
C
C:
C
C
A
(11)
On the other hand, the inertia tensor calculated in the CM frame has the components
I 11 =
I
12
Z b=2
Z b=2
³
´
M Z b=2
1
2
2
dx
dy
dz
¢
y
+
z
= M b2 = I 22 = I 33 (12)
3
¡ b=2
¡ b=2
¡ b=2
b
6
Z b=2
Z b=2
M Z b=2
= ¡ 3
dx
dy
dz ¢ x y = 0 = I 23 = I 31
¡ b=2
¡ b=2
¡ b=2
b
(13)
and thus the CM inertia matrix for the cube is
0
M b2
I =
6
B
B
B
B
B
B
@
1
0
0
0
1
0
0
0
1
1
C
C
C
C:
C
C
A
(14)
The displacement vector ½ from the CM point to the corner O is given as
½=¡
b
(bx + by + bz) ;
2
so that ½2 = 3b2=4. By using the Parallel-Axis Theorem (7), the inertia tensor
0
M
³
½2 1 ¡ ½ ½
´
M b2
=
4
B
B
B
B
B
B
@
2
¡1
¡1
¡1
2
¡1
¡1
¡1
2
1
C
C
C
C
C
C
A
when added to the CM inertia tensor (14), yields the inertia tensor (11)
1.4
Principal Axes of Inertia
In general, the CM inertia tensor I can be made into a diagonal tensor with components
given by the eigenvalues I1 , I2, and I3 of the inertia tensor. These components (known as
principal moments of inertia) are the three roots of the cubic polynomial
I 3 ¡ Tr(I) I 2 + Ad(I) I ¡ Det(I) = 0;
5
(15)
obtained from Det(I ¡ I 1) = 0, with coe±cients
Tr(I) = I 11 + I 22 + I 33;
Ad(I) = ad11 + ad22 + ad33;
Det(I) = I 11 ad11 ¡ I 12 ad12 + I 13 ad13;
where adij is the determinant of the two-by-two matrix obtained from I by removing the
ith-row and j th-column from the inertia matrix I.
Each principal moment of inertia Ii represents the moment of inertia calculated about
the principal axis of inertia with unit vector bei. The unit vectors (be1; be2 ; be3 ) are related by
a sequence of rotations to the Cartesian CM unit vectors (bx1 ; bx2; bx3 ) by the relation
bei
= Rij xbj ;
(16)
where Rij are components of the rotation matrix R. By denoting as I0 the diagonal inertia
tensor calculated in the body frame of reference (along the principal axes), we ¯nd
0
I0 = R ¢ I ¢ RT
1
I1 0 0
B
= @ 0 I2 0 C
A;
0 0 I3
(17)
where RT denotes the transpose of R, i.e., (RT )ij = Rji . In the body frame, the inertia
tensor is, therefore, expressed in dyadic form as
I0 = I1 be1 be1 + I2 be2 be2 + I3 be3 be3;
(18)
and the rotational kinetic energy (6) is
0
Krot
=
´
1
1³
! ¢ I0 ¢ ! =
I1 !21 + I2 !22 + I3 !32 :
2
2
(19)
An object is said to be a spherical top if its three principal moments of inertia are equal
(I1 = I2 = I3), i.e., the three roots of the cubic polynomial (15) are triply degenerate. The
object is said to be a symmetric top if two of its principal moments of inertia are equal
(I1 = I2 6= I3), i.e., I3 is a single root and I1 = I2 are doubly-degenerate roots of the cubic
polynomial (15). Lastly, when the three roots (I1 6= I2 6= I3) are all single roots of the
cubic polynomial (15), the object is said to be an asymmetric top.
Before proceeding further, we consider the example of a dumbbell composed of two
equal point masses m placed at the ends of a massless rod of total length 2 b and rotating
about the z-axis with angular frequency !.
6
Here, the positions of the two masses are expressed as
r§ = § b [sin µ (cos ' bx + sin ' by) + cos µ bz ] ;
so that the CM inertia tensor is
0
I =
B
B
B
2 m b2 B
B
B
@
1 ¡ cos2 ' sin2 µ
¡ cos ' sin ' sin2 µ
¡ cos ' cos µ sin µ
¡ cos ' sin ' sin2 µ
1 ¡ sin2 ' sin2 µ
¡ sin ' cos µ sin µ
¡ cos ' cos µ sin µ
¡ sin ' cos µ sin µ
1 ¡ cos2 µ
1
C
C
C
C:
C
C
A
(20)
After some tedious algebra, we ¯nd Tr(I) = 4 mb2, Ad(I) = (2 mb2 )2, and Det(I) = 0, and
thus the cubic polynomial (15) has the single root I3 = 0 and the double root I1 = I2 =
2 mb2, which makes the dumbbell a symmetric top.
The root I3 = 0 clearly indicates that one of the three principal axes is the axis of
symmetry of the dumbbell (be3 = rb). The other two principal axes are located on the
b From these choices,
plane perpendicular to the symmetry axis (i.e., be1 = µb and be2 = ').
we easily recover the rotation matrix R
0
1
cos ' cos µ sin ' cos µ ¡ sin µ
B
C
R = R2(¡ µ) ¢ R3 (') = @ ¡ sin '
cos '
0 A;
cos ' sin µ sin ' sin µ
cos µ
so that, using the spherical coordinates (r; µ; '), we ¯nd
b
e1
b
e2
b
e3
b
= cos µ (cos ' bx + sin ' by ) ¡ sin µ bz = µ;
b
= ¡ sin ' bx + cos ' by = ';
= sin µ (cos ' bx + sin ' by) + cos µ bz = rb:
Indeed, the principal moment of inertia about the rb-axis is zero, while the principal moments
b and '-axes
b
of inertia about the perpendicular µare equally given as 2 mb2.
7
2
Angular Momentum
2.1
Euler Equations
The canonical angular momentum of a discrete particle distribution is
X
L =
r ® £ p® =
®
(
=
X
r® £ f m® (V + ! £ r® ) g
®
X
)
m® (r® 1 ¡ r ® r® )
¢ ! = I ¢ !;
(21)
®
P
where ® m® r® = 0 was used and the de¯nition of the CM inertia tensor (4) was introduced. Since the two vectors L and ! are in general not colinear, we ¯nd ! £ L 6= 0.
The time derivative of the angular momentum (21) in the ¯xed (LAB) frame is given
as
Ã
dL
dt
!
Ã
=
f
dL
dt
!
+ ! £ L = N;
r
where N represents the external torque applied to the system and (dL=dt)r denotes the
rate of change of L in the rotating frame. By choosing the body frame as the rotating
frame, we ¯nd
Ã
dL
dt
!
= I ¢ !_ = (I1 !_ 1 ) be1 + (I2 !_ 2 ) be2 + (I3 !_ 3) be3;
(22)
r
while
! £ L = ¡ be1 f !2 !3 (I2 ¡ I3) g ¡ be2 f !3 !1 (I3 ¡ I1 ) g ¡ be3 f !1 !2 (I1 ¡ I2 ) g : (23)
Thus the time evolution of the angular momentum
scribed in terms of
I1 !_ 1 ¡ !2 !3 (I2 ¡ I3)
I2 !_ 2 ¡ !3 !1 (I3 ¡ I1)
I3 !_ 3 ¡ !1 !2 (I1 ¡ I2)
in the body frame of reference is de9
= N1 >
=
= N2 > ;
= N3 ;
(24)
which are known as the Euler equations. Lastly, we note that the rate of change of the
rotational kinetic energy (6) is expressed as
dKrot
= ! ¢ I ¢ !_ = ! ¢ (¡ ! £ L + N) = N ¢ !:
dt
(25)
We note that in the absence of external torque (N = 0), not only is kinetic energy conserved
P
but also L2 = 3i=1 (Ii!i )2, as can be veri¯ed from Eq. (24).
8
2.2
Euler Equations for a Force-Free Symmetric Top
As an application of the Euler equations (24) we consider the case of the dynamics of
a force-free symmetric top, for which N = 0 and I1 = I2 6= I3. Accordingly, the Euler
equations (24) become
9
I1 !_ 1 = !2 !3 (I1 ¡ I3) >
=
I1 !_ 2 = !3 !1 (I3 ¡ I1) ;
(26)
>
;
I3 !_ 3 = 0
The last Euler equation states that if I3 6= 0, we have !_ 3 = 0 or that !3 is a constant of
motion. Next, after de¯ning the frequency
µ
¶
I3
!p = !3
¡1 ;
(27)
I1
which may be positive (I3 > I1 ) or negative (I3 < I1), the ¯rst two Euler equations yield
!_ 1(t) = ¡ !p !2(t) and !_ 2(t) = !p !1(t):
(28)
The general solutions for !1 (t) and !2(t) are
!1(t) = !0 cos(!pt + Á0 ) and !2(t) = !0 sin(!pt + Á 0);
(29)
where !0 is a constant and Á0 is an initial phase associated with initial conditions for !1(t)
and !2(t). Since !3 and !20 = !12(t) + !22 (t) are constant, then the magnitude of the angular
velocity !,
q
! = !12 + !22 + !23 ;
is also a constant. Thus the angle ® between ! and be3 is constant, with
!3 = ! cos ® and
q
!21 + !22 = !0 = ! sin ®:
Since the magnitude of ! is also constant, the !-dynamics simply involves precession
motion of ! about the be3-axis with a precession frequency !p ; as a result of precession, the
vector ! spans the body cone
9
with !p > 0 if I3 > I1 (for a pancake-shaped or oblate symmetric top) or !p < 0 if I3 < I1
(for a cigar-shaped or prolate symmetric top).
For example, to a good approximation, Earth is an oblate spheroid with
I1 =
³
´
1
2
M a2 + c2 = I2 and I3 = M a 2 > I1;
5
5
where 2 c = 12; 714 km is the Pole-to-Pole distance and 2 a = 12; 756 km is the equatorial
diameter, so that
I3
a 2 ¡ c2
¡ 1 = 2
= 0:003298::: = ²:
I1
a + c2
The precession frequency (27) of the rotation axis of Earth is, therefore, !p = ² !3, where
!3 = 2¼ rad/day is the rotation frequency of the Earth, so that the precession motion
repeats itself every ² ¡1 days or 303 days; the actual period is 430 days and the di®erence is
due to the non-rigidity of Earth and the fact that the Earth is not a pure oblate spheroid.
A slower precession motion of approximately 26,000 years is introduced by the combined
gravitational e®ect of the Sun and the Moon on one hand, and the fact that the Earth's
rotation axis is at an angle 23.5o to the Ecliptic plane (on which most planets move).
The fact that the symmetric top is force-free implies that its rotational kinetic energy
is constant [see Eq. (25)] and, hence, L ¢ ! is constant while ! £ L ¢ be3 = 0 according to
Eq. (23). Since L itself is constant in magnitude and direction in the LAB (or ¯xed) frame,
we may choose the bz-axis to be along L (i.e., L = L bz). If at a given instant, !1 = 0, then
!2 = !0 = ! sin ® and !3 = ! cos ®. Likewise, we may write L1 = I1 !1 = 0, and
L2 = I2 !2 = I1 ! sin ® = L sin µ;
L3 = I3 !3 = I3 ! cos ® = L cos µ;
where L ¢ ! = L ! cos µ, with µ represents the space-cone angle. From these equations, we
¯nd the relation between the body-cone angle ® and the space-cone angle µ to be
µ
tan µ =
I1
I3
¶
tan ®;
(30)
which shows that µ > ® for I3 < I1 and µ < ® for I3 > I1.
2.3
Euler Equations for a Force-Free Asymmetric Top
We now consider the general case of an asymmetric top moving under force-free conditions.
To facilitate our discussion, we assume that I1 > I2 > I3 and thus Euler's equations (24)
for a force-free asymmetric top are
9
>
I1 !_ 1 = !2 !3 (I2 ¡ I3) =
I2 !_ 2 = ¡ !3 !1 (I1 ¡ I3 ) ;
>
I3 !_ 3 = !1 !2 (I1 ¡ I2) ;
10
(31)
where we henceforth assume that !1 < 0 so that !_ 2 is positive. As previously mentioned,
the Euler equations (31) have two constants of the motion: kinetic energy
K =
´
1³
I1 !12 + I2 !22 + I3 !23 ;
2
and the squared magnitude of the angular momentum
L2 = I12 !12 + I22 !22 + I33 !32:
To proceed with a general solution of Eqs. (31), we introduce the constants
¾ = 2 I1K ¡ L2 and ½ = L2 ¡ 2 I3K;
from which we obtain expressions for !1 and !3 in terms of !2:
!1 =
v
u
u½
¡t
v
u
u ¾ ¡ I2 (I1 ¡ I2 ) !22
¡ I2 (I2 ¡ I3 ) !22
and !3 = t
:
I1 (I1 ¡ I3)
I3 (I1 ¡ I3 )
(32)
When we substitute these expressions in the Euler equation for !2, we easily obtain
!_ 2 = ®
q
(!21 ¡ !22) (!23 ¡ !22);
(33)
where ® is a positive dimensionless constant de¯ned as
® =
sµ
¶
I2 ¶ µ I2
1¡
¡1 ;
I1
I3
while the constant frequencies ! 1 and ! 3 are de¯ned as
!21
2 I1 K ¡ L 2
L2 ¡ 2 I3K
2
=
and !3 =
:
I2(I1 ¡ I2)
I2 (I2 ¡ I3 )
We immediately note that the evolution of !2 is characterized by the two frequencies ! 1
and !3 , which also represent the turning points at which !_ 2 vanishes.
Next, by introducing a dimensionless frequency u = !2 =! 1 (here, we assume that
!1 > ! 3) and a dimensionless time ¿ = ®!3 t, the Euler equation (33) can now be
integrated to yield
¿ =
Z u
0
q
ds
(1 ¡ s2 ) (1 ¡ k2 s 2)
=
Z £(u)
0
p
dµ
;
1 ¡ k2 sin2 µ
(34)
where k 2 = ! 21=!23 > 1, £(u) = arcsin u, and we assume that !2(t = 0) = 0; compare
Eq. (34) with the solution of the pendulum equation. The turning points for !2 are now
represented in terms of turning points for u as !2 = !1 ! u = 1 and !2 = !3 ! u =
k¡1 = !3=!1 < 1.
11
3
3.1
Symmetric Top with One Fixed Point
Eulerian Angles as generalized Lagrangian Coordinates
To describe the physical state of a rotating object with principal moments of inertia
(I1; I2; I3), we need the three Eulerian angles ('; µ; Ã) in the body frame of reference.
The Eulerian angle ' is associated with the rotation of the ¯xed-frame unit vectors
(bx; by; bz) about the z-axis.
We thus obtain the new unit vectors (bx0; by 0; bz0 ) de¯ned as
0
1
b
x0
B 0 C
b A
@ y
zb0
z0
= R3 (')
}|
1{
0
1
bx
cos ' sin ' 0
B
C B
C
= @ ¡ sin ' cos ' 0 A ¢ @ yb A
bz
0
0
1
(35)
The rotation matrix R3(') has the following properties associated with a general rotation
matrix Ri (®), where a rotation of axes about the xi-axis is performed through an arbitrary
angle ®. First, the matrix Ri(¡ ®) is the inverse matrix of Ri(®), i.e.,
Ri (¡ ®) ¢ Ri(®) = 1 = Ri (®) ¢ Ri (¡ ®):
Next, the determinant of Ri (®) is
Det[Ri(®)] = + 1:
Lastly, the eigenvalues of Ri(®) are + 1, ei® , and e¡ i® .
The Eulerian angle µ is associated with the rotation of the unit vectors (bx0 ; by0 ; bz0) about
the x0-axis.
12
We thus obtain the new unit vectors (bx00; by 00 ; bz00 ) de¯ned as
0
1
b
x00
B b00 C
@ y A
zb00
= R 1(µ)
z0
}|
1{
0
1
b
1
0
0
x0
B
C B b0 C
= @ 0 cos µ sin µ A ¢ @ y A
b
0 ¡ sin µ cos µ
z0
(36)
The Eulerian angle à is associated with the rotation of the unit vectors (bx00 ; by00 ; bz00) about
the z 00-axis.
We thus obtain the body-frame unit vectors (be1; be2; be3) de¯ned as
0
1
b
e1
B b C
@ e2 A
b
e3
= R 3(Ã)
z
0
}|
1{
0
1
b
cos à sin à 0
x00
B
C B b00 C
= @ ¡ sin à cos à 0 A ¢ @ y A
b
0
0
1
z00
(37)
Hence, the relation between the ¯xed-frame unit vectors (bx; by; bz) and the body-frame
unit vectors (be1; be2; be3 ) involves the matrix R = R3(Ã) ¢ R1(µ) ¢ R3('), such that
bei
= Rij bxj ;
13
(38)
or
b
e1
b
e2
b
e3
3.2
= cos à (cos ' bx + sin ' by ) ¡ sin à f cos µ (sin ' bx ¡ cos ' by) ¡ sin µ bz g ;
= ¡ sin à (cos ' bx + sin ' by ) ¡ cos à f cos µ (sin ' bx ¡ cos ' by) ¡ sin µ bz g ;
= sin µ (sin ' bx ¡ cos ' by) + cos µ bz:
Angular Velocity in terms of Eulerian Angles
The angular velocity ! represented in the three Figures above is expressed as
! = '_ bz + µ_ bx0 + Ã_ be3:
The unit vectors bz and xb0 are written in terms of the body-frame unit vectors (be1; be2 ; be3) as
b
z
b0
x
= sin µ (sin à be1 + cos à be2) + cos µ be3;
= bx00 = cos à be1 ¡ sin à be2:
The angular velocity can, therefore, be written exclusively in the body frame of reference
as
! = !1 be1 + !2 be2 + !3 be3;
(39)
where
!1 = '_ sin µ sin à + µ_ cos Ã
!2 = '_ sin µ cos à ¡ µ_ sin Ã
!3 = Ã_ + '_ cos µ
9
>
=
>
;
:
(40)
Note that all three components are independent of ' (i.e., @!i=@' = 0), while the derivatives with respect to à and Ã_ are
@!1
@!2
@!3
= ! 2;
= ¡ !1; and
= 0;
@Ã
@Ã
@Ã
and
@!1
@!2
@!3
=
0
=
and
= 1:
@ Ã_
@ Ã_
@ Ã_
The relations (40) can be inverted to yield
'_ = csc µ (sin à !1 + cos à !2);
µ_ = cos à !1 ¡ sin à !2 ;
Ã_ = !3 ¡ cot µ (sin à !1 + cos à !2 ):
14
3.3
Rotational Kinetic Energy of a Symmetric Top
The rotational kinetic energy (6) for a symmetric top can be written as
³
´o
1 n
I3 !23 + I1 !12 + !22
;
Krot =
2
_ Ã)
_ as
or explicitly in terms of the Eulerian angles ('; µ; Ã) and their time derivatives (';
_ µ;
¾
´2
³
´
1 ½ ³_
I3 Ã + '_ cos µ + I1 µ_2 + '_ 2 sin2 µ
:
(41)
2
We now brie°y return to the case of the force-free symmetric top for which the Lagrangian
_ ';
_ = Krot . Since ' and à are ignorable coordinates, i.e., the force-free
is simply L(µ; µ;
_ Ã)
Lagrangian (41) is independent of ' and Ã, their canonical angular momenta
Krot =
@L
= I3 (Ã_ + '_ cos µ) cos µ + I1 sin2 µ ';
_
@ '_
@L
pà =
= I3 (Ã_ + '_ cos µ) = I3 !3
_
@Ã
are constants of the motion. By inverting these relations, we obtain
p' =
p' ¡ pà cos µ
(p' ¡ pà cos µ) cos µ
and Ã_ = !3 ¡
;
2
I1 sin µ
I1 sin2 µ
and the rotational kinetic energy (41) becomes
'_ =
Krot
1
=
2
(
_2
I1 µ +
I3 !23
(p ¡ pà cos µ)2
+ '
I1 sin2 µ
(42)
(43)
(44)
)
:
(45)
Hence, the motion of a force-free symmetric top can be described in terms of solutions of
the Euler-Lagrange equation for the Eulerian angle µ
d
dt
Ã
!
@L
= '_ sin µ (I1 cos µ '_ ¡ pÃ)
@µ
(p ¡ pà cos µ) (pà ¡ p' cos µ)
= ¡ '
:
(46)
I1 sin µ
sin2 µ
Once µ(t) is solved for given values of the principal moments of inertia I1 = I2 and I3
and the invariant canonical angular momenta p' and pÃ, the functions '(t) and Ã(t) are
determined from the time integration of Eqs. (44).
3.4
@L
@ µ_
= I1 µÄ =
Lagrangian Dynamics of a Symmetric Top with One Fixed
Point
We now consider the case of a spinning symmetric top of mass M and principal moments of
inertia (I1 = I2 6= I3) with one ¯xed point O moving in a gravitational ¯eld with constant
acceleration g.
15
The rotational kinetic energy of the symmetric top is given by Eq. (41) while the potential
energy for the case of a symmetric top with one ¯xed point is
U(µ) = M gh cos µ;
(47)
where h is the distance from the ¯xed point O and the center of mass (CM) of the symmetric
top. The Lagrangian for the symmetric top with one ¯xed point is
_ ';
_ = 1
L(µ; µ;
_ Ã)
2
½
I3
³
Ã_ + '_ cos µ
´2
+ I1
³
´ ¾
µ_2 + '_ 2 sin2 µ
¡ M gh cos µ:
(48)
Once again the canonical angular momenta p' and pà are constants of the motion since the
Lagrangian (48) is independent of the Eulerian angles ' and Ã, respectively. The solution
for µ(t) is then most easily obtained by considering the energy equation
1
E =
2
(
(p ¡ pà cos µ)2
I1 µ_ 2 + I3 !23 + '
I1 sin2 µ
)
+ M gh cos µ:
(49)
simply obtained by adding the rotational kinetic energy (45) and the gravitational potential
energy (47); here p' and pà = I3 !3 are constants of the motion.
Since the total energy E is itself a constant of the motion, we may de¯ne a new energy
constant
1
E 0 = E ¡ I3 !32;
2
and an e®ective potential energy
(p' ¡ pà cos µ)2
V (µ) =
+ Mgh cos µ;
2I1 sin2 µ
16
(50)
so that Eq. (49) becomes
E0 =
1
I1 µ_ 2(t) + V (µ);
2
(51)
which can be formally solved as
t(µ) = §
Z
q
dµ
(2=I1 ) [E0 ¡ V (µ)]
:
(52)
A simpler formulation for this problem is obtained as follows. First, we de¯ne the
following quantities
!2 =
2 M gh
2 E0
p'
pÃ
; ² =
; ® =
; and ¯ =
;
2
I1
I1 !
I1 !
I1 !
(53)
so that Eq. (52) becomes
¿(u) = §
Z
du
q
(1 ¡ u2)(² ¡ u) ¡ (® ¡ ¯ u)2
= §
Z
du
q
(1 ¡ u2)[² ¡ W (u)]
;
(54)
where ¿(u) = ! t(u), u = cos µ, and the energy equation (51) becomes
2 Ã
1
4
² =
(1 ¡ u2)
du
d¿
3
!2
+ (® ¡
¯ u)2 5
Ã
2 ¡1
+ u = (1 ¡ u )
du
d¿
!2
+ W (u):
(55)
We note that the e®ective potential W (u) is in¯nite at u = §1 and has a single minimum
at u = u0 (or µ = µ0 ) de¯ned by the quartic equation
Ã
W 0 (u0) = 1 + 2u0
® ¡ ¯ u0
1 ¡ u20
!2
Ã
¡ 2¯
® ¡ ¯ u0
1 ¡ u20
!
= 0:
(56)
This equation has four roots: two roots are complex roots, a third root is always greater
than one for ® > 0 and ¯ > 0 (which is unphysical since u = cos µ < 1), while the fourth
root is less than one for ® > 0 and ¯ > 0; hence, this root is the only physical root
corresponding to a single minimum for the e®ective potential W (u) (see Figure below).
17
We now investigate the motion of the symmetric top at the minimum angle µ 0 for which
² = W(u0 ) and u(u
_ 0) = 0. For this purpose, we note that the dimensionless azimuthal
frequency
®¡¯u
d'
=
= º(u)
d¿
1 ¡ u2
has two solutions when evaluated at the potential minimum u = u0 (or µ = µ0):
¯
º(u0) =
2u0
s
Ã
!
2u
1 ¡ 20 ;
¯
1 §
Hence, the precession frequency '_ 0 = º(u0) ! at µ = µ 0 has a slow component and a fast
component
2
('_ 0)slow =
I3 !3
6
41 ¡
2 I1 cos µ0
2
( '_ 0)f ast =
I3 !3
6
41 +
2 I1 cos µ0
v
u
u
t
v
u
u
t
1 ¡ 2
1 ¡ 2
µ
µ
I1 !
I3 ! 3
I1 !
I3 ! 3
¶2
¶2
3
cos µ 0 7
5;
3
7
cos µ0 5 :
We further note that these solutions require that the radicand be positive, or
I3 ! 3 > 2 I1 ! u0 ;
if u0 > 0 (or µ 0 < ¼=2); no conditions are applied to !3 for the case u0 < 0 (or µ0 > ¼=2)
since the radicand is strictly positive in this case. We further note that for µ0 < ¼=2 (or
cos µ 0 > 0) the two precession frequencies ('_ 0)slow and ( '_ 0)f ast have the same sign while
for µ 0 > ¼=2 (or cos µ0 < 0) the two precession frequencies have opposite signs ('_ 0 )slow < 0
and ('_ 0 )fast > 0.
Next, we investigate the case with two turning points u1 < u0 and u2 > u1 (or µ1 > µ2 )
where ² = W (u) (see the Figure below). Hence, the µ-dynamics oscillates between µ1 and
µ2 and this phenomenon is called nutation.
18
The turning points u1 and u2 are roots of the function
F (u) = (1 ¡ u2) [² ¡ W (u)] = u3 ¡ (² + ¯ 2) u2 ¡ (1 ¡ 2 ®¯) u + (² ¡ ®2):
(57)
Although a third root u3 exists for F (u) = 0, it is unphysical since u3 > 1.
Since the azimuthal frequencies at the turning points are expressed as
d'1
® ¡ ¯ u1
d'2
® ¡ ¯ u2
=
and
=
;
2
d¿
1 ¡ u1
d¿
1 ¡ u22
where ® ¡¯ u1 > ® ¡¯ u2 , we see that three scenarios for nutation appear (here, we assume
that both ® and ¯ are positive). In the ¯rst scenario, where ® > ¯ u2 , the precession
frequency d'=d¿ is strictly positive for u1 < u < u2 and nutation proceeds monotonically.
In the second scenario, where ® = ¯ u2, the precession frequency d'=d¿ is positive for
u1 < u < u2 and vanishes at u = u2 ; nutation in this scenario exhibits a cusp at µ2. In
the third scenario, where ® < ¯ u2 , the precession frequency d'=d¿ reverses its sign at
ur = ®=¯ or µ2 < µr = arccos(®=¯) < µ1.
3.5
Stability of the Sleeping Top
Let us consider the case where a symmetric top with one ¯xed point is launched with
initial conditions µ0 6= 0 and µ_0 = '_ 0 = 0, with Ã_ 0 6= 0. In this case, the invariant canonical
momenta are
pà = I3 Ã_ 0 and p' = pà cos µ 0:
These initial conditions (u0 = ®=¯; u_ 0 = 0), therefore, imply from Eq. (55) that ² = u0 and
that the energy equation (55) now becomes
Ã
du
d¿
!2
=
h
(1 ¡ u2 ) ¡ ¯ 2 (u0 ¡ u)
19
i
(u0 ¡ u):
(58)
Next, we consider the case of the sleeping top for which an additional initial condition
is µ0 = 0 (and u0 = 1). Thus Eq. (58) becomes
Ã
du
d¿
!2
=
³
´
1 + u ¡ ¯ 2 (1 ¡ u)2 :
(59)
The sleeping top has the following equilibrium points (where u_ = 0): u1 = 1 and u2 = ¯2 ¡1.
We now investigate the stability of the equilibrium point u1 = 1 by writing u = 1 ¡± (with
± ¿ 1) so that Eq. (59) becomes
q
d±
= 2 ¡ ¯ 2 ±:
d¿
The solution of this equation is exponential (and, therefore, u1 is unstable) if ¯2 < 2
or oscillatory (and, therefore, u1 is stable) if ¯2 > 2. Note that in the latter case, the
condition ¯ 2 > 2 implies that the second equilibrium point u2 = ¯2 ¡ 1 > 1 is unphysical.
We, therefore, see that stability of the sleeping top requires a large spinning frequency !3 ;
in the presence of friction, the spinning frequency slows down and ultimately the sleeping
top becomes unstable.
20