Download Chapter 9 Rigid Body Motion in 3D - RIT

Chapter 9 Rigid Body Motion in 3D
Rigid body rotation in 3D is a complicated problem requiring the introduction of tensors.
Upon completion of this chapter we will be able to describe such things as the motion of
a top, or of a bicycle.
1
Rotation of body about an arbitrary axis—Moments and
Products of Inertia, Inertia Tensor, Angular Momentum
and Kinetic Energy
Consider an arbitrary rigid object and a set of xyz axes. Our first task will be to find the
moment of inertia about an axis passing through the origin but oriented in an arbitrary
direction, not one of the coordinate axes. The line is bidirectional, but once we set the
object into rotation the vector ω
~ will define a unique direction.
In Chapter 1 we discussed direction cosines. Angles α, β, γ are the angles between a vector
and the xyz axes (pages 16, 17 of the text) so that Ax = A cos α etc.
Refer to Figure 9.1.1 that shows the object, the z-axis of the coordinate system, a rotation
axis with ω
~ , and a small piece of the object labeled mi a distance ~r from the origin, with
moment arm ri⊥ from the axis. It has a velocity ~vi that is perpendicular to both ~ri and ω
~.
The scalar moment of inertia of the object about the rotation axis is by definition
X
2
I=
mi ri⊥
(1)
where we have used a summation, but we could easily move to continuous mass distributions
and do an integral.
Now we want to express this in terms of the cartesian coordinates xi , yi , zi and the direction
cosines.
1
Figure 1: Rigid body rotating about a general axis, not one of the coordinate axes. Unit
vector for axis set by angular velocity is n̂ One infinitesimal piece of object is shown.
Define the unit vector n̂ as ω
~ = ωn̂. We can recognize that
ri⊥ = |ri sin θi | = |~ri × n̂|
(2)
where θi is the angle between the axis of rotation n̂ and the radius vector ~ri .
The position of the infinitesimal mass is ~ri = xi î + yi ĵ + zi k̂. The unit vector can be written
in terms of the cartesian unit vectors and the direction cosines
n̂ = î cos α + ĵ cos β + k̂ cos γ
(3)
and using this and some algebra
2
ri⊥
= |~ri × n̂|2
= (yi cos γ − zi cos β)2 + (zi cos α − xi cos γ)2 + (xi cos β − yi cos α)2
= (yi2 + zi2 ) cos2 α + (zi2 + x2i ) cos2 β + (x2i + yi2 ) cos2 γ
−2yi zi cos β cos γ − 2zi xi cos γ cos α − 2xi yi cos α cos β
Hence the moment of inertia relative to the axis of rotation is
X
X
X
I =
mi (yi2 + zi2 ) cos2 α +
mi (zi2 + x2i ) cos2 β +
mi (x2i + yi2 ) cos2 γ
X
X
X
−2
mi yi zi cos β cos γ − 2
zi xi cos γ cos α − 2
xi yi cos α cos β
(4)
(5)
Wow. A seemingly small change results in a really messy expression. We can recognize
the terms in the first line as containing moments of inertia about the coordinate axes, and
modify our notation to have double subscripts:
X
Ixx =
mi (yi2 + zi2 )
moment of inertia about x−axis
(6)
X
Iyy =
mi (zi2 + x2i )
moment of inertia about y−axis
(7)
X
Izz =
mi (x2i + yi2 )
moment of inertia about z−axis
(8)
2
The sums involving products will be called the products of inertia. Our text includes the
negative sign in the definition—other texts define the products as positive numbers.
X
Ixy = Iyx = −
mi xi yi
(9)
X
Iyz = Izy = −
mi yi zi
(10)
X
Ixz = Izx = −
mi xi zi
(11)
Hence we can write
I = Ixx cos2 α + Iyy cos2 β + Izz cos2 γ
+2Ixy cos α cos β + 2Iyz cos β cos γ + 2Ixz cos α cos γ
(12)
It is much more compact to cast this into tensor/vector notation.
Represent the unit vector n̂ as a column vector,


cos α
n̂ =  cos β 
cos γ
(13)
The transpose of this vector is a row vector
˜ = (cos α
n̂
cos β
cos γ)
(14)
For the moment of inertia tensor use


Ixx Ixy Ixz
I =  Iyx Iyy Iyz 
Izx Izy Izz
(15)
where we know relations like Ixy = Iyx . The scalar moment of inertia can be found by
simple matrix multiplication
˜
I = n̂In̂
(16)
E.g. Moment of inertia tensor of a rectangular lamina Consider a rectangular lamina lying in the x-y plane with length a in the x and 2a in the y. Find the moment
of inertia tensor relative to these coordinates. We can write the surface density
ρ = m/2a2 . The moments of inertia about the coordinate axes can be written from
previous work, Ixx = (1/3)m(2a)2 = (4/3)ma2 , Iyy = (1/3)ma2 and from the perpendicular axis theorem, Izz = Ixx + Iyy = (5/3)ma2 .
3
Figure 2: A uniform rectangular lamina of dimensions a by 2a.
Since z = 0 for the lamina, two products of inertia Ixz = Iyz = 0. So now we just
need to find Ixy .
Z a Z 2a
ρxy dx dy
(17)
Ixy = −
0
0
Z a
Z 2a
m
= − 2
x dx
y dy
(18)
2a 0
0
1
= − ma2
(19)
2
So the tensor is
 4

2
− 12 ma2
0
3 ma
0 
I =  − 21 ma2 13 ma2
(20)
5
2
0
0
3 ma
Find the scalar moment of inertia for the rectangular lamina with an axis at 37◦
to the x-axis
The direction cosines are cos α = cos 37◦ = 0.80, cos β = cos 53◦ = 0.60, cos γ = 0 so
need to evaluate
 4



− 12 0
0.80
3
1
0  ma2  0.60  = 0.493 ma2
I = (0.80 0.60 0)  − 21
(21)
3
0
0 53
0
1.1
Angular Momentum Vector and Dyad Products
We now introduce a new vector product. Recall that the dot product of two vectors gave
a scalar and the cross product gave a vector. The dyad product 1 of two vectors will give
a tensor via


Ax Bx Ax By Ax Bz
~B
~ ≡  Ay Bx Ay By Ay Bz 
(22)
A
Az Bx Az By Az Bz
1
See Wikipedia, http://en.wikipedia.org/wiki/Dyadic_tensor for more discussion.
4
A unit tensor (be sure you can distinguish one 1 and I) will be


1 0 0
1 =  0 1 0  = îî + ĵ ĵ + k̂ k̂
0 0 1
(23)
The product of a unit tensor with a vector returns the vector: 1~
ω=ω
~ where putting the
tensor next to the vector impies the usual matrix-vector multiplication. Note that the text
often (but not always I think) puts a “·” between the tensor and vector. I will usually
reserve the dot for dot products of vectors.
Some dyad properties–these are shown in the text.
• Dyad product with vector (yields a vector): (~a~b) ~c = ~a (~b · ~c)
g
• Commutation rule: ~a ~b = (~b ~a)
~˜a)(~˜b~c) = (d~ · ~a)(~b · ~c)
~˜ a~b)~c = (d~
• Dyad between vectors (yields a scalar): d(~
Now we want to write the relation between angular momentum vector and the angular
velocity vector. In Chapter 8, with a fixed axis and generally symmetrical objects, this
~ = I~
was just L
ω . The general case is considerably more complicated.
We will do it with summations. We know that ~vi = ω
~ × ~ri . The angular momentum
is
X
X
~ =
L
~ri × mi~vi =
[mi~ri × (~
ω × ~ri )]
(24)
and expanding the triple cross product as discussed in Chapter 1 (BAC - CAB)
X
X
~ =
L
mi ri2 ω
~−
mi~ri (~ri · ω
~)
hX
X
i
=
mi ri2 1 −
(mi~ri~ri ) ω
~
(25)
(26)
This takes a bit to process, but with a little work you should see that it works.
Now we write the relation between angular momentum and angular velocity in tensor
form:


 2
 2
xi xi yi xi zi
xi + yi2 + zi2
0
0
X
X
~ =
ω
0
0
x2i + yi2 + zi2
~
~−
mi  xi yi yi2 yi zi  ω
L
mi 
2
2
2
2
x i zi yi zi zi
0
0
xi + yi + zi
 2

2
yi + zi −xi yi
−xi zi
X
=
mi  −yi xi x2i + zi2 −yi zi  ω
~ = I~
ω
(27)
2
2
−xi zi
−yi zi xi + yi
This should tell you that The direction of the angular velocity and the direction of the
angular momentum are not, in general, the same!
5
E.g. Find angle between angular momentum and angular velocity for the rect~
angular lamina rotating about an axis at 37◦ to the x-axis. First compute L:
 4





− 12 0
0.80
0.767
3
1
~ =  −1
0  ma2  0.60  ω =  −0.20  ma2 ω
L
(28)
2
3
5
0
0
0 3
0
q
~˜ L)
~ is 0.793ma2 ω and since L
~ ·ω
~ = Lω cos φ we can find
The magnitude of this ( L
◦
φ = 51 .
Be sure to look at the examples in the text, 9.1.1 and 9.1.2, for a square lamina and rotation
about (a) the x-axis and (b) the diagonal. In the case of the diagonal, the angle is just 0,
and we will shortly describe this as one of the principal axes of the object.
1.2
Rotational Kinetic Energy Using Tensors
First consider an object in pure rotation, and use the relation ~vi = ω
~ × ~ri
Trot =
X1
2
mi~vi · ~vi =
1X
(~
ω × mi~ri ) · ~vi
2
(29)
~ × B)
~ ·C
~ =A
~ · (B
~ × C)
~ from chapter 1. Hence
But (A
Trot =
=
=
=
1X
ω
~ · (mi~ri × ~vi )
2
1 X
ω
~·
(mi~ri × ~vi )
2
1
~
ω
~ ·L
2
1˜
ω
~ Iω
2
(30)
(31)
(32)
(33)
E.g. Kinetic Energy of Rectangular Plate about 37◦ Evaluating this gives
1
Trot = (0.493)ma2 ω 2
2
(34)
and this is just what we get using the scalar moment of inertia.
2
Principal Axes of a Rigid Body
The origin of our coordinates may be fixed in a problem, but the orientation of the axes is
usually our choice. We can always find some orientation in which the products of inertia
6
are all zero and thus the moment of inertia tensor is diagonal. In this case the diagonal
elements are called the principal moments, and the axes are called the principal axes. We
will use the following notation when we have the axes oriented to be principal axes:
Ixx ≡ I1 ωx ≡ ω1 î ≡ ê1
Iyy ≡ I2 ωy ≡ ω2 ĵ ≡ ê2
Izz ≡ I3 ωz ≡ ω3 k̂ ≡ ê3
and

I1 0 0
I =  0 I2 0 
0 0 I3
(35)

(36)
In some cases one or more of the principal axes may be found by inspection. In more
general cases we must diagonalize the inertia tensor, thereby determining the principal
moments and the principal axes. The rigid object can still be rotated about any axis, not
just the principle axes.
Once we have found the principal axes and moments, it is easy to determine the scalar
moment of inertia about an axis other than the principal axes, and to find the angular
momentum and rotational kinetic energy about this new axis.
˜ = (cos α cos β cos γ). Then the scalar
Suppose we rotate abut an axis with normal n̂
moment of inertia is



I1 0 0
cos α
˜
I = n̂In̂
= (cos α cos β cos γ)  0 I2 0   cos β 
0 0 I3
cos γ
= I1 cos2 α + I2 cos2 β + I3 cos2 γ
(37)
The angular velocity is in the same direction as n̂ and so the angular momentum is



I1 0 0
ω1
~ = I~
L
ω =  0 I 2 0   ω2 
0 0 I3
ω3


I1 ω1
=  I2 ω2  = ê1 I1 ω1 + ê2 I2 ω2 + ê3 I3 ω3
(38)
I3 ω3
Finally the rotational kinetic energy is

Trot
1˜
= ω
~ I~
ω =
2
=


I1 0 0
ω1
ω2 ω3 )  0 I2 0   ω2 
0 0 I3
ω3
1
I1 ω12 + I2 ω22 + I3 ω32
2
1
(ω1
2
7
(39)
2.1
Symmetry
Often we can see the principal axes by inspection. Here we assume the origin is at the
center of mass.
Rectangular Block about the cm We expect the principal axes to be parallel to the
sides of the block.
If the sides are a, b and c, from Chapter 8 we have the principal moments
m
m
m
I1 = (b2 + c2 )
I2 = (a2 + c2 )
I3 = (a2 + b2 )
(40)
12
12
12
Notice that for a cube each moment is ma2 /6. Example 9.2.2 shows that the scalar
moment of inertia of the cube about any other axis through the center of mass of the
cube is also ma2 /6.
Laminar Objects If the laminar object lies in the x − y plane, then the z axis must be
a principal axis since Ixz = Iyz = 0. If there is also an axis of symmetry, such as for
a ping-pong paddle or tennis racquet, then that is another principal axis. The third
principal axis is orthogonal to these two.
Figure 3: A model ping-pong paddle. Handle has mass m/2 and length 2a, head has mass
m/2 and radius a. The paddle is laminar.
Consider a model paddle consisting of circular lamina of mass m/2 and radius a
attached to a thin rod of mass m/2 and length 2a. The center of mass is at the point
where the rod meets the circle (see Figure 9.2.1). Call axis 1 the axis of symmetry
and axis 3 normal to the lamina. About axis 1 we can write the total inertia as
1m 2 1
I1 = Irod + Icircle = 0 +
a = ma2
(41)
4 2
8
About axis 2,
1m
1m 2 m 2
31
2
I2 = Irod + Icircle =
(2a) +
a + a = ma2
(42)
3 2
4 2
2
24
8
and from the perpendicular axis theorem,
I3 =
2.2
17
ma2
12
(43)
Dynamic Balancing
Suppose an object is rotating about one of its principal axes, say axis 1. Then ω2 = ω3 = 0
and the angular momentum is just
~ = ê1 I1 ω1 = I1 ω
L
~
(44)
This tells us that for rotation about a principal axis, the angular momentum and angular
velocity are parallel, while for rotation about an axis that is not a principal axis, angular
momentum and angular velocity are NOT parallel.
This has direct consequence in balancing rotating systems such as automobile tires. Static
balancing occurs when the center of mass lies on the axis of rotation. Thus the gravitational
force will not cause the object to rotate. However if we set the object into rotation it may
wobble terribly unless it is dynamically balanced so that the axis of rotation is along a
principal axis.
When the wheel is NOT dynamically balanced we can view the situation as having ω
~
~ in some other direction, so that as the wheel rotates the angular
along the axle, but L
momentum changes direction, tracing out a cone. Whenever the angular momentum is
changing we know that there must be a net torque and it will be at right angles to the axis
of rotation.
2.3
Finding Principal Axes and Principal Moments I: One Axis Known
For a general inertia tensor we can find the principal moments and axes by diagonalizing
the tensor. We start with a simpler case where one of the principal axes is known, call it
I3 . The tensor looks like


Ixx Ixy 0
 Ixy Iyy 0 
(45)
0
0 I3
What we want to find is the orientation of the other principal axes, 1 and 2, relative to the
starting x − y axes. The angle sought is just
tan θ =
9
ωy
ωx
(46)
For rotation about principal axis 1, the angular
parallel and related by the principal moment, so



ωx
~ = I1 ω
L
~ = I1  ωy  = I~
ω=
ωz
momentum and angular velocity will be


Ixx Ixy 0
ωx
Ixy Iyy 0   ωy 
0
0 I3
ωz
(47)
Doing the matrix multiplication and equating terms we get
Ixx ωx + Ixy ωy = I1 ωx
(48)
Ixy ωx + Iyy ωy = I1 ωy
(49)
Eliminate ωy by introducing θ to get
Ixx + Ixy tan θ = I1
(50)
Ixy + Iyy tan θ = I1 tan θ
(51)
(Ixx − Iyy ) tan θ = Ixy (1 − tan2 θ)
(52)
and eliminating I1 we get
Using the trig identity tan 2θ = 2 tan θ/(1 − tan2 θ) results in
tan 2θ =
2Ixy
Ixx − Iyy
(53)
Find the principal axes for our rectangle relative to coordinate axes at the corner.
We have Ixx = 4ma2 /3, Iyy = ma2 /3, Ixy = −ma2 /2. Putting these into our expression we get
tan 2θ = −1
(54)
◦
2θ = −45 , 135
◦
◦
(55)
◦
θ = −22.5 (or 157.5 ), 67.5
◦
(56)
So the 1-axis is at -22◦ and the 2-axis is at 67.5◦ relative to the original x-axis. Notice
that the 2-axis is not the body diagonal, at 63.4◦ .
Balancing a tire Suppose that we have a solid disk of radius a and mass m. It has a
symmetry axis that we will call the x-axis. We know that this is a principal axis for
the disk with Ixx = ma2 /2, Iyy = ma2 /4, Ixy = 0. The axle for the disk is mounted
at a slight angle θ away from the symmetry axis. What we will do is show that by
adding two small masses m0 at distances b from the disk we can move the principal
axis to align with the axle. Refer to Figure 9.2.4 for the visual.
10
Figure 4: A wheel is made from a solid disk of radius a and mass m and lies in the yz
plane. It’s axle is crooked, tilted an angle θ to the x-axis. By adding symmetrical placed
masses m0 a distance b from the wheel, the principle axis can be made to coincide with the
axle, leading to a wheel that is balanced both statically and dynamically.
For the combined system (disk plus masses) we can write
1
ma2 + 2m0 a2
2
1
=
ma2 + 2m0 b2
4
= −[(−b)m0 a + (b)m0 (−a)] = 2abm0
Ixx =
(57)
Iyy
(58)
Ixy
tan 2θ =
2Ixy
4abm0
=
Ixx − Iyy
ma2 /4 + 2m0 (a2 − b2 )
(59)
(60)
For small angles and m0 m,
bm0
(61)
am
For values of m = 10 kg, a = 18 cm, b = 5 cm and θ = 1◦ , we find the required
balancing masses to be m0 = 76 grams.
θ≈8
2.4
Principal Axes by Diagonalizing a Matrix
Suppose we have an inertial tensor that has all non-zero elements relative to an xyz set
of coordinates. We want to find the principal axes (and their orientations relative to xyz)
and the principal moments of inertia. This is an example of an eigenfunction problem that
will arise extensively in quantum mechanics.
11
When we have rotation about a principal axis êi , angular momentum and angular velocity
are parallel, I~
ω and ω
~ are parallel. We can write the requirement for component i as
Iêi = λi êi
(I − λi 1)êi = 0
(62)
(63)
where the λi are the principal moments. For this to be true, the determinant of the matrix
in parentheses must be zero
Ixx − λ
Ixy
Ixz Iyy − λ
Iyz = 0
(64)
|I − λi 1| = Ixy
Ixz
Iyz
Izz − λ
This is a cubic equation in λ whose solutions are the principal moments.
The next task is to find the unit vectors that describe the principal axes relative to the
original xyz coordinates. If ê1 is the unit vector of the axis that has a principal moment
λ1 , we can write


Ixx − λ1
Ixy
Ixy
Iyy − λ1
Ixz
Iyz
(I − λ1 1)ê1 = 0


Ixz
cos α
  cos β  = 0
Iyz
Izz − λ1
cos γ
(65)
(66)
and the three component equations can be solved for the direction cosines.
Like most matrix efforts, if the matrix is given in numbers it is easier to let a computer
do the work. An online resource is http://www.math.ubc.ca/~israel/applet/mcalc/
matcalc.html or search on “(ubc java matrix”).
If we use the rectangle and diagonalize we get principal moments I1 = 0.126 ma2 , I2 =
1.540 ma2 , I3 = 1.667 ma2 with corresponding unit vectors
ê˜1 = (−0.3827 − 0.9239 0)
ê˜2 = (0.9239 − 0.3827 0)
ê˜3 = (0 0 1)
(67)
(68)
(69)
From the unit vectors we can get the angles of the principal axes relative to the original
axes. Remember that inverse cosines are double valued, ±θ, so some adjusting and choosing
is necessary. I get α = −112.5◦ , β = 157.5◦ γ = 90◦ for the first vector, α = −22.5◦ , β =
−112.5◦ γ = 90◦ for the second, and α = 90◦ , β = 90◦ γ = 0◦ for the third.
The origin of the coordinate system is determined by us. If we change the coordinate
system we will change the inertia tensor, the principal moments and the principal axes.
12
For example, if we consider the center of the rectangle as the origin,


1/3
0
0
0  ma2
I =  0 1/12
0
0
5/12
(70)
and the xyz axes are the principal axes.
3
Euler’s Equations For Rigid Body Motion
Newton’s Second Law for rotation is, in an inertial frame,
~
~ = dL
N
dt
(71)
~ can be easily expressed if we use principal axes. We then imagine
For a rigid body L
rotation about an axis that is NOT a principal axis. When the object rotates about this
axis by some amount, and if the axes are inertial (fixed), then the inertia tensor will
change.
To make the mathematics tractable, we use two sets of coordinates. One is an inertial or
fixed set, the other rotates with the body and we choose the principal axes for the body set.
The body coordinates are a non-inertial reference frame like those discussed in Chapter 5.
In that chapter we determined the following relation for any vector. The subscript “fixed”
means evaluate the derivative for variables referred to a fixed inertial reference system.
The subscript “rotating” means evaluate the derivative relative to variables measured in
the rotating reference system.
!
!
~
~
dL
dL
~˙ rot + ω
~
~ =L
~ ×L
(72)
=
+ω
~ ×L
dt
dt
f ixed
rotating
~ = I~
By choosing the body coordinates to be the principal axes we can use L
ω , with a
˙~
˙
~
diagonal inertia tensor, hence Lrot = Iω
~ and ω
~ ×L=ω
~ × I~
ω . In tensor form this is

 
 

N1
I1 ω̇1
ω2 ω3 (I3 − I2 )
 N2  =  I2 ω̇2  +  ω3 ω1 (I1 − I3 ) 
(73)
N3
I3 ω̇3
ω1 ω2 (I2 − I1 )
The second term on the right is obtained by the usual method of cross products using
~˜ rot = (I1 ω1 I2 ω2 I3 ω3 ).
L
13
3.1
Rigid Body About Fixed Axis in Uniform Rotation
Neither the size nor direction of ω
~ change, so the time derivatives are zero. Euler’s equations
become
N1 = ω2 ω3 (I3 − I2 )
(74)
N2 = ω3 ω1 (I1 − I3 )
(75)
N3 = ω1 ω2 (I2 − I1 )
(76)
These are the components of torque that the axle must supply to the system to maintain
the uniform rotation.
Example. What torque must be supplied to rotate the rectangle about an axis through
the short side? We use the principal axes, and note that the angles between the axis of
rotation and the principal axes are 112.5◦ , 22.5◦ , 90◦ so that the angular velocity in the
frame of the principal axes is


−0.3827
(77)
ω
~ =  0.9239  ω
0
Recalling that I1 = 0.126 ma2 and I2 = 1.540 ma2 , we find N1 = N2 = 0, N3 = 0.500 ma2 ω 2 .
If we want to think in terms of force, the ~r is in the 1-2 plane, the torque is along the 3
axis, so the required force must be in the 1-2 plane.
4
Rigid Body in Free Rotation and Poinsot’s Ellipsoids
In Chapter 7 we showed that the motion of a system can be decomposed into translational
motion of the center of mass plus rotational motion about the center of mass. We are going
to apply Euler’s equations to a rigid body on which no torques act. An example of such a
situation is throwing an object in the air. Neglecting air resistance, the only force acting
on the object is the weight and it acts at the center of mass2 .If we use the center of mass
as the origin of our body coordinate system, there is no torque due to this force.
~
Since there is no torque, in the fixed reference frame the angular momentum vector L
is constant both in direction and magnitude. On the rotating body frame, however, the
angular momentum is fixed in magnitude but not direction. We can write this (body
frame) as
~ ·L
~ = L21 + L22 + L23 = constant
L
(78)
2
Well technically at the center of gravity, but in a uniform gravitational field this is the same thing.
14
In the body frame the tip of the angular momentum vector lies on the surface of a sphere.
If we look at the angular velocity in the body frame,
I12 ω12 + I22 ω22 + I32 ω32 = L2 = constant
(79)
The angular velocity vector thus lies on the surface of an ellipsoid. The lengths of the semimajor axes in the three principal axes are inversely proportional to the inertia component,
in the ratio
1 1 1
:
:
(80)
I1 I2 I3
Since there is no torque, the rotational kinetic energy in the body frame is also constant,
~ = 2Trot = const or in terms of the angular velocity
and we can write ω
~ ·L
I1 ω12 + I2 ω22 + I3 ω32 = 2Trot = constant
(81)
This shows that the angular velocity also lies on the surface an ellipsoid, the Poinsot
Ellipsoid or Inertial Ellipsoid, with semi-major axes in the ratio
1
1
1
√ :√ :√
I1
I2
I3
(82)
The angular momentum vector must therefore lie on two different ellipsoids, one determined by angular momentum, one determined by kinetic energy. Since there is an angular
momentum, the ellipsoids must intersect at at least one point. In general the angular velocity vector lies on a path that is the intersection of the two ellipsoids. One example is
shown in Figure 9.4.1 in the text, where the angular velocity follows a circular path about
axis 3. The path is given the rather obscure name polhode.
A special case is when the object rotates about a principal axis, say ω
~ = ω1 ê1 . In this case
the two ellipsoids intersect at a point on the 1-axis.
4.1
The Tennis Racquet Theorem
Consider a tennis racquet. The principal axes are easily seen to be along the handle,
through the center of mass normal to the racquet, and through the center of mass in the
plane of the racquet. See Figure 9.4.2.
If the racquet is spun about one of the principal axes and launched into a torque-free
environment, the rotation will be stable providing the rotation axis is along a principal
axis associated with either the maximum or the minimum principal moment. The axis
with the intermediate moment will be unstable. This is seen very clearly if you spin a
racquet and launch it into the air.
15
The text discusses this intersections of the ellipsoids qualitatively. First I will try to justify
the text figure 9.4.2. Suppose I1 : I2 : I : 3 = 5 : 4 : 3. Then the ratio of ellipsoid axes from
angular momentum is (0.20:0.25:0.33) and the ratio from kinetic energy is (0.44:0.50:0.58).
If rotation is about the 1-axis, maximum moment, I will rescale the angular momentum
ratio so that the first term is equal to the first term in the kinetic energy ratio, resulting is
(0.44:0.55:0.73). The ellipsoids intersect at a single point since on ellipsoid lies completely
outside the other. Doing this for the 3-axis, the axis with minimum moment, the angular
momentum ratio becomes (0.35:0.43:0.58), and again the ellipsoids intersect at a single
point. However for the 2-axis, intermediate moment, the ratio becomes (0.40:0.50:0.66),
and now the ellipsoids must intersect along a curve rather than just a point.
For the axes with minimum and maximum moments, a slight disturbance from rotation
around the axis will result in a restoring movement, with oscillations about the stable
point. For the axis with intermediate moment, a slight disturbance from rotation about
the axis will result in unstable equilibrium.
For more details, consider the model of the tennis racquet, from Classical Mechanics,
Barger and Olsson.
Figure 5: Model of a tennis racquet. The head is a ring, the handle a thin rod. The left
vertical axis is used to find the center of mass. The right axis passes through the center of
mass and is used for moments of inertia.
Consider a circular head that is a ring of mass ma = 0.15 kg and radius a = 0.13 m.
Attached is a handle that is a thin rod of mass m` = 0.18 kg and length ` = 0.38 m. First
find the center of mass. Measured from the center of the head this is
R=
m` (a + `/2)
= 0.175 meters
ma + m`
(83)
Now find the principal moments of inertia relative to the center-of-mass origin. Call the
16
axis through the handle the 1-axis.
1
I1 = ma a2 = 1.27 × 10−3 kg m2
2
(84)
Call the 2-axis the one in the plane of the racquet. The parallel axis theorem must be
used
1
1
2
2
2
I2 =
ma a + ma R +
m` ` + m` [a + `/2 − R] = 11.81 × 10−3 kg m2
(85)
2
12
and from the perpendicular axis theorem, I3 = I1 + I2 = 13.08 × 10−3 kg m2
Now we can write Euler’s equations, one from each component. Remember that the torque
is zero.
I1 ω̇1 + ω3 ω2 (I3 − I2 ) = 0
(86)
I2 ω̇2 + ω1 ω3 (I1 − I3 ) = 0
(87)
I3 ω̇3 + ω2 ω1 (I2 − I1 ) = 0
(88)
Since the racquet is a laminar object with I3 = I1 + I2 these can be simplified to
ω̇3 +
ω̇1 + ω3 ω2 = 0
(89)
ω̇2 − ω1 ω3 = 0
I2 − I1
ω1 ω2 = 0
I2 + I1
(90)
(91)
For the tennis racquet model I2 I1 , so the last equation is approximately
ω̇3 + ω1 ω2 = 0
(92)
At this point we have 3 coupled differential equations for the angular velocity of the tennis
racquet.
4.1.1
Axis 1 or Axis 3
Suppose that the racquet is launched in perfect alignment with one of the principal axes,
say 1. Then ω2 = ω3 = 0 and Euler’s equations say that ω1 = constant. This is true for
all three principal axes.
Now let us suppose that there is a slight disturbance from the perfect alignment. Since ω2 ω3
is small the first Euler equation is still ω̇1 ≈ 0, ω1 ≈ constant. The other two equations
can be combined by defining a complex quantity for the angular velocity transverse to axis
1 (I will denote complex with ă)
ω̆T = ω3 + iω2
(93)
17
Using this the other two Euler equations can be combined into
˘ T − iω1 ω̆T = 0
ω̇
(94)
ω̆T = Aei(ω1 t+α)
(95)
ω2 = A sin(ω1 t + α)
(96)
ω3 = A cos(ω1 t + α)
(97)
with solution
Hence
Since by our initial assumption the disturbance is small, A is small and the transverse
angular velocity components trace out a circle around the main angular velocity ω1 , or in
other words the situation is stable, slight disturbances do not lead to radically different
orientations.
Looking at the Euler equations for this situation, notice that 1 and 3 are in similar positions with similar signs. Thus rotation about the principal axes with either maximum or
minimum moment is shown to be stable.
4.1.2
Axis 2
For rotation about the principal axis with intermediate moment of inertia, we combine the
other two Euler equations into
(ω̇1 + ω̇3 ) + (ω1 + ω3 )ω2 = 0
(98)
(ω̇1 − ω̇3 ) − (ω1 − ω3 )ω2 = 0
(99)
(ω1 + ω3 ) = Ae−ω2 t
(100)
+ω2 t
(101)
with solutions
(ω1 − ω3 ) = Be
or
ω1 =
ω3 =
1
Ae−ω2 t + Be+ω2 t
2
1
Ae−ω2 t − Be+ω2 t
2
(102)
(103)
These solutions show that the disturbances grow in size with time. The solutions shown
are only valid for small transverse angular velocities, but it is clear that about this axis
that the racquet will tumble.
18
5
Analysis of a football
The text does the analysis of the football, using the symmetry axis as the 3-axis and
defining
Is = I3
(104)
I = I1 = I2
(105)
I ω̇1 + ω3 ω2 (Is − I) = 0
(106)
I ω̇2 + ω1 ω3 (I − Is ) = 0
(107)
Is ω̇3 = 0
(108)
Euler’s equations become
The latter equation tells us that ω3 = constant. Define a new constant
Ω = ω3
Is − I
I
(109)
so that the first two Euler equations can be written
ω̇1 + Ωω2 = 0
(110)
ω̇2 − Ωω1 = 0
(111)
ω̈1 + Ωω̇2 = 0
(112)
ω̈1 + Ω2 ω1 = 0
(113)
ω1 = ω0 cos(Ωt + α)
(114)
Differentiate the first of these
and use the second to yield
which is simple harmonic motion
Solving for the other angular velocity
ω2 = ω0 sin(Ωt + α)
(115)
As discussed in the text we can view the football as having an angular velocity ω
~ that is
constant in magnitude. It has a component ω3 = ω cos α along the symmetry axis, where
cos α is the direction cosine for the angular velocity and the 3-axis. The projection onto
the 1-2 plane is ω0 = ω sin α which has a constant magnitude but rotates around the 3-axis
with angular velocity Ω. Figure 9.5.2 shows this.
The circular path that the angular velocity traces out represents the intersection of the
two ellipsoids discussed earlier.
19
5.1
General 3-D Object
Solving Euler’s equations for a more complicated object can be done also as is illustrated
on pages 387-390 of the text. Specifically, the text looks at an ellipsoid
x2 y 2 z 2
+
+
=1
9
4
1
(116)
Using Mathematica to evaluate the integrals,
I1 = π
I2 = 2π
I3 = 8.168
(117)
Euler’s equations become
A1 ω2 ω3 = ω̇1
(118)
A2 ω1 ω3 = ω̇2
(119)
A3 ω2 ω1 = ω̇3
(120)
Using initial conditions of ω1 (0) = ω2 (0) = ω3 (0) = 1 rad/s and numerically solving the
differential equations results in the phase plot shown in Figure 9.5.4.
Refer to that section for more details.
6
More General Rotation and Eulerian Angles
We have looked at two cases so far of applications of Euler’s Equations: rotation about a
~ = 0). To most
fixed axis at constant angular velocity (ω
~˙ = 0) and torque free motions (N
easily deal with the more general case we need to discuss the Eulerian angles.
The Eulerian angles provide a way to transform from the fixed frame to the body frame
via three rotations (recall the rotation matrices from Chapter 1). Call the fixed frame
Oxyz and the rotating body frame O123. A third frame, Ox0 y 0 z 0 , also rotating, will be the
intermediary.
The z 0 axis will coincide with the 3-axis, and the x0 axis will lie in the xy plane and will
define the line of nodes. Starting from the fixed axis, rotate around the z axis by an angle
φ to get to the x0 axis. Next rotate about the x0 axis through an angle θ to bring the z
axis to z 0 . Finally rotate about the z 0 axis through an angle ψ to get to the 123 set of axes.
The angles φ, θ, ψ are called the Eulerian angles.
There are two planes to consider: The xy plane in which x and y axes lie, and the x0 y 0
plane in which axes 1 and 2 also lie. The x0 axis lies at the intersection of the two planes,
and is called the line of nodes .
20
Next we look at how to transform from the Oxyz frame to the O123 frame using the
Eulerian angles. Recall the rotation matrices from Chapter 1. We can write






cos φ sin φ 0
1
0
0
cos ψ sin ψ 0
λφ =  − sin φ cos φ 0  λθ =  0 cos θ sin θ  λψ =  − sin ψ cos ψ 0 
0
0
1
0 − sin θ cos θ
0
0
1
(121)
The three rotations can then be written as the product λ = λψ λθ λφ which is


cos φ cos ψ − sin φ cos θ sin ψ
sin φ cos ψ + cos φ cos θ sin ψ sin θ sin ψ
 − cos φ sin ψ − sin φ cos θ cos ψ − sin φ sin ψ + cos φ cos θ cos ψ sin θ cos ψ 
sin φ sin θ
− cos φ sin θ
cos θ
(122)
Now suppose that the body is rotating about some arbitrary axis with angular velocity ω
~
NOT one of the principal axes. We set about determining the angular velocity components
in the body frame using the Eulerian angles.
In a small time dt the rigid body rotates through an angle
~ + dθ~ + dψ
~
dβ~ = ω
~ dt = dφ
(123)
and thus the angular velocity can be written as
~˙ + θ~˙ + ψ
~˙
ω
~ =φ
(124)
The meaning of the time derivatives is
~˙
φ
˙
θ~
~˙
ψ
Rotation about the z axis of the fixed (inertial) system
Rotation about the line of nodes, the x0 axis (intermediate rotating system)
Rotation about the 3 axis of the body (rotating) system
So to get the components of ω
~ we need the components of the time derivatives of the vector
Eulerian angles. These directions of these derivatives are shown on Figure 9.6.2. We would
like to get the derivatives in all three reference frames, Oxyz, Ox0 y 0 z 0 , and O123.
Careful inspection of Figure 9.6.2 lets us write the derivatives in terms of the Ox0 y 0 z 0
coordinates (note error in text Equation 9.6.3)
φ̇x0 = 0
θ̇x0 = θ̇
ψ̇x0 = 0
(125)
φ̇y0 = φ̇ sin θ
θ̇y0 = 0
ψ̇y0 = 0
(126)
φ̇z 0 = φ̇ cos θ
θ̇z 0 = 0
ψ̇z 0 = ψ̇
(127)
21
Thus the components of angular velocity in the intermediate system are
ωx0
= θ̇
(128)
ωy0
= φ̇ sin θ
(129)
ωz 0
= φ̇ cos θ + ψ̇
(130)
Now for the components in the body, O123 system
φ̇1 = φ̇ sin θ sin ψ
φ̇2 = φ̇ sin θ cos ψ
φ̇3 = φ̇ cos θ
θ̇1 = θ̇ cos ψ
ψ̇1 = 0
θ̇2 = −θ̇ sin ψ
θ̇3 = 0
ψ̇2 = 0
ψ̇3 = ψ̇
(131)
(132)
(133)
making the angular velocity components in the body system
ω1 = φ̇ sin θ sin ψ + θ̇ cos ψ
(134)
ω2 = φ̇ sin θ cos ψ − θ̇ sin ψ
(135)
ω3 = φ̇ cos θ + ψ̇
(136)
Similarly we could get the components relative to the fixed reference system, and if I did
this right the result is
6.1
ωx = θ̇ cos ψ − ψ̇ sin θ cos φ
(137)
ωy = θ̇ sin ψ − ψ̇ sin θ sin φ
(138)
ωz = φ̇ + ψ̇ cos θ
(139)
Free Rotation (No Torques) Of a Symmetric Object
Let’s restate the meaning of the angles: θ is the angle between the z axis and both the z 0
and 3 axes; φ is the angle between the x and x0 axes; ψ is the angle between the line of
nodes and the 1 axis measured in the 12 plane. Note that x0 is perpendicular to the 3, z
and z 0 axes!
For free rotation, the angular momentum in the fixed coordinates is constant in magnitude
~ is along the z
and direction. For convenience, orient the fixed coordinates so that L
axis. This will be called the invariable line. Referring to Figure 9.6.1 we can write the
components in the intermediate system
Lx0 = 0
Ly0 = L sin θ
Lz 0 = L cos θ
(140)
We restrict ourselves to a body with rotational symmetry about the 3-axis, I3 = Is , I1 =
I2 = I as for the football. For such symmetric objects, the 1 and 2 axes can be rotated
22
around the 3 axis and still result in a diagonal inertia tensor. We will choose the 1 axis to
be coincident with the x0 axis, meaning ψ = 0. Hence the body and intermediate frames
are identical, and we can write
Lx0 = Iωx0
Ly0 = Iωy0
Lz 0 = Is ωz 0
(141)
Since ωx0 = θ̇ and Lx0 = 0, we have ωx0 = 0, θ̇ = 0. Thus we see that ω
~ lies in the y 0 z 0
0
~ (along z axis), ω
plane with an angle between it and the z axis being called α. Thus L
~
and the 3-axis lie in the same plane.
A useful visualization is the Java applet at http://faculty.ifmo.ru/butikov/Applets/
Precession.html (programmed initially in Easy Java Simulations). If you open this,
initially turn off the “Show Point Trace”. Start the simulation and you will see two red
vectors on a dark background (or two blue vectors on a white background.). The vertical
one is the angular velocity of the rotation of the symmetry axis (the precession) while the
inclined one is the angular velocity of the object in the body frame. The vector sum, in
yellow on the dark background (red on the white background), is the combined angular
velocity, and is the angular velocity of the object in the space frame. Note that it is NOT
constant in direction, but only in magnitude.
The angle between the red (blue) vectors, i.e. between the z and 3-axes, is θ. The angle
between vertical and the yellow(red) vector, i.e. between z and ω
~ , is α
Here is Marion’s description (Classical Dynamics of Particles and Systems). In addition to
~
constant angular momentum in the fixed frame, the rotational kinetic energy Trot = ω
~ · L/2
is constant. The dot product can be interpreted as proportional to the projection of the
angular velocity along the direction of the angular momentum. So the projection of ω
~
onto the z axis is constant, and the angular velocity ω
~ precesses around the z-axis making
a constant angle α with the angular momentum. This is the same thing that we got
before.
View the situation from the fixed frame. We can visualize a space cone traced out by the
angular velocity as it precesses around the z-axis. This is the fixed cone shown on the right
of the simulation.
In the body frame, the inertia tensor is diagonalized, and a body cone is traced out by the
angular velocity rotating about the 3-axis. You can look at the simulation and consider
the view from a frame fixed in the body to see this. Since ω
~ lies on both cones, it must lie
on the intersection of the cones, and we visualize the body cone rolling around the space
cone as is suggested in Figure 9.6.4 and shown in the simulation.
Back to Fowles and Cassiday. We have chosen our coordinates so that ω
~ lies in the y 0 z 0
and 23 planes. The inertia tensor is diagonal in both these frames. Thus
ωx0 = 0
ωy0 = ω sin α
23
ωz 0 = ω cos α
(142)
L x0 = 0
Ly0 = L sin θ = Iω sin α
Lz 0 = L cos θ = Is ω cos α
(143)
~ and 3—and α—the angle
and we can get a relation between angles θ—the angle between L
between ω
~ and 3.
I
tan α
(144)
tan θ =
Is
For prolate objects like a rod, α < θ while for oblate objects like a coin, α > θ. This
means that the space cone can engulf the body cone (prolate case) or the space cone can
be outside the body cone (oblate case). See Figure 9.6.4. You can adjust the aspect
ratio in the simulation between 0.5 (oblate) and 5 (prolate). Prolateness of 1 refers to a
sphere.
We end up with three useful angular velocities: ω
~ of the body about the rotation axis
~
(yellow line in the simulation), Ω of the angular velocity vector about the body symmetry
axis 3, and φ̇ of the symmetry axis (3) about the space axis z, the invariable line. When
we did the football we determined
Is
Ω=
− 1 ω cos α
(145)
I
The φ̇ for this symmetric situation is the angular rate of precession of both the body
~
symmetry axis, 3, and the angular velocity, ω
~ , about the invariable line (z axis or L.)
This appears as a wobble of an imperfectly thrown frisbee or football. Earlier we had
φ̇ = ωy0 / sin θ and ωy0 = ω sin α, so we can combine these to get
φ̇ = ω
sin α
sin θ
With some trig identities and algebra we get
2
Is
2
− 1 cos α
φ̇ = ω 1 +
I2
(146)
(147)
giving the wobble rate in terms of the angular speed ω of the body and the inclination α
between the body 3-axis and the angular velocity.
E.g. Precession of a Frisbee A highly oblate object like a Frisbee is approximately a
laminar object, so Is = 2I, Is /I = 2. If the disk is tossed into the air with an angular
velocity inclined to the body symmetry axis, 3, by an angle α, then
Ω = ω cos α
(148)
and
φ̇ = ω
p
1 + 3 cos2 α ≈ 2ω
for small angles. The wobble rate is twice the angular speed of rotation.
24
(149)
This had an interesting historical impact on Richard Feynman3 .
“I was in the cafeteria and some guy, fooling around, throws a plate in the
air. As the plate went up in the air I saw it wobble, and I noticed the red
medallion of Cornell on the plate going around. It was pretty obvious to
me that the medallion went around faster than the wobbling. I had nothing
to do, so I start figuring out the motion of the rotating plate. I discovered
that when the angle is very slight, the medallion rotates twice as fast as
the wobble rate—two to one. It came out of a complicated equation! I
went on to work out equations for wobbles. Then I thought about how the
electron orbits start to move in relativity. Then there’s the Dirac equation
in electrodynamics. And then quantum electrodynamics. And before I
knew it . . . the whole business that I got the Nobel prize for came from
that piddling around with the wobbling plate.”
E.g. Free Precession of Earth The earth is a slightly oblate spheroid, Is /I ≈ 1.00327.
The axis of rotation of the earth is inclined by α ≈ 0.200 = 0.97 × 10−6 rad to the
symmetry axis.
We then have Ω = 0.00327ω.
We know that ω = 2π/(1 day) so the period of the precession of the the earth’s axis of
rotation about the pole is predicted to be 2π/Ω = 305 days. In fact the observed value
is 440 days, attributed to the fact that the Earth is not a perfect oblate spheroid,
and that is is not a rigid object.
The wobble of the Earth’s body axis is φ̇ = 1.00327ω yielding a period of 0.997 days.
7
Mr. Euler, Meet Mr. Lagrange
7.1
Free Rotation of a General Rigid Body
Consider the O123 coordinate system in which the inertia tensor is diagonal. Then
T =
3
X
Ii ωi2
V = constant
(150)
1
In Chapter 10 we used L = T − V for the Lagrangian. In this chapter L represents angular
momentum, so I will use Λ = T − V . In the torque free case, the Lagrangian Λ = T . Let’s
3
Feynman R P 1985 Surely You Are Joking, Mr. Feynman! (New York: W W Norton) see pp 157—158
for a discussion of the rotating plate motion
25
choose the Euler angles as the generalized coordinates. Then for the Euler angle ψ,
∂T
d ∂T
=
∂ψ
dt ∂ ψ̇
(151)
The ωi can be expressed as functions of the Euler angles as we did earlier. Hence we can
write for the ψ coordinate,
3
3
X
∂T ∂ωi
d X ∂T ∂ωi
=
∂ωi ∂ψ
dt
∂ωi ∂ ψ̇
1
(152)
1
From the expression for kinetic energy we have
∂T
= Ii ωi
∂ωi
(153)
Now let’s evaluate the other partials using the previous expressions for the angular velocities
in terms of the Euler angles, Equations ??, ??, ??.
∂ω1
= φ̇ sin θ cos ψ − θ̇ sin ψ = ω2
∂ψ
∂ω2
= −φ̇ sin θ sin ψ − θ̇ cos ψ = −ω1
∂ψ
∂ω3
=0
∂ψ
∂ω2
=0
∂ ψ̇
∂ω2
=0
∂ ψ̇
∂ω3
=1
∂ ψ̇
(154)
(155)
(156)
Putting these values into Equation ?? we get
I1 ω1 ω2 + I2 ω2 (−ω1 ) =
d
I 3 ω3
dt
(157)
or
(I1 − I2 )ω1 ω2 − I3 ω̇3 = 0
(158)
Since the designation of the 3-axis is arbitrary, similar relations hold for the other Euler
angle Lagrangians,
X
(Ii − Ij )ωi ωj −
Ik ω̇k ijk = 0
(159)
k
where the permutation symbol ijk is
ijk = 0 any two subscripts equal
(160)
= +1 any even permutation 123, 231, 312
(161)
= −1 any odd permutation, 132, 213, 321
(162)
These are just the Euler equations for a torque free environment.
26
7.2
Rotation with Applied Torque
In the case of applied torque we start with
!
~
d
L
~ =
=
N
dt
f ixed
~
dL
dt
!
~
+ω
~ ×L
(163)
body
and take components along the 3-axis
N3 = L̇3 + ω1 L2 − ω2 L1
(164)
In the body frame the inertia is diagonalized so Li = Ii ωi and we get the general Euler
equations that we can write in the somewhat obscure form
X
(Ii − Ij )ωi ωj −
( Ik ω̇k − Nk ) ijk = 0
(165)
k
If i = j this is 0=0. For i 6= j, If k equals either i or j, so the summation is non-zero
only for a different value of k and the sign of the last terms depends on the perturbation
symbol. So in reality Equation ?? can be written as
(Ii − Ij )ωi ωj − ( Ik ω̇k − Nk ) ijk = 0
8
(166)
Motion of a Symmetrical Top
We now turn to the motion of a rotationally symmetric top, I1 = I2 = I, I3 = Is , that
rotates about a tip fixed in location , but with a uniform gravitational field. In Chapter 8
we discussed possible choices of an axis of rotation. The center of mass is always a good
choice, and is what we have used thus far. When an object rotates about a fixed point,
that point is also a good choice of origin, and is what we will use here.
Figure 9.7.1 shows the top in a tilted position. The inertial system has a vertical z-axis,
and the body 3-axis is tilted by an angle θ. As before the x0 axis is perpendicular to z, z 0
and 3 axes. Also due to the symmetry of the top the inertia tensor is diagonal in both the
body and intermediate frames of reference. Call the distance from the tip of the top to
the center of mass `, measured along the 3-axis, hence we can write the components of the
gravitational torque.
Nx0 = mg` sin θ
Ny0 = Nz 0 = 0.
27
(167)
Figure 6:
Earlier we expressed the angular velocities in the primed frame in terms of the Euler angles,
and using this we get
Lx0
= I θ̇
(168)
Ly0
= I φ̇ sin θ
(169)
Lz 0
= Is (φ̇ cos θ + ψ̇) ≡ Is S
(170)
where we have introduced a quantity S = φ̇ cos θ + ψ̇ that is the spin of the top about the
z 0 = 3 axis.
Using Lagrangian formulation, we can write
1
1
(171)
T = I(ω12 + ω22 ) + Is ω32
2
2
Using the expressions for the angular velocity components in terms of the Euler angles,
Equations ??, ??, ??, and doing the algebra we see this becomes
1
1
T = I(φ̇2 sin2 θ + θ̇2 ) + Is (φ̇ cos θ + ψ̇)2
2
2
(172)
V = mg` cos θ
(173)
The potential energy is
so the Lagrangian is
1
1
(174)
Λ = I(φ̇2 sin2 θ + θ̇2 ) + Is (φ̇ cos θ + ψ̇)2 − mg` cos θ
2
2
Notice that φ and ψ are ignorable coordinates meaning that the conjugate angular momenta
are
∂Λ
pφ =
= (I sin2 θ + Is cos2 θ)φ̇ + Is ψ̇ cos θ = constant
(175)
∂ φ̇
∂Λ
pψ =
= Is (ψ̇ + φ̇ cos θ) = Is S ≡ Is ω3 = constant
(176)
∂ ψ̇
28
Thus the spin, as defined by Fowles and Cassiday, is constant. The spin is the angular
velocity about the body symmetry axis. Now referring to Figure 9.6.2 we can see that
pφ ≡ Lz , the angular momentum about the fixed z-axis and a little more algebra shows
that this can be written
Lz ≡ pφ = I φ̇ sin2 θ + Is S cos θ
(177)
Likewise we can recognize that pψ = Is S ≡ Lz 0 , the angular momentum component about
the body symmetry axis.
Since the momenta are constants of the motion, we can use the above equations to solve
for ψ̇ and φ̇. Use Equation ?? and solve for
Lz 0 − Is φ̇ cos θ
Is
(178)
Lz − Lz 0 cos θ
I sin2 θ
(179)
Lz 0
(Lz − Lz 0 cos θ) cos θ
−
Is
I sin2 θ
(180)
ψ̇ =
Put this into Equation ?? and solve for
φ̇ =
Then put this into Equation ?? to get
ψ̇ =
For the θ generalized coordinate we write
∂L
= I θ̇
∂ θ̇
∂L
∂θ
(181)
= I φ̇2 sin θ cos θ − Is (φ̇ cos θ + ψ̇)φ̇ sin θ + mg` sin θ
(182)
= I φ̇2 sin θ cos θ − Is S φ̇ sin θ + mg` sin θ
(183)
So the equation of motion is
I θ̈ = I φ̇2 sin θ cos θ − Is S φ̇ sin θ + mg` sin θ
(184)
The expression for φ̇ can be inserted to give a differential equation in one variable. Once
θ(t) is found, the other Euler angle functions can be found. We will be able to deduce
much of the top’s motion without solving this equation.
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8.1
Steady Precession of A Top
Consider first a horizontal top, θ = 90◦ = constant. The equation of motion then reduces
to
mg` = Is S φ̇
(185)
So φ̇ = constant. The precession rate, φ̇ should decrease if the spin is increased, but
increase if the mass is increased. We’ll see if we can demo this.
If we consider the case of arbitrary angle, but still want steady precession, θ̈ = 0, then the
equation of motion becomes
mg` = Is S φ̇ − S φ̇2 cos θ
(186)
This is a quadratic in φ̇ so there are two possible rates of steady precession. Usually (Fowles
and Cassaday say) the slower will occur, but with the correct initial conditions, the fast
one may occur. The precession rate is
p
Is S ± Is2 S 2 − 4mg`I cos θ
φ̇ =
(187)
2I cos θ
For precession to occur, this must be real hence
Is2 S 2 ≥ 4mg`I cos θ
(188)
If a there is friction in the bearings so that the top’s spin slows, once it reaches this
threshold the top will begin to fall, and eventually topple over.
E.g. Rotating Top Consider a simple top made from a spindle of negligible mass with
a solid disk of mass 2.00 kg and radius 5.0 cm mounted 4.0 cm above the tip of the
top.
(a) Find the minimum spin of the top so that it can remain in its vertical position
without toppling. Is = 0.5(2.00)(0.05)2 = 2.5 × 10−3 kg m2 , using the parallel
axis theorem I = (0.25(2.00)(0.05)2 + 2.00(0.04)2 ) = 4.45 × 10−3 kg m2 , ` = 0.04
m. Hence the required spin is
s
4mg`I
S≥
= 47.3 rad/s = 7.52 rev/sec = 451 rpm
(189)
Is2
This is a period of 0.132 s
(b) If the axis is set horizontally and the spin is 500 rpm, find the precession rate
and period. S = 52.4 rad/s
φ̇ =
mg`
= 5.98 rad/s = 57.1 rpm
Is S
The period is then 1.05 s.
30
(190)
(c) If the top has spin of 500 rpm and is released at an initial angle of 30◦ , find the
two rates of precession and the two periods. The precession rates are 7.75 rad/s
with period 0.81 s and 26.2 rad/s with period 0.24 s.
9
Nutation and the Energy Equation
The normal force at the tip of the top is assumed to do no work (no friction) so energy is
also a constant of motion
1
1
E =
I(φ̇2 sin2 θ + θ̇2 ) + Is (φ̇ cos θ + ψ̇)2 + mg` cos θ = constant
(191)
2
2
1
1
I(φ̇2 sin2 θ + θ̇2 ) + Is S 2 + mg` cos θ = constant
(192)
=
2
2
where the second equation has inserted the spin. However we already know that the spin
S is a constant of motion, so we can look at the quantity
1
1
E 0 = E − Is S 2 = I(φ̇2 sin2 θ + θ̇2 ) + mg` cos θ = constant
2
2
From Equation ?? we can rewrite the equation as
1
E 0 = I θ̇2 + V (θ)
2
(193)
(194)
where
(Lz − Lz 0 cos θ)2
+ mg` cos θ
(195)
2I sin2 θ
What we have done is to reduce the three dimensional problem to an equivalent onedimensional problem in θ. This is the same treatment that we used in the central-force
problem, where we looked at radial variations by introducing an effective potential.
V (θ) =
Figure 9.8.2 shows the general shape of the effective potential. Note that it has a minimum
and is concave up. This implies some sort of a restoring torque will cause the top to bob
up and down (in θ) in a pattern called nutation.
The minimum of the potential is the situation of steady precession.
10
And then ...
We have discussed only a small portion of rotational problems. Even for the symmetrical
top there are other things that we could discuss such as a top whose point of contact with
a table makes a circle as the top spins, or a Tippy Top, http://www.youtube.com/watch?
v=xu_Dp9IfgSU.
31