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 ofa top, or of a bicycle.

1 Rotation of body about an arbitrary axis—Moments andProducts of Inertia, Inertia Tensor, Angular Momentumand Kinetic Energy

Consider an arbitrary rigid object and a set of xyz axes. Our first task will be to find themoment of inertia about an axis passing through the origin but oriented in an arbitrarydirection, not one of the coordinate axes. The line is bidirectional, but once we set theobject 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 vectorand the xyz axes (pages 16, 17 of the text) so that Ax = A cosα etc.

Refer to Figure 9.1.1 that shows the object, the z-axis of the coordinate system, a rotationaxis with ~ω, and a small piece of the object labeled mi a distance ~r from the origin, withmoment 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

I =∑

mir2i⊥ (1)

where we have used a summation, but we could easily move to continuous mass distributionsand do an integral.

Now we want to express this in terms of the cartesian coordinates xi, yi, zi and the directioncosines.

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Figure 1: Rigid body rotating about a general axis, not one of the coordinate axes. Unitvector for axis set by angular velocity is n One infinitesimal piece of object is shown.

Define the unit vector n as ~ω = ωn. We can recognize that

ri⊥ = |ri sin θi| = |~ri × n| (2)

where θi is the angle between the axis of rotation n and the radius vector ~ri.

The position of the infinitesimal mass is ~ri = xii+yij+zik. The unit vector can be writtenin terms of the cartesian unit vectors and the direction cosines

n = i cosα+ j cosβ + k cos γ (3)

and using this and some algebra

r2i⊥ = |~ri × n|2

= (yi cos γ − zi cosβ)2 + (zi cosα− xi cos γ)2 + (xi cosβ − yi cosα)2

= (y2i + z2i ) cos2 α+ (z2i + x2i ) cos2 β + (x2i + y2i ) cos2 γ

−2yizi cosβ cos γ − 2zixi cos γ cosα− 2xiyi cosα cosβ (4)

Hence the moment of inertia relative to the axis of rotation is

I =∑

mi(y2i + z2i ) cos2 α+

∑mi(z

2i + x2i ) cos2 β +

∑mi(x

2i + y2i ) cos2 γ

−2∑

miyizi cosβ cos γ − 2∑

zixi cos γ cosα− 2∑

xiyi cosα cosβ (5)

Wow. A seemingly small change results in a really messy expression. We can recognizethe terms in the first line as containing moments of inertia about the coordinate axes, andmodify our notation to have double subscripts:

Ixx =∑

mi(y2i + z2i ) moment of inertia about x−axis (6)

Iyy =∑

mi(z2i + x2i ) moment of inertia about y−axis (7)

Izz =∑

mi(x2i + y2i ) moment of inertia about z−axis (8)

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The sums involving products will be called the products of inertia. Our text includes thenegative sign in the definition—other texts define the products as positive numbers.

Ixy = Iyx = −∑

mixiyi (9)

Iyz = Izy = −∑

miyizi (10)

Ixz = Izx = −∑

mixizi (11)

Hence we can write

I = Ixx cos2 α+ Iyy cos2 β + Izz cos2 γ

+2Ixy cosα cosβ + 2Iyz cosβ cos γ + 2Ixz cosα cos γ (12)

It is much more compact to cast this into tensor/vector notation.

Represent the unit vector n as a column vector,

n =

cosαcosβcos γ

(13)

The transpose of this vector is a row vector

˜n = (cosα cosβ cos γ) (14)

For the moment of inertia tensor use

I =

Ixx Ixy IxzIyx Iyy IyzIzx Izy Izz

(15)

where we know relations like Ixy = Iyx. The scalar moment of inertia can be found bysimple matrix multiplication

I = ˜nIn (16)

E.g. Moment of inertia tensor of a rectangular lamina Consider a rectangular lam-ina lying in the x-y plane with length a in the x and 2a in the y. Find the momentof inertia tensor relative to these coordinates. We can write the surface densityρ = m/2a2. The moments of inertia about the coordinate axes can be written fromprevious work, Ixx = (1/3)m(2a)2 = (4/3)ma2, Iyy = (1/3)ma2 and from the per-pendicular axis theorem, Izz = Ixx + Iyy = (5/3)ma2.

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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 justneed to find Ixy.

Ixy = −∫ a

0

∫ 2a

0ρxy dx dy (17)

= − m

2a2

∫ a

0x dx

∫ 2a

0y dy (18)

= −1

2ma2 (19)

So the tensor is

I =

43ma

2 −12ma

2 0−1

2ma2 1

3ma2 0

0 0 53ma

2

(20)

Find the scalar moment of inertia for the rectangular lamina with an axis at 37

to the x-axis

The direction cosines are cosα = cos 37 = 0.80, cosβ = cos 53 = 0.60, cos γ = 0 soneed to evaluate

I = (0.80 0.60 0)

43 −1

2 0−1

213 0

0 0 53

ma2

0.800.60

0

= 0.493ma2 (21)

1.1 Angular Momentum Vector and Dyad Products

We now introduce a new vector product. Recall that the dot product of two vectors gavea scalar and the cross product gave a vector. The dyad product 1 of two vectors will givea tensor via

~A~B ≡

AxBx AxBy AxBzAyBx AyBy AyBzAzBx AzBy AzBz

(22)

1See Wikipedia, http://en.wikipedia.org/wiki/Dyadic_tensor for more discussion.

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A unit tensor (be sure you can distinguish one 1 and I) will be

1 =

1 0 00 1 00 0 1

= ii+ jj + kk (23)

The product of a unit tensor with a vector returns the vector: 1~ω = ~ω where putting thetensor next to the vector impies the usual matrix-vector multiplication. Note that the textoften (but not always I think) puts a “·” between the tensor and vector. I will usuallyreserve 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)

• Commutation rule: ~a~b = (~b~a)

• Dyad between vectors (yields a scalar): ~d(~a~b)~c = ( ~d~a)(~b~c) = (~d · ~a)(~b · ~c)

Now we want to write the relation between angular momentum vector and the angularvelocity vector. In Chapter 8, with a fixed axis and generally symmetrical objects, thiswas just ~L = I~ω. The general case is considerably more complicated.

We will do it with summations. We know that ~vi = ~ω × ~ri. The angular momentumis

~L =∑

~ri ×mi~vi =∑

[mi~ri × (~ω × ~ri)] (24)

and expanding the triple cross product as discussed in Chapter 1 (BAC - CAB)

~L =∑

mir2i ~ω −

∑mi~ri(~ri · ~ω) (25)

=[(∑

mir2i 1)−∑

(mi~ri~ri)]~ω (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 tensorform:

~L =∑

mi

x2i + y2i + z2i 0 00 x2i + y2i + z2i 00 0 x2i + y2i + z2i

~ω −∑

mi

x2i xiyi xizixiyi y2i yizixizi yizi z2i

~ω

=∑

mi

y2i + z2i −xiyi −xizi−yixi x2i + z2i −yizi−xizi −yizi x2i + y2i

~ω = I~ω (27)

This should tell you that The direction of the angular velocity and the direction of theangular momentum are not, in general, the same!

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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:

~L =

43 −1

2 0−1

213 0

0 0 53

ma2

0.800.60

0

ω =

0.767−0.20

0

ma2ω (28)

The magnitude of this (

√~L~L) is 0.793ma2ω and since ~L · ~ω = Lω cosφ we can find

φ = 51.

Be sure to look at the examples in the text, 9.1.1 and 9.1.2, for a square lamina and rotationabout (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 =∑ 1

2mi~vi · ~vi =

1

2

∑(~ω ×mi~ri) · ~vi (29)

But ( ~A× ~B) · ~C = ~A · ( ~B × ~C) from chapter 1. Hence

Trot =1

2

∑~ω · (mi~ri × ~vi) (30)

=1

2~ω ·∑

(mi~ri × ~vi) (31)

=1

2~ω · ~L (32)

=1

2~ωIω (33)

E.g. Kinetic Energy of Rectangular Plate about 37 Evaluating this gives

Trot =1

2(0.493)ma2ω2 (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 isusually our choice. We can always find some orientation in which the products of inertia

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are all zero and thus the moment of inertia tensor is diagonal. In this case the diagonalelements are called the principal moments, and the axes are called the principal axes. Wewill use the following notation when we have the axes oriented to be principal axes:

Ixx ≡ I1 ωx ≡ ω1 i ≡ e1Iyy ≡ I2 ωy ≡ ω2 j ≡ e2Izz ≡ I3 ωz ≡ ω3 k ≡ e3

(35)

and

I =

I1 0 00 I2 00 0 I3

(36)

In some cases one or more of the principal axes may be found by inspection. In moregeneral cases we must diagonalize the inertia tensor, thereby determining the principalmoments and the principal axes. The rigid object can still be rotated about any axis, notjust the principle axes.

Once we have found the principal axes and moments, it is easy to determine the scalarmoment of inertia about an axis other than the principal axes, and to find the angularmomentum and rotational kinetic energy about this new axis.

Suppose we rotate abut an axis with normal ˜n = (cosα cosβ cos γ). Then the scalarmoment of inertia is

I = ˜nIn = (cosα cosβ cos γ)

I1 0 00 I2 00 0 I3

cosαcosβcos γ

= I1 cos2 α+ I2 cos2 β + I3 cos2 γ (37)

The angular velocity is in the same direction as n and so the angular momentum is

~L = I~ω =

I1 0 00 I2 00 0 I3

ω1

ω2

ω3

=

I1ω1

I2ω2

I3ω3

= e1I1ω1 + e2I2ω2 + e3I3ω3 (38)

Finally the rotational kinetic energy is

Trot =1

2~ωI~ω =

1

2(ω1 ω2 ω3)

I1 0 00 I2 00 0 I3

ω1

ω2

ω3

=

1

2

(I1ω

21 + I2ω

22 + I3ω

23

)(39)

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2.1 Symmetry

Often we can see the principal axes by inspection. Here we assume the origin is at thecenter of mass.

Rectangular Block about the cm We expect the principal axes to be parallel to thesides of the block.

If the sides are a, b and c, from Chapter 8 we have the principal moments

I1 =m

12(b2 + c2) I2 =

m

12(a2 + c2) I3 =

m

12(a2 + b2) (40)

Notice that for a cube each moment is ma2/6. Example 9.2.2 shows that the scalarmoment of inertia of the cube about any other axis through the center of mass of thecube is also ma2/6.

Laminar Objects If the laminar object lies in the x − y plane, then the z axis must bea principal axis since Ixz = Iyz = 0. If there is also an axis of symmetry, such as fora ping-pong paddle or tennis racquet, then that is another principal axis. The thirdprincipal axis is orthogonal to these two.

Figure 3: A model ping-pong paddle. Handle has mass m/2 and length 2a, head has massm/2 and radius a. The paddle is laminar.

Consider a model paddle consisting of circular lamina of mass m/2 and radius aattached to a thin rod of mass m/2 and length 2a. The center of mass is at the pointwhere the rod meets the circle (see Figure 9.2.1). Call axis 1 the axis of symmetryand axis 3 normal to the lamina. About axis 1 we can write the total inertia as

I1 = Irod + Icircle = 0 +1

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m

2a2 =

1

8ma2 (41)

About axis 2,

I2 = Irod + Icircle =1

3

m

2(2a)2 +

(1

4

m

2a2 +

m

2a2)

=31

24ma2 (42)

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and from the perpendicular axis theorem,

I3 =17

12ma2 (43)

2.2 Dynamic Balancing

Suppose an object is rotating about one of its principal axes, say axis 1. Then ω2 = ω3 = 0and the angular momentum is just

~L = e1I1ω1 = I1~ω (44)

This tells us that for rotation about a principal axis, the angular momentum and angularvelocity are parallel, while for rotation about an axis that is not a principal axis, angularmomentum and angular velocity are NOT parallel.

This has direct consequence in balancing rotating systems such as automobile tires. Staticbalancing occurs when the center of mass lies on the axis of rotation. Thus the gravitationalforce will not cause the object to rotate. However if we set the object into rotation it maywobble terribly unless it is dynamically balanced so that the axis of rotation is along aprincipal axis.

When the wheel is NOT dynamically balanced we can view the situation as having ~ωalong the axle, but ~L in some other direction, so that as the wheel rotates the angularmomentum changes direction, tracing out a cone. Whenever the angular momentum ischanging we know that there must be a net torque and it will be at right angles to the axisof 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 diagonalizingthe tensor. We start with a simpler case where one of the principal axes is known, call itI3. The tensor looks like Ixx Ixy 0

Ixy Iyy 00 0 I3

(45)

What we want to find is the orientation of the other principal axes, 1 and 2, relative to thestarting x− y axes. The angle sought is just

tan θ =ωyωx

(46)

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For rotation about principal axis 1, the angular momentum and angular velocity will beparallel and related by the principal moment, so

~L = I1~ω = I1

ωxωyωz

= I~ω =

Ixx Ixy 0Ixy Iyy 00 0 I3

ωxωyωz

(47)

Doing the matrix multiplication and equating terms we get

Ixxωx + Ixyωy = I1ωx (48)

Ixyωx + Iyyωy = I1ωy (49)

Eliminate ωy by introducing θ to get

Ixx + Ixy tan θ = I1 (50)

Ixy + Iyy tan θ = I1 tan θ (51)

and eliminating I1 we get

(Ixx − Iyy) tan θ = Ixy(1− tan2 θ) (52)

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 expres-sion 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. Noticethat 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 asymmetry axis that we will call the x-axis. We know that this is a principal axis forthe disk with Ixx = ma2/2, Iyy = ma2/4, Ixy = 0. The axle for the disk is mountedat a slight angle θ away from the symmetry axis. What we will do is show that byadding two small masses m′ at distances b from the disk we can move the principalaxis to align with the axle. Refer to Figure 9.2.4 for the visual.

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Figure 4: A wheel is made from a solid disk of radius a and mass m and lies in the yzplane. It’s axle is crooked, tilted an angle θ to the x-axis. By adding symmetrical placedmasses m′ a distance b from the wheel, the principle axis can be made to coincide with theaxle, leading to a wheel that is balanced both statically and dynamically.

For the combined system (disk plus masses) we can write

Ixx =1

2ma2 + 2m′a2 (57)

Iyy =1

4ma2 + 2m′b2 (58)

Ixy = −[(−b)m′a+ (b)m′(−a)] = 2abm′ (59)

tan 2θ =2Ixy

Ixx − Iyy=

4abm′

ma2/4 + 2m′(a2 − b2)(60)

For small angles and m′ m,

θ ≈ 8bm′

am(61)

For values of m = 10 kg, a = 18 cm, b = 5 cm and θ = 1, we find the requiredbalancing masses to be m′ = 76 grams.

2.4 Principal Axes by Diagonalizing a Matrix

Suppose we have an inertial tensor that has all non-zero elements relative to an xyz setof 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 thatwill arise extensively in quantum mechanics.

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When we have rotation about a principal axis ei, angular momentum and angular velocityare parallel, I~ω and ~ω are parallel. We can write the requirement for component i as

Iei = λiei (62)

(I− λi1)ei = 0 (63)

where the λi are the principal moments. For this to be true, the determinant of the matrixin parentheses must be zero

|I− λi1| =

∣∣∣∣∣∣Ixx − λ Ixy IxzIxy Iyy − λ IyzIxz Iyz Izz − λ

∣∣∣∣∣∣ = 0 (64)

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 theoriginal xyz coordinates. If e1 is the unit vector of the axis that has a principal momentλ1, we can write

(I− λ11)e1 = 0 (65) Ixx − λ1 Ixy IxzIxy Iyy − λ1 IyzIxz Iyz Izz − λ1

cosαcosβcos γ

= 0 (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 computerdo 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 = 0.126ma2, I2 =1.540ma2, I3 = 1.667ma2 with corresponding unit vectors

˜e1 = (−0.3827 − 0.9239 0) (67)

˜e2 = (0.9239 − 0.3827 0) (68)

˜e3 = (0 0 1) (69)

From the unit vectors we can get the angles of the principal axes relative to the originalaxes. Remember that inverse cosines are double valued, ±θ, so some adjusting and choosingis 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 coordinatesystem we will change the inertia tensor, the principal moments and the principal axes.

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For example, if we consider the center of the rectangle as the origin,

I =

1/3 0 00 1/12 00 0 5/12

ma2 (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,

~N =d~L

dt(71)

For a rigid body ~L can be easily expressed if we use principal axes. We then imaginerotation about an axis that is NOT a principal axis. When the object rotates about thisaxis by some amount, and if the axes are inertial (fixed), then the inertia tensor willchange.

To make the mathematics tractable, we use two sets of coordinates. One is an inertial orfixed 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 inthe rotating reference system.(

d~L

dt

)fixed

=

(d~L

dt

)rotating

+ ~ω × ~L = ~Lrot + ~ω × ~L (72)

By choosing the body coordinates to be the principal axes we can use ~L = I~ω, with a

diagonal inertia tensor, hence ~Lrot = I~ω and ~ω × ~L = ~ω × I~ω. In tensor form this is N1

N2

N3

=

I1ω1

I2ω2

I3ω3

+

ω2ω3(I3 − I2)ω3ω1(I1 − I3)ω1ω2(I2 − I1)

(73)

The second term on the right is obtained by the usual method of cross products using~Lrot = (I1ω1 I2ω2 I3ω3).

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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 equationsbecome

N1 = ω2ω3(I3 − I2) (74)

N2 = ω3ω1(I1 − I3) (75)

N3 = ω1ω2(I2 − I1) (76)

These are the components of torque that the axle must supply to the system to maintainthe uniform rotation.

Example. What torque must be supplied to rotate the rectangle about an axis throughthe short side? We use the principal axes, and note that the angles between the axis ofrotation and the principal axes are 112.5, 22.5, 90 so that the angular velocity in theframe of the principal axes is

~ω =

−0.38270.9239

0

ω (77)

Recalling that I1 = 0.126ma2 and I2 = 1.540ma2, we findN1 = N2 = 0, N3 = 0.500ma2ω2.If we want to think in terms of force, the ~r is in the 1-2 plane, the torque is along the 3axis, 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 translationalmotion of the center of mass plus rotational motion about the center of mass. We are goingto apply Euler’s equations to a rigid body on which no torques act. An example of such asituation is throwing an object in the air. Neglecting air resistance, the only force actingon the object is the weight and it acts at the center of mass2.If we use the center of massas 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 ~Lis constant both in direction and magnitude. On the rotating body frame, however, theangular momentum is fixed in magnitude but not direction. We can write this (bodyframe) as

~L · ~L = L21 + L2

2 + L23 = constant (78)

2Well technically at the center of gravity, but in a uniform gravitational field this is the same thing.

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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,

I21ω21 + I22ω

22 + I23ω

23 = L2 = constant (79)

The angular velocity vector thus lies on the surface of an ellipsoid. The lengths of the semi-major axes in the three principal axes are inversely proportional to the inertia component,in the ratio

1

I1:

1

I2:

1

I3(80)

Since there is no torque, the rotational kinetic energy in the body frame is also constant,and we can write ~ω · ~L = 2Trot = const or in terms of the angular velocity

I1ω21 + I2ω

22 + I3ω

23 = 2Trot = constant (81)

This shows that the angular velocity also lies on the surface an ellipsoid, the PoinsotEllipsoid or Inertial Ellipsoid, with semi-major axes in the ratio

1√I1

:1√I2

:1√I3

(82)

The angular momentum vector must therefore lie on two different ellipsoids, one deter-mined by angular momentum, one determined by kinetic energy. Since there is an angularmomentum, the ellipsoids must intersect at at least one point. In general the angular ve-locity vector lies on a path that is the intersection of the two ellipsoids. One example isshown in Figure 9.4.1 in the text, where the angular velocity follows a circular path aboutaxis 3. The path is given the rather obscure name polhode.

A special case is when the object rotates about a principal axis, say ~ω = ω1e1. In this casethe 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 theplane of the racquet. See Figure 9.4.2.

If the racquet is spun about one of the principal axes and launched into a torque-freeenvironment, the rotation will be stable providing the rotation axis is along a principalaxis associated with either the maximum or the minimum principal moment. The axiswith the intermediate moment will be unstable. This is seen very clearly if you spin aracquet and launch it into the air.

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The text discusses this intersections of the ellipsoids qualitatively. First I will try to justifythe text figure 9.4.2. Suppose I1 : I2 : I : 3 = 5 : 4 : 3. Then the ratio of ellipsoid axes fromangular 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 momentumratio 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 completelyoutside the other. Doing this for the 3-axis, the axis with minimum moment, the angularmomentum ratio becomes (0.35:0.43:0.58), and again the ellipsoids intersect at a singlepoint. 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 rotationaround the axis will result in a restoring movement, with oscillations about the stablepoint. For the axis with intermediate moment, a slight disturbance from rotation aboutthe 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 leftvertical axis is used to find the center of mass. The right axis passes through the center ofmass 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. Firstfind the center of mass. Measured from the center of the head this is

R =m`(a+ `/2)

ma +m`= 0.175meters (83)

Now find the principal moments of inertia relative to the center-of-mass origin. Call the

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axis through the handle the 1-axis.

I1 =1

2maa

2 = 1.27× 10−3 kgm2 (84)

Call the 2-axis the one in the plane of the racquet. The parallel axis theorem must beused

I2 =

(1

2maa

2 +maR2

)+

(1

12m``

2 +m`[a+ `/2−R]

)= 11.81× 10−3 kgm2 (85)

and from the perpendicular axis theorem, I3 = I1 + I2 = 13.08× 10−3 kgm2

Now we can write Euler’s equations, one from each component. Remember that the torqueis zero.

I1ω1 + ω3ω2(I3 − I2) = 0 (86)

I2ω2 + ω1ω3(I1 − I3) = 0 (87)

I3ω3 + ω2ω1(I2 − I1) = 0 (88)

Since the racquet is a laminar object with I3 = I1 + I2 these can be simplified to

ω1 + ω3ω2 = 0 (89)

ω2 − ω1ω3 = 0 (90)

ω3 +

(I2 − I1I2 + I1

)ω1ω2 = 0 (91)

For the tennis racquet model I2 I1, so the last equation is approximately

ω3 + ω1ω2 = 0 (92)

At this point we have 3 coupled differential equations for the angular velocity of the tennisracquet.

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 forall three principal axes.

Now let us suppose that there is a slight disturbance from the perfect alignment. Since ω2ω3

is small the first Euler equation is still ω1 ≈ 0, ω1 ≈ constant. The other two equationscan be combined by defining a complex quantity for the angular velocity transverse to axis1 (I will denote complex with a)

ωT = ω3 + iω2 (93)

17

Using this the other two Euler equations can be combined into

˘ωT − iω1ωT = 0 (94)

with solutionωT = Aei(ω1t+α) (95)

Hence

ω2 = A sin(ω1t+ α) (96)

ω3 = A cos(ω1t+ α) (97)

Since by our initial assumption the disturbance is small, A is small and the transverseangular velocity components trace out a circle around the main angular velocity ω1, or inother words the situation is stable, slight disturbances do not lead to radically differentorientations.

Looking at the Euler equations for this situation, notice that 1 and 3 are in similar posi-tions with similar signs. Thus rotation about the principal axes with either maximum orminimum moment is shown to be stable.

4.1.2 Axis 2

For rotation about the principal axis with intermediate moment of inertia, we combine theother two Euler equations into

(ω1 + ω3) + (ω1 + ω3)ω2 = 0 (98)

(ω1 − ω3)− (ω1 − ω3)ω2 = 0 (99)

with solutions

(ω1 + ω3) = Ae−ω2t (100)

(ω1 − ω3) = Be+ω2t (101)

or

ω1 =1

2

(Ae−ω2t +Be+ω2t

)(102)

ω3 =1

2

(Ae−ω2t −Be+ω2t

)(103)

These solutions show that the disturbances grow in size with time. The solutions shownare only valid for small transverse angular velocities, but it is clear that about this axisthat the racquet will tumble.

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5 Analysis of a football

The text does the analysis of the football, using the symmetry axis as the 3-axis anddefining

Is = I3 (104)

I = I1 = I2 (105)

Euler’s equations become

Iω1 + ω3ω2(Is − I) = 0 (106)

Iω2 + ω1ω3(I − Is) = 0 (107)

Isω3 = 0 (108)

The latter equation tells us that ω3 = constant. Define a new constant

Ω = ω3Is − II

(109)

so that the first two Euler equations can be written

ω1 + Ωω2 = 0 (110)

ω2 − Ωω1 = 0 (111)

Differentiate the first of theseω1 + Ωω2 = 0 (112)

and use the second to yieldω1 + Ω2ω1 = 0 (113)

which is simple harmonic motion

ω1 = ω0 cos(Ωt+ α) (114)

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 isconstant in magnitude. It has a component ω3 = ω cosα along the symmetry axis, wherecosα is the direction cosine for the angular velocity and the 3-axis. The projection ontothe 1-2 plane is ω0 = ω sinα which has a constant magnitude but rotates around the 3-axiswith angular velocity Ω. Figure 9.5.2 shows this.

The circular path that the angular velocity traces out represents the intersection of thetwo ellipsoids discussed earlier.

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5.1 General 3-D Object

Solving Euler’s equations for a more complicated object can be done also as is illustratedon pages 387-390 of the text. Specifically, the text looks at an ellipsoid

x2

9+y2

4+z2

1= 1 (116)

Using Mathematica to evaluate the integrals,

I1 = π I2 = 2π I3 = 8.168 (117)

Euler’s equations become

A1ω2ω3 = ω1 (118)

A2ω1ω3 = ω2 (119)

A3ω2ω1 = ω3 (120)

Using initial conditions of ω1(0) = ω2(0) = ω3(0) = 1 rad/s and numerically solving thedifferential 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 afixed axis at constant angular velocity (~ω = 0) and torque free motions ( ~N = 0). To mosteasily 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 framevia three rotations (recall the rotation matrices from Chapter 1). Call the fixed frameOxyz and the rotating body frame O123. A third frame, Ox′y′z′, also rotating, will be theintermediary.

The z′ axis will coincide with the 3-axis, and the x′ axis will lie in the xy plane and willdefine the line of nodes. Starting from the fixed axis, rotate around the z axis by an angleφ to get to the x′ axis. Next rotate about the x′ axis through an angle θ to bring the zaxis to z′. Finally rotate about the z′ 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 x′y′

plane in which axes 1 and 2 also lie. The x′ 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 theEulerian angles. Recall the rotation matrices from Chapter 1. We can write

λφ =

cosφ sinφ 0− sinφ cosφ 0

0 0 1

λθ =

1 0 00 cos θ sin θ0 − sin θ cos θ

λψ =

cosψ sinψ 0− sinψ cosψ 0

0 0 1

(121)

The three rotations can then be written as the product λ = λψλθλφ which is cosφ cosψ − sinφ cos θ sinψ sinφ cosψ + cosφ cos θ sinψ sin θ sinψ− cosφ sinψ − sinφ cos θ cosψ − sinφ sinψ + cosφ cos θ cosψ sin θ cosψ

sinφ sin θ − cosφ sin θ cos θ

(122)

Now suppose that the body is rotating about some arbitrary axis with angular velocity ~ωNOT one of the principal axes. We set about determining the angular velocity componentsin the body frame using the Eulerian angles.

In a small time dt the rigid body rotates through an angle

d~β = ~ωdt = d~φ+ d~θ + d~ψ (123)

and thus the angular velocity can be written as

~ω = ~φ+ ~θ + ~ψ (124)

The meaning of the time derivatives is

~φ Rotation about the z axis of the fixed (inertial) system

~θ Rotation about the line of nodes, the x′ axis (intermediate rotating system)

~ψ 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 vectorEulerian angles. These directions of these derivatives are shown on Figure 9.6.2. We wouldlike to get the derivatives in all three reference frames, Oxyz, Ox′y′z′, and O123.

Careful inspection of Figure 9.6.2 lets us write the derivatives in terms of the Ox′y′z′

coordinates (note error in text Equation 9.6.3)

φx′ = 0 θx′ = θ ψx′ = 0 (125)

φy′ = φ sin θ θy′ = 0 ψy′ = 0 (126)

φz′ = φ cos θ θz′ = 0 ψz′ = ψ (127)

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Thus the components of angular velocity in the intermediate system are

ωx′ = θ (128)

ωy′ = φ sin θ (129)

ωz′ = φ cos θ + ψ (130)

Now for the components in the body, O123 system

φ1 = φ sin θ sinψ θ1 = θ cosψ ψ1 = 0 (131)

φ2 = φ sin θ cosψ θ2 = −θ sinψ ψ2 = 0 (132)

φ3 = φ cos θ θ3 = 0 ψ3 = ψ (133)

making the angular velocity components in the body system

ω1 = φ sin θ sinψ + θ cosψ (134)

ω2 = φ sin θ cosψ − θ sinψ (135)

ω3 = φ cos θ + ψ (136)

Similarly we could get the components relative to the fixed reference system, and if I didthis right the result is

ωx = θ cosψ − ψ sin θ cosφ (137)

ωy = θ sinψ − ψ sin θ sinφ (138)

ωz = φ+ ψ cos θ (139)

6.1 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′

and 3 axes; φ is the angle between the x and x′ axes; ψ is the angle between the line ofnodes and the 1 axis measured in the 12 plane. Note that x′ is perpendicular to the 3, zand z′ axes!

For free rotation, the angular momentum in the fixed coordinates is constant in magnitudeand direction. For convenience, orient the fixed coordinates so that ~L is along the zaxis. This will be called the invariable line. Referring to Figure 9.6.1 we can write thecomponents in the intermediate system

Lx′ = 0 Ly′ = L sin θ Lz′ = 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

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around the 3 axis and still result in a diagonal inertia tensor. We will choose the 1 axis tobe coincident with the x′ axis, meaning ψ = 0. Hence the body and intermediate framesare identical, and we can write

Lx′ = Iωx′ Ly′ = Iωy′ Lz′ = Isωz′ (141)

Since ωx′ = θ and Lx′ = 0, we have ωx′ = 0, θ = 0. Thus we see that ~ω lies in the y′z′

plane with an angle between it and the z′ axis being called α. Thus ~L (along z axis), ~ωand 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 redvectors on a dark background (or two blue vectors on a white background.). The verticalone is the angular velocity of the rotation of the symmetry axis (the precession) while theinclined one is the angular velocity of the object in the body frame. The vector sum, inyellow on the dark background (red on the white background), is the combined angularvelocity, and is the angular velocity of the object in the space frame. Note that it is NOTconstant in direction, but only in magnitude.

The angle between the red (blue) vectors, i.e. between the z and 3-axes, is θ. The anglebetween 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 toconstant angular momentum in the fixed frame, the rotational kinetic energy Trot = ~ω ·~L/2is constant. The dot product can be interpreted as proportional to the projection of theangular velocity along the direction of the angular momentum. So the projection of ~ωonto the z axis is constant, and the angular velocity ~ω precesses around the z-axis makinga constant angle α with the angular momentum. This is the same thing that we gotbefore.

View the situation from the fixed frame. We can visualize a space cone traced out by theangular velocity as it precesses around the z-axis. This is the fixed cone shown on the rightof the simulation.

In the body frame, the inertia tensor is diagonalized, and a body cone is traced out by theangular velocity rotating about the 3-axis. You can look at the simulation and considerthe view from a frame fixed in the body to see this. Since ~ω lies on both cones, it must lieon the intersection of the cones, and we visualize the body cone rolling around the spacecone 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′z′

and 23 planes. The inertia tensor is diagonal in both these frames. Thus

ωx′ = 0 ωy′ = ω sinα ωz′ = ω cosα (142)

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Lx′ = 0 Ly′ = L sin θ = Iω sinα Lz′ = L cos θ = Isω cosα (143)

and we can get a relation between angles θ—the angle between ~L and 3—and α—the anglebetween ~ω and 3.

tan θ =I

Istanα (144)

For prolate objects like a rod, α < θ while for oblate objects like a coin, α > θ. Thismeans that the space cone can engulf the body cone (prolate case) or the space cone canbe outside the body cone (oblate case). See Figure 9.6.4. You can adjust the aspectratio in the simulation between 0.5 (oblate) and 5 (prolate). Prolateness of 1 refers to asphere.

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 symmetryaxis 3, and φ of the symmetry axis (3) about the space axis z, the invariable line. Whenwe did the football we determined

Ω =

(IsI− 1

)ω cosα (145)

The φ for this symmetric situation is the angular rate of precession of both the bodysymmetry 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 hadφ = ωy′/ sin θ and ωy′ = ω sinα, so we can combine these to get

φ = ωsinα

sin θ(146)

With some trig identities and algebra we get

φ = ω

[1 +

(I2sI2− 1

)cos2 α

](147)

giving the wobble rate in terms of the angular speed ω of the body and the inclination αbetween the body 3-axis and the angular velocity.

E.g. Precession of a Frisbee A highly oblate object like a Frisbee is approximately alaminar object, so Is = 2I, Is/I = 2. If the disk is tossed into the air with an angularvelocity inclined to the body symmetry axis, 3, by an angle α, then

Ω = ω cosα (148)

andφ = ω

√1 + 3 cos2 α ≈ 2ω (149)

for small angles. The wobble rate is twice the angular speed of rotation.

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This had an interesting historical impact on Richard Feynman3.

“I was in the cafeteria and some guy, fooling around, throws a plate in theair. As the plate went up in the air I saw it wobble, and I noticed the redmedallion of Cornell on the plate going around. It was pretty obvious tome that the medallion went around faster than the wobbling. I had nothingto do, so I start figuring out the motion of the rotating plate. I discoveredthat when the angle is very slight, the medallion rotates twice as fast asthe wobble rate—two to one. It came out of a complicated equation! Iwent on to work out equations for wobbles. Then I thought about how theelectron orbits start to move in relativity. Then there’s the Dirac equationin electrodynamics. And then quantum electrodynamics. And before Iknew it . . . the whole business that I got the Nobel prize for came fromthat 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 ≈ 0.2′′ = 0.97 × 10−6 rad to thesymmetry 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 ofrotation about the pole is predicted to be 2π/Ω = 305 days. In fact the observed valueis 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 φ = 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∑1

Iiω2i V = constant (150)

In Chapter 10 we used L = T −V for the Lagrangian. In this chapter L represents angularmomentum, so I will use Λ = T − V . In the torque free case, the Lagrangian Λ = T . Let’s

3Feynman R P 1985 Surely You Are Joking, Mr. Feynman! (New York: W W Norton) see pp 157—158for a discussion of the rotating plate motion

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choose the Euler angles as the generalized coordinates. Then for the Euler angle ψ,

∂T

∂ψ=

d

dt

∂T

∂ψ(151)

The ωi can be expressed as functions of the Euler angles as we did earlier. Hence we canwrite for the ψ coordinate,

3∑1

∂T

∂ωi

∂ωi∂ψ

=d

dt

3∑1

∂T

∂ωi

∂ωi

∂ψ(152)

From the expression for kinetic energy we have

∂T

∂ωi= Iiωi (153)

Now let’s evaluate the other partials using the previous expressions for the angular velocitiesin terms of the Euler angles, Equations ??, ??, ??.

∂ω1

∂ψ= φ sin θ cosψ − θ sinψ = ω2

∂ω2

∂ψ= 0 (154)

∂ω2

∂ψ= −φ sin θ sinψ − θ cosψ = −ω1

∂ω2

∂ψ= 0 (155)

∂ω3

∂ψ= 0

∂ω3

∂ψ= 1 (156)

Putting these values into Equation ?? we get

I1ω1ω2 + I2ω2(−ω1) =d

dtI3ω3 (157)

or(I1 − I2)ω1ω2 − I3ω3 = 0 (158)

Since the designation of the 3-axis is arbitrary, similar relations hold for the other Eulerangle Lagrangians,

(Ii − Ij)ωiωj −∑k

Ikωkεijk = 0 (159)

where the permutation symbol εijk is

εijk = 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.

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7.2 Rotation with Applied Torque

In the case of applied torque we start with

~N =

(d~L

dt

)fixed

=

(d~L

dt

)body

+ ~ω × ~L (163)

and take components along the 3-axis

N3 = L3 + ω1L2 − ω2L1 (164)

In the body frame the inertia is diagonalized so Li = Iiωi and we get the general Eulerequations that we can write in the somewhat obscure form

(Ii − Ij)ωiωj −∑k

( Ikωk −Nk) εijk = 0 (165)

If i = j this is 0=0. For i 6= j, If k equals either i or j, so the summation is non-zeroonly for a different value of k and the sign of the last terms depends on the perturbationsymbol. So in reality Equation ?? can be written as

(Ii − Ij)ωiωj − ( Ikωk −Nk) εijk = 0 (166)

8 Motion of a Symmetrical Top

We now turn to the motion of a rotationally symmetric top, I1 = I2 = I, I3 = Is, thatrotates about a tip fixed in location , but with a uniform gravitational field. In Chapter 8we discussed possible choices of an axis of rotation. The center of mass is always a goodchoice, 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 x′ axis is perpendicular to z, z′

and 3 axes. Also due to the symmetry of the top the inertia tensor is diagonal in both thebody and intermediate frames of reference. Call the distance from the tip of the top tothe center of mass `, measured along the 3-axis, hence we can write the components of thegravitational torque.

Nx′ = mg` sin θ Ny′ = Nz′ = 0. (167)

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Figure 6:

Earlier we expressed the angular velocities in the primed frame in terms of the Euler angles,and using this we get

Lx′ = Iθ (168)

Ly′ = Iφ sin θ (169)

Lz′ = Is(φ cos θ + ψ) ≡ IsS (170)

where we have introduced a quantity S = φ cos θ + ψ that is the spin of the top about thez′ = 3 axis.

Using Lagrangian formulation, we can write

T =1

2I(ω2

1 + ω22) +

1

2Isω

23 (171)

Using the expressions for the angular velocity components in terms of the Euler angles,Equations ??, ??, ??, and doing the algebra we see this becomes

T =1

2I(φ2 sin2 θ + θ2) +

1

2Is(φ cos θ + ψ)2 (172)

The potential energy isV = mg` cos θ (173)

so the Lagrangian is

Λ =1

2I(φ2 sin2 θ + θ2) +

1

2Is(φ cos θ + ψ)2 −mg` cos θ (174)

Notice that φ and ψ are ignorable coordinates meaning that the conjugate angular momentaare

pφ =∂Λ

∂φ= (I sin2 θ + Is cos2 θ)φ+ Isψ cos θ = constant (175)

pψ =∂Λ

∂ψ= Is(ψ + φ cos θ) = IsS ≡ Isω3 = constant (176)

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Thus the spin, as defined by Fowles and Cassiday, is constant. The spin is the angularvelocity about the body symmetry axis. Now referring to Figure 9.6.2 we can see thatpφ ≡ Lz, the angular momentum about the fixed z-axis and a little more algebra showsthat this can be written

Lz ≡ pφ = Iφ sin2 θ + IsS cos θ (177)

Likewise we can recognize that pψ = IsS ≡ Lz′ , the angular momentum component aboutthe body symmetry axis.

Since the momenta are constants of the motion, we can use the above equations to solvefor ψ and φ. Use Equation ?? and solve for

ψ =Lz′ − Isφ cos θ

Is(178)

Put this into Equation ?? and solve for

φ =Lz − Lz′ cos θ

I sin2 θ(179)

Then put this into Equation ?? to get

ψ =Lz′

Is− (Lz − Lz′ cos θ) cos θ

I sin2 θ(180)

For the θ generalized coordinate we write

∂L

∂θ= Iθ (181)

∂L

∂θ= Iφ2 sin θ cos θ − Is(φ cos θ + ψ)φ sin θ +mg` sin θ (182)

= Iφ2 sin θ cos θ − IsSφ sin θ +mg` sin θ (183)

So the equation of motion is

Iθ = Iφ2 sin θ cos θ − IsSφ sin θ +mg` sin θ (184)

The expression for φ can be inserted to give a differential equation in one variable. Onceθ(t) is found, the other Euler angle functions can be found. We will be able to deducemuch 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 reducesto

mg` = IsSφ (185)

So φ = constant. The precession rate, φ should decrease if the spin is increased, butincrease 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, θ = 0, then theequation of motion becomes

mg` = IsSφ− Sφ2 cos θ (186)

This is a quadratic in φ so there are two possible rates of steady precession. Usually (Fowlesand Cassaday say) the slower will occur, but with the correct initial conditions, the fastone may occur. The precession rate is

φ =IsS ±

√I2sS

2 − 4mg`I cos θ

2I cos θ(187)

For precession to occur, this must be real hence

I2sS2 ≥ 4mg`I cos θ (188)

If a there is friction in the bearings so that the top’s spin slows, once it reaches thisthreshold 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 witha solid disk of mass 2.00 kg and radius 5.0 cm mounted 4.0 cm above the tip of thetop.

(a) Find the minimum spin of the top so that it can remain in its vertical positionwithout toppling. Is = 0.5(2.00)(0.05)2 = 2.5 × 10−3 kg m2, using the parallelaxis theorem I = (0.25(2.00)(0.05)2+2.00(0.04)2) = 4.45×10−3 kg m2, ` = 0.04m. Hence the required spin is

S ≥

√4mg`I

I2s= 47.3 rad/s = 7.52 rev/sec = 451 rpm (189)

This is a period of 0.132 s

(b) If the axis is set horizontally and the spin is 500 rpm, find the precession rateand period. S = 52.4 rad/s

φ =mg`

IsS= 5.98 rad/s = 57.1 rpm (190)

The period is then 1.05 s.

30

(c) If the top has spin of 500 rpm and is released at an initial angle of 30, find thetwo rates of precession and the two periods. The precession rates are 7.75 rad/swith 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 isalso a constant of motion

E =1

2I(φ2 sin2 θ + θ2) +

1

2Is(φ cos θ + ψ)2 +mg` cos θ = constant (191)

=1

2I(φ2 sin2 θ + θ2) +

1

2IsS

2 +mg` cos θ = constant (192)

where the second equation has inserted the spin. However we already know that the spinS is a constant of motion, so we can look at the quantity

E′ = E − 1

2IsS

2 =1

2I(φ2 sin2 θ + θ2) +mg` cos θ = constant (193)

From Equation ?? we can rewrite the equation as

E′ =1

2Iθ2 + V (θ) (194)

where

V (θ) =(Lz − Lz′ cos θ)2

2I sin2 θ+mg` cos θ (195)

What we have done is to reduce the three dimensional problem to an equivalent one-dimensional problem in θ. This is the same treatment that we used in the central-forceproblem, where we looked at radial variations by introducing an effective potential.

Figure 9.8.2 shows the general shape of the effective potential. Note that it has a minimumand is concave up. This implies some sort of a restoring torque will cause the top to bobup 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 symmetricaltop there are other things that we could discuss such as a top whose point of contact witha table makes a circle as the top spins, or a Tippy Top, http://www.youtube.com/watch?v=xu_Dp9IfgSU.

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