Assignment 5

Goldstein 5.6 (a) Show that the angular momentum of the torque-free symmetrical top rotates in thebody coordinates about the symmetry axis with an angular frequency Ω. Show also that the symmetry axisrotates in space about the fixed direction of the angular momentum with the angular frequency

φ =I3ω3

I1 cos θ

where φ is the Euler angle of the line of nodes with respect to the angular momentum as the space z axis.

(b) Using the results of Exrcise 15, Chapter 4, show that ~ω rotates in space about the angular momentumwith the same frequency φ, but that the angle θ′ between ~ω and ~L is given by

sin θ′ =Ωφ

sin θ′′

where θ′′ is the inclination of ω to the symmetry axis. Using the data given in Section 5.6, show thereforethat Earth’s rotation axis and the axis of angular momentum are never more than 1.5 cm apart on Earth’ssurface.

(a). From class or the text, we know that the components of the angular velocity in the body frame are

ω′1 = A cos(Ωt− α

)ω′2 = A sin

(Ωt− α

)ω′3 = const

where A is just an amplitude, Ω is given by ω′3(I3 − I1)/I1 and α is just a phase shift that we can set to zerowithout loss of generality. Note that the magnitude of the angular velocity is given by |ω| =

√A2 + ω′23 .

Because ~L = I~ω and we use the principal axes, we can immediately write down the angular momentumcomponents:

L′1 = I1A cos ΩtL′2 = I1A sinΩtL′3 = I3ω

′3

As a result we can see that the 3 component of the angular momentum is constant while the other twocomponents trace out a circle perpendicular to the symmetry axis. Hence, ~L rotates about the symmetryaxis with the same frequency, Ω, as does the angular velocity.

If we now think about the problem in the inertial frame for a bit, the angular momentum must be afixed, constant vector in that frame. The other axes (symmetry and rotation) must then move about it.To show this, recall the equations for the angular velocity in the body axes expressed in terms of the Eulerangles:

ω′1 = φ sin θ sinψ + θ cosψ

ω′2 = φ sin θ cosψ + θ sinψ

ω′3 = φ cos θ + ψ

We now want to equate these expression to those we found in terms of t above. First note that there is nonutation for a torque-free symmetrical top. Hence, we have θ = 0 and θ is a constant. Thus,

A cos Ωt = φ sin θ sinψ

A sinΩt = φ sin θ cosψ

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Combining these equations in clever ways, we get

0 = φ sin θ cos(ψ + Ωt

)A2 = φ2 sin2 θ

Provided φ is not a constant (a rather boring case), our solutions for these become

ψ =π

2− Ωt

and that φ is a constant given by

φ =A

sin θ,

which, in turn, can be integrated to give φ = φ t+ φ0.Using these relations, it is straightforward to show

ω′3 = φ cos θ + ψ

= φ cos θ − Ω

= φ cos θ − ω′3I3 − I1I1

Rearranging a bit gives, then, the answer for φ:

φ =I3 ω

′3

I1 cos θ

where we have used ω′3 instead of ω3 for the component of the angular velocity along the symmetry axis inthe body frame.

(b). If we consider the angular velocity in the inertial frame, we have

ω1 = θ cosφ+ ψ sin θ sinφ = −Ω sin θ sinφ

ω2 = θ sinφ− ψ sin θ cosφ = Ω sin θ cosφ

ω3 = ψ cos θ + φ = −Ω cos θ + φ

What this says is that in the inertial frame, the component of the angular velocity along the angularmomentum is a constant and the other two components trace out a circle perpendicular to the angularmomentum with frequency φ (because φ = φ t+ φ0).

In the inertial frame, the angular velocity and the angular momentum vectors make an angle of θ′ witheach other. This angle can be described as

sin θ′ =

√ω2

1 + ω22

|ω|

=Ω sin θ|ω|

.

A similar angle exists in the body frame between the angular velocity vector and the symmetry axis of thebody. We will call this θ′′. It can be expressed as

sin θ′′ =

√ω′21 + ω′22|ω|

=φ sin θ|ω|

2

Taking the ratio of these two quantities, we find

sin θ′ =Ωφ

sin θ′′.

In order to consider the spatial separation of the Earth’s rotation axis (the axis that ~ω lies along andthe axis of angular momentum, we need to put the above quantity in terms of the principal moments ofinertia, namely

Ωφ

= ω′3I3 − I1I1

· I1 cos θI3 ω′3

=I3 − I1I3

cos θ

From the text, we are told two pertinent pieces of information. One is that

I3 − I1I1

= 0.00327

The other is that Earth’s rotation axis is moving about the pole (axis of symmetry) by approximately 10m. This latter is a statement that θ′′ is very small and that 2 sin θ′′ ≈ 2θ′′ ≈ (10m)/Re where Re is theradius of the earth and we threw in the 2 to account for back and forth motion about the pole. Hence, forthe motion of the angular momentum axis about the rotation axis, we get (in meters)

2Reθ′ ≈ 2Re sin θ′ ≈ 0.0326 cos θ

which is no more than 3 cm total. In terms of the distance the Earth’s rotation axis is away from the axisof angular moemntum, they are never more than 1.5 cm apart on the Earth’s surface.

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Goldstein 5.17 A uniform right circular cone of height h, half-angle α, and density ρ rolls on its sidewithout slipping on a uniform horizontal plane in such a manner that it returns to its original position in atime τ . Find expressions for the kinetic energy and the components of the angular momentum of the cone.

This problem has a considerable computational component. However, the key idea is really that theconstraint of rolling on a plane means that the instantaneous axis of rotation is the line of contact betweenthe cone and the plane. Hence, in the inertial frame, there is no component of the angular velocity in thevertical (call it z) direction. In the body frame (principal axes), we have

ω′1 = φ sin θ sinψ + θ cosψ

ω′2 = φ sin θ cosψ + θ sinψ

ω′3 = φ cos θ + ψ

We also know that θ + α = π/2 and θ = 0.The no-slip condition means in the inertial frame we have

ω3 = ψ cos θ + φ = 0

This is equivalent to φ = −ψ sinα. We have what amounts to a holonomic constraint (it could be integratedif we chose to). The angular velocity now becomes

ω′1 = φ cosα sinψ

ω′2 = φ cosα cosψ

ω′3 = −φ cos2 αsinα

In the inertial frame, the center of mass is located on the axis of symmetry of the cone. We will call thedistance from the point of the cone to the center of mass L. (We will have to calculate this.) The velocityof the center of mass we will call vcm. It is related to the rotational velocity of the cone about the (inertialframe’s) z-axis which we take to be perpendicular to the plane on which the cone rolls and which is centeredat the tip of the cone as it rolls. This rotational velocity is given by φ so that

vcm = φ L cosα

as we need the horizontal distance L cosα from the z axis to the center of mass.To construct the kinetic energy and angular momentum, we need the principal moments of inertia. It

is also necessary to find the center of mass of our system. The following will only sketch their derivations asI trust you know how to integrate.

The center of mass along the z axis of a cone is given by

zcm =∫ρ0zdV∫ρ0dV

where the integral in the denominator, of course, is the total mass: M = 13πρ0h

3 tan2 α. We could also writethis partially in terms of the radius of the base of the cone, rb = h tanα. The other integral need only recallsome techniques from multiple integrals:∫

ρ0zdV = ρ0

∫ ∫ ∫z ·RdRdϕdz

= 2πρ0

∫ (h−z) tan α

0

∫ h

0

RdR z dz

=π

12ρ0h

4 tan2 α

4

so we have zcm = h/4. Of course, the way we have calculated this center of mass distance is in a coordinatesystem centered at the base of the cone. To get our earlier defined L, we subtract this from the height ofthe cone, hence L = 3h/4. Nonetheless, we continue with the principal moments of inertia. We have

I1 =∫ρ0

(y2 + z2

)RdRdϕdz

= ρ0

∫ (h−z) tan α

0

∫ 2π

0

∫ h

0

(z2 +R2 sin2 φ

)RdRdϕdz

=π

60ρ0 h

5 tan2 α(2 + 3 tan2 α

)=

160Mh2

(2 + 3 tan2 α

)I3 =

∫ρ0

(x2 + y2

)RdRdϕdz

=310Mh2 tan2 α

Again, we have calculated this in a convenient coordinate system centered on the base of the cone. To getthe principal moments of inertia in a coordinate system centered on the center of mass, we must use theparallel axis theorem and shift the moment of inertia by the moment of inertia of the whole mass rotatingaround an axis paralell to the original. As I3cm = I3, we need only do this for I1:

I1 = I1cm +(h4)2M

As a result, we get

I1cm =380Mh2

(1 + 4 tan2 α

).

The components of the angular momentum then are

L′1 = I1cmω′1 =

380Mh2

(1 + 4 tan2 α

)φ cosα sinψ

L′2 = I1cmω′2 =

380Mh2

(1 + 4 tan2 α

)φ cosα cosψ

L′3 = I3ω′3 = − 3

10Mh2 sinα φ

while the kinetic energy becomes

T =12Mv2

cm +12I1cm

(ω′1

2 + ω′22)

+12I3 ω

′32

=340Mh2φ2

(1 + 5 cos2 α

)

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Goldstein 5.18 (a) A bar of negligible weight and length l has equal mass points m at the two ends. Thebar is made to rotate uniformly about an axis passing through the center of the bar and making an angle θwith the bar. From Euler’s equations find the components along the principal axes of the bar of the torquedriving the bar.

(b) From the fundamental torque equation (1.26) find the components of the torque along axes fixed in space.Show that these components are consistent with those found in part (a).

There are a couple of ways to do this. One is to simply construct the angular velocity, moments of inertiaand angular momentum from what is given. However, we are asked to work from the Euler equations. Tothis end, we work in the body frame and with respect to principal axes. In this case, the principal momentsof inertia are quite simple. They are Ixx = Iyy = I1 = 2m(l/2)2 and Izz = 0.

Now, the rotation is uniform. So going back to the inertial frame for a moment, we have that θ = 0and ~ω has a component only in the z direction. From the form of ~ω in terms of Euler angles in the inertialframe,

ω1 = θ cosφ+ ψ sin θ sinφ

ω2 = θ sinφ− ψ sin θ cosφ

ω3 = ψ cos θ + φ

we conclude that ψ = 0 as well and that ω3 = φ which, by our assumption of uniform rotation, must be aconstant and which we will call (somewhat surprisingly) ω.

Returning to the body axes, the angular velocity components expressed there in terms of Euler anglesbecome

ω′1 = φ sin θ sinψ + θ cosψ = φ sin θ sinψ

ω′2 = φ sin θ cosψ − θ sinψ = φ sin θ cosψ

ω′3 = ψ cos θ + φ = φ cos θ

which are, of course, all constants.The Euler equations for rigid body rotation in the body axes are now

I1 ω′1 + ω′2ω

′3

(I3 − I2

)= N ′

1

I2 ω′2 + ω′3ω

′1

(I1 − I3

)= N ′

2

I3 ω′3 + ω′1ω

′2

(I2 − I1

)= N ′

3

The derivatives vanish as does the difference between I1 and I2. Using our forms for the angular velocitycomponents, we have

N ′1 = −1

2ml2ω2 sin θ cos θ cosψ

N ′2 =

12ml2ω2 sin θ cos θ sinψ

N ′3 = 0

Given the symmetry in our problem, the angle ψ merely sets the relative x′-y′ axes in the plane perpendicularto z′, so we can set this value arbitrarily. We let ψ = π/2 in which case the component of the torque in thex′ direction vanishes and our torque is perpendicular to the symmetry axis of our rotating system.

(b) In the inertial frame we can construct the angular momentum with respect to the space axes. Inparticular we have

~L = ~r1 × ~p1 + ~r2 × ~p2

where the vectors are the position and momentum of the particles. Of course, ~r2 = −~r1 and ~p2 = −~p1 so

~L = 2~r × ~p

6

where~r =

12l(ı sin θ cosωt+ sin θ sinωt+ k cos θ

)~p =

12mlω

(−ı sin θ sinωt+ sin θ cosωt

)As a result, the angular momentum is

~L = −12ml2ω sin θ

[ı cos θ cosωt+ cos θ sinωt− k sin θ

]The torque is thus

~N =d~L

dt=

12ml2ω2 sin θ cos θ

[ı sinωt− cosωt

]which is, indeed, consistent with part (a). Here our torque is rotating yet remains perpendicular ( ~N · ~r = 0)to the rod and the angular velocity.

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Goldstein 5.25 (a) Express in terms of Euler’s angles the constraint conditions for a uniform sphere rollingwithout slipping on a flat horizontal surface. Show that they are nonholonomic.

(b) Set up the Lagrangian equations for this problem by the method of Lagrange multipliers. Show thatthe translational and rotational parts of the kinetic energy are separately conserved. Are there any otherconstants of the motion?

For a general sphere moving through space, there are six degrees of freedom counting the three transla-tional degrees of freedom of the center of mass and the three Euler angles describing the sphere’s orientation.Once it starts rolling, the height of the sphere is constrained to be a constant. In addition, there is theconstraint of no slipping. In this regard this amounts to a pair of constraints such that there is only asingle generalized coordinate (angle) which describes the orientation of the sphere about the vertical axisconnecting the center of mass and the point of contact. Thus there are only three degrees of freedom in ourproblem and we need to account for three constraints. One we have, namely z = a, the radius of the sphere.The others can be found as follows. The velocity of the center of mass, ~vcm, must be parallel to the plane.The angular velocity, ~ω, will then lie in a plane perpendicular to ~vcm. We can describe this in the inertialframe as

~ω = ω‖(ı cosα+ sinα

)+ k ω⊥

where ω‖ and ω⊥ are those components of ~ω parallel and perpendicular to the plane. The angle α is justto allow a decomposition of the parallel part in the plane itself. With these introduced coordinates, ~vcm

becomes~vcm = |vcm|

(ı sinα− cosα

)which is, of course, perpendicular to ~ω.

Our constraint is that the point of contact on the sphere has zero instantaneous velocity:

~vcontact = 0 = ~vcm + ~ω × ~r

where ~r = −k a. Working this out, we get

0 =(|vcm| − aω‖

)(ı sinα− cosα

)and hence vcm = aω‖.

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