Solutions to Goldstein Rigid body

1. Assignment 5
Goldstein 5.6 (a) Show that the angular momentum of the torque-free symmetrical top rotates in the
body coordinates about the symmetry axis with an angular frequency
. Show also that the symmetry axis
rotates in space about the fixed direction of the angular momentum with the angular frequency
˙
= I3!3
I1 cos
where 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 ~! rotates in space about the angular momentum
with the same frequency ˙
, but that the angle 0 between ~! and ~L
is given by
sin 0 =
˙
sin 00
where 00 is the inclination of ! to the symmetry axis. Using the data given in Section 5.6, show therefore
that Earth’s rotation axis and the axis of angular momentum are never more than 1.5 cm apart on Earth’s
surface.
(a). From class or the text, we know that the components of the angular velocity in the body frame are
!01
= A cos
t −
!02
t −
= A sin
!03
= const
where A is just an amplitude,
is given by !03
(I3 −I1)/I1 and is just a phase shift that we can set to zero
without loss of generality. Note that the magnitude of the angular velocity is given by |!| =
p
A2 + !02
3 .
Because ~L
= I~! and we use the principal axes, we can immediately write down the angular momentum
components:
L0
1 = I1A cos
t
L0
2 = I1A sin
t
L0
3 = I3!03
As a result we can see that the 3 component of the angular momentum is constant while the other two
components trace out a circle perpendicular to the symmetry axis. Hence, ~L
rotates about the symmetry
axis with the same frequency,
, as does the angular velocity.
If we now think about the problem in the inertial frame for a bit, the angular momentum must be a
fixed, 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 Euler
angles:
!01
= ˙
sin sin + ˙ cos
!02
= ˙
sin cos + ˙ sin
!03
= ˙
cos + ˙
We now want to equate these expression to those we found in terms of t above. First note that there is no
nutation for a torque-free symmetrical top. Hence, we have ˙ = 0 and is a constant. Thus,
A cos
t = ˙
sin sin
A sin
t = ˙
sin cos
1

2. Combining these equations in clever ways, we get
0 = ˙
sin cos
+
t
A2 = ˙
2 sin2
Provided is not a constant (a rather boring case), our solutions for these become
=
2 −
t
and that ˙
is a constant given by
˙
= A
sin
,
which, in turn, can be integrated to give = ˙
t + 0.
Using these relations, it is straightforward to show
!03
= ˙
cos + ˙
= ˙
cos −
= ˙
cos − !03
I3 − I1
I1
Rearranging a bit gives, then, the answer for ˙
:
˙
= I3 !03
I1 cos
where we have used !03
instead of !3 for the component of the angular velocity along the symmetry axis in
the body frame.
(b). If we consider the angular velocity in the inertial frame, we have
!1 = ˙ cos + ˙
sin sin = −
sin sin
!2 = ˙ sin − ˙
sin cos =
sin cos
!3 = ˙
cos + ˙
= −
cos + ˙
What this says is that in the inertial frame, the component of the angular velocity along the angular
momentum is a constant and the other two components trace out a circle perpendicular to the angular
momentum with frequency ˙
(because = ˙
t + 0).
In the inertial frame, the angular velocity and the angular momentum vectors make an angle of 0 with
each other. This angle can be described as
sin 0 =
p
!2
1 + !2
2
|!|
=
sin
|!|
.
A similar angle exists in the body frame between the angular velocity vector and the symmetry axis of the
body. We will call this 00. It can be expressed as
sin 00 =
p
!02
1 + !02
2
|!|
=
˙
sin
|!|
2

3. Taking the ratio of these two quantities, we find
sin 0 =
˙
sin 00.
In order to consider the spatial separation of the Earth’s rotation axis (the axis that ~! lies along and
the axis of angular momentum, we need to put the above quantity in terms of the principal moments of
inertia, namely
˙
= !03
I3 − I1
I1
·
I1 cos
I3 !03
= I3 − I1
I3
cos
From the text, we are told two pertinent pieces of information. One is that
I3 − I1
I1
= 0.00327
The other is that Earth’s rotation axis is moving about the pole (axis of symmetry) by approximately 10
m. This latter is a statement that 00 is very small and that 2 sin 00 200 (10m)/Re where Re is the
radius of the earth and we threw in the 2 to account for back and forth motion about the pole. Hence, for
the motion of the angular momentum axis about the rotation axis, we get (in meters)
2Re0 2Re sin 0 0.0326 cos
which is no more than 3 cm total. In terms of the distance the Earth’s rotation axis is away from the axis
of angular moemntum, they are never more than 1.5 cm apart on the Earth’s surface.
3

4. Goldstein 5.17 A uniform right circular cone of height h, half-angle , and density rolls on its side
without slipping on a uniform horizontal plane in such a manner that it returns to its original position in a
time . 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 the
constraint of rolling on a plane means that the instantaneous axis of rotation is the line of contact between
the cone and the plane. Hence, in the inertial frame, there is no component of the angular velocity in the
vertical (call it z) direction. In the body frame (principal axes), we have
!01
= ˙
sin sin + ˙ cos
!02
= ˙
sin cos + ˙ sin
!03
= ˙
cos + ˙
We also know that + = /2 and ˙ = 0.
The no-slip condition means in the inertial frame we have
!3 = ˙
cos + ˙
= 0
This is equivalent to ˙
= − ˙
sin . We have what amounts to a holonomic constraint (it could be integrated
if we chose to). The angular velocity now becomes
!01
= ˙
cos sin
!02
= ˙
cos cos
!03
= −˙
cos2
sin
In the inertial frame, the center of mass is located on the axis of symmetry of the cone. We will call the
distance from the point of the cone to the center of mass L. (We will have to calculate this.) The velocity
of the center of mass we will call vcm. It is related to the rotational velocity of the cone about the (inertial
frame’s) z-axis which we take to be perpendicular to the plane on which the cone rolls and which is centered
at the tip of the cone as it rolls. This rotational velocity is given by ˙
so that
vcm = ˙
Lcos
as we need the horizontal distance Lcos 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 as
I trust you know how to integrate.
The center of mass along the z axis of a cone is given by
zcm =
R
R0zdV
0dV
where the integral in the denominator, of course, is the total mass: M = 1
30h3 tan2 . We could also write
this partially in terms of the radius of the base of the cone, rb = h tan . The other integral need only recall
some techniques from multiple integrals:
Z
0zdV = 0
Z Z Z
z · RdR d' dz
= 20
Z (h−z) tan
0
Z h
0
RdR z dz
=
12 0h4 tan2
4

5. so we have zcm = h/4. Of course, the way we have calculated this center of mass distance is in a coordinate
system centered at the base of the cone. To get our earlier defined L, we subtract this from the height of
the cone, hence L = 3h/4. Nonetheless, we continue with the principal moments of inertia. We have
I1 =
Z
0
y2 + z2
RdR d' dz
= 0
Z (h−z) tan
0
Z 2
0
Z h
0
z2 + R2 sin2
RdR d' dz
=
60 0 h5 tan2
2 + 3 tan2
=
1
60 Mh2
2 + 3 tan2
I3 =
Z
0
x2 + y2
RdR d' dz
=
3
10 Mh2 tan2
Again, we have calculated this in a convenient coordinate system centered on the base of the cone. To get
the principal moments of inertia in a coordinate system centered on the center of mass, we must use the
parallel axis theorem and shift the moment of inertia by the moment of inertia of the whole mass rotating
around an axis paralell to the original. As I3cm = I3, we need only do this for I1:
I1 = I1cm +
h
4
2
M
As a result, we get
I1cm =
3
80 Mh2
1 + 4 tan2
.
The components of the angular momentum then are
L0
1 = I1cm!01
=
3
80 Mh2
1 + 4 tan2
˙
cos sin
L0
2 = I1cm!02
=
3
80 Mh2
1 + 4 tan2
˙
cos cos
3
3
L0
= I3!= ˙ −
Mh2 sin 3 10 while the kinetic energy becomes
0T =
1
2 Mv2
cm +
1
2 I1cm
!01
2 + !02
2
+
1
2 I3 !03
2
=
2
1 + 5 cos2
3
40 Mh2˙
5

6. Goldstein 5.18 (a) A bar of negligible weight and length l has equal mass points m at the two ends. The
bar is made to rotate uniformly about an axis passing through the center of the bar and making an angle
with the bar. From Euler’s equations find the components along the principal axes of the bar of the torque
driving 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 inertia
and angular momentum from what is given. However, we are asked to work from the Euler equations. To
this end, we work in the body frame and with respect to principal axes. In this case, the principal moments
of 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 ˙ = 0
and ~! has a component only in the z direction. From the form of ~! in terms of Euler angles in the inertial
frame,
!1 = ˙ cos + ˙
sin sin
!2 = ˙ sin − ˙
sin cos
!3 = ˙
cos + ˙
we conclude that ˙
= 0 as well and that !3 = ˙
which, by our assumption of uniform rotation, must be a
constant and which we will call (somewhat surprisingly) !.
Returning to the body axes, the angular velocity components expressed there in terms of Euler angles
become
!01
= ˙
sin sin + ˙ cos = ˙
sin sin
!02
= ˙
sin cos − ˙ sin = ˙
sin cos
!03
= ˙
cos + ˙
= ˙
cos
which are, of course, all constants.
The Euler equations for rigid body rotation in the body axes are now
I1 !˙ 01
+ !02
!03
I3 − I2
= N01
I2 !˙ 02
+ !03
!01
I1 − I3
= N02
I3 !˙ 03
+ !01
!02
I2 − I1
= N03
The derivatives vanish as does the difference between I1 and I2. Using our forms for the angular velocity
components, we have
N01
= −
1
2 ml2!2 sin cos cos
N02
=
1
2 ml2!2 sin cos sin
N03
= 0
Given the symmetry in our problem, the angle merely sets the relative x0-y0 axes in the plane perpendicular
to z0, so we can set this value arbitrarily. We let = /2 in which case the component of the torque in the
x0 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. In
particular 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

7. where
~r =
1
2 l
ˆı sin cos !t + ˆ| sin sin !t + ˆk cos
~p =
1
2 ml!
−ˆı sin sin !t + ˆ| sin cos !t
As a result, the angular momentum is
~L
= −
1
2 ml2! sin
h
ˆı cos cos !t + ˆ| cos sin !t − ˆk sin
i
The torque is thus
~N
= d~L
dt
=
1
2 ml2!2 sin cos
h
ˆı sin !t − ˆ| cos !t
i
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.
7

8. Goldstein 5.25 (a) Express in terms of Euler’s angles the constraint conditions for a uniform sphere rolling
without 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 that
the translational and rotational parts of the kinetic energy are separately conserved. Are there any other
constants 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 the
constraint of no slipping. In this regard this amounts to a pair of constraints such that there is only a
single generalized coordinate (angle) which describes the orientation of the sphere about the vertical axis
connecting the center of mass and the point of contact. Thus there are only three degrees of freedom in our
problem 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, ~!, will then lie in a plane perpendicular to ~vcm. We can describe this in the inertial
frame as
~! = !k
ˆı cos + ˆ| sin
+ ˆk !?
where !k and !? are those components of ~! parallel and perpendicular to the plane. The angle is just
to allow a decomposition of the parallel part in the plane itself. With these introduced coordinates, ~vcm
becomes
ˆı sin − ˆ| cos
~vcm = |vcm|
which is, of course, perpendicular to ~!.
Our constraint is that the point of contact on the sphere has zero instantaneous velocity:
~vcontact = 0 = ~vcm + ~! ×~r
where ~r = −ˆk a. Working this out, we get
0 =
|vcm| − a!k
ˆı sin − ˆ| cos
and hence vcm = a!k.
8