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Unformatted text preview: Chapter 26 Symmetric Tops In this chapter we will use Hamilton’s equations to further study the motion of
symmetric tops. We will ﬁrst introduce a useful set of generalized coordinates
for the description of the rotational motion of rigid bodies, the socalled Euler
angles. Imagine a rigid body with its three principal axes originally aligned with
a space—ﬁxed orthonormal frame {m,y,z}. Call this orientation the standard
orientation. An arbitrary orientation of the rigid body can be reached from the
standard orientation by making three successive rotations as follows (Fig. 26.1): 1. Rotate about the z—axis by an angle gt; call the new axes {e’1’,e’2,e’3}.
2. Rotate about the cf; axis by an angle 6; call the new axes {e’1,e’2, B3}. 3. Rotate about the 53 axis by an angle «p; call the ﬁnal axes {61, 62, 63}. The three angles (15,6 and 1!) (0 3 gb < 27r, 0 S 6 3 7r, 0 3 1b < 277) are called
the Euler angles specifying the orientation of a rigid body. Using the Euler
angles, the angular velocity to can be written as wzgzlz+9e§+rbeg. (26.1) Let us consider a symmetric top with 63 as the symmetry axis (I 1 = 12). Be—
cause of the axial symmetry any two orthogonal axes on the 6162 plane together
with 63 will constitute a set of principal axes. We will choose {61,653,63} as a
set of principal axes with respect to which components of to will be calculated.
From Fig. 26.1 it is easily seen that z:(zeﬂ)eﬂ+(ze3)e3=—sin663+c056e3. (26.2)
Eq.(26.1) then implies
w : (—qlsinﬁ) e’l +ée§ + (d) + qlcosQ) 63 . (26.3) 127 128 CHAPTER 26. SYMJVIETRIC TOPS It follows from (21.9) that [3
2 _I1'22 '2
Trot——((/) sm 9+0 )+ 2 (1/) + q'scos 6)2 . (26.4) Suppose the symmetric top is ﬁxed at a point on the symmetry axis lower
than the center of mass (CM) (Fig. 26.2) and the only external force is the
gravitational force acting at the center of mass. From (20.10) and (20.11) the total kinetic energy is given by 1 T = —2MV2 + Trot , (26.5) Where M is the mass of the top and V is the velocity of the center of mass. As
the top moves the CM moves on the surface of a sphere of radius 7". Hence %MV2 2 $662692 + sin2 66?) . (26.6)
Combining the last three equations we have
1 1 ) ‘9 ‘9 . 9 I3 ‘ ' 9
T: 5(I +JVIT“)(0‘ +q5‘ sm‘ 9) + 3(1/2+¢cos<9)“ . (26.7)
Realizing that the potential energy is given by
U : Mgr cosﬁ , (26.8)
and deﬁning
I’EIl+Mr2, (26.9)
the Lagrangian can be written as
I] .9 "J . '7 ' ' 9
L = 3(0‘ + (b‘ sm‘ 6) + 3% + ¢cos€)‘ —— Mgr c056. (2610)
The canonical momenta are then given by
(91; I . . 2 3 . .
p¢=6—¢.S=I¢>s1n 0+I c059(w+¢c036), (26.11)
(9L 
= —T = 1’0 , 26.12
. m 80 < >
p1), 8L : I301) + 6cos 0) . (26.13) :8? i
A
93
cm
a r
3
ME};
1173.26 2. 129 From these equations one can solve for (6,9 and to yield  p¢ —— 10,), c050
= ——— , 26.14
if) I’ sin29 ( )
 P9
_p¢ p¢—p¢cos6
’3’ Substituting these expressions in and (26.8) for T and U, respectively,
we then have the following expression for the Hamiltonian: (1% ‘W COS 9)2 P3 P311
H:T+U=————+——+——+J\/Igrcos6. (26.17)  2
2I’sm 6’ 21’ 213 Exercise 26.1 Verify the above expression for the Hamiltonian. A glance at the Hamiltonian reveals that gt and it are cyclic coordinates
(8H/6qﬁ : arr/aw = 0). Thus, by Hamilton’s equations, (9H 8H
' :——: ' :——:O 26.18
which imply
19¢ = constant , pd, : constant . (26.19) Since BH/Bt = O, the total energy E is also conserved (of. Theorem 24.3).
We have thus found three constants of motion, and according to Def. 24.1, the
problem is integrable. It is then reduced to a one—dimensional one (in 0) and
we can write 9
P5
21’
where U5 f f is an effective potential deﬁned by H: + Ugff (6) , (26.20) (19¢ — pg, cos 0)2 P30
————— + —— + JVIgr c030. (26.21) U6” (6) 21' 51112 0 213 The equation of motion for 0 is given by the Hamilton equation 8H
' = — .22
P9 60 a (26 )
which, by (26.12) and (26.20), is equivalent to,
2
I’ d—6 — ———dUE”(6) (26.23) alt2 ' d6 This equation can in principle be integrated directly. We will, however, not
solve it explicitly, but will instead attempt to gain a qualitative understanding 130 CHAPTER 26. SYJVIMETRIC TOPS Fri; 6 i f3» of the solutions by examining the shape of the effective potential U8 f f(6), a
sketch of which is given in Fig. 26.3 for the case p¢ 75 :l:p¢. (For this case the
vertical lines 6 = 0 and 6 = 7r are asymptotes to the potential energy curve). E 2 Energy conservation is expressed by the following equation: 2
P_9
2I’ Obviously the classically allowed values of 6 must be in the region where E Z Ueff(6). If E = Ueff(60), then 6 : 60 = constant. Eqs.(26.f4) and (2619) then 5 q: imply that = constant, and the motion of the top is a steady precession of the all is, A symmetry axis 63 around the vertical direction with constant angular speed and inclined at constant angle 6 = 60 (Fig. 26.4). This case was already treated by an approximate method in Chapter 21 [recall the discussion surrounding 5;; /
0 (21.25)]. + Ueff((9) : E = constant . (26.24) If E > Ueff(60), the motion is periodic with 6 assuming values in the range
61 g 6 S 62. The top “nods” at the same time as (1) changes in value. The time
evolution of 6 and (15 can then be represented by tracing the path of the CM
(or any point on the symmetry axis except the ﬁxed point) on the surface of a
sphere (since any point on the surface of a sphere can be coordinatized by 6 and
gb). [For the steady precession case discussed above, this path is just a latitude
circle (with constant 6)]. . From (26.14) we see that ((5 vanishes only when c056 2 13¢ /p¢. We thus
consider the following three cases according to the value of cos”1(p¢, / 19¢). (1) cos—1 ¢ [61, 62 Since is never zero, it cannot change in sign. P111
Thus gb increases (or decreases) monotonically as 6 oscillates between 61 and 62 (see Fig. 26.5). (2) 61 < cos“1 <13qu ) < 62. In ghis case vanishes at a value of 6 inter— Flfl‘ 25'5
psi
mediate between 61 and @L Hence ¢ both increases and decreases depending on the value of 6. A possible motion is indicated in Fig. 26.6. 131 (3) (a) 61 = cos“1 (b) 02 = cos‘1 A possible motion for (a) is
1/1 ¢
shown in Fig. 26.7 and one for (b) is shown in Fig. 26.8. Fig. 26.7, 26.8 a
.x/ C) :5) 3 _ g3}. :26. ‘7 ~ 16% ...
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 Spring '08
 LAM
 mechanics

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