416 lines
No EOL
18 KiB
Text
416 lines
No EOL
18 KiB
Text
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§35
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Eulerian angles
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111
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Let us now express the components of the angular velocity vector S along
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the moving axes X1, X2, X3 in terms of the Eulerian angles and their derivatives.
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To do this, we must find the components along those axes of the angular
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velocities 6, b, 4. The angular velocity è is along the line of nodes ON, and
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its components are 1 = O cos 4/5, = - O sin 4/5, = 0. The angular velo-
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city is along the Z-axis; its component along the x3-axis is 03 = cos 0, and
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in the x1x2-plane sin A. Resolving the latter along the X1 and X2 axes, we
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have 01 = sin 0 sin 4/s, O2 = sin 0 cos 4. Finally, the angular velocity is
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is along the x3-axis.
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Collecting the components along each axis, we have
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S21 = 0 COS 4,
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Q2 = sin 0 cosy-osiny,
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(35.1)
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S23 = o cos0+4. =
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If the axes X1, X2, X3 are taken to be the principal axes of inertia of the body,
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the rotational kinetic energy in terms of the Eulerian angles is obtained by
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substituting (35.1) in (32.8).
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For a symmetrical top (I1 = I2 # I3), a simple reduction gives
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Trot =
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(35.2)
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This expression can also be more simply obtained by using the fact that the
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choice of directions of the principal axes X1, X2 is arbitrary for a symmetrical
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top. If the X1 axis is taken along the line of nodes ON, i.e. 4 = 0, the compo-
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nents of the angular velocity are simply
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O2 = o sin A,
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(35.3)
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As a simple example of the use of the Eulerian angles, we shall use them
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to determine the free motion of a symmetrical top, already found in $33.
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We take the Z-axis of the fixed system of co-ordinates in the direction of the
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constant angular momentum M of the top. The x3-axis of the moving system
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is along the axis of the top; let the x1-axis coincide with the line of nodes at
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the instant considered. Then the components of the vector M are, by
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formulae (35.3), M1 = I1 = I, M2 = IS2 = sin 0, M3 = I3Q3
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= I3( cos 0+4). Since the x1-axis is perpendicular to the Z-axis, we have
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M1 = 0, M2 = M sin 0, M3 = M cos 0. Comparison gives
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0=0,
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I = M,
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=
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(35.4)
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The first of these equations gives 0 = constant, i.e. the angle between the
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axis of the top and the direction of M is constant. The second equation gives
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the angular velocity of precession = M/I1, in agreement with (33.5).
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Finally, the third equation gives the angular velocity with which the top
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rotates about its own axis: S3 = (M/I3) cos 0.
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112
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Motion of a Rigid Body
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§35
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PROBLEMS
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PROBLEM 1. Reduce to quadratures the problem of the motion of a heavy symmetrical
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top whose lowest point is fixed (Fig. 48).
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SOLUTION. We take the common origin of the moving and fixed systems of co-ordinates
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at the fixed point O of the top, and the Z-axis vertical. The Lagrangian of the top in a gravita-
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tional field is L = (02 +02 sin ²0 + 1/3(1- cos 0)2-ugl - cos 0, where u is the mass
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of the top and l the distance from its fixed point to the centre of mass.
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Z
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X3
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x2
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a
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ug
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Y
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x1
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N
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FIG. 48
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The co-ordinates 4 and are cyclic. Hence we have two integrals of the motion:
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P4 = = cos 0) = constant = M3
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(1)
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= = cos 0 = constant III M2,
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(2)
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where I'1 = I1+ul2; the quantities P4 and Po are the components of the rotational angular
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momentum about O along the X3 and Z axes respectively. The energy
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E = cos 0
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(3)
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is also conserved.
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From equations (1) and (2) we find
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=
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0)/I'1
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sin 20,
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(4)
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(5)
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Eliminating b and of from the energy (3) by means of equations (4) and (5), we obtain
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E' =
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where
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E'
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=
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(6)
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§35
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Eulerian angles
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113
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Thus we have
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t=
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(7)
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this is an elliptic integral. The angles 4 and are then expressed in terms of 0 by means of
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integrals obtained from equations (4) and (5).
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The range of variation of 0 during the motion is determined by the condition E' Ueff(0).
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The function Uett(8) tends to infinity (if M3 # M2) when 0 tends to 0 or II, and has a minimum
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between these values. Hence the equation E' = Ueff(0) has two roots, which determine the
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limiting values 01 and O2 of the inclination of the axis of the top to the vertical.
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When 0 varies from 01 to O2, the derivative o changes sign if and only if the difference
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M-M3 cos 0 changes sign in that range of 0. If it does not change sign, the axis of the top
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precesses monotonically about the vertical, at the same time oscillating up and down. The
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latter oscillation is called nutation; see Fig. 49a, where the curve shows the track of the axis
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on the surface of a sphere whose centre is at the fixed point of the top. If does change sign,
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the direction of precession is opposite on the two limiting circles, and so the axis of the top
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describes loops as it moves round the vertical (Fig. 49b). Finally, if one of 01, O2 is a zero of
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M2-M3 cos 0, of and è vanish together on the corresponding limiting circle, and the path
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of the axis is of the kind shown in Fig. 49c.
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O2
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O2
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O2
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(a)
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(b)
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(c)
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FIG. 49
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PROBLEM 2. Find the condition for the rotation of a top about a vertical axis to be stable.
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SOLUTION. For 0 = 0, the X3 and Z axes coincide, so that M3 = Mz, E' = 0. Rotation
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about this axis is stable if 0 = 0 is a minimum of the function Ueff(9). For small 0 we have
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Ueff 22 whence the condition for stability is M32 > 41'1ugl or S232
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> 41'1ugl/I32.
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PROBLEM 3. Determine the motion of a top when the kinetic energy of its rotation about
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its axis is large compared with its energy in the gravitational field (called a "fast" top).
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SOLUTION. In a first approximation, neglecting gravity, there is a free precession of the
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axis of the top about the direction of the angular momentum M, corresponding in this case
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to the nutation of the top; according to (33.5), the angular velocity of this precession is
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Sunu = M/I' 1.
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(1)
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In the next approximation, there is a slow precession of the angular momentum M about
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the vertical (Fig. 50). To determine the rate of this precession, we average the exact equation
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of motion (34.3) dM/dt = K over the nutation period. The moment of the force of gravity
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on the top is K=uln3xg, where n3 is a unit vector along the axis of the top. It is evident
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from symmetry that the result of averaging K over the "nutation cone" is to replace n3 by
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its component (M/M) cos a in the direction of M, where a is the angle between M and the
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axis of the top. Thus we have dM/dt = -(ul/M)gxM cos a. This shows that the vector M
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114
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Motion of a Rigid Body
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§36
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precesses about the direction of g (i.e. the vertical) with a mean angular velocity
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Spr (ul/M)g cos a
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(2)
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which is small compared with Senu
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Spr
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in
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no
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a
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FIG. 50
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In this approximation the quantities M and cos a in formulae (1) and (2) are constants,
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although they are not exact integrals of the motion. To the same accuracy they are related
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to the strictly conserved quantities E and M3 by M3 = M cos a,
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§36. Euler's equations
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The equations of motion given in §34 relate to the fixed system of co-
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ordinates: the derivatives dP/dt and dM/dt in equations (34.1) and (34.3)
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are the rates of change of the vectors P and M with respect to that system.
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The simplest relation between the components of the rotational angular
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momentum M of a rigid body and the components of the angular velocity
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occurs, however, in the moving system of co-ordinates whose axes are the
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principal axes of inertia. In order to use this relation, we must first transform
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the equations of motion to the moving co-ordinates X1, X2, X3.
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Let dA/dt be the rate of change of any vector A with respect to the fixed
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system of co-ordinates. If the vector A does not change in the moving system,
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its rate of change in the fixed system is due only to the rotation, so that
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dA/dt = SxA; see §9, where it has been pointed out that formulae such as
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(9.1) and (9.2) are valid for any vector. In the general case, the right-hand
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side includes also the rate of change of the vector A with respect to the moving
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system. Denoting this rate of change by d'A/dt, we obtain
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dAdd
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(36.1)
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§36
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Euler's equations
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115
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Using this general formula, we can immediately write equations (34.1) and
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(34.3) in the form
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=
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K.
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(36.2)
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Since the differentiation with respect to time is here performed in the moving
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system of co-ordinates, we can take the components of equations (36.2) along
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the axes of that system, putting (d'P/dt)1 = dP1/dt, ..., (d'M/dt)1 = dM1/dt,
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..., where the suffixes 1, 2, 3 denote the components along the axes x1, x2, X3.
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In the first equation we replace P by V, obtaining
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(36.3)
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=
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If the axes X1, X2, X3 are the principal axes of inertia, we can put M1 = I,
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etc., in the second equation (36.2), obtaining
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=
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I2 = K2,
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}
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(36.4)
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I3 = K3.
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These are Euler's equations.
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In free rotation, K = 0, so that Euler's equations become
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= 0,
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}
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(36.5)
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= 0.
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As an example, let us apply these equations to the free rotation of a sym-
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metrical top, which has already been discussed. Putting I1 = I2, we find from
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the third equation SQ3 = 0, i.e. S3 = constant. We then write the first two
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equations as O = -wS2, Q2 = wS1, where
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=
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(36.6)
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is a constant. Multiplying the second equation by i and adding, we have
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= so that S1+iD2 = A exp(iwt), where A is a
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constant, which may be made real by a suitable choice of the origin of time.
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Thus
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S1 = A cos wt
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Q2 = A sin wt.
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(36.7)
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116
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Motion of a Rigid Body
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§37
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This result shows that the component of the angular velocity perpendicular
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to the axis of the top rotates with an angular velocity w, remaining of constant
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magnitude A = Since the component S3 along the axis of the
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top is also constant, we conclude that the vector S rotates uniformly with
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angular velocity w about the axis of the top, remaining unchanged in magni-
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tude. On account of the relations M1 = , M2 = I2O2, M3 = I3O3 be-
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tween the components of S and M, the angular momentum vector M evidently
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executes a similar motion with respect to the axis of the top.
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This description is naturally only a different view of the motion already
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discussed in §33 and §35, where it was referred to the fixed system of co-
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ordinates. In particular, the angular velocity of the vector M (the Z-axis in
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Fig. 48, $35) about the x3-axis is, in terms of Eulerian angles, the same as
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the angular velocity - 4. Using equations (35.4), we have
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cos
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or - is = I23(I3-I1)/I1, in agreement with (36.6).
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§37. The asymmetrical top
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We shall now apply Euler's equations to the still more complex problem
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of the free rotation of an asymmetrical top, for which all three moments of
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inertia are different. We assume for definiteness that
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I3 > I2 I.
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(37.1)
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Two integrals of Euler's equations are known already from the laws of
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conservation of energy and angular momentum:
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= 2E,
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(37.2)
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= M2,
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where the energy E and the magnitude M of the angular momentum are given
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constants. These two equations, written in terms of the components of the
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vector M, are
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(37.3)
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M2.
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(37.4)
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From these equations we can already draw some conclusions concerning
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the nature of the motion. To do so, we notice that equations (37.3) and (37.4),
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regarded as involving co-ordinates M1, M2, M3, are respectively the equation
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of an ellipsoid with semiaxes (2EI1), (2EI2), (2EI3) and that of a sphere
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of radius M. When the vector M moves relative to the axes of inertia of the
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top, its terminus moves along the line of intersection of these two surfaces.
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Fig. 51 shows a number of such lines of intersection of an ellipsoid with
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§37
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The asymmetrical top
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117
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spheres of various radii. The existence of an intersection is ensured by the
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obviously valid inequalities
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2EI1 < M2 < 2EI3,
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(37.5)
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which signify that the radius of the sphere (37.4) lies between the least and
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greatest semiaxes of the ellipsoid (37.3).
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x1
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X2
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FIG. 51
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Let us examine the way in which these "paths"t of the terminus of the
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vector M change as M varies (for a given value of E). When M2 is only slightly
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greater than 2EI1, the sphere intersects the ellipsoid in two small closed curves
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round the x1-axis near the corresponding poles of the ellipsoid; as M2 2EI1,
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these curves shrink to points at the poles. When M2 increases, the curves
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become larger, and for M2 = 2EI2 they become two plane curves (ellipses)
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which intersect at the poles of the ellipsoid on the x2-axis. When M2 increases
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further, two separate closed paths again appear, but now round the poles on
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the
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x3-axis; as M2 2EI3 they shrink to points at these poles.
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First of all, we may note that, since the paths are closed, the motion of the
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vector M relative to the top must be periodic; during one period the vector
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M describes some conical surface and returns to its original position.
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Next, an essential difference in the nature of the paths near the various
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poles of the ellipsoid should be noted. Near the x1 and X3 axes, the paths lie
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entirely in the neighbourhood of the corresponding poles, but the paths which
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pass near the poles on the x2-axis go elsewhere to great distances from those
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poles. This difference corresponds to a difference in the stability of the rota-
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tion of the top about its three axes of inertia. Rotation about the x1 and X3
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axes (corresponding to the least and greatest of the three moments of inertia)
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t The corresponding curves described by the terminus of the vector Ca are called polhodes.
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118
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Motion of a Rigid Body
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§37
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is stable, in the sense that, if the top is made to deviate slightly from such a
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state, the resulting motion is close to the original one. A rotation about the
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x2-axis, however, is unstable: a small deviation is sufficient to give rise to a
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motion which takes the top to positions far from its original one.
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To determine the time dependence of the components of S (or of the com-
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ponents of M, which are proportional to those of (2) we use Euler's equations
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(36.5). We express S1 and S3 in terms of S2 by means of equations (37.2)
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and (37.3):
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S21 =
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(37.6)
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Q32 =
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and substitute in the second equation (36.5), obtaining
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dSQ2/dt (I3-I1)21-23/I2
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= V{[(2EI3-M2-I2(I3-I2)22]
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(37.7)
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Integration of this equation gives the function t(S22) as an elliptic integral.
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In reducing it to a standard form we shall suppose for definiteness that
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M2 > 2EI2; if this inequality is reversed, the suffixes 1 and 3 are interchanged
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in the following formulae. Using instead of t and S2 the new variables
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(37.8)
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S = S2V[I2(I3-I2)/(2EI3-M2)],
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and defining a positive parameter k2 < 1 by
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(37.9)
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we obtain
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ds
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the origin of time being taken at an instant when S2 = 0. When this integral
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is inverted we have a Jacobian elliptic function S = sn T, and this gives O2
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as a function of time; S-1(t) and (33(t) are algebraic functions of 22(t) given
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by (37.6). Using the definitions cn T = V(1-sn2r), dn T =
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we find
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Superscript(2) = [(2EI3-M2/I1(I3-I1)] CNT,
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O2 =
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(37.10)
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O3 = dn T.
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These are periodic functions, and their period in the variable T is 4K,
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where K is a complete elliptic integral of the first kind:
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=
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(37.11)
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§37
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The asymmetrical top
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119
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The period in t is therefore
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T =
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(37.12)
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After a time T the vector S returns to its original position relative to the
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axes of the top. The top itself, however, does not return to its original position
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relative to the fixed system of co-ordinates; see below.
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For I = I2, of course, formulae (37.10) reduce to those obtained in §36
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for a symmetrical top: as I I2, the parameter k2 0, and the elliptic
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functions degenerate to circular functions: sn -> sin T, cn T cos
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T,
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dn T -> 1, and we return to formulae (36.7).
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When M2 = 2EI3 we have Superscript(1) = S2 = 0, S3 = constant, i.e. the vector S
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is always parallel to the x3-axis. This case corresponds to uniform rotation of
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the top about the x3-axis. Similarly, for M2 = 2EI1 (when T III 0) we have
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uniform rotation about the x1-axis.
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Let us now determine the absolute motion of the top in space (i.e. its
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motion relative to the fixed system of co-ordinates X, Y, Z). To do so, we
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use the Eulerian angles 2/5, o, 0, between the axes X1, X2, X3 of the top and the
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axes X, Y, Z, taking the fixed Z-axis in the direction of the constant vector M.
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Since the polar angle and azimuth of the Z-axis with respect to the axes
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x1, X2, X3 are respectively 0 and 1/77 - is (see the footnote to $35), we obtain on
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taking the components of M along the axes X1, X2, X3
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M sin 0 sin y = M1 = ,
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M sin A cos is = M2 = I2O2,
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(37.13)
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M cos 0 = M3 = I3S23.
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Hence
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cos 0 = I3S3/M,
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tan / =
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(37.14)
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and from formulae (37.10)
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COS 0 = dn T,
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(37.15)
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tan 4 = cn r/snt,
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which give the angles 0 and is as functions of time; like the components of the
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vector S, they are periodic functions, with period (37.12).
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The angle does not appear in formulae (37.13), and to calculate it we
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must return to formulae (35.1), which express the components of S in terms
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of the time derivatives of the Eulerian angles. Eliminating O from the equa-
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tions S1 = sin 0 sin 4 + O cos 2/5, S2 = sin 0 cos 4-0 - sin 2/5, we obtain
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& = (Superscript(2) sin 4+S2 cos 4)/sin 0, and then, using formulae (37.13),
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do/dt =
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(37.16)
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The function (t) is obtained by integration, but the integrand involves
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elliptic functions in a complicated way. By means of some fairly complex
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120
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Motion of a Rigid Body
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§37
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transformations, the integral can be expressed in terms of theta functions;
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we shall not give the calculations, but only the final result.
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The function (t) can be represented (apart from an arbitrary additive
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constant) as a sum of two terms:
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$(t) = (11(t)++2(t),
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(37.17)
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one of which is given by
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(37.18)
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where D01 is a theta function and a a real constant such that
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sn(2ixK) = iv[I3(M2-2I1)/I1(2EI3-M2]
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(37.19)
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K and Tare given by (37.11) and (37.12). The function on the right-hand side
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of (37.18) is periodic, with period 1T, so that 01(t) varies by 2n during a time
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T. The second term in (37.17) is given by
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(37.20)
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This function increases by 2nr during a time T'. Thus the motion in is a
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combination of two periodic motions, one of the periods (T) being the same
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as the period of variation of the angles 4 and 0, while the other (T') is incom-
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mensurable with T. This incommensurability has the result that the top does
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not at any time return exactly to its original position.
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PROBLEMS
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PROBLEM 1. Determine the free rotation of a top about an axis near the x3-axis or the
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x1-axis.
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SOLUTION. Let the x3-axis be near the direction of M. Then the components M1 and M2
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are small quantities, and the component M3 = M (apart from quantities of the second and
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higher orders of smallness). To the same accuracy the first two Euler's equations (36.5) can
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be written dM1/dt = DoM2(1-I3/I2), dM2/dt = QOM1(I3/I1-1), where So = M/I3. As
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usual we seek solutions for M1 and M2 proportional to exp(iwt), obtaining for the frequency w
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(1)
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The values of M1 and M2 are
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cos wt, sin wt,
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(2)
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where a is an arbitrary small constant. These formulae give the motion of the vector M
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relative to the top. In Fig. 51, the terminus of the vector M describes, with frequency w,
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a small ellipse about the pole on the x3-axis.
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To determine the absolute motion of the top in space, we calculate its Eulerian angles.
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In the present case the angle 0 between the x3-axis and the Z-axis (direction of M) is small,
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t These are given by E. T. WHITTAKER, A Treatise on the Analytical Dynamics of Particles
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and Rigid Bodies, 4th ed., Chapter VI, Dover, New York 1944. |