93 lines
4.3 KiB
Markdown
93 lines
4.3 KiB
Markdown
---
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title: 33-angular-momentum-of-a-rigid-body
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---
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Angular momentum of a rigid body
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105
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the notation being as in Problem 7. The instantaneous axis of rotation is the generator OA
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which passes through the point where the cone touches the plane. The centre of mass is at a
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distance a sin a from this axis, and so l = V/a sin a = O/sin a. The components of the
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vector Ca along the principal axes of inertia are, if the x2-axis is taken perpendicular to the
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axis of the cone and to the line OA, O sin a = 6, 0, N cos a = è cot a. The kinetic energy
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is therefore
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T cot2a
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= 314h282(sec2x+5)/40.
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Z
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0
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Y
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A
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FIG. 43
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PROBLEM 9. Find the kinetic energy of a homogeneous ellipsoid which rotates about one
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of its axes (AB in Fig. 44) while that axis itself rotates about a line CD perpendicular to it
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and passing through the centre of the ellipsoid.
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SOLUTION. Let the angle of rotation about CD be 0, and that about AB (i.e. the angle
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between CD and the x1-axis of inertia, which is perpendicular to AB) be . Then the com-
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ponents of S along the axes of inertia are é cos , 0 sin , b, if the x3-axis is AB. Since the
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centre of mass, at the centre of the ellipsoid, is at rest, the kinetic energy is
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=
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D
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B
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D
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A
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Do
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A
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1a
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C
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of
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8
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FIG. 44
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FIG. 45
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PROBLEM 10. The same as Problem 9, but for the case where the axis AB is not perpendicu-
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lar to CD and is an axis of symmetry of the ellipsoid (Fig. 45).
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SOLUTION. The components of Ca along the axis AB and the other two principal axes of
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inertia, which are perpendicular to AB but otherwise arbitrary, are è cos a cos , è cos a X
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sin , o to sin a. The kinetic energy is T = 11102 a)2.
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$33. Angular momentum of a rigid body
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The value of the angular momentum of a system depends, as we know, on
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the point with respect to which it is defined. In the mechanics of a rigid body,
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106
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Motion of a Rigid Body
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§33
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the most appropriate point to choose for this purpose is the origin of the
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moving system of co-ordinates, i.e. the centre of mass of the body, and in
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what follows we shall denote by M the angular momentum SO defined.
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According to formula (9.6), when the origin is taken at the centre of mass
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of the body, the angular momentum M is equal to the "intrinsic" angular
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momentum resulting from the motion relative to the centre of mass. In the
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definition M = Emrxv we therefore replace V by Sxr:
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M = =
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or, in tensor notation,
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Mi = OK
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Finally, using the definition (32.2) of the inertia tensor, we have
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(33.1)
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If the axes X1, X2, X3 are the same as the principal axes of inertia, formula
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(33.1) gives
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M2 = I2DQ,
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M3 = I303. =
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(33.2)
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In particular, for a spherical top, where all three principal moments of inertia
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are equal, we have simply
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M = IS,
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(33.3)
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i.e. the angular momentum vector is proportional to, and in the same direc-
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tion as, the angular velocity vector. For an arbitrary body, however, the
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vector M is not in general in the same direction as S; this happens only
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when the body is rotating about one of its principal axes of inertia.
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Let us consider a rigid body moving freely, i.e. not subject to any external
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forces. We suppose that any uniform translational motion, which is of no
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interest, is removed, leaving a free rotation of the body.
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As in any closed system, the angular momentum of the freely rotating body
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is constant. For a spherical top the condition M = constant gives C = con-
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stant; that is, the most general free rotation of a spherical top is a uniform
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rotation about an axis fixed in space.
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The case of a rotator is equally simple. Here also M = IS, and the vector
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S is perpendicular to the axis of the rotator. Hence a free rotation of a rotator
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is a uniform rotation in one plane about an axis perpendicular to that plane.
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The law of conservation of angular momentum also suffices to determine
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the more complex free rotation of a symmetrical top. Using the fact that the
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principal axes of inertia X1, X2 (perpendicular to the axis of symmetry (x3)
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of the top) may be chosen arbitrarily, we take the x2-axis perpendicular to
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the plane containing the constant vector M and the instantaneous position
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of the x3-axis. Then M2 = 0, and formulae (33.2) show that Q2 = 0. This
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means that the directions of M, St and the axis of the top are at every instant
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in one plane (Fig. 46). Hence, in turn, it follows that the velocity V = Sxr
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of every point on the axis of the top is at every instant perpendicular to that
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