127 lines
6.1 KiB
Text
127 lines
6.1 KiB
Text
The equations of motion of a rigid body
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107
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plane. That is, the axis of the top rotates uniformly (see below) about the
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direction of M, describing a circular cone. This is called regular precession
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of the top. At the same time the top rotates uniformly about its own axis.
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M
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n
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x3
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22pr
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x1
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FIG. 46
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The angular velocities of these two rotations can easily be expressed in
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terms of the given angular momentum M and the angle 0 between the axis
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of the top and the direction of M. The angular velocity of the top about its
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own axis is just the component S3 of the vector S along the axis:
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Q3 = M3/I3 = (M/I3) cos 0.
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(33.4)
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To determine the rate of precession Spr, the vector S must be resolved into
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components along X3 and along M. The first of these gives no displacement
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of the axis of the top, and the second component is therefore the required
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angular velocity of precession. Fig. 46 shows that Spr sin 0 = Q1, and, since
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S21 = M1/I1 = (M/I1) sin 0, we have
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Spr r=M/I1.
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(33.5)
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$34. The equations of motion of a rigid body
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Since a rigid body has, in general, six degrees of freedom, the general
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equations of motion must be six in number. They can be put in a form which
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gives the time derivatives of two vectors, the momentum and the angular
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momentum of the body.
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The first equation is obtained by simply summing the equations p = f
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for each particle in the body, p being the momentum of the particle and f the
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108
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Motion of a Rigid Body
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§34
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force acting on it. In terms of the total momentum of the body P =
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and total force acting on it F = Ef, we have
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dP/dt = F.
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(34.1)
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Although F has been defined as the sum of all the forces f acting on the
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various particles, including the forces due to other particles, F actually
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includes only external forces: the forces of interaction between the particles
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composing the body must cancel out, since if there are no external forces
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the momentum of the body, like that of any closed system, must be conserved,
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i.e. we must have F = 0.
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If U is the potential energy of a rigid body in an external field, the force
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F is obtained by differentiating U with respect to the co-ordinates of the
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centre of mass of the body:
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F = JUIR.
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(34.2)
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For, when the body undergoes a translation through a distance SR, the radius
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vector r of every point in the body changes by SR, and so the change in the
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potential energy is
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SU = (U/dr) Sr = RR Couldr = SR SR.
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It may be noted that equation (34.1) can also be obtained as Lagrange's
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equation for the co-ordinates of the centre of mass, (d/dt)àL/JV = aL/JR,
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with the Lagrangian (32.4), for which
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OL/OV=,MV=P, 0L/JR = JU/OR = F.
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Let us now derive the second equation of motion, which gives the time
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derivative of the angular momentum M. To simplify the derivation, it is
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convenient to choose the "fixed" (inertial) frame of reference in such a way
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that the centre of mass is at rest in that frame at the instant considered.
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We have M = (d/dt) Erxp = Eixp+. Our choice of the frame of
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reference (with V = 0) means that the value of i at the instant considered is
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the same as V = i. Since the vectors V and p = mv are parallel, ixp = 0.
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Replacing p by the force f, we have finally
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dM/dt = K,
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(34.3)
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where
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K = .
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(34.4)
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Since M has been defined as the angular momentum about the centre of
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mass (see the beginning of $33), it is unchanged when we go from one inertial
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frame to another. This is seen from formula (9.5) with R = 0. We can there-
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fore deduce that the equation of motion (34.3), though derived for a particular
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frame of reference, is valid in any other inertial frame, by Galileo's relativity
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principle.
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The vector rxf is called the moment of the force f, and so K is the total
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torque, i.e. the sum of the moments of all the forces acting on the body. Like
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§34
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The equations of motion of a rigid body
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109
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the total force F, the sum (34.4) need include only the external forces: by
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the law of conservation of angular momentum, the sum of the moments of
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the internal forces in a closed system must be zero.
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The moment of a force, like the angular momentum, in general depends on
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the choice of the origin about which it is defined. In (34.3) and (34.4) the
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moments are defined with respect to the centre of mass of the body.
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When the origin is moved a distance a, the new radius vector r' of each
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point in the body is equal to r-a. Hence K = Erxf = Er'xf+ Eaxf or
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K = K'+axF.
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(34.5)
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Hence we see, in particular, that the value of the torque is independent of
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the choice of origin if the total force F = 0. In this case the body is said to
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be acted on by a couple.
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Equation (34.3) may be regarded as Lagrange's equation (d/dt) OL/OS
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= 0L/dd for the "rotational co-ordinates". Differentiating the Lagrangian
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(32.4) with respect to the components of the vector S2, we obtain
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= IikOk = Mi. The change in the potential energy resulting from an
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infinitesimal rotation SO of the body is SU = - Ef.Sr = -
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= So. Erxf = -K.SO, whence
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K =-20/00, =
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(34.6)
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so that aL/dd = 00/08 = K.
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Let us assume that the vectors F and K are perpendicular. Then a vector a
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can always be found such that K' given by formula (34.5) is zero and
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K a x F.
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(34.7)
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The choice of a is not unique, since the addition to a of any vector parallel
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to F does not affect equation (34.7). The condition K' = 0 thus gives a straight
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line, not a point, in the moving system of co-ordinates. When K is perpendi-
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cular to F, the effect of all the applied forces can therefore be reduced to that
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of a single force F acting along this line.
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Such a case is that of a uniform field of force, in which the force on a particle
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is f = eE, with E a constant vector characterising the field and e characterising
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the properties of a particle with respect to the field. Then F = Ee,
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K = erxE. Assuming that # 0, we define a radius vector ro such that
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(34.8)
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Then the total torque is simply
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=roxF
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(34.9)
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Thus, when a rigid body moves in a uniform field, the effect of the field
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reduces to the action of a single force F applied at the point whose radius
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vector is (34.8). The position of this point is entirely determined by the
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t For example, in a uniform electric field E is the field strength and e the charge; in a
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uniform gravitational field E is the acceleration g due to gravity and e is the mass m.
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110
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Motion of a Rigid Body
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