186 lines
8.9 KiB
Text
186 lines
8.9 KiB
Text
the latter being in the same vertical plane as AB. Determine the reactions of the planes and
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the tensions in the strings.
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SOLUTION. The tensions TA and TB are from A to D and from B to C respectively. The
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reactions RA and RB are perpendicular to the corresponding planes. The solution of the
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equations of equilibrium gives RB = P, TB = 1P cot a, RA = TB sin B, TA = TB cos B.
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PROBLEM 4. Two rods of length l and negligible weight are hinged together, and their ends
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are connected by a string AB (Fig. 54). They stand on a plane, and a force F is applied
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at the midpoint of one rod. Determine the reactions.
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RC
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C
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PA
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F
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1
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RB
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T
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T
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A
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B
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FIG. 54
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SOLUTION. The tension T acts at A from A to B, and at B from B to A. The reactions RA
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and RB at A and B are perpendicular to the plane. Let Rc be the reaction on the rod AC at
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the hinge; then a reaction - -Rc acts on the rod BC. The condition that the sum of the moments
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of the forces RB, T and - -Rc acting on the rod BC should be zero shows that Rc acts along
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BC. The remaining conditions of equilibrium (for the two rods separately) give RA = 1F,
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RB = 1F, Rc = 1F cosec a, T = 1F cot a, where a is the angle CAB.
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§39. Motion in a non-inertial frame of reference
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Up to this point we have always used inertial frames of reference in discuss-
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ing the motion of mechanical systems. For example, the Lagrangian
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L = 1mvo2- U,
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(39.1)
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and the corresponding equation of motion m dvo/dt = - au/dr, for a single
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particle in an external field are valid only in an inertial frame. (In this section
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the suffix 0 denotes quantities pertaining to an inertial frame.)
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Let us now consider what the equations of motion will be in a non-inertial
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frame of reference. The basis of the solution of this problem is again the
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principle of least action, whose validity does not depend on the frame of
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reference chosen. Lagrange's equations
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(39.2)
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are likewise valid, but the Lagrangian is no longer of the form (39.1), and to
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derive it we must carry out the necessary transformation of the function Lo.
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§39
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Motion in a non-inertial frame of reference
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127
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This transformation is done in two steps. Let us first consider a frame of
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reference K' which moves with a translational velocity V(t) relative to the
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inertial frame K0. The velocities V0 and v' of a particle in the frames Ko and
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K' respectively are related by
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vo = v'+ V(t).
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(39.3)
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Substitution of this in (39.1) gives the Lagrangian in K':
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L' = 1mv2+mv.+1mV2-U
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Now V2(t) is a given function of time, and can be written as the total deriva-
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tive with respect to t of some other function; the third term in L' can there-
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fore be omitted. Next, v' = dr'/dt, where r' is the radius vector of the par-
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ticle in the frame K'. Hence
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mV(t)+v'= mV.dr/'dt = d(mV.r')/dt-mr'.dV/dt.
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Substituting in the Lagrangian and again omitting the total time derivative,
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we have finally
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L' =
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(39.4)
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where W = dV/dt is the translational acceleration of the frame K'.
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The Lagrange's equation derived from (39.4) is
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(39.5)
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Thus an accelerated translational motion of a frame of reference is equivalent,
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as regards its effect on the equations of motion of a particle, to the application
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of a uniform field of force equal to the mass of the particle multiplied by the
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acceleration W, in the direction opposite to this acceleration.
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Let us now bring in a further frame of reference K, whose origin coincides
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with that of K', but which rotates relative to K' with angular velocity Su(t).
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Thus K executes both a translational and a rotational motion relative to the
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inertial frame Ko.
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The velocity v' of the particle relative to K' is composed of its velocity
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V
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relative to K and the velocity Sxr of its rotation with K: v' = Lxr
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(since the radius vectors r and r' in the frames K and K' coincide). Substitut-
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ing this in the Lagrangian (39.4), we obtain
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L = +mv.Sx+1m(xr)2-mW.r-
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(39.6)
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This is the general form of the Lagrangian of a particle in an arbitrary, not
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necessarily inertial, frame of reference. The rotation of the frame leads to the
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appearance in the Lagrangian of a term linear in the velocity of the particle.
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To calculate the derivatives appearing in Lagrange's equation, we write
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128
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Motion of a Rigid Body
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§39
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the total differential
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dL = mv.dv+mdv.Sxr+mv.Sxdr+
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=
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v.dv+mdv.xr+mdr.vxR+
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The terms in dv and dr give
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0L/dr X Q-mW-dU/0r. - -
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Substitution of these expressions in (39.2) gives the required equation of
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motion:
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mdv/dt = (39.7)
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We see that the "inertia forces" due to the rotation of the frame consist
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of three terms. The force mrxo is due to the non-uniformity of the rotation,
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but the other two terms appear even if the rotation is uniform. The force
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2mvxs is called the Coriolis force; unlike any other (non-dissipative) force
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hitherto considered, it depends on the velocity of the particle. The force
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mSX(rxS) is called the centrifugal force. It lies in the plane through r and
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S, is perpendicular to the axis of rotation (i.e. to S2), and is directed away
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from the axis. The magnitude of this force is mpO2, where P is the distance
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of the particle from the axis of rotation.
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Let us now consider the particular case of a uniformly rotating frame with
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no translational acceleration. Putting in (39.6) and (39.7) S = constant,
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W = 0, we obtain the Lagrangian
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L
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=
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(39.8)
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and the equation of motion
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mdv/dt = -
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(39.9)
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The energy of the particle in this case is obtained by substituting
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p =
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(39.10)
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in E = p.v-L, which gives
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E =
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(39.11)
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It should be noticed that the energy contains no term linear in the velocity.
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The rotation of the frame simply adds to the energy a term depending only
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on the co-ordinates of the particle and proportional to the square of the
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angular velocity. This additional term - 1m(Sxr)2 is called the centrifugal
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potential energy.
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The velocity V of the particle relative to the uniformly rotating frame of
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reference is related to its velocity V0 relative to the inertial frame Ko by
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(39.12)
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§39
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Motion in a non-inertial frame of reference
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129
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The momentum p (39.10) of the particle in the frame K is therefore the same
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as its momentum Po = MVO in the frame K0. The angular momenta
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M = rxpo and M = rxp are likewise equal. The energies of the particle
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in the two frames are not the same, however. Substituting V from (39.12) in
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(39.11), we obtain E = 1mv02-mvo Sxr+U = 1mvo2 + mrxvo S.
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The first two terms are the energy E0 in the frame K0. Using the angular
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momentum M, we have
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E = E0 n-M.S.
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(39.13)
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This formula gives the law of transformation of energy when we change to a
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uniformly rotating frame. Although it has been derived for a single particle,
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the derivation can evidently be generalised immediately to any system of
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particles, and the same formula (39.13) is obtained.
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PROBLEMS
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PROBLEM 1. Find the deflection of a freely falling body from the vertical caused by the
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Earth's rotation, assuming the angular velocity of this rotation to be small.
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SOLUTION. In a gravitational field U = -mg. r, where g is the gravity acceleration
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vector; neglecting the centrifugal force in equation (39.9) as containing the square of S, we
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have the equation of motion
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v = 2vxSu+g.
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(1)
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This equation may be solved by successive approximations. To do so, we put V = V1+V2,
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where V1 is the solution of the equation V1 = g, i.e. V1 = gt+ (Vo being the initial velocity).
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Substituting V = V1+v2in (1) and retaining only V1 on the right, we have for V2 the equation
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V2 = 2v1xSc = 2tgxSt+2voxS. Integration gives
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(2)
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where h is the initial radius vector of the particle.
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Let the z-axis be vertically upwards, and the x-axis towards the pole; then gx = gy = 0,
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n sin 1, where A is the latitude (which for definite-
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ness we take to be north). Putting V0 = 0 in (2), we find x = 0, =-1t300 cos A. Substitu-
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tion of the time of fall t 22 (2h/g) gives finally x = 0,3 = - 1(2h/g)3/2 cos A, the negative
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value indicating an eastward deflection.
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PROBLEM 2. Determine the deflection from coplanarity of the path of a particle thrown
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from the Earth's surface with velocity Vo.
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SOLUTION. Let the xx-plane be such as to contain the velocity Vo. The initial altitude
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h = 0. The lateral deviation is given by (2), Problem 1: y =
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or, substituting the time of flight t 22 2voz/g, y =
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PROBLEM 3. Determine the effect of the Earth's rotation on small oscillations of a pendulum
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(the problem of Foucault's pendulum).
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SOLUTION. Neglecting the vertical displacement of the pendulum, as being a quantity
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of the second order of smallness, we can regard the motion as taking place in the horizontal
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xy-plane. Omitting terms in N°, we have the equations of motion x+w2x = 20zy, j+w2y
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= -20zx, where w is the frequency of oscillation of the pendulum if the Earth's rotation is
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neglected. Multiplying the second equation by i and adding, we obtain a single equation
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130
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Motion of a Rigid Body
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§39
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+2i02s+w28 = 0 for the complex quantity $ = xtiy. For I2<<, the solution of this
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equation is
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$ = exp(-is2t) [A1 exp(iwt) +A2 exp(-iwt)]
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or
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xtiy = (xo+iyo) exp(-is2zt),
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where the functions xo(t), yo(t) give the path of the pendulum when the Earth's rotation is
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neglected. The effect of this rotation is therefore to turn the path about the vertical with
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angular velocity Qz.
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CHAPTER VII
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THE CANONICAL EQUATIONS
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