72 lines
2.4 KiB
Markdown
72 lines
2.4 KiB
Markdown
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section: The Equations of Motion
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Find the Lagrangian for each of the following systems when placed in a uniform gravitational field (acceleration g).
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TODO figures
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§5
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The Lagrangian for a system of particles
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11
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PROBLEM 1. A coplanar double pendulum (Fig. 1).
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'!!!!!!!
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pl4
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m,
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2
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L2
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M2
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FIG. 1
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SOLUTION. We take as co-ordinates the angles 01 and O2 which the strings l1 and l2 make
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with the vertical. Then we have, for the particle M1, T1 = U = -migh cos 01. In
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order to find the kinetic energy of the second particle, we express its Cartesian co-ordinates
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x2, y2 (with the origin at the point of support and the y-axis vertically downwards) in terms
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of the angles 01 and O2: X2 = l1 sin 01 sin O2, y2 = l1 cos 1+12 cos O2. Then we find
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T2 = tm2(x22++22)
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=
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Finally
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L = +m2) m2/1/20102 cos(01-02)+(m1+m2)gl cos 01 + m2gl2 cos O2.
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PROBLEM 2. A simple pendulum of mass M2, with a mass M1 at the point of support which
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can move on a horizontal line lying in the plane in which m2 moves (Fig. 2).
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x
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m,
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l
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M2
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FIG. 2
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SOLUTION. Using the co-ordinate x of M1 and the angle between the string and the
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vertical, we have
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L = cos b) +m2gl cos p.
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PROBLEM 3. A simple pendulum of mass m whose point of support (a) moves uniformly
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on a vertical circle with constant frequency y (Fig. 3), (b) oscillates horizontally in the plane
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of motion of the pendulum according to the law x = a cos yt, (c) oscillates vertically accord-
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ing to the law y = a cos yt.
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SOLUTION. (a) The co-ordinates of m are x = a cos rt+l sin , y = -a sin rt+l cos o.
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The Lagrangian is
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L = sin(-t) +mgl cos ;
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here terms depending only on time have been omitted, together with the total time derivative
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of mlay cos(-t).
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12
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The Equations of Motion
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§5
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(b) The co-ordinates of m are x = a cos yt+l sin , y = l cos p. The Lagrangian is (omit-
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ting total derivatives)
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L = st sin +mgl cos .
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(c) Similarly
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L = cos st cos +mgl cos .
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ly
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m
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FIG. 3
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PROBLEM 4. The system shown in Fig. 4. The particle M2 moves on a vertical axis and the
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whole system rotates about this axis with a constant angular velocity S2.
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A
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a
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m
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m,
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a
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m2
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FIG. 4
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SOLUTION. Let 0 be the angle between one of the segments a and the vertical, and $ the
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angle of rotation of the system about the axis; ; b = S. For each particle M1, the infinitesimal
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displacement is given by dl-2 = a2 sin² 0 do2. The distance of M2 from the point
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of support A is 2a cos 0, and so dl2 = -2a sin 0 d0. The Lagrangian is
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L = m1a2(j2 + Q2 sin20) +2m2a202 sin20+2(m1 +m2)ga cos 0.
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