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