HW2_prob4_GPS22_G20

# HW2_prob4_GPS22_G20 - GPS Problem 1.22(2nd ed 1.20 Planar...

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GPS Problem 1.22 (2 nd ed. 1.20) Planar Double Pendulum Because the motion of each mass point is con f ned to the same plane, the system has only two degrees of freedom. Use the two polar angles, θ 1 and θ 2 , de f ned in Fig. 1.4, as the two generalized coordinates. The angles will be measured from a y axis that points downwards so that θ 1 and θ 2 are positive as showninF ig .1 .4 . The kinetic energy of the pendulum is T = m 1 2 ( · x 1 2 + · y 1 2 )+ m 2 2 ( · x 2 2 + · y 2 2 ) . (1) In the polar coordinate system we have x 1 = l 1 sin θ 1 , (2) y 1 = l 1 cos θ 1 , (3) x 2 = x 1 l 2 sin θ 2 , (4) y 2 = y 1 + l 2 cos θ 2 , (5) from which it follows that · x 1 = l 1 · θ 1 cos θ 1 , (6) · y 1 = l 1 · θ 1 sin θ 1 , (7) · x 2 = · x 1 l 2 · θ 2 cos θ 2 , (8) · y 2 = · y 1 l 2 · θ 2 sin θ 2 . (9) First note that · x 2 2 + · y 2 2 = · x 1 2 + · y 1 2 + μ l 2 · θ 2 2 2 l 2 · θ 2 ( · x 1 cos θ 2 + · y 1 sin θ 2

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HW2_prob4_GPS22_G20 - GPS Problem 1.22(2nd ed 1.20 Planar...

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