Exercise 11.6. Consider the double pendulum from Example 12.1.5 with equal masses and rods of equal length (so the mass

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Exercise 11.6. Consider the double pendulum from Example 12.1.5 with equal masses and rods of equal length (so the mass

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Exercise 11 6 Consider The Double Pendulum From Example 12 1 5 With Equal Masses And Rods Of Equal Length So The Mass 1
Exercise 11 6 Consider The Double Pendulum From Example 12 1 5 With Equal Masses And Rods Of Equal Length So The Mass 1 (141.14 KiB) Viewed 92 times
Exercise 11.6. Consider the double pendulum from Example 12.1.5 with equal masses and rods of equal length (so the mass ratio and length ratio are both equal to 1). The two angles 0₁ (t) and 02 (t) that describe the state of the system at time t satisfy the pair of autonomous second-order non-linear ODE's given by (11.7.1) 2 cos(0₁-0₂)] Cos (0₁ 0₂) 1 ¹₂] [81] = (a) Express this as a 4-dimensional first-order ODE system x' = f(x) for a vector x(t) = [-2 sin 0₁ (0₂)2 sin(0₁-0₂)] - sin 0₂ + (01)² sin(0₁ – 0₂) - [0₁(t)] 0₂ (t) x3 (t) [x4(t)] and function f: R4 → R4. You will find it convenient to express the last two entries in f in the form M (01₁, 0₂) ¹v (01,02) for the 2 x 2 matrix M (0₁, 0₂) = [COB(02² - cos(01-0₂) cos(01-0₂) 1 appearing in the second-order system (11.7.1) and some vector-valued function v: R² → R²; there is no need to write out the inverse of M(01,02) explicitly, but do briefly explain why M (01,02) is always invertible. (b) Find the stationary points of the first-order system in (a), where the first two coordinates are kept in the interval of values 0 ≤ 01,02 < 27. Interpret what these 4-vectors correspond to in terms of the double pendulum (both position and velocities!).
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