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5.30 Consider an inverted pendulum that is composed of a weightless rigid rod of length L to which a mass m and a linear spring of spring constant k are attached at its free end. The pendulum is initially vertical. The unstretched length of the spring is L. The rota-tion of the pendulum’s pivot has a damping c, and the pendulum is driven by a moment
M(t). The governing equation describing the angular motion is38
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– sin u + b a1
b sin u
5 – 4 cos u
t = tA
a = 1c/m2
and t is the time.
If M = 0, b = 10, a = 0.1, u(0) = p/4, and du(0)/dt = 0, then plot the rotation u as a function of t for 1,000 equally spaced values of t from 0 … t … 50 and in a sep-arate figure, plot u(t) versus du(t)/dt, which is called the phase portrait.
The oscillations of the height Z of the separation between the fluid levels in two rectan-gular prismatic reservoirs connected by a long pipeline can be determined from39
+ signum1dZ>dt2 pa
+ qZ = 0
See, for example, D. M. Etter, Engineering Problem Solving with MATLAB, Prentice Hall, Upper Sad-dle River, NJ, 1997, pp. 220–221.
Wilson and Turcotte, Advanced Mathematics, p. 279.
D. N. Roy, Applied Fluid Mechanics, Ellis Horwood Limited, Chichester, UK, 1988, pp. 290–293.
If p = 0.375 m – 1, q = 7.4 * 10 – 4 s- 2 and the initial conditions are Z(0) = Zn m and dZ(0)/dt = 0 m/s, then determine the value of the first occurrence of tn, n = 1, 2, for which Z(tn) = 0 when Z1 = 10 m and Z2 = 50 m. Use interp1 to determine tn. The quantity signum is determined with sign. Suggestion: Use an appropriate combination of min and find to determine the index of the first value of t at which Z is negative.Then take a small range of values of t around this value over which interp1 should perform the inter-polation. [Answers: for Z1 = 10 m, t1 = 114.2692s andfor Z2 = 510m, t2 = 276.1428 s.]
Consider Eq. (5.3) and its numerical solution to a step input; that is, h(t) = u(t). Determine the value of j that makes the following quantity a minimum:
f(j) = aN 1y(tn) – 122
n = 1
where y(tn), n = 1, 2, Á , N are the values of y at the times tn over the range
0 … t … 35 that are determined by ode45.
Determine the solution to the following system of nonlinear ordinary differential equations:
= – a
+ gLb y2 –
= 1 + e sin 7(vt + 9p/8)
= 7ev sin 6(vt + 9p/8) cos(vt + 9p/8)
The initial conditions are y1(0) = – 1 and y2(0) = 1 and the constants have the follow-ing values: e = 0.16, g = 0.4, and v = 0.97. Plot y1(t) versus y2(t).