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辅导案例-MEC4428-Assignment 2

By May 15, 2020No Comments

MONASH UNIVERSITY Department of Mechanical and Aerospace Engineering MEC4428 — Advanced Dynamics — Assignment 2: Rigid Body Dynamics — The Motion of a Com- pound Pendulum due: Thursday 10am – 17th October 2018, assignment box building 31 A correct worked derivation of the governing equations with accompanying extracts of the function file, and the commands to solve for and plot the solutions will earn 4 marks contributing to your final score. A Compound Pendulum Write a MATLAB code to determine the motion of the two component pendulum shown in the figure below. The two parts are both rigid bodies, connected by a pivot at point D. The first link is pinned at the point O. Specifically: 1. plot the trajectory of point C, the centre of the square, 2. plot the phase portraits (θ˙ versus θ and φ˙ versus φ), 3. plot the total energy – this should be a constant as there is no dissipation, 4. finally, determine the approximate governing equations for small angles (i.e., θ → 0, φ → pi). Compare the solution from the original set of equations with the approximate set for initial conditions θ(0) = pi/12, φ(0) = pi − pi/12, θ˙(0) = φ˙(0) = 0. You have choice of how to derive the equations. You can use Angular Momentum Balance for the disk about the point D and for the whole system about O. You could also substitute one of these equations with total energy conservation for the two masses. It is also possible to use either virtual work or the Lagrange equations. In any case you should get a system of 4 first-order differential equations for θ˙, θ¨, φ˙, φ¨, which will then be solved using your favourite MATLAB ODE solver (typically ODE45). Assume the following parameters and initial conditions. I have put up some MATLAB code for a double pendulum (two rods) that you might like to modify for the current problem. • m1 = m2 = m = 1 • L = 2, H = 1 (IS = mH2/6; IR = mL2/12) • d = 1/6 (the length CD) • g = 1 (Note this is not a mistake – the problem has been non-dimensionalised…) • θ(0) = pi/2, φ(0) = pi/2 • θ˙(0) = φ˙(0) = 0 1 HR m, R, IS O C G θ φ D m, L, I 2

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