ME240_CA1_W08_Example - ME 240: Introduction to Dynamics...

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Unformatted text preview: ME 240: Introduction to Dynamics and Vibrations Mechanical Engineering Department The University of Michigan Computer Assignment #1 Supplemental Document Originally prepared by Akira Saito and modified here by Todd Lillian 1 Introduction In this document, we present how to solve a second-order ordinary differential equation using Mat- lab . Sample Matlab codes are provided for solving a simple example problem that is similar to computer assignment #1. Modifying the Matlab code in this document may be a good starting point for working on your assignment. Example Problem: The equation of motion of a simple pendulum shown in Fig. 1 can be expressed as a second-order, nonlinear ordinary differential equation of the following form: mL + mg sin = 0 (1) where is the angle defined in Fig. 1 , denotes the second derivative of with respect to time t , m is the mass and is set to (1/9.8) kg, g is gravity and is set to 9.8 m/s 2 , and L is the length and is set to 9.8 m. Considering the following initial conditions: ( t = 0) = 4 , ( t = 0) = 0 (2) answer the following questions by solving Eq. ( 1 ) using Matlab : (a) Obtain the value of when the pendulum first hits = 0 . (b) Plot tension T as a function of . T is defined as T = mg cos + mL 2 Light, Rigid rod Figure 1: Simple Pendulum 1 2 State Equations In order to make Eq. ( 1 ) more suitable for a numerical solution, we transform this equation into a special form called state equations . In general the state equations have the following form: x = f ( x , u ) (3) where x contains all the state variables , f is a state function , and u contains inputs . If you have n state variables, Eq. ( 3 ) can be written more explicitly as, x 1 x 2 . . . x n = f 1 ( x , u ) f 2 ( x , u ) . . . f n ( x , u ) (4) Now lets transform Eq. ( 1 ) into the form of Eq. ( 4 ). We define as the first state variable, x 1 , and therefore x 1 = . The natural choice for the second state variable is , therefore x 2 = . We note that that the second time derivative of is the first time derivative of . Let us re-write the definitions using the new notation: x 1 x 2 = (5) Now considering the time derivative of Eq. ( 5 ) and using Eq. ( 1 ), =- g L sin , we get the following: x 1 x 2 = = x 2- g L sin x 1 (6) It is noted that we have two components in the state functions: f 1 ( x , u ) = f 1 ([ x 1 ; x 2 ] , ) = x 2 (7) f 2 ( x , u ) = f 2 ([ x 1 ; x 2 ] , ) =- g L sin x 1 (8) The next step is to write a Matlab code to solve these equations....
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ME240_CA1_W08_Example - ME 240: Introduction to Dynamics...

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