examples - System Dynamics & Response: 4600340...

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Unformatted text preview: System Dynamics & Response: 4600340 March 23, 2004 Contents 1 Mechanical Systems 2 1.1 Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Transient Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2 Electrical Systems 32 2.1 Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 2.2 Transient Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3 Fluid and Thermal Systems 39 3.1 Modeling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.2 Transient Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 4 Laplace Transforms 42 5 Other 49 1 1 Mechanical Systems 1.1 Modeling Problem 1: In the system shown to the right, the bar has mass m and length . The block of mass 2 m is suspended from the bar by an inextensible cable. a) What is the degree-of-freedom for this system? b) Derive the equations of motion for this system for small rotations of the bar; c) Find the transfer function between the forcing F ( s ) and the rotation of the bar ( s ). You may assume that the system is in static equilibrium when in the horizontal position with no forcing (equivalently, you may ne- glect gravity). 3 2 3 ( m, ) k b x 2 m k k f ( t ) G x 1 z y O Solution: a) This is a two degree-of-freedom system. To uniquely determine the configuration of the system we must specify both the rotation of the bar and the displacmeent of the block (or the equivalent of these two). b) We begin the modeling by identifying the five coordinates as shown in the figure. As a two degree-of-freedom system, there must exist three relationships among these. For small rotations of the bar: x 1 = 2 3 , x 2 = 3 , z = y- x 1 The free-body diagram of the block is shown to the right. Also, the angu- lar acceleration of the bar and trans- lational acceleration of the block are: B/ F = k , F a G =- y . Also, the moment of inertia of the bar about O is: I O = m 2 12 + m 6 2 = m 2 9 .- k x 2 - b x 2 k x 1 - k z k z - f ( t ) F R 2 Therefore, applying angular momentum balance on the bar about O : X M O = I O B/ F , (- k x 1 + k z ) 2 3- ( k x 2 + b x 2 ) 3 k = m 2 9 k , and linear momentum balance on the block: X F = m F a G , k z- f ( t ) =- m y . Finally, eliminating x 1 , x 2 , and z from these equations, we obtain the equation of motion as: m 2 9 + b 2 9 + ( k 2 ) - 2 k 3 y = , m y- 2 k 3 + k y = f ( t ) ....
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This note was uploaded on 04/10/2008 for the course ME 3514 taught by Professor J.renno during the Fall '08 term at Virginia Tech.

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examples - System Dynamics & Response: 4600340...

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