mynotes1 - Part 1 System Modelling Imagine a rotating mass...

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Part 1 - 1 Part 1 - System Modelling Imagine a rotating mass supported by two bearings (as shown below). There is friction in the bearings and, over time, this will cause the mass to stop rotating. We will assume that the friction in the bearings is viscous friction . This means that the torque due to friction will be proportional to the speed of rotation. ω b T = where T = torque due to friction b = friction coefficient ω = angular velocity Newton&s Second Law tells us that the change in angular velocity (the angular acceleration) will depend upon the torque applied and the moment of inertia of the rotating mass. The negative sign in the formula below reflects the fact that the torque slows down the rotation. J T / = & where J = moment of inertia of the rotating mass. Combining the two equations gives a single equation that describes the behaviour of the system. 0 ) / ( = + J b &
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Part 1 - 2 This differential equation can easily be solved to produce an equation for ω in terms of time t . t J b e ) / ( 0 = ω where ω 0 is the angular velocity at time t = 0 This is an example of &exponential decay±. The angular velocity decays exponentially, approaching zero as time approaches infinity but never (in theory) actually quite stopping. The diagram below illustrates the process. It was created using ω 0 = 5 rads/sec, b = 2 N-m/(rad/sec), and J = 20 kg m 2 . In practice, of course, the rotating mass would in fact stop at some point. The discrepancy between theory and reality is a consequence of our simplified friction model. Real friction is not strictly linear. Not let us consider another system: one which, on first blush, would seem to have very little in common with a rotating mass. This time the system will consist of a capacitor (initially charged), a switch (initially open), and a resistor. We want to know what will happen when the switch is closed.
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Part 1 - 3 Once the switch is closed, the capacitor will discharge through the resistor. The defining equation for the capacitor is q C V c ) / 1 ( = where V C is the voltage across the capacitor C is the capacitance of the capacitor q is the charge on the capacitor For the resistor iR V R = where i is the current through the resistor R is the resistance of the resistor The current through the resistor is equal to the rate at which the charge on the capacitor is changing, and the sum of the voltage drops around the circuit must be zero. 0 = + = R C V V q i & Putting everything together, we get 0 ) / 1 ( = + q RC q & Solving this to get an equation for the charge on the capacitor as a function of time produces t RC e q q ) / 1 ( 0 = where q 0 is the charge on the capacitor at time t = 0
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Part 1 - 4 The charge decays exponentially. If this seems very familiar, it should. Though the two examples we&ve looked at come from different branches of engineering, the underlying equations (and hence their solutions) are essentially identical.
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This note was uploaded on 07/16/2009 for the course SYSC 3600 taught by Professor John bryant during the Winter '08 term at Carleton CA.

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mynotes1 - Part 1 System Modelling Imagine a rotating mass...

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