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The mass associated with the equivalent model therefore depends upon the amount of mechanical advantage provided by the lever formed by the rod and the pivot There are two ways to calculate the equivalent mass of a system The rst is to balance reaction forces as in the example above This method has an advantage in that the equations of motion are developed concurrently, though it is the more tedious of the two approaches The other method is to consider the kinetic energy of the original model and the lumped model Masses and inertias store energy in the form of kinetic energy, so for the lumped model to represent the original, the kinetic energy of the two systems must be the same This works not only for masses but also for inertias, as demonstrated in the following example

=-k/dt,

Figure 118a illustrates a simple model of a geared system with an inertial load Each gear and the load are assumed to have inertia and the desired model is an equivalent inertia re ected to the input shaft (Fig 118b) The kinetic energy of the original system is

What is the average value of the function g( x) = sin x over the interval [0, 2 ]

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The equivalent inertia can be obtained by equating the two expressions for the kinetic energy and incorporating the kinematics of the system In this case, the kinematic relationship is simply the gear ratio or Rpinion w out = w Rgear in This gives Rpinion 2 1 1 1 2 2 J eq w in = J pinion w in + ( J gear + J load ) w in 2 2 2 Rgear so Rpinion Jeq = J pinion + ( Jgear + Jload ) Rgear This would be the inertia felt by a motor or person turning the input shaft Conversely, if the motor were attached to the output shaft, the elements in Fig 118 would have an equivalent inertia of Rgear Jeq = Jgear + Jload + J Rpinion pinion Because of the analogy to mechanical advantage, it is not surprising that the equivalent inertia of a system depends upon the reference point chosen

1 [ cos x] 2

2 2 2

In system modeling, the term spring refers not only to machine elements such as coil, torsional, or leaf springs, but to any component of the system that de ects in response to load In other words, every part of the system that is not assumed to be perfectly rigid acts as a spring for the purposes of modeling Springs selected from a catalog and compliant elements share several properties Most fundamentally, they are described by some relationship between de ection and the force applied at either end of the spring This is most commonly modeled as a linear relationship F = Kd (1140)

60 60 = 65e-2k -= e -2k , and switching to logarithms ' 65 60 - 2 k = ln-, 65

1 [ 1 ( 1) ] = 0 2

keeping in mind that the force exerted by the spring always acts to oppose the compression or extension The value K is known as the spring stiffness or spring rate and has units of force/distance with lbf/in and N m being common In Eq 1140, the de ection d is measured from the free or unstretched length, Lf, of the spring and F is the force exerted on the spring (Fig 119)

There are several reasons why a linear spring model is a useful approximation First, systems described by combinations of masses, linear springs, and linear dampers possess linear equations of motion These are more readily analyzed than general nonlinear forms The linear description is also compact, requiring only a single parameter (the rate of stiffness) to describe the spring From a practical standpoint, linearity proves to be a reasonable assumption for most physical springs, at least within a small range of de ection If more accuracy is desired, however, springs can instead be represented as a general nonlinear function such as: F = fs (d ) (1141)

We see that this answer is consistent with our intuition: the function g( x) = sin x takes positive values and negative values with equal weight over the interval [0, 2 ] The average is intuitively equal to zero And that is the actual computed value

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