# Troublshooting your pc for dummies - Bourg D.M.

ISBN 0-596-00006-5

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typedef struct _PointMass {

float mass;

Vector designPosition;

Vector correctedPosition;

} PointMass;

// Assume that _NUMELEMENTS has been defined PointMass Elements[_NUMELEMENTS];

Here’s some code that illustrates how to calculate the total mass and combined center of gravity of the elements:

int i; float TotalMass;

Vector CombinedCG;

Vector First,foment;

TotalMass = 0;

for(i=0; i<_NUMELEMENTS; i++)

TotalMass+= Elementsfl].mass;

FirstMoment = Vector(o, 0, 0); '

for(i=0; i< NUMELEMENTS; i++) .

{

FirstMoment += Eletnent[i].mass * Element[i].designPosition;

}

CombinedCG = FirstMoment / TotalMass;

Now that the combined center of gravity location has been found, you can calculate the relative position of each point mass as follows:

for(i=0; i<_NUMELEMENTS; i++)

{

Element[i].correctedPosition = Elementfi].designPosition -

CombinedCG;

}

To calculate the mass moment of inertia, you need to take the second moment of each elemental mass making up the body about each coordinate axis. Here, the lever is not the distance to the elemental mass centroid along the coordinate axis, as in the calculation for center of mass; it is the perpendicular distance from the coordinate axis, about which we want to calculate the moment of inertia, to the elemental mass centroid. The second moment is then the product of the mass times this distance squared.

Referring to Figure 1-2 for an arbitrary body in three dimensions, when calculating moment of inertia about the x-axis, Ixx, this distance, r, will be in the yz-plane such that r 2 = y2 + zl ? Similarly for the moment of inertia about the y-axis, lyy, ry2 = z2 + *2, and for the moment of inertia about the z-axis, Izz, r\ = x2 + y2.

8 | Chapter 1: Basic Concepts

Figure 1-2. Arbitrary Body ip 3D

The equations for mass moment of inertia about the coordinate axes in 3D are:

4x = j r2dm- J (y2 + z2) dm

Iyy — j r2 dm - j (z2 + x2) dm

Izz = j r2 dm = J {x2 + y1) dm

Let’s look for a moment at a common situation that arises in practice. Say you are given the moment of inertia, Ia, of a body about an axis, called the neutral axis, passing through the center of mass of the body but you want to know the moment of inertia, I, about an axis some distance from, but parallel to, this neutral axis. In this case, you can use the transfer of axes, or parallel axis theorem, to determine the moment of inertia about this new axis. The formula to use is

/ = /0+ md2

where m is the mass of the body and d is the perpendicular distance between the parallel axes.

There is an important practical observation to make here: the new moment of inertia is a function of the distance separating the axes squared. This means that in cases in which Jo is known to be relatively small and d relatively large, you can safely ignore I0, since the md2 term will dominate. You must use your judgment here, of course. This formula for transfer of axes also indicates that the moment of inertia of a body will be at its minimum when calculated about an axis passing through the body’s center of gravity. The body’s moment of inertia about any parallel axis will always increase by an amount md2 when calculated about an axis that does not pass through the body’s center of mass.

In practice, calculating mass moment of inertia for all but the simplest shapes of uniform density is a complicated endeavor, so we will often approximate the moment of inertia

Mass, Center of Mass, and Moment of Inertia j 9

of a body about axes passing through its center of mass by using simple formulas for basic shapes that approximate the object. Further, we will break down complicated bodies into smaller components and take advantage of the fact ihatiiA, may be negligible for certain components, considering its md2 contribution to the total body’s moment of inertia.

Figures 1-3 through 1-7 show some simple solid geometries for which you can easily calculate mass moments of inertia. The mass moment of inertia formulas for each of these simple geometries of homogenous density about the three coordinate axes are shown in the figure captions. Similar formulas for other basic geometries can readily be found in college-level dynamics texts (see the bibliography at the end of this book for a few sources).

Figure 1-3. Circular Cylinder; lxx = lyy = (l/4)mr2 + (1/12)m(2; ?? = (l/2)mr2

As you can see, these formulas are relatively simple to implement. The trick here is to break up a complex body into a number of smaller, simpler representative geometries whose combination will approximate the complex body’s inertia properties. This exercise is largely a matter of judgment considering the desired level of accuracy

ID I Chapter 1: Basic Concepts

Figure 1-5. Rectangular Cylinder; l„ = (l/12)m(a2 +12); Iyy = (l/12)m(i>2 +11); Ia = (1/12) m(a2 + b2) '? .

J Z-— f t • V Y

Kgwre J-6. Sphere; lxx = Iyy = Iu = (2/5)mr2

> Z-— : iV; Y

Figure 1-7. Spherical Shell; lxx = lyy = lu = {2/3)mr2 A

Mass, Center of Mass, and Moment of Inertia I 11

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