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(1130)

14 34 4 4

Similarly, Eq 1121 can be directly used to obtain the moment of inertia of a body about a given point Moments of inertia of simple geometries can often be found in tables, Popov (1976), Fenster and Gould (1985) As a rst approximation, complex objects can often be broken into one or more simple shapes and the moments of inertia (about the center of mass of the combination of shapes) summed Moments of inertia can also be obtained experi-

= 40

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mentally While there are several ways to accomplish this, one of the more common methods is to calculate the natural frequency of a pendulum made from the object (an interesting application of this to moment of inertia calculation for passenger cars is described by Garrott and Monk (1988) The following simple example describes the direct use of the formulas for determining the mass, center of mass, and moment of inertia of a at plate

=ek+l,

EXAMPLE Consider the square plate with sides of length s, a thickness t, and a density r shown in Fig 116 The mass of the plate is given by

Use the method of cylindrical shells to calculate the volume enclosed when the curve y = x 2 , 0 x 3, is rotated about the x-axis (Figure 820)

The coordinates of the center of gravity of the plate are given by ycg = xcg = rst x 2 1 0 xr(st )dx = m 2 m x=

how to write pdf file in asp.net c#

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We reverse, in our analysis, the roles of the x- and y-axes Of course y ranges from 0 to 9 For each position y in that range, there is a segment stretching from x = y to x = 3 Thus it has length 3 y Then the cylinder generated when this segment (thickened to a strip of width y) is rotated about the x-axis has volume V ( y) = 2 y [3

which, of course, describe the center of the plate The moment of inertia about the center of the plate is

1132 Moving Moments of Inertia to Other Points Sometimes the moment of inertia about the center of mass is known, but the moment of inertia about another point on the body is required While the moment of inertia about the desired point could be calculated directly, it is often easier to simply move the moment of inertia If the distance between the center of mass and the point O where the moment of inertia is desired is given by d, the two moments of inertia are related by the parallel-axis theorem Jo = Jcg + md 2 (1131)

Solution:

2 y [3

As with the previous results, planar motion is assumed (though an analogous result holds for spatial motion with proper de nition of the distance d and the axes of rotation) This relationship can be proven easily by substituting the expression xo = xcg + x into Eq 1121 and realizing that d2 = x2 + y2

(1132)

= 2

(1133)

= 2 = 2 = 2 =

Continuing with the example of the at plate in Fig 116, suppose that the plate is pinned to rotate around point O From the discussion above, the kinetic energy can be described in terms of the angular velocity and the moment of inertia about point O To derive the moment of inertia about point O from the moment of inertia about the center of mass, the parallel axis theorem can be used 1 s = Jcg + ms 2 Jo = Jcg + m 2 2 As a check, the moment of inertia about O can be derived using the basic formula:

y = ln(2x2 + 1)

x =0

y5/2 3y2 2 5/2

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how to write pdf file in asp.net c#

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how to write pdf file in asp.net c#

how to write pdf file in asp.net c#