Moment of Inertia. Terminology. Definitions Moment of inertia of a body with mass, m, about the x axis: Transfer Theorem - 1. ( )dm. = y 2 + z 2.

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1 Terinology Moent of Inertia ME 202 Moent of inertia (MOI) = second ass oent Instead of ultiplying ass by distance to the first power (which gives the first ass oent), we ultiply it by distance to the second power. 2 Definitions Moent of inertia of a body with ass,, about the x axis: ( )d I x = y 2 + z 2 Moent of inertia of a body with ass,, about the y axis: ( )d I y = x 2 + z 2 Moent of inertia of a body with ass,, about the z axis: ( )d = x 2 + y 2 3 Transfer Theore - 1 We can transfer the oent of inertia fro one axis to another, provided that the two axes are parallel. In other words, if we know the oent of inertia about one axis, we can copute it about any other axis parallel to the first axis. 4

2 Transfer Theore - 2 If the oent of inertia of a body with ass about an axis x through the ass center is I G x, and the perpendicular distance fro the x axis to the (parallel) axis x is d, then the oent of inertia of the body about the x axis is I x = I G x + d! 2 transfer ter Transfer Theore - 3 The oent of inertia to which the transfer ter is added is always the one for an axis through the ass center. The oent of inertia about an axis through the ass center is saller than the oent of inertia about any other parallel axis. 5 6 Transfer Theore - 4 We can transfer fro any axis to a parallel axis through the ass center by subtracting the transfer ter. I G x = I x d 2 Radius of Gyration The radius of gyration about the x axis is denoted by. k x ( )d I x = y 2 + z 2 By definition, the radius of gyration of a ass about the x axis is k x = I x = k x 2 7 8

3 Coposite Masses Since the oent of inertia is an integral, and since the integral over a su of several asses equals the su of the integrals over the individual asses, we can find the oent inertia of a coposite ass by adding the oents of inertia of its parts. (Regions with no ass can be subtracted.) γ lb / ft For a straight, slender, unifor bar: I G = 1 12 L2 I A = I G + d 2 Find IA (about the axis through A that is noral to the plane of the figure). Into how any parts should we divide the structure? Use two bars, each with length I A = 1 12 L2 + L 2 2 L ft L2 + 2L = L = γ L g L2 36 I A 1 γ L 3 36 g 1 ( 3 lb / ft) ( 3 ft) 3 = 2.17 slug ft ft / sec For a right, unifor, circular cone: I G 10 r 2 = ρv V = 1 3 πr 2 h Find Iz. Into how any parts should we divide the structure? 11 ρ = 200 kg / 3 Is the issing peak the sae size as the hole? h peak 200 = h peak = Use three parts: large, full cone; sall issing peak of large cone; hole. = I large I peak I hole 10 large R peak r hole r 2 ρ = 200 kg / 3 10 large R peak r hole r 2 12 For a right, unifor, circular cone: I G 10 r 2 = ρv ( ) 10 ρ V large R2 V peak r 2 V hole r 2 ( ) = π 10 ρ R4 h large r 4 h peak r 4 h hole ( ) = π 10 ρ R4 h large r 4 h peak + h hole V = 1 3 πr 2 h

4 ρ = 200 kg / 3 ( ) = π 10 ρ R4 h large r 4 h peak + h hole = π ( 200 kg / 3) kg 2 ( ) ( ) Rod: 3 kg/ Plate with hole: 12 kg/ 2 Find IG (about an axis noral to the plane of the figure). First, find G. Second, copute the MOI. Into how any parts should we divide the structure? Three pieces: two rods and one plate

5 Moent of Inertia 2 Although the FMM is a vector, the MOI is not. Mass and the square of a distance are both scalars. The diensions of the MOI are ass ties the square of length. Typical units are slug ties foot squared or kilogra ties eter squared. 3 In these integrals, each eleent of ass, d, is ultiplied by the square of its distance fro the axis about which the MOI is coputed. Results of these integrals depend on the shape of the body and the distribution of the ass within it. For a body with what is called spherical syetry, all three of the MOI are equal. A unifor sphere and a unifor cube are both aong the bodies that have spherical syetry. 4 Coputing the MOI using the integrals on the previous slide can be tedious. But any books contain tables showing the MOI for various three-diensional bodies. The MOI are typically given about three perpendicular axes that pass through the bodies ass centers. The transfer theore (aka the parallel axis theore) akes it easy to use the result fro a table to copute the MOI about an axis that does not pass through a body s ass center but is parallel to an axis for which the MOI is given in the table. 5 You should eorize this equation. And you should understand that it can be used between any two parallel axes, provided that one of the passes through the ass center of the body in question. Page 1 of 4

6 6 You should eorize the fact that the MOI about a given axis through the ass center is the sallest MOI that can be coputed for any axis parallel to the given axis. 7 This is siply a rearrangeent of the equation on slide 5, showing how to use the transfer theore to copute the MOI about an axis through the ass center when we already know the MOI about a parallel axis that does not pass through the ass center. 8 The position of a body s ass center is where we would find all of the body s ass if we considered the body to be a particle. In a siilar way, the radius of gyration is where we would find all of the body s ass if we considered the body to be an infinitely thin cylinder with radius k. For the MOI about the x axis, the cylinder would be parallel to and centered on the x axis. Since all of the cylinder s ass would be at the sae distance fro the axis, the squared distance in the first integral on slide 3 is the radius of gyration and can be taken outside the integral. That leads to the final equation on this slide. If we have any two of the quantities ass, radius of gyration and MOI, we can copute the third using this definition. 9 Because a MOI is an integral, the techniques used to locate a body s ass center or to copute the FMM for a body ade up of several siple shapes can also be applied to copute the body s MOI. The MOI of a hole can be subtracted. But adding and subtracting are correct only when all of the MOI of the body s pieces are coputed about the sae axis. Page 2 of 4

7 10 We cannot account for the exact shape of the region where the rods intersect, because we have no inforation about that. Two equations are needed for this calculation. One is the MOI for a thin, unifor rod about an axis perpendicular to the rod and passing through the rod s ass center. The other equation is the parallel axis theore. Each rod s MOI is coputed using the two equations, and the two MOI are added. Note that the ass of each rod is its weight per unit length divided by g and ultiplied by L. In the final equation, we ust expand the lb unit to slug-ft/sec 2 before siplifying. 11 A siple calculation shows that the conical hole is not the sae size as the issing peak of the larger cone. Forulas for a right, circular cone can be found in books or at web sites. In the final equation, the radius of the base of the large cone is denoted by R, and the radius of the bases of the issing peak and hole is denoted by r. It reains to substitute for the asses and volues, siplify and copute. These operations are shown on the next two slides. 12 The final equation shows that each ter is proportional to the fourth power of a radius. 13 This is the nuerical result for the final equation fro the previous slide. Page 3 of 4

8 14 For any coposite shapes, it is necessary to find the ass center before coputing the MOI. If the ass center s location were known, it would be a siple atter to transfer the MOI for each piece of the body to the ass center. But in transferring the MOI for each piece to the body s overall ass center, the transfer distance depends on the location of the ass center. Because the location is not given, we ust first find it. That requires that we copute the FMM of each piece relative to a coon point, add the FMM and divide by the total ass. Page 4 of 4