6161103 10.9 mass moment of inertiaPresentation Transcript
10.9 Mass Moment of Inertia Mass moment of inertia of a body is the property that measures the resistance of the body to angular acceleration Mass moment of inertia is defined as the integral of the second moment about an axis of all the elements of mass dm which compose the bodyExample Consider rigid body
10.9 Mass Moment of InertiaFor body’s moment of inertia about the z axis, I = ∫ r 2 dm mHere, the moment arm r is the perpendiculardistance from the axis to the arbitrary elementdmSince the formulation involves r, the value of I isunique for each axis z about which it iscomputedThe axis that is generally chosen for analysis,passes through the body’s mass center G
10.9 Mass Moment of InertiaMoment of inertia computed about this axiswill be defined as IGMass moment of inertia is always positiveIf the body consists of material having avariable density ρ = ρ(x, y, z), the elementmass dm of the body may be expressed asdm = ρ dVUsing volume element for integration, I = ∫ r 2 ρdV V
10.9 Mass Moment of InertiaIn the special case of ρ being a constant, I = ρ ∫ r 2 dV VWhen element volume chosen for integration hasdifferential sizes in all 3 directions, dV = dx dy dzMoment of inertia of the body determined bytriple integrationSimplify the process to single integrationby choosing an element volume witha differential size or thickness in 1direction such as shell or disk elements
10.9 Mass Moment of InertiaProcedure for Analysis Consider only symmetric bodies having surfaces which are generated by revolving a curve about an axisShell Element For a shell element having height z, radius y and thickness dy, volume dV = (2πy)(z)dy
10.9 Mass Moment of InertiaProcedure for AnalysisShell Element Use this element to determine the moment of inertia Iz of the body about the z axis since the entire element, due to its thinness, lies at the same perpendicular distance r = y from the z axisDisk Element For disk element having radius y, thickness dz, volume dV = (πy2) dz
10.9 Mass Moment of InertiaProcedure for AnalysisDisk Element Element is finite in the radial direction and consequently, its parts do not lie at the same radial distance r from the z axis To perform integration using this element, determine the moment of inertia of the element about the z axis and then integrate this result
10.9 Mass Moment of InertiaExample 10.11Determine the mass moment of inertia of thecylinder about the z axis. The density of thematerial is constant.
10.9 Mass Moment of InertiaSolutionShell Element For volume of the element, dV = (2πr )(h )dr For mass, dm = ρdV = ρ (2πrh dr ) Since the entire element lies at the same distance r from the z axis, for the moment of inertia of the element, dI z = r 2 dm = ρ 2πhr 3
10.9 Mass Moment of InertiaSolution Integrating over entire region of the cylinder, R ρπ 4 I z = ∫ r dm = ρ 2πh ∫ r dr = 2 3 R h m 0 2 For the mass of the cylinder R m = ∫ dm = ρ 2πh ∫ rdr = ρπhR 2 m 0 So that 1 I z = mR 2 2
10.9 Mass Moment of InertiaExample 10.12A solid is formed by revolving the shaded areaabout the y axis. If the density of the material is5 Mg/m3, determine the mass moment of inertiaabout the y axis.
10.9 Mass Moment of InertiaSolutionDisk Element Element intersects the curve at the arbitrary point (x, y) and has a mass dm = ρ dV = ρ (πx2)dy Although all portions of the element are not located at the same distance from the y axis, it is still possible to determine the moment of inertia dIy about the y axis
10.9 Mass Moment of InertiaSolution In the previous example, it is shown that the moment of inertia for a cylinder is I = ½ mR2 Since the height of the cylinder is not involved, apply the about equation for a disk dI y = 1 2 1 [( ) ] (dm) x 2 = ρ πx 2 dy x 2 2 For moment of inertia for the entire solid, 5π 1 5π 1 Iy = 2 ∫ 0 x 4 dy = 2 ∫ 0 y 8 dy = 0.873Mg.m 2 = 873kg.m 2
10.9 Mass Moment of InertiaParallel Axis Theorem If the moment of inertia of the body about an axis passing through the body’s mass center is known, the moment of inertia about any other parallel axis may be determined by using parallel axis theorem Considering the body where the z’ axis passes through the mass center G, whereas the corresponding parallel z axis lie at a constant distance d away
10.9 Mass Moment of InertiaParallel Axis Theorem Selecting the differential mass element dm, which is located at point (x’, y’) and using Pythagorean theorem, r 2 = (d + x’)2 + y’2 For moment of inertia of body about the z axis, m m [ 2 ] I = ∫ r 2 dm = ∫ (d + x) + y 2 dm m ( 2 2 ) = ∫ x + y dm + 2d ∫ x dm + d m 2 ∫ dm m First integral represent IG
10.9 Mass Moment of InertiaParallel Axis Theorem Second integral = 0 since the z’ axis passes through the body’s center of mass Third integral represents the total mass m of the body For moment of inertia about the z axis, I = IG + md2
10.9 Mass Moment of InertiaRadius of Gyration For moment of inertia expressed using k, radius of gyration, I I = mk 2 or k= m Note the similarity between the definition of k in this formulae and r in the equation dI = r2 dm which defines the moment of inertia of an elemental mass dm of the body about an axis
10.9 Mass Moment of InertiaComposite Bodies If a body is constructed from a number of simple shapes such as disks, spheres, and rods, the moment of inertia of the body about any axis z can be determined by adding algebraically the moments of inertia of all the composite shapes computed about the z axis Parallel axis theorem is needed if the center of mass of each composite part does not lie on the z axis
10.9 Mass Moment of InertiaExample 10.13If the plate has a density of 8000kg/m3 and athickness of 10mm, determine its mass moment ofinertia about an axis perpendicular to the page andpassing through point O.
10.9 Mass Moment of InertiaSolution The plate consists of 2 composite parts, the 250mm radius disk minus the 125mm radius disk Moment of inertia about O is determined by computing the moment of inertia of each of these parts about O and then algebraically adding the results
10.9 Mass Moment of InertiaSolutionDisk For moment of inertia of a disk about an axis perpendicular to the plane of the disk, 1 2 I G = mr 2 Mass center of the disk is located 0.25m from point O md = ρ dVd = 8000[π (0.25)2 (0.01)] = 15.71kg (I O )d = 1 md rd2 + md d 2 2 1 = (15.71)(0.25)2 + (15.71)(0.25)2 = 1.473kg.m 2 2
10.9 Mass Moment of InertiaSolutionHole [ ] mh = ρ hVh = 8000 π (0.125) (0.01) = 3.93kg 2 1 (I O )h = mh rh2 + mh d 2 2 1 = (3.93)(0.125)2 + (3.93)(0.25)2 = 0.276kg.m 2 2 For moment of inertia of plate about point O, I O = (I O )d − (I O )h = 1.473 − 0.276 = 1.20kg.m 2
10.9 Mass Moment of InertiaExample 10.14The pendulum consists of twothin robs each having a mass of100kg. Determine thependulum’s mass moment ofinertia about an axis passingthrough (a) the pin at point O,and (b) the mass center G of thependulum.
10.9 Mass Moment of InertiaSolutionPart (a) For moment of inertia of rod OA about an axis perpendicular to the page and passing through the end point O of the rob, 1 IO = ml 2 3 Hence, 1 1 (IOA )O = ml 2 = (100)(3)2 = 300kg.m2 3 3 Using parallel axis theorem, 1 2 IG = ml 12 (IOA )O = 1 ml 2 + md 2 = 1 (100)(3)2 + (100)(1.5)2 = 300kg.m2 12 12
10.9 Mass Moment of InertiaSolution For rod BC, 1 1 (I BC )O = ml 2 + md 2 = (100)(3) + (100)(3) 2 2 12 12 = 975kg.m 2 For moment of inertia of pendulum about O, I O = 300 + 975 = 1275kg.m 2
10.9 Mass Moment of InertiaSolutionPart (b) Mass center G will be located relative to pin at O For mass center, ∑ ~m 1.5m(100kg ) + 3m(100kg ) y y= = = 2.25m ∑m 100kg + 100kg Mass of inertia IG may be computed in the same manner as IO, which requires successive applications of the parallel axis theorem in order to transfer the moments of inertias of rod OA and BC to G
10.9 Mass Moment of InertiaSolution Apply the parallel axis theorem for IO, I O = I G + md 2 ; 125kg.m 2 = I G + (200)(2.25) 2 I G = 262.5kg.m 2