Chapter 2 Centroids and Moments of Inertia

Transcription

1 Chpter Centroids nd Moments of Inerti.1 Centroids nd Center of Mss.1.1 First Moment nd Centroid of Set of Points The position vector of point P reltive to point is r P nd sclr ssocited with P is s, for emple, the mss m of prticle situted t P. Thefirst moment of point P with respect to point is the vector M = sr P. The sclr s is clled the strength of P.Thesetofn points P i, i = 1,,...,n,is{S},Fig..1 {S} = {P 1, P,...,P n } = {P i } i=1,,...,n. The strengths of the points P i re s i, i = 1,,...,n, thtis,n sclrs, ll hving the sme dimensions, nd ech ssocited with one of the points of {S}. The centroid of the set {S} is the point C with respect to which the sum of the first moments of the points of {S} is equl to zero. The centroid is the point defining the geometric center of the sstem or of n object. The position vector of C reltive to n rbitrril selected reference point is r C,Fig..1b. The position vector of P i reltive to is r i. The position vector of P i reltive to C is r i r C. The sum of the first moments of the points P i with respect to C is n i=1 s i(r i r C ).IfC is to be centroid of {S}, this sum is equl to zero: n i=1 n n s i (r i r C )= i r i r C s i =. i=1s i=1 D.B. Mrghitu nd M. Dupc, dvnced Dnmics: nlticl nd Numericl Clcultions with MTLB, DI 1.17/ , Springer Science+Business Medi, LLC 1 73

2 74 Centroids nd Moments of Inerti P 1 (s 1 ) P (s ) P i (s i ) {S} b r C C r i P i (s i ) {S} P n (s n ) Fig..1 () Set of points nd (b) centroid of set of points The position vector r C of the centroid C, reltive to n rbitrril selected reference point, isgivenb r C = n i=1 n i=1 If n i=1 s i =, the centroid is not defined. The centroid C of set of points of given strength is unique point, its loction being independent of the choice of reference point. The Crtesin coordintes of the centroid C( C, C, z C ) of set of points P i, i = 1,...,n, of strengths s i, i = 1,...,n, re given b the epressions C = n i=1 n i=1 s i i s i, C = n s i r i i=1 n i=1 s i. s i i s i, z C = n s i z i i=1 n i=1 The plne of smmetr of set is the plne where the centroid of the set lies, the points of the set being rrnged in such w tht corresponding to ever point on one side of the plne of smmetr there eists point of equl strength on the other side, the two points being equidistnt from the plne. set{s } of points is clled subset of set {S} if ever point of {S } is point of {S}. The centroid of set {S} m be locted using the method of decomposition: Divide the sstem {S} into subsets. Find the centroid of ech subset. ssign to ech centroid of subset strength proportionl to the sum of the strengths of the points of the corresponding subset. Determine the centroid of this set of centroids. s i.

3 .1 Centroids nd Center of Mss Centroid of Curve, Surfce, or Solid The position vector of the centroid C of curve, surfce, or solid reltive to point is r dτ τ r C =, (.1) dτ τ where τ is curve, surfce, or solid; r denotes the position vector of tpicl point of τ, reltive to ; nddτ is the length, re, or volume of differentil element of τ. Ech of the two limits in this epression is clled n integrl over the domin τ (curve, surfce, or solid). The integrl τ dτ gives the totl length, re, or volume of τ, thtis, dτ = τ. τ The position vector of the centroid is r C = 1 τ r dτ. τ Let ı, j, k be mutull perpendiculr unit vectors (Crtesin reference frme) with the origin t. The coordintes of C re C, C, z C nd It results tht r C = C ı + C j + z C k. C = 1 dτ, C = 1 dτ, z C = 1 z dτ. (.) τ τ τ τ τ τ The coordintes for the centroid of curve L, Fig.., is determined b using three sclr equtions C = L dl dl zdl L, C =, z C = L. (.3) dl dl dl L L L

4 76 Centroids nd Moments of Inerti C r dl L d C dv C V z Fig.. Centroid of curve L, n re, nd volume V The centroid of n re,fig..,is d d zd C =, C =, z C =, (.4) d d d nd similrl, the centroid of volume V,Fig..,is dv dv zdv V C = V, C =, z C = V. (.5) dv dv dv V V V For curved line in the plne, the centroidl position is given b C = dl L, C = dl L, (.6) where L is the length of the line. Note tht the centroid C is not generll locted long the line. curve mde up of simple curves is considered. For ech simple curve, the centroid is known. The line segment, L i, hs the centroid C i with coordintes Ci, Ci, i = 1,...,n. For the entire curve, C = n i=1 Ci L i, C = L n i=1 Ci L i, where L = L n i=1 L i.

5 .1 Centroids nd Center of Mss 77 Fig..3 Mss center position vector P n z P m C m n r r C r n r 1 P 1 m Mss Center of Set of Prticles The mss center of set of prticles {S} = {P 1,P,...,P n } = {P i } i=1,,...,n is the centroid of the set of points t which the prticles re situted with the strength of ech point being tken equl to the mss of the corresponding prticle, s i = m i, i = 1,,...,n. For the sstem of n prticles in Fig..3, one cn write ( n i=1m i ) nd the mss center position vector is r C = where M is the totl mss of the sstem. r C = n i=1 n i=1 m i r i m i r i, M, (.7).1.4 Mss Center of Curve, Surfce, or Solid To stud problems concerned with the motion of mtter under the influence of forces, tht is, dnmics, it is necessr to locte the mss center. The position vector of the mss center C of continuous bod τ, curve, surfce, or solid, reltive to point is rρ dτ τ r C = = 1 rρ dτ, (.8) ρ dτ m τ τ

6 78 Centroids nd Moments of Inerti or using the orthogonl Crtesin coordintes C = 1 ρ dτ, C = 1 ρ dτ, z C = 1 zρ dτ, m τ m τ m τ where ρ is the mss densit of the bod: mss per unit of length if τ is curve, mss per unit re if τ is surfce, nd mss per unit of volume if τ is solid; r is the position vector of tpicl point of τ, reltive to ;dτ is the length, re, or volume of differentil element of τ; m = τ ρ dτ is the totl mss of the bod; nd C, C, z C re the coordintes of C. If the mss densit ρ of bod is the sme t ll points of the bod, ρ = constnt, the densit, s well s the bod, re sid to be uniform. The mss center of uniform bod coincides with the centroid of the figure occupied b the bod. The densit ρ of bod is its mss per unit volume. The mss of differentil element of volume dv is dm = ρ dv. Ifρ is not constnt throughout the bod nd cn be epressed s function of the coordintes of the bod, then ρdv ρdv zρdv τ C = τ, C =, z C = τ. (.9) ρdv ρdv ρdv τ τ τ The centroid of volume defines the point t which the totl moment of volume is zero. Similrl, the center of mss of bod is the point t which the totl moment of the bod s mss bout tht point is zero. The method of decomposition cn be used to locte the mss center of continuous bod: Divide the bod into number of simpler bod shpes, which m be prticles, curves, surfces, or solids; holes re considered s pieces with negtive size, mss, or weight. Locte the coordintes Ci, Ci, z Ci of the mss center of ech prt of the bod. Determine the mss center using the equtions C = n i=1 n i=1 dτ τ dτ τ, C = n i=1 n i=1 dτ τ dτ τ, z C = n i=1 n i=1 zdτ τ dτ τ, (.1) where τ is curve, re, or volume, depending on the centroid tht is required. Eqution (.1) cn be simplif s C = n i=1 n i=1 Ci τ i τ i, C = n i=1 n i=1 Ci τ i τ i, z C = n i=1 n i=1 z Ci τ i τ i, (.11) where τ i is the length, re, or volume of the ith object, depending on the tpe of centroid.

7 .1 Centroids nd Center of Mss 79 Fig..4 Plnr surfce of re d Fig..5 Centroid nd centroidl coordintes for plnr surfce C centroid C.1.5 First Moment of n re plnr surfce of re nd reference frme in the plne of the surfce re shown in Fig..4. The first moment of re bout the -is is M = d, (.1) nd the first moment bout the -is is M = d. (.13) The first moment of re gives informtion of the shpe, size, nd orienttion of the re. Theentirere cn be concentrted t position C( C, C ), the centroid, Fig..5. The coordintes C nd C re the centroidl coordintes. To compute the centroidl coordintes, one cn equte the moments of the distributed re with tht of the concentrted re bout both es d C = d, = C = = M, (.14) C = d, = C = d = M. (.15) The loction of the centroid of n re is independent of the reference es emploed, tht is, the centroid is propert onl of the re itself.

8 8 Centroids nd Moments of Inerti Fig..6 Plne re with is of smmetr d d C is of smmetr If the es hve their origin t the centroid, C, then these es re clled centroidl es. The first moments bout centroidl es re zero. ll es going through the centroid of n re re clled centroidl es for tht re, nd the first moments of n re bout n of its centroidl es re zero. The perpendiculr distnce from the centroid to the centroidl is must be zero. Finding the centroid of bod is gretl simplified when the bod hs is of smmetr. Figure.6 shows plne re with the is of smmetr colliner with the is. The re cn be considered s composed of re elements in smmetric pirs such s shown in Fig..6. The first moment of such pir bout the is of smmetr is zero. The entire re cn be considered s composed of such smmetric pirs nd the coordinte C is zero C = 1 d =. Thus, the centroid of n re with one is of smmetr must lie long the is of smmetr. The is of smmetr then is centroidl is, which is nother indiction tht the first moment of re must be zero bout the is of smmetr. With two orthogonl es of smmetr, the centroid must lie t the intersection of these es. For such res s circles nd rectngles, the centroid is esil determined b inspection. If bod hs single plne of smmetr, then the centroid is locted somewhere on tht plne. If bod hs more thn one plne of smmetr, then the centroid is locted t the intersection of the plnes. In mn problems, the re of interest cn be considered formed b the ddition or subtrction of simple res. For simple res, the centroids re known b inspection. The res mde up of such simple res re composite res. For composite res C = i Ci i nd C = i Ci i, (.16)

9 .1 Centroids nd Center of Mss 81 w F R w() L Fig..7 Distributed lod w F = 1 w (b ) w (b ) 3 b 1 (b ) 3 Fig..8 Tringulr distributed lod where Ci nd Ci re the centroidl coordintes to simple re i (with proper signs) nd is the totl re. The centroid concept cn be used to determine the simplest resultnt of distributedloding. In Fig..7, the distributedlod w() is considered. The resultnt force F R of the distributed lod w() loding is given s L F R = w() d. (.17) From the eqution bove, the resultnt force equls the re under the loding curve. The position, C,ofthesimplest resultnt lod cn be clculted from the reltion L F R C = w() d = C = L w() d F R. (.18) The position C is ctull the centroidl coordinte of the loding curve re. Thus, the simplest resultnt force of distributed lod cts t the centroid of the re under the loding curve. For the tringulr distributed lod shown in Fig..8, one cn replce the distributed loding b force F equl to ( 1 ) (w )(b ) t position ( 1 3) (b ) from the right end of the distributed loding.

10 8 Centroids nd Moments of Inerti.1.6 Center of Grvit The center of grvit is point which loctes the resultnt weight of sstem of prticles or bod. The sum of moments due to individul prticle weight bout n point is the sme s the moment due to the resultnt weight locted t the center of grvit. The sum of moments due to the individul prticles weights bout center of grvit is equl to zero. Similrl, the center of mss is point which loctes the resultnt mss of sstem of prticles or bod. The center of grvit of bod is the point t which the totl moment of the force of grvit is zero. The coordintes for the center of grvit of bod cn be determined with ρ gdv ρ gdv zρ gdv V C = V, C =, z C = V. (.19) ρ gdv ρ gdv ρ gdv V V The ccelertion of grvit is g, g = 9.81 m/s or g = 3. ft/s.ifg is constnt throughout the bod, then the loction of the center of grvit is the sme s tht of the center of mss. V.1.7 Theorems of Guldinus Pppus The theorems of Guldinus Pppus re concerned with the reltion of surfce of revolution to its generting curve, nd the reltion of volume of revolution to its generting re. Theorem.1. Consider coplnr generting curve nd n is of revolution in the plne of this curve in Fig..9. The surfce of revolution developed b rotting the generting curve bout the is of revolution equls the product of the length of the generting L curve times the circumference of the circle formed b the centroid of the generting curve C in the process of generting surfce of revolution = π C L. (.) The generting curve cn touch but must not cross the is of revolution. Proof. n element dl of the generting curve is considered in Fig..9. For single revolutionof the genertingcurve bout the -is, the line segment dl trces n re d = π dl. For the entire curve, this re, d, becomes the surfce of revolution,, given s = π dl = π C L,

11 .1 Centroids nd Center of Mss 83 L dl Generting curve C C is of revolution Fig..9 Surfce of revolution developed b rotting the generting curve bout the is of revolution L 1 C 1 C L ( C, C ) L 4 C L C 3 C 4 3 C1 C1 Fig..1 Composed generting curve where L is the length of the curve nd C is the centroidl coordinte of the curve. The circumferentil length of the circle formed b hving the centroid of the curve rotte bout the -is is π C, q.e.d. The surfce of revolution is equl to π times the first moment of the generting curve bout the is of revolution. If the generting curve is composed of simple curves, L i, whose centroids re known, Fig..1, the surfce of revolution developed b revolving the composed generting curve bout the is of revolution is = π ( 4 i=1l i Ci ), (.1) where Ci is the centroidl coordinte to the ith line segment L i.

12 84 Centroids nd Moments of Inerti Generting plne surfce d C is of revolution C d Fig..11 Volume of revolution developed b rotting the generting plne surfce bout the is of revolution Theorem.. Consider generting plne surfce nd n is of revolution coplnr with the surfce Fig..11. The volume of revolution V developed b rotting the generting plne surfce bout the is of revolution equls the product of the re of the surfce times the circumference of the circle formed b the centroid of the surfce C in the process of generting the bod of revolution V = π C. (.) The is of revolution cn intersect the generting plne surfce onl s tngent t the boundr or hve no intersection t ll. Proof. The plne surfce is shown in Fig..11. The volume generted b rotting n element d of this surfce bout the -is is dv = π d. The volume of the bod of revolution formed from is then V = π d = π C. Thus, the volume V equls the re of the generting surfce times the circumferentil length of the circle of rdius C, q.e.d. The volume V equls π times the first moment of the generting re bout the is of revolution.

13 . Moments of Inerti 85. Moments of Inerti..1 Introduction sstemofn prticle P i, i = 1,,...,n is considered. The mss of the prticle P i is m i s shown in Fig..1. The position vector of the prticle P i is r i = i ı + i j + z i k. The moments of inerti of the sstem bout the plnes, z, ndz re I = m i z i, i I z = m i i, i I z = m i i. (.3) i The moments of inerti of the sstem bout,, ndz es re ), I = = m i ( i + z i i I = B = m i (z i + i i I zz = C = m i ( i + i i The moment of inerti of the sstem bout the origin is I = m i ( i + i + z i i ), ). (.4) ). (.5) The products of inerti of the sstem bout the es, z, ndz re I z = D = m i i z i, i I z = E = m i z i i, i I = F = m i i i. (.6) i z i z r i m 1 m P 1 ( ) P ( ) P i (m i ) i Fig..1 Prticle P i with the mss m i i

14 86 Centroids nd Moments of Inerti Fig..13 Rigid bod in spce with mss m nd differentil volume dv z dv r Between the different moments of inerti, one cn write the reltions I = I + I z + I z = 1 (I + I + I zz ), nd I = I z + I z. For continuous domin D, the previous reltions become I = z dm, I z = dm, I z = dm, D D D ( I = + z ) ( dm, I = + z ) ( dm, I zz = + ) dm, D D D ( I = + + z ) dm, D I = dm, I z = zdm, I z = zdm. (.7) D D D The infinitesiml mss element dm cn hve the vlues dm = ρ v dv, dm = ρ d, dm = ρ l dl, where ρ v, ρ,ndρ l re the volume densit, re densit, nd length densit. Moment of Inerti bout n rbitrr is For rigid bod with mss m, densit ρ, ndvolumev, s shown in Fig..13, the moments of inerti re defined s follows: I = ρ ( + z ) dv, I = ρ ( z + ) dv, I zz = ρ ( + ) dv, (.8) V V V

15 . Moments of Inerti 87 Fig..14 Rigid bod nd n rbitrr is Δ of unit vector u Δ dm z r d θ Δ u Δ nd the products of inerti I = I = ρ dv, I z = I z = ρ zdv, I z = I z = ρ zdv. (.9) V V V The moment of inerti given in (.8) is just the second moment of the mss distribution with respect to Crtesin is. For emple, I is the integrl of summtion of the infinitesiml mss elements ρ dv, ech multiplied b the squre of its distnce from the -is. The effective vlue of this distnce for certin bod is known s its rdius of grtion with respect to the given is. The rdius of grtion corresponding to I jj is defined s Ijj k j = m, where m is the totl mss of the rigid bod nd where the smbol j cn be replced b, or z. Theinerti mtri of rigid bod is represented b the mtri [I]= I I I z I I I z I z I z I zz. Moment of Inerti bout n rbitrr is Consider the rigid bod shown in Fig..14. The reference frme,, z hs the origin t. The direction of n rbitrr is Δ through is defined b the unit vector u Δ u Δ = cosαı + cosβ J + cosγk, where cosα, cosβ,cosγ re the direction cosines. The moment of inerti bout the Δ is, for differentil mss element dm of the bod, is b definition I Δ = d dm, D

16 88 Centroids nd Moments of Inerti where d is the perpendiculr distnce from dm to Δ. The position of the mss element dm is locted using the position vector r nd then d = r sinθ, which represents the mgnitude of the cross product u Δ r. The moment of inerti cn be epressed s I Δ = u Δ r dm = (u Δ r) (u Δ r)dm. D D If the position vector is r =ı + J + zk,then u Δ r =(zcosβ cosγ)ı +(cosγ zcosα)j +(cosα cosβ )k. fter substituting nd performing the dot-product opertion, one cn write the moment of inerti s [ I Δ = (zcosβ cosγ) +(cosγ zcosα) +(cosα cosβ ) ] dm D = cos ( α + z ) dm + cos ( β z + ) dm + cos ( γ + ) dm D D D cosα cosβ dm cosβ cosγ zdm cosγ cosα zdm. D The moment of inerti with respect to the Δ is is I Δ = I cos α + I cos β + I zz cos γ D I cosα cosβ I z cosβ cosγ I z cosγ cosα. (.3) D.. Trnsltion of Coordinte es The defining equtions for the moments nd products of inerti, s given b (.8) nd(.9), do not require tht the origin of the Crtesin coordinte sstem be tken t the mss center. Net, one cn clculte the moments nd products of inerti for given bod with respect to set of prllel es tht do not pss through the mss center. Consider the bod shown in Fig..15. The mss center is locted t the origin C of the primed sstem z. The coordinte of with respect to the unprimed sstem z is ( c, c,z c ). n infinitesiml volume element dv is locted t (,,z) in the unprimed sstem nd t (,,z ) in the primed sstem. These coordintes re relted b the equtions = + c, = + c, z = z + z c. (.31)

17 . Moments of Inerti 89 Fig..15 Rigid bod nd centroidl es z : = + c, = + c, z = z + z c z z dv Centroid ( c, c, z c ) C z The moment of inerti bout the -is cn be written in terms of primed coordintes b using (.8)nd(.31) [ ( I = ρ ) ( + c + z ) ] + z c dv V = I C + c V ρ dv + z c V ρz dv + m ( c + z c), (.3) where m is the totl mss of the rigid bod, nd the origin of the primed coordinte sstem ws chosen t the mss center. ne cn write V ρ dv = V ρ dv = V ρz dv =, (.33) nd therefore, the two integrls on the right-hnd side of (.3) re zero. In similr w, one cn obtin I nd I. The results re summrized s follows: I = I C + m( c + z c), I = I C + m( c + z c), I zz = I Cz z + m( c + c), (.34) or, in generl, I kk = I Ck k + md, (.35) where d is the distnce between given unprimed is nd prllel primed is pssing through the mss center C. Eqution (.35) represents the prllel es

18 9 Centroids nd Moments of Inerti Fig..16 Rottion of coordinte es z z dv r Theorem. The products of inerti re obtined in similr mnner, using (.9)nd(.31) I = ρ ( )( + c ) + c dv V = I C + c V ρ dv + c V ρ dv + m c c. The two integrls on the previous eqution re zero. The other products of inerti cn be clculted in similr mnner, nd the results cn be written s follows: I = I C + m c c, I z = I C z + m c z c, I z = I C z + m c z c. (.36) Equtions (.34) nd(.36) shows tht trnsltion of es w from the mss center results in n increse in the moments of inerti. The products of inerti m increse or decrese, depending upon the prticulr cse...3 Principl es Net, the chnges in the moments nd product of inerti of rigid bod due to rottion of coordinte es re considered, s shown in Fig..16. The origin of the coordinte es is locted t the fied point. In generl, the origin is not the mss center C of the rigid bod. From the definitions of the moments of inerti given in (.8), it results tht the moments of inerti cnnot be negtive. Furthermore, I + I + I zz = ρr dv, (.37) V

19 . Moments of Inerti 91 where r is the squre of the distnce from the origin, r = + + z. The distnce r corresponding to n mss element ρdv of the rigid bod does not chnge with rottion of es from z to z (Fig..16). Therefore, the sum of the moments of inerti is invrint with respect to coordinte sstem rottion. In terms of mtri nottion, the sum of the moments of inerti is just the sum of the elements on the principl digonl of the inerti mtri nd is known s the trce of tht mtri. So the trce of the inerti mtri is unchnged b coordinte rottion becuse the trce of n squre mtri is invrint under n orthogonl trnsformtion. Net, the products of inerti re considered. coordinte rottion of es cn result in chnge in the signs of the products of inerti. 18 rottion bout the -is, for emple, reverses the signs of I nd I z, while the sign of I z is unchnged. This occurs becuse the directions of the positive nd z es re reversed. n the other hnd, 9 rottion bout the -is reverses the sign of I z.it cn be seen tht the moments nd products of inerti vr smoothl with chnges in the orienttion of the coordinte sstem becuse the direction cosines vr smoothl. Therefore, n orienttion cn lws be found for which given product of inerti is zero. It is lws possible to find n orienttion of the coordinte sstem reltive to given rigid bod such tht ll products of inerti re zero simultneousl, tht is, the inerti mtri is digonl. The three mutull orthogonl coordinte es re known s principl es in this cse, nd the corresponding moments of inerti re the principl moments of inerti. The three plnes formed b the principl es re clled principl plnes. If I is principl moment of inerti, then I stisfies the cubic chrcteristic eqution I I I I z I I I I z I z I z I zz I =. (.38) Eqution (.38) is used to determine the ssocited principl moments of inerti. Suppose tht u 1, u, u 3 re mutull perpendiculr unit vectors ech prllel to principl is of the rigid bod reltive to. The principl moments of inerti ssocited to u 1, u, u 3 for the rigid bod reltive to re I 1, I,ndI 3. The inerti mtri, in this cse, is I 1 I = I. I 3 When the point under considertion is the mss center of the rigid bod, one speks of centrl principl moments of inerti.

20 9 Centroids nd Moments of Inerti 1 Δ z z 1 1 Fig..17 Ellipsoid of inerti..4 Ellipsoid of Inerti The ellipsoid of inerti for given bod nd reference point is plot of the moment of inerti of the bod for ll possible is orienttions through the reference point. This grph in spce hs the form of n ellipsoid surfce. Consider rigid bod in rottionl motion bout n is Δ. Theellipsoid of inerti with respect to n rbitrr point is the geometricl locus of the points Q,whereQis the etremit of the vector Q with the module Q = 1 nd where I Δ is the moment of inerti IΔ bout the instntneous is of rottion Δ, s shown in Fig..17. The segment Q is clculted with Q = 1 = 1 IΔ k m, where k is the rdius of grtion of the bod bout the given is nd m is the totl mss. For Crtesin sstem of es, the eqution of the ellipsoid surfce centered t is I + I + I zz z + I + I z z+ I z z= 1. (.39) The rdius of grtion of given rigid bod depends upon the loction of the is reltive to the bod nd is not depended upon the position of the bod in spce. The ellipsoid of inerti is fied in the bod nd rottes with it. The 1, 1,ndz 1 es re ssumed to be the principl es of the ellipsoid, s shown in Fig..17.For the principl es, the eqution of the inerti ellipsoid, (.39), tkes the following simple form I I 1 + I 3 z 3 = 1. (.4) The previous eqution is of the sme form s (.39) for the cse where the 1, 1, nd z 1 es re the principl es of the rigid bod, nd ll products of inerti vnish.

21 . Moments of Inerti 93 Fig..18 Moments of inerti for re bout nd es d r Therefore, I 1, I,ndI 3 re the principl moments of inerti of the rigid bod, nd furthermore, the principl es of the bod coincide with those of the ellipsoid of inerti. From (.4), the lengths of the principl semies of the ellipsoid of inerti re l 1 = 1 I1, l = 1 I, l 3 = 1 I3. From the prllel-is theorem, (.35), one cn remrk tht the minimum moment of inerti bout the mss center is lso the smllest possible moment of inerti for the given bod with respect to n reference point. If the point is the sme s the mss center ( C), the ellipsoid is nmed principl ellipsoid of inerti...5 Moments of Inerti for res The moment of inerti (second moment) of the re bout nd es, see Fig..18, denoted s I nd I, respectivel, re I = d, (.41) I = d. (.4) The second moment of re cnnot be negtive. The entire re m be concentrted t single point (k,k ) to give the sme second moment of re for given reference. The distnces k nd k re clled the rdii of grtion. Thus, k = I = d = k = d = I,

22 94 Centroids nd Moments of Inerti Fig..19 re,, with n is of smmetr d C d Centroid d k = I = d = k = = I. (.43) This point (k,k ) depends on the shpe of the re nd on the position of the reference. The centroid loction is independent of the reference position. The product of inerti for n re is defined s I = d. (.44) This quntit m be positive or negtive nd reltes n re directl to set of es. If the re under considertion hs n is of smmetr, the product of re for this is is zero. Consider the re in Fig..19, which is smmetricl bout the verticl is. The plnr Crtesin frme is. The centroid is locted somewhere long the smmetricl is. Two differentil element of res tht re positioned s mirror imges bout the -is re shown in Fig..19. The contribution to the product of re of ech elementl re is d, but with opposite signs, nd so the result is zero. The entire re is composed of such elementl re pirs, nd the product of re is zero. The product of inerti for n re I is zero (I = ) if either the -or-is is n is of smmetr for the re. Trnsfer Theorem or Prllel-is Theorem The -is in Fig.. is prllel to n is, nd it is t distnce b from the is.theis is going through the centroidc of the re, nd it is centroidl is. The second moment of re bout the -is is I = d = ( + b) d,

23 . Moments of Inerti 95 Fig.. re nd centroidl is C d C Centroid b where the distnce = + b. Crring out the opertions I = d + b d + b. The first term of the right-hnd side is b definition I I C = d. The second term involves the first moment of re bout the is, nd it is zero becuse the is is centroidl is d =. The second moment of the re bout n is I is equl to the second moment of the re bout prllel is t centroid I C plus b,whereb is the perpendiculr distnce between the is for which the second moment is being computed nd the prllel centroidl is I = I C + b. With the trnsfer theorem, the second moments or products of re bout n is cn be computed in terms of the second moments or products of re bout prllel set of es going through the centroid of the re in question. In hndbooks, the res nd second moments bout vrious centroidl es re listed for mn of the prcticl configurtions, nd using the prllel-is theorem (or Hugens Steiner theorem), second moments cn be clculted for es not t the centroid. In Fig..1 re shown two references, one t the centroid C nd the other rbitrr but positioned prllel reltive to. The coordintes of the centroid C( C, C ) of re mesured from the reference, re c nd b, C = c, C = b. The centroid coordintes must hve the proper signs. The product of re bout the noncentroidl es is I = d = ( + c)( + b) d,

24 96 Centroids nd Moments of Inerti Fig..1 Centroidl es prllel to reference es : C nd C -( -c) d c C Centroid b -( -b) or I = d + c d + b d + bc. The first term of the right-hnd side is b definition I I = d. The net two terms of the right-hnd side re zero since nd re centroidl es d = nd d =. Thus, the prllel-is theorem for products of re is s follows. The product of re for n set of es I is equl to the product of re for prllel set of es t centroid I C plus cb,wherec nd b re the coordintes of the centroid of re, I = I C + cb. With the trnsfer theorem, the second moments or products of re cn be found bout n is in terms of second moments or products of re bout prllel set of es going through the centroid of the re. PolrMomentofre In Fig..18, there is reference ssocited with the origin. Summing I nd I, I + I = d + d = ( + ) d = r d, where r = +. The distnce r is independent of the orienttion of the reference, nd the sum I + I is independent of the orienttion of the coordinte sstem.

25 . Moments of Inerti 97 Fig.. Reference nd reference rotted with n ngle α d α Therefore, the sum of second moments of re bout orthogonl es is function onl of the position of the origin for the es. The polr moment of re bout the origin is I = I + I. (.45) The polr moment of re is n invrint of the sstem. The group of terms I I I is lso invrint under rottion of es. The polr rdius of grtion is I k =. (.46) Principl es In Fig.., nre is shown with reference hving its origin t. nother reference with the sme origin is rotted with n ngle α from (counterclockwise s positive). The reltions between the coordintes of the re elements d for the two references re = cosα + sinα, = sinα + cosα. The second moment I cn be epressed s I = ( ) d = ( sin α + cosα) d = sin α d sinα cosα d + cos α d = I sin α + I cos α I sinα cosα. (.47) Using the trigonometric identities cos α = 1 + cosα, sin α = 1 cosα, sinα = sinα cosα,

26 98 Centroids nd Moments of Inerti Fig..3 Principl is of re α (.47) becomes I = I + I + I I cosα I sinα. (.48) Replcing α with α + π/in(.48) nd using the trigonometric reltions the second moment I is cos(α + π)= cosα, sin(α + π)= sinα, I = I + I The product of re I is computed in similr mnner I = I I cosα + I sinα. (.49) d = I I sinα + I cosα. (.5) If I, I,ndI re known for reference with n origin, then the second moments nd products of re for ever set of es t cn be computed. Net, it is ssumed tht I, I,ndI re known for reference. The sum of the second moments of re is constnt for n reference with origin t.theminimum second moment of re corresponds to n is t right ngles to the is hving the mimum second moment, s shown in Fig..3. This prticulr set of es if clled principl is of re nd the corresponding moments of inerti with respect to these es re clled principl moments of inerti. The second moments of re cn be epressed s functions of the ngle vrible α. The mimumsecond momentm be determinedb setting the prtil derivtive of I with respect to α equl to zero. Thus, or I α =(I I )( sin α) I cosα =, (.51) (I I )sin α I cosα =,

27 . Moments of Inerti 99 where α is the vlue of α which defines the orienttion of principl es. Hence, tnα = I I I. (.5) The ngle α corresponds to n etreme vlue of I (i.e., to mimum or minimum vlue). There re two roots for α, which re π rdins prt, tht will stisf the previous eqution. Thus, nd α 1 = tn 1 α = tn 1 I I I = α 1 = 1 tn 1 I I I, I + π = α = 1 I I I tn 1 + π I I. This mens tht there re two es orthogonl to ech other hving etreme vlues for the second moment of re t. ne of the es is the mimum second moment of re, nd the minimum second moment of re is on the other is. These es re the principl es. With α = α, the product of re I becomes I = I I sinα + I cosα. (.53) For α = α 1, the sine nd cosine epressions re I (I I ) sinα 1 =, cosα 1 =. (I I ) + 4I (I I ) + 4I For α = α, the sine nd cosine epressions re I I I sinα =, cosα =. (I I ) + 4I (I I ) + 4I Eqution (.53)ndα = α 1 give I I = (I I I I )[ ] (I I ) + 4I 1/ + I [ (I I ) + 4I] 1/ =. In similr w, (.53)ndα = α give I =. The product of re corresponding to the principl es is zero.

28 1 Centroids nd Moments of Inerti C M M C d l Fig..4 Emple.1 The mimum or minimum moment of inerti for the re re I 1, = I m,min = I + I ± (I I ) + I. (.54) If I is principl moment of inerti, then I stisfies the qudrtic chrcteristic eqution I I I I I =. (.55) I.3 Emples Emple.1. Find the position of the mss center for nonhomogeneous stright rod, with the length = l (Fig..4). The liner densit ρ of the rod is liner function with ρ = ρ t nd ρ = ρ 1 t. Solution reference frme is selected with the origin t nd the -is long the rectiliner rod (Fig..4). Let M(, ) be n rbitrril given point on the rod, nd let MM be n element of the rod with the length d nd the mss dm = ρ d. The densit ρ is liner function of given b dm = dm L l ρ = ρ()=ρ + ρ 1 ρ. l The center mss of the rod, C, hs c s bsciss. The mss center c, with respect to point,is l l ( ρ + ρ ) 1 ρ d ρ + ρ 1 L l l c = ( ) = 6. ρ d = ρ d l ρ + ρ 1 ρ l d ρ + ρ 1 l

29 .3 Emples 11 Fig..5 Emple. The mss center c is c = ρ + ρ 1 3(ρ + ρ 1 ) l. In the specil cse of homogeneous rod, the densit nd the position of the mss center re given b ρ = ρ 1 nd c = l. The MTLB progrm is sms rho rhol L rho = rho + (rhol - rho)*/l; m = int(rho,,, L); M = simplif(int(*rho,,, L)); C = simplif(m/m); fprintf( m = %s \n, chr(m)) fprintf( M = %s \n, chr(m)) fprintf( C = M/m = %s \n, chr(c)) The MTLB sttement int(f,,, b) is the definite integrl of f with respect to its smbolic vrible from to b. Emple.. Find the position of the centroid for homogeneous circulr rc. The rdius of the rc is R, nd the center ngle is α rdins s shown in Fig..5. Solution The is of smmetr is selected s the -is ( C =). Let MM be differentil element of rc with the length ds = Rdθ. The mss center of the differentil element of rc MM hs the bsciss

30 1 Centroids nd Moments of Inerti ( = Rcos θ + dθ ) R cos(θ). The bsciss C of the centroid for homogeneous circulr rc is clculted from (.8) α α c Rdθ = Rdθ. (.56) α α Becuse = Rcos(θ) nd with (.56), one cn write α α c Rdθ = R cos(θ)dθ. (.57) α α From (5.19), fter integrtion, it results c αr = R sin (α), or c = Rsin(α). α For semicirculr rc when α = π, the position of the centroid is c = R π, nd for the qurter-circulr α = π 4, c = π R. Emple.3. Find the position of the mss center for the re of circulr sector. The center ngle is α rdins, nd the rdius is R s shown in Fig..6. Solution The origin is the verte of the circulr sector. The -is is chosen s the is of smmetr nd C =. Let MNPQ be surfce differentil element with the re d = ρdρdθ. The mss center of the surfce differentil element hs the bsciss ( = ρ + dρ ) ( cos θ + dθ ) ρ cos(θ). (.58) Using the first moment of re formul with respect to, (.15), the mss center bsciss C is clculted s C ρdρdθ = ρdρdθ. (.59)

31 .3 Emples 13 Fig..6 Emple.3 Equtions (.58)nd(.59)give C ρdρdθ = ρ cos(θ)dρdθ, or C R α R α ρdρ dθ = ρ dρ cos(θ)dθ. (.6) α α From (.6), fter integrtion, it results ( ( ) 1 1 C )(α)= R 3 R3 (sin(α)), or C = Rsin(α). (.61) 3α For semicirculr re, α = π,the-coordinte to the centroid is C = 4R 3π. For the qurter-circulr re, α = π,the-coordinte to the centroid is C = 4R 3π.

32 14 Centroids nd Moments of Inerti d = sin() b d = d d = sin() d d d Fig..7 Emple.4 Emple.4. Find the coordintes of the mss center for homogeneous plnr plte locted under the curve of eqution = sin from = to =. Solution verticl differentil element of re d = d =(sin)(d) is chosen s shown in Fig..7. The -coordinte of the mss center is clculted from (.15): c (sin )d = (sin )d or c { cos } = c (1 cos )= The integrl (sin )d is clculted with (sin )d, (sin )d. (.6) (sin )d = {( cos )} ( cos )d = {( cos )} + {sin } = sin cos. (.63) or Using (.6)nd (.63) fter integrtion, it results sin cos c =. 1 cos The -coordinte of the mss center, C, cn be clculted using the differentil element of re d = dd, s shown in Fig..7b. The re of the figure is = = dd = sin d {} sin = sin dd = d d The first moment of the re bout the -is is (sin )d = { cos } = 1 cos.

33 .3 Emples 15 M = d = = sin d {} sin = dd = sin d d (sin )d = sin cos. The -coordinte of the mss center is C = M /. The-coordinte of the mss center is C = M /, where the first moment of the re bout the -is is sin sin M = d = dd = d d { } sin sin = d = d = 1 sin d. The integrl sin d is clculted with sin d = sin d( cos )={sin ( cos )} + cos d or The coordinte C is = sin (cos )+ sin d = C = M (1 sin )d= sin (cos )+ sin cos. = sin cos 4(1 cos). The MTLB progrm is given b sms % d = d d % = int d d ; << <<sin() % = int d ; <<sin() =int(1,,,sin()) % = int d ; ; << =int(,,,) % M = int d d ; << <<sin() % Q = int d ; <<sin() Q=int(1,,,sin()) % M = int Q d ; << M=int(*Q,,,) sin d,

34 16 Centroids nd Moments of Inerti Fig..8 Emple.5 % M = int d d ; << <<sin() % Q = int d ; <<sin() % M = int Q d ; << Q=int(,,,sin()) M=int(Q,,,) C=M/; C=M/; prett(c) prett(c) Emple.5. Find the position of the mss center for homogeneous plnr plte ( = 1 m), with the shpe nd dimensions given in Fig..8. Solution The plte is composed of four elements: the circulr sector re 1, the tringle, the squre 3, nd the circulr re 4 to be subtrcted. Using the decomposition method, the positions of the mss center i, the res i, nd the first moments with respect to the es of the reference frme M i nd M i, for ll four elements re clculted. The results re given in the following tble: i i i i M i = i i M i = i i Circulr sector 1 8 π 18π Tringle Squre Circulr sector π π 9π 54π π (π + 6) 18(8 3π) 3 9(8 6π) 3

35 .3 Emples 17 The nd coordintes of the mss center C re c = i i i = c = i i i The MTLB progrm is sms (8 3π) =.311 =.311 m, π + 6 = 8 6π = = m. π + 6 1=; 1=-8*/pi; 1=18*pi*ˆ; 11=1*1; 11=1*1; fprintf( 1 1 = %s \n, chr(11)) fprintf( 1 1 = %s \n, chr(11)) =-*; =*; =18*ˆ; =*; =*; fprintf( = %s \n, chr()) fprintf( = %s \n, chr()) 3=3*; 3=3*; 3=36*ˆ; 33=3*3; 33=3*3; fprintf( 3 3 = %s \n, chr(33)) fprintf( 3 3 = %s \n, chr(33)) 4=6*-8*/pi; 4=4; 4=-9*pi*ˆ; 44=simplif(4*4); 44=simplif(4*4); fprintf( 4 4 = %s \n, chr(44)) fprintf( 4 4 = %s \n, chr(44)) C=(1*1+*+3*3+4*4)/(1++3+4); C=simplif(C);

36 18 Centroids nd Moments of Inerti Fig..9 Emple.6 C=(1*1+*+3*3+4*4)/(1++3+4); C=simplif(C); fprintf( C = %s = %s \n, chr(c), chr(vp(c,6))) fprintf( C = %s = %s \n, chr(c), chr(vp(c,6))) Emple.6. The homogeneous plte shown in Fig..9 is delimitted b the htched re. The circle, with the center t C 1, hs the unknown rdius r.thecircle, with the center t C, hs the given rdius = m,(r < ). The position of the mss center of the htched re, C, is locted t the intersection of the circle, with the rdius r nd the center t C 1, nd the positive -is. Find the rdius r. Solution The -is is smmetr is for the plnr plte, nd the origin of the reference frme is locted t C 1 = (, ).The-coordinte of the mss center of the htched re is C =. The coordintes of the mss center of the circle with the center t C nd rdius re C ( r, ). The re of the of circle C 1 is 1 = πr,ndthere of the circle C is = π. The totl re is given b = 1 + = π ( r ). The coordinte C of the mss center is given b C = C C = C = C ( r)π = π ( r )

37 .3 Emples 19 z b + z = R z dv dz R z C R z R Fig..3 Emple.7 or C ( r ) = ( r). If C = r, the previous eqution gives r + r =, with the solutions r = ± 5. Becuse r >, the correct solution is r = ( 5 ) 1.6 =.6()=1.4 m. Emple.7. Locte the position of the mss center of the homogeneous volume of hemisphere of rdius R with respect to its bse, s shown in Fig..3. Solution The reference frme is selected s shown in Fig..3, nd the z-is is the smmetr is for the bod: C = nd C =. Using the sphericl coordintes, z = ρ sinϕ, nd the differentil volume element is dv = ρ cosϕdρdθdϕ. Thez coordinte of the mss center is clculted from z C ρ cosϕdρdθdϕ = ρ 3 sin ϕ cosϕdρdθdϕ, or z C R π π/ R π π/ ρ dρ dθ cosϕdϕ = ρ 3 dρ dθ sinϕ cosϕdϕ,

38 11 Centroids nd Moments of Inerti or z C = R R π ρ 3 dρ ρ dρ From (.64), fter integrtion, it results π π/ dθ π/ dθ z C = 3R 8. sin ϕ cosϕdϕ. (.64) cosϕdϕ nother w of clculting the position of the mss center z C is shown in Fig..3b. The differentil volume element is V dv = π dz = π (R z )dz, nd the volume of the hemisphere of rdius R is R ( R V = dv= π (R z )dz=π R R ) dz z dz =π The coordinte z C is clculted from the reltion ( R R ) zdv π R zdz z 3 dz V z C = = = π V V V = π R4 4V = π ( ) R4 3 4 π R 3 = 3R 8. sms rho thet phi R rel % dv = rhoˆ cos(phi) drho dthet dphi % <rho<r <thet<pi <phi<pi/ Ir = int(rhoˆ, rho,, R); It = int(1, thet,, *pi); Ip = int(cos(phi), phi,, pi/); V = Ir*It*Ip; fprintf( V = %s \n, chr(v)) (R 3 )= R3 3 ) (R R R4 4 % dmz = rho rhoˆ cos(phi) drho dthet dphi % <rho<r <thet<pi <phi<pi/ Mr = int(rhoˆ3, rho,, R); Mt = int(1, thet,, *pi); Mp = int(sin(phi)*cos(phi), phi,, pi/); Mz = Mr*Mt*Mp; fprintf( Mz = %s \n, chr(mz)) π R3. 3

39 .3 Emples 111 Fig..31 Emple.8 z r dz h z R zc = Mz/V; fprintf( zc = Mz/V = %s \n, chr(zc)) fprintf( nother method \n ) sms z R rel % ˆ + zˆ = Rˆ % dv = pi ˆ dz = pi (Rˆ-zˆ) dz V = int(pi*(rˆ-zˆ), z,, R); fprintf( V = %s \n, chr(v)) M = int(z*pi*(rˆ-zˆ), z,, R); fprintf( M = %s \n, chr(m)) zc = M/V; fprintf( zc = M/V = %s \n, chr(zc)) Emple.8. Find the position of the mss center for homogeneous right circulr cone, with the bse rdius R nd the height h, s shown in Fig..31. Solution The reference frme is shown in Fig..31. Thez-is is the smmetr is for the right circulr cone. B smmetr, C = nd C =. The volume of the thin disk differentil volume element is dv = π r dz. From geometr r R = h z h,

40 11 Centroids nd Moments of Inerti Fig..3 Emple.9 C B r r I P or ( r = R 1 z ). h The z coordinte of the centroid is clculted from or z C πr h ( 1 z h z C = h h ) dz = πr h ( z 1 z ) dz h ( 1 z ) = h 4. dz h The MTLB progrm is sms z h rel V = int((1-z/h)ˆ, z,, h); M = int(z*(1-z/h)ˆ, z,, h); zc=m/v; fprintf( V = %s \n,chr(v)) fprintf( M = %s \n,chr(m)) fprintf( zc = %s \n,chr(zc)) ( z 1 h) z dz, Emple.9. The densit of squre plte with the length is given s ρ = kr, where k = constnt nd r is the distnce from the origin to current point P(, ) on the plte s shown in Fig..3.FindthemssM of the plte. Solution The mss of the plte is given b M = ρ dd.

41 .3 Emples 113 The densit is ρ = k +, (.65) where ρ = kr = k k = ρ. (.66) Using (.65)nd(.66), the densit is ρ = ρ +. (.67) Using (.67), the mss of the plte is M = ρ = ρ ρ d [ [ + d + + ( ln + ) ] + d ( )] + + ln + = + d { ( = ρ + d ) } ln + d { [ = ρ + + ( ln + ) ]} + { [ ( 1 + ρ + ) ln = 1 [ ( ρ + ) ln = ρ 3 [ + ( ln 1 + )]. ln ( + + ) ]} ln ( + + ) ] Emple.1. Revolving the circulr re of rdius R through 36 bout the -is, complete torus is generted. The distnce between the center of the circle nd the -is is d, s shown in Fig..33. Find the surfce re nd the volume of the obtined torus. Solution Using the Guldinus Pppus formuls = π C L,

42 114 Centroids nd Moments of Inerti Fig..33 Emple.1 C R C d Fig..34 Emple.11 C C V = π C, nd with C = d therendthevolumere =(πd)(πr)=4π Rd, V =(πd) ( πr ) = π R d. Emple.11. Find the position of the mss center for the semicirculr re shown in Fig..34. Solution Rotting the semicirculr re with respect to -is, sphere is obtined. The volume of the sphere is given b V = 4πR3 3. The re of the semicirculr re is = πr.

43 .3 Emples 115 Fig..35 Emple.1 R 1 h Using the second Guldinus Pppus theorem, the position of the mss center is C = V π = 4πR 3 3 π πr 4πR 3 = 3 π R = 4R 3π. Emple.1. Find the first moment of the re with respect to the -is for the surfce given in Fig..35. Solution The composite re is considered to be composed of the semicirculr re nd the tringle. The mss center of the semicirculr surfce is given b 1 = 3 Rsin π π = 4R 3π. The mss center of the tringle is given b = b 3.

44 116 Centroids nd Moments of Inerti Fig..36 Emple.13 d d h C u v α b The first moment of re with respect to -is for the totl surfce is given b M = i=1 i i = 4R πr 3π h Rh 3, or M = R ( R h ). 3 Emple.13. rectngulr plnr plte with the sides b = 1 m nd h = m is shown in Fig..36. () Find the product of inerti nd the moments of inerti with respect to the es of the reference frme with the origin t. (b) Determine the product of inerti nd the moments of inerti with respect to the centroidl es tht re locted t the mss center C of the rectngle nd re prllel to its sides. (c) nother reference uv with the sme origin is rotted with n ngle α = 45 from (counterclockwise s positive). Find the inerti mtri of the plte with respect to uv es. (d) Find the principl moments nd the principl directions with the reference frme with the origin t. Solution () The differentil element of re is d = dd. The product of inerti of the rectngle bout the es is

45 .3 Emples 117 h b I = d = b h dd = d d = b h = b h = 1 ( ) = 1m The moment of inerti of the rectngle bout -is is h b b h I = d = dd = d d = b h3 3 = bh3 3 = (1)3 =.666 m 4. 3 The moment of inerti of the rectngle bout -is is h b b h I = d = dd = d d = b3 3 h = hb3 3 = ()13 =.666 m 4. 3 The moment of inerti of the rectngle bout z-is (the polr moment bout )is I = I zz = I + I = ( b + h ) = 1() ( 1 + ) = 3.33 m The inerti mtri of the plne figure with respect to es is represented b [I] = = I I I z I I I z I z I z I zz =. bh 3 3 b h 4 b h 4 hb (b + h ) (b) The product of inerti of the rectngle bout the es is h/ b/ h/ b/ I = d = dd = d d =. h/ b/ h/ b/ The sme results is obtined using the prllel-is theorem or I = I + b h, I = I bh h 4 (bh)=b 4 b h =. 4

46 118 Centroids nd Moments of Inerti The moment of inerti of the rectngle bout is is I = h/ b/ h/ b/ d = dd = d d h/ b/ h/ b/ { } 3 b/ = {} h/ h/ 3 b/ = bh3 1 = (1)3 1 =.666 m4. Using the prllel-is theorem, the moment of inerti of the rectngle bout is is ( ) b I = I = hb3 3 hb3 4 = hb3 1 = ()13 1 =.166 m4. The moment of inerti of the rectngle bout z is (the centroid polr moment) is I C = I z z = I + I = 1 (b + h )= 1() 1 (1 + )=.833 m 4. The inerti mtri of the plne figure with respect to centroidl es is represented b bh 3 I I I z 1 [I C ]= I I I z = hb 3 I z I z I 1 z z 1 (b + h ).666 = (c) The moment of inerti of the rectngle bout u-is is I uu = I + I + I I cosα I sinα = + cos(45 ) (1)sin(45 )=.666 m 4. The moment of inerti of the rectngle bout v-is is I vv = I + I I I cosα + I sinα = cos(45 )+(1)sin(45 )=.666 m 4.

47 .3 Emples 119 The product of inerti of the rectngle bout uv es is I uv = I I sinα + I cosα = sin(45 )+(1)cos(45 )=1m 4. The polr moment of inerti of the rectngle bout is I = I zz = I uu + I vv = I + I = = 3.33 m 4. The inerti mtri of the plne figure with respect to uv es is [I α ]= I uu I uv I vu I vv I zz = (d) The mimum or minimum moment of inerti for the re is. I 1, = I m,min = I (I + I I ± I 1 = I m = I (I + I I + ) + I, ) + I = (.666 ) = 3.8 m 4, I = I min = I + I (I I ) + I = (.666 ) =.5 m 4. The polr moment of inerti of the rectngle bout is I = I zz = I 1 + I = I uu + I vv = I + I = = 3.33 m 4. The principl directions re obtined from tnα = I I I,

48 1 Centroids nd Moments of Inerti or α = 1 tn 1 The principl directions re The MTLB progrm for this emple is sms b h rel list={b, h}; listn={1, }; % m I I I = 1 tn 1 (1) =.5. α 1 =.5 nd α = α 1 + π/ = % d = d d % = int d d ; <<b <<h % = int d ; <<b % = int d ; <<h =int(1,,,b); =int(1,,,h); =*; % I = int d d ; <<b <<h % I1 = int d ; <<b % I = int d ; <<h I1=int(,,, b); I=int(,,, h); I=I1*I; In=subs(I, list, listn); fprintf( I = %s = %g (mˆ4) \n,chr(i), In); % I = int ˆ d d ; <<b <<h % I1 = int d ; <<b % I = int ˆ d ; <<h I1=int(1,,, b); I=int(ˆ,,, h); I=I1*I; In=subs(I, list, listn); fprintf( I = %s = %g (mˆ4) \n,chr(i), In); % I = int ˆ d d ; <<b <<h % I1 = int d ; <<b % I = int ˆ d ; <<h

49 .3 Emples 11 I1=int(ˆ,,, b); I=int(1,,, h); I=I1*I; In=subs(I, list, listn); fprintf( I = %s = %g (mˆ4) \n,chr(i), In); I=I+I; In=In+In; fprintf( I = %s = %g (mˆ4) \n,chr(i), In); % Ip = int ˆ d d ; -b/<<b/ ; -h/<<h/ % I1p = int d ; -b/<<b/ % Ip = int ˆ d ; -h/<<h/ I1p=int(1,, -b/, b/); Ip=int(ˆ,, -h/, h/); Ip=I1p*Ip; Ipn=subs(Ip, list, listn); fprintf( Ip = %s = %g (mˆ4) \n,chr(ip), Ipn); Ip=I-(b*h)*(b/)ˆ; Ipn=subs(Ip, list, listn); fprintf( Ip = %s = %g (mˆ4) \n,chr(ip), Ipn); Ip=I-(b*h)*(-b/)*(-h/); Ipn=subs(Ip, list, listn); fprintf( Ip = %s = %g (mˆ4) \n,chr(ip), Ipn); IC=Ip+Ip; ICn=Ipn+Ipn; fprintf( IC = %s = %g (mˆ4) \n,chr(ic), ICn); fprintf( \n ); lph=pi/4; fprintf( lph = %g (degree) \n, lph*18/pi); fprintf( \n ); I=... (I+I)/+(I-I)/*cos(*lph) -I*sin(*lph); In=subs(I, list, listn); fprintf( Iuu45 = %g (mˆ4) \n, In);

50 1 Centroids nd Moments of Inerti I=... (I+I)/-(I-I)/*cos(*lph) +I*sin(*lph); In=subs(I, list, listn); fprintf( Ivv45 = %g (mˆ4) \n, In); I=(I-I)/*sin(*lph)+I*cos(*lph); In=subs(I, list, listn); fprintf( Iuv45 = %g (mˆ4) \n, In); I=In+In; fprintf( I45 = Iuu45+Ivv45 = %g (mˆ4) \n, I); fprintf( \n ); I1=(I+I)/+sqrt((I-I)ˆ/4+Iˆ); I1n=subs(I1, list, listn); I=(I+I)/-sqrt((I-I)ˆ/4+Iˆ); I1n=subs(I1, list, listn); In=subs(I, list, listn); fprintf( I1 = Im = %g (mˆ4) \n, I1n); fprintf( I = Imin = %g (mˆ4) \n, In); fprintf( I1+I = %g (mˆ4) \n, I1n+In); fprintf( \n ); tnlph=simplif(*i/(i-i)); fprintf( tn ( lph) = %s \n,chr(tnlph)); lphn=tn(*in/(in-in)); fprintf( lph = %g (degrees) \n, lphn/*18/pi); Emple.14. Determine the moment of inerti for the slender rod, shown in Fig..37, with respect to es of reference with the origin t the end nd with respect to centroidl es. The length of the rod is l, the densit is ρ, nd the crosssectionl re i. Epress the results in terms of the totl mss, m, of the rod. Solution The mss of the rod is m = ρ l, nd the densit will be ρ = m/(l). The differentil element of mss is dm = ρ d. The moment of inerti of the slender rod bout the -orz-es is I = I zz = l l m ρ d = l d = m l l d = ml 3. The -is is smmetr is nd tht is wh I =. The moment of inerti of the slender rod bout the centroidl es or z is clculted with the prllel-is theorem

51 .3 Emples 13 Fig..37 Emple.14 z z d C l dr b r dα d r dr dα d R α R α R 1 Fig..38 Emple.15 ( ) l I = I z z = I m = ml 3 ml 4 = ml 1. Emple.15. Find the polr moment of inerti of the plnr flwheel shown in Fig..38. The rdii of the wheel re R 1 nd R (R 1 < R ). Clculte the moments of inerti of the re of circle with rdius R bout dimetrl is nd bout the polr is through the center s shown in Fig..38b.

52 14 Centroids nd Moments of Inerti Solution The polr moment of inerti is given b the eqution I = r d, where r is the distnce from the pole to n rbitrr point on the wheel, nd the differentil element of re is d = r dα dr. The polr moment of inerti is I = R π R 1 R π r 3 dr dα = r 3 dr R 1 Thereofthewheelis = π(r R 1 ) nd dα = R4 R4 1 4 (π)= R4 R4 1 π. I = R + R 1. (.68) If R 1 = ndr = R, the polr moment of inerti of the circulr re of rdius R is, Fig..38b, I = R = π R4. B smmetr for the circulr re, shown in Fig..38b, the moment of inerti bout dimetrl is is I = I nd I = I + I = I = I = I / = π R 4 /4. The results cn be obtined using the integrtion I = d = = R4 4 R π (r sinα) r dr dα = { 1 α sinα } π = πr4 4. π R 4 4 (sinα) dα Emple.16. Find the moments of inerti nd products of inerti for the re showninfig..39, with respect to the es nd with respect to the centroidl es tht pss through the mss center C. Find the principl moments of inerti for the re nd the principl directions. Solution The plte is composed of two element re: the rectngulr re 1 nd the rectngulr re, Fig..39b. The nd coordintes of the mss center C re C = C C1 1 = (/)( )+()( ) = 5 4 C = C C1 1 = ()( )+(/)( ) = 3 4.

53 .3 Emples 15 Fig..39 Emple.16 3 b 1 1 C 1 1 C C The product of inerti for the re shown in Fig..39 bout es is given b I = = ( 4 ) 4 dd + 3 dd = ( 9 )( ) + = The sme result is obtined if prllel-is theorem is used: 3 d d + d d I = I C C1 C1 1 + I C + C C ( ) ( = + ()( )+ +() ( ) )=3 4. The product of inerti for the re bout z nd z es re I z = I z =.

54 16 Centroids nd Moments of Inerti The moment of inerti of the figure with respect to -is is I = I C ( C1 ) 1 + I C +( C ) = ()3 1 + ( )+ ()3 1 ( + ( ) )= The moment of inerti of the figure with respect to -is is I = I C ( C1 ) 1 + I C +( C ) = ()3 1 ( ) + ( )+ ()3 +() ( )= The moment of inerti of the re with respect to z-is is I zz = I + I = = The inerti mtri of the plne figure is represented b the mtri [I]= I I I z I I I z I z I z I zz = Using the prllel-is theorem, the moments of inerti of the re with respect to the centroidl es z re ( I = I ( C ) = I = I ( C ) = 84 3 ( 5 4 I z z = I + I = = 54 6, I = I ( C )( C ) = 3 4 I z = I z =. ( 3 4 The centroidl inerti mtri of the plne figure is ) (4 )= 134 1, ) (4 )= 374 1, )( 5 4 )(4 )= 34 4,

55 .3 Emples 17 [ I ] = I I I z I I I z I z I z I z z = The principl moments of inerti for the re re I 1, = I m,min = I + I ± ( I I = 54 1 ± /4 = 1 ± ) + I I 1 = 14 nd I = The invrint of the sstem is I + I = I 1 + I. The principl directions re obtined from tnα = I 34 I I = = 3 4 = α 1 = 1 tn 1 ( 3/4)= nd α = π + 1 tn 1 ( 3/4)= Emple.17. Find the inerti mtri of the re delimited b the curve = p, from = to = s shown in Fig.4, bout the es of the Crtesin frme with the origin t. Clculte the centroidl inerti mtri. Solution From Fig..4, when =, thevlueof-coordinte is = b nd b = p = p = b /. The epression of the function is = p= b. The differentil element of re is d = dd, nd the re of the figure is = dd = = p p d {} p p = p dd = d d p p d= p 1/ d= b { } 3/ 3/ = 4b 3.

56 18 Centroids nd Moments of Inerti b = p = p b b C C Fig..4 Emple.17 The moment of inerti of the re with respect to -is is or I = p dd = p dd = d p d p { } 3 p = d 3 = (p) 3/ d p 3 { } = 3 (p)3/ 3/ d = 5/ 3 (p)3/ = 4b3 5/ 15 = 4b 3 I = b 5. The moment of inerti of the re with respect to -is is I = dd = = p p dd = p d d p p d {} p = (p) 1/ d { = b 5/ d = b 7/ 7/ } = 43 b 7 = 4b 3 3 7, b 5,

57 .3 Emples 19 or I = 3 7. The moment of inerti of the re with respect to z-is is I zz = I + I = b = 7 ( b The product of inerti of the re with respect to es is p p I = dd = dd = d p d p { } p = d =. p The products of inerti of the re with respect to z nd z es re I z = I z =. The inerti mtri of the plne figure is b I I I 5 z [I]= I I I z 3 =. 7 I z I z I zz ( ) b The first moment of the re with respect to -is is p p M = dd = dd = d p d p p = d {} p = p d { } = b 3/ d = b 5/ = b 5/ 5/ 5/ = 4b 5. The -coordinte of the mss center, Fig.4b, is C = M = 4b 3 5 4b = 3 5. The first moment of the re with respect to -is is M = nd C = M / =. Using the prllel-is theorem, the moments of inerti of the re with respect to the centroidl es z,fig.4b, re ).

58 13 Centroids nd Moments of Inerti I = I d = I = b 5, I = I ( C ) = I ( 3 5 ) = = ), ( I z z = I + I = b b 175 = I = I ( C )() =, I z = I z =. The centroidl inerti mtri of the plne figure is [ I ] = I I I z I I I z I z I z I z z = b ( b )..4 Problems.1 Find the -coordinte of the centroid of the indicted region where = m nd k = π/8m 1 (Fig..41).. Find the -coordinte of the centroid of the shded region shown in the figure. The region is bounded b the curves = nd =. ll coordintes m be treted s dimensionless (Fig..4)..3 The region shown is bounded b the curves = b nd = k 3,whereb = k 3. Find the coordintes of the centroid (Fig..43)..4 Determinethe centroid of the re where (, ). Use integrtion (Fig..44)..5 Locte the center of grvit of the volume where = /b, = b = 1m. The mteril is homogeneous (Fig..45). = cos(k) Fig..41 Problem.1

59 .4 Problems 131 Fig..4 Problem. = = Fig..43 Problem.3 = k 3 Fig..44 Problem.4 = m (,) = k Fig..45 Problem.5 ( ) = b b

60 13 Centroids nd Moments of Inerti Fig..46 Problem.6 b ( b ) = Fig..47 Problem.7 b c d.6 Locte the centroid of the prboloid, shown in Fig..46 defined b the eqution / =(/b) where = b = 5 m. The mteril is homogeneous..7 Determine the loction of the centroid C oftherewhere = 6m,b = 6m, c = 7m,ndd = 7 m, s shown in Fig The solid object, shown in Fig..48 it consists of solid hlf-circulr bse with n etruded ring. The bottom of the object is hlf-circulr in shpe with thickness, t, equl to 1 mm. The hlf-circle hs rdius, R, equl to 3 mm. The coordinte es re ligned so tht the origin is t the bottom of the object t the center point of the hlf-circle. This hlf-ring hs n inner rdius, r, of mm, nd n outer rdius, R, of 3 mm. The etrusion height for the