1 CHAPTER 2 MOMENT OF INERTIA. m m,, K such that the distance of the mass m i from the axis is r i. The quantity m r i i

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1 CHAPTER MOMENT OF INERTIA. Definition of Moment of Inerti Consider stright line (the "is") nd set of point msses m m,, K such tht the, m distnce of the mss m i from the is is r i. The quntity m r i i is the second moment of the i th mss with respect to (or "bout") the is, nd the sum m i r i is the second moment of mss of ll the msses with respect to the is. Aprt from some subtleties encountered in generl reltivity, the word "inerti" is synonymous with mss - the inerti of body is merely the rtio of n pplied force to the resulting ccelertion. Thus m i r i cn lso be clled the second moment of inerti. The second moment of inerti is discussed so much in mechnics tht it is usully referred to s just "the" moment of inerti. In this chpter we shll consider how to clculte the (second) moment of inerti for different sizes nd shpes of body, s well s certin ssocited theorems. But the question should be sked: "Wht is the purpose of clculting the squres of the distnces of lots of prticles from n is, multiplying these squres by the mss of ech, nd dding them ll together? Is this merely pointless mke-work eercise in rithmetic? Might one just s well, for ll the good it does, clculte the sum m i ri? Does m i r i hve ny physicl significnce?". Mening of Rottionl Inerti. If force cts of body, the body will ccelerte. The rtio of the pplied force to the resulting ccelertion is the inerti (or mss) of the body. If torque cts on body tht cn rotte freely bout some is, the body will undergo n ngulr ccelertion. The rtio of the pplied torque to the resulting ngulr ccelertion is the rottionl inerti of the body. It depends not only on the mss of the body, but lso on how tht mss is distributed with respect to the is. Consider the system shown in figure II..

2 F O r m τ & θ O r m FIGURE II. A prticle of mss m is ttched by light (i.e. zero or negligible mss) rm of length r to point t O, bout which it cn freely rotte. A force F is pplied, nd the mss consequently undergoes liner ccelertion F/m. The ngulr ccelertion is then & θ F. mr. Also, the torque is τ Fr. The rtio of the pplied torque to the ngulr ccelertion is therefore mr. Thus the rottionl inerti is the second moment of inerti. Rottionl inerti nd (second) moment of inerti re one nd the sme thing, ecept tht rottionl inerti is physicl concept nd moment of inerti is its mthemticl representtion.. Moments of inerti of some simple shpes. A student my well sk: "For how mny different shpes of body must I commit to memory the formuls for their moments of inerti?" I would be tempted to sy: "None". However, if ny re to be committed to memory, I would suggest tht the list to be memorized should be limited to those few bodies tht re likely to be encountered very often (prticulrly if they cn be used to determine quickly the moments of inerti of other bodies) nd for which it is esier to remember the formuls thn to derive them. With tht in mind I would recommend lerning no more thn five. In the following, ech body is supposed to be of mss m nd rottionl inerti I.. A rod of length l bout n is through the middle, nd t right ngles to the rod: I ml... A uniform circulr disc of rdius bout n is through the centre nd perpendiculr to the

3 plne of the disc: I m... A uniform right-ngled tringulr lmin bout one of its shorter sides - i.e. not the hypotenuse. The other not-hypotenuse side is of length : I 6 m.. 4. A uniform solid sphere of rdius bout n is through the centre. I 5 m A uniform sphericl shell of rdius bout n is through the centre. I m..5 I shll now derive the first three of these by clculus. The derivtions for the spheres will be left until lter.. Rod, length l (Figure II.) FIGURE II. δ The mss of n element δ t distnce from the middle of the rod is mδ l nd its second moment of inerti is m δ. l The moment of inerti of the entire rod is m l m l d d l l l ml.

4 . Disc, rdius. (Figure II.) 4 FIGURE II. r The re of n elementl nnulus, rdii r, r + δ r is πrδr. The re of the entire disc is π. πrδrm mrδr Therefore the mss of the nnulus is. π nd its second moment of inerti is mr δ r. m The moment of inerti of the entire disc is. r dr m. Right-ngled tringulr lmin. (Figure II.4) FIGURE II.4 b y b( / ) δ

5 5 The eqution to the hypotenuse is y b( / ). The re of the elementl strip is yδ b( / ) δ nd the re of the entire tringle is b/. Therefore the mss of the elementl strip is m( ) δ nd its second moment of inerti is m ( ) δ. The second moment of inerti of the entire tringle is the integrl of this from to, which is m /6. Uniform circulr lmin bout dimeter. For the ske of one more bit of integrtion prctice, we shll now use the sme rgument to show tht the moment of inerti of uniform circulr disc bout dimeter is m /4. However, we shll see lter tht it is not necessry to resort to integrl clculus to rrive t this result, nor is it necessry to commit the result to memory. In little while it will become immeditely pprent nd ptently obvious, with no clcultion, tht the moment of inerti must be m /4. However, for the time being, let us hve some more clculus prctice. See figure II.5. FIGURE II.5

6 6 The disc is of rdius, nd the re of the elementl strip is yδ. But y nd re relted through the eqution to the circle, which is y ( ) / ( ) / δ. The re of the whole disc is /. Therefore the re of the strip is π, so the mss of the strip is ( ) δ m / m ( ) δ. The second moment of inerti bout the y-is is π π m / ( ) δ. For the entire disc, we integrte from to +, or, if you prefer, π from to nd then double it. The result m /4 should follow. If you need hint bout how to do the integrtion, let cosθ (which it is, nywy), nd be sure to get the limits of integrtion with respect to θ right. The moment of inerti of uniform semicirculr lmin of mss m nd rdius bout its bse, or dimeter, is lso m /4, since the mss distribution with respect to rottion bout the dimeter is the sme..4 Rdius of gyrtion. The second moment of inerti of ny body cn be written in the form mk. Thus, for the rod, the disc (bout n is perpendiculr to its plne), the tringle nd the disc (bout dimeter), k hs the vlues l.866l,.77,.48,.5 6 respectively. k is clled the rdius of gyrtion. If you were to concentrte ll the mss of body t its rdius of gyrtion, its moment of inerti would remin the sme..5 Prllel nd Perpendiculr Aes Theorems O m 5 m 6 m m 4 m m FIGURE II.6

7 7 In figure II.6, the two unbroken lines represent two fied coordinte es. I hve drwn severl point msses m, m m K distributed in plne. The -coordinte of mss m i is i. The dshed, line is moveble, nd it -coordinte is, so tht the distnce of m i this line is i. The moment of inerti of the system of msses bout the dshed line is + I m ( ) + m ( ) + m ( ) Now imgine wht hppens if the dshed line is moved to the right. The moment of inerti decreses nd decreses - nd decreses. But eventully the line finds itself to the right of m 4, nd then of m 5, nd then of m 6. After tht is by no mens obvious tht the moment of inerti is going to continue to decrese. Indeed, by this time it is cler tht t some point I is going to go through minimum nd then strt to increse gin s more nd more of the msses find themselves to the left of the dshed line. Just where is the dshed line when the moment of inerti is minimum? I ll leve you to differentite eqution.5. with respect to, nd hence show tht I is lest when m + m + m m + m + m +... Tht is, the moment of inerti is lest when is pssing through the centre of mss.. Tht is, the moment of inerti is lest for n In figure II.6b, the line CC psses through the centre of mss; the moment of inerti is lest bout this line. The line AA is t distnce from CC, nd the moment of inerti is greter bout AA thn bout CC. The Prllel Aes Theorem tells us by how much. m 5 A m 6 C m m 4 A m C FIGURE II.6b m

8 8 Let us mesure distnces from CC, so tht the distnce of m i from CC is i nd the distnce of m i from AA is i +. It is cler tht I m i CC i nd tht I AA mi ( i + ) mi + mi i + mi..5. The first term on the right hnd side is I CC. The sum in the second term is the first moment of mss bout the centre of mss, nd is zero. The sum in the third term is the totl mss. We therefore rrive t the Prllel Aes Theorem: I I + M..5.4 AA CC In words, the moment of inerti bout n rbitrry is is equl to the moment of inerti bout prllel is through the centre of mss plus the totl mss times the squre of the distnce between the prllel es. The theorem holds lso for msses distributed in three-dimensionl spce. The Perpendiculr Aes Theorem, on the other hnd, holds only for msses distributed in plne, or for plne lmins. y FIGURE II.7 m 4 z m Figure II.7 shows some point msses distributed in the y plne, the z is being perpendiculr to the plne of the pper. The moments of inerti bout the, y nd z es re denoted respectively by A, B nd C. The distnce of m i from the z is is ( ). inerti of the msses bout the z is is m i y i / + Therefore the moment of C m ( + y )..5.5 Tht is to sy: C A + B..5.6 i i i m

9 9 This is the Perpendiculr Aes Theorem. Note gin very crefully tht, unlike the prllel es theorem, this theorem pplies only to plne lmins nd to point msses distributed in plne. Emples of the Use of the Prllel nd Perpendiculr Aes Theorems. From section. we know the moments of inerti of discs, rods nd tringulr lmins. We cn mke use of the prllel nd perpendiculr es theorems to write down the moments of inerti of most of the following emples lmost by sight, with no clculus. Hoop nd discs, rdius. m m m m Rods, length l. m m 4 m 5 m 4 ml 4 ml Rectngulr lmins, sides nd b; > b. m mb ( ) m + b m b ( + b )

10 Squre lmins, side. m m m m Tringulr lmins. m 6 m 6 b m( + b ) 6( + b) θ m 6 ( ) m + b + c c b b c

11 I 6 6 m m ( + cot θ) mb ( + tn θ) mc ( sin θ) 6 6 ( b + c ) m( c ) m( + b ) Three-dimensionl solid figures. Spheres, cylinders, cones Sphere, mss m, rdius. y The volume of n elementl cylinder of rdii, + δ, height y is ( ) / δ. 4π yδ 4π Its mss is m / second moment of inerti is ( ) δ. sphere is / ( ) δ m / ( ) δ. 4π m Its π 4 The second moment of inerti of the entire ( ) / d. m 5 m The moment of inerti of uniform solid hemisphere of mss m nd rdius bout dimeter of its bse is lso 5 m, becuse the distribution of mss round the is is the sme s for complete sphere. Problem: A hollow sphere is of mss M, eternl rdius nd internl rdius. Its rottionl inerti is.5 M. Show tht is given by the solution of nd clculte to four significnt figures. (Answer.686.)

12 Solid cylinder, mss m, rdius, length l δ l l The mss of n elementl disc of thickness δ is m δ. Its moment of inerti bout its dimeter l is mδ m δ. Its moment of inerti bout the dshed is through the centre of the 4 l 8l m δ mδ m( + 4 ) δ cylinder is +. The moment of inerti of the entire cylinder 8l l 8l l m( + 4 ) d bout the dshed is is m( 4 + l ). 8l In similr mnner it cn be shown tht the moment of inerti of uniform solid tringulr prism of mss m, length l, cross section n equilterl tringle of side bout n is through its centre nd perpendiculr to its length is m ( + l ). 6 Solid cone, mss m, height h, bse rdius. y h y h The mss of the elementl disc of thickness δ is

13 π y δ m π h my δ. h Its second moment of inerti bout the is of the cone is But y nd re relted through y 4 my δ my δ y. h h, so the moment of inerti of the elementl disk is h 4 m δ 5. h The moment of inerti of the entire cone is m h 4 d 5 h m. The following, for solid cone of mss m, height h, bse rdius, re left s n eercise: m m ( + 4h ) ( + h ).7 Three-dimensionl hollow figures. Spheres, cylinders, cones. Hollow sphericl shell, mss m, rdius.

14 4 sin θ δθ θ The re of the elementl zone is π sin θδθ. Its mss is π sin θδθ m msin θδθ. 4π Its moment of inerti is m sin θ δθ sin θ m sin θ δ θ. The moment of inerti of the entire sphericl shell is m sin θdθ m π This result cn be used to clculte, by integrtion, the moment of inerti sphere. Or, if you strt with 5. 5 m of solid m for solid sphere, you cn differentite to find the result m for hollow sphere. Write the moment of inerti for solid sphere in terms of its density rther thn its mss. Then dd lyer d nd clculte the increse di in the moment of inerti. We cn lso use the moment of inerti for hollow sphere ( m ) to clculte the moment of inerti of nonuniform solid sphere in which the density vries s ρ ρ(r). For emple, if ρ ρ r ( / ) moment of inerti is, see if you cn show tht the mss of the sphere is.467ρ m. A much esier method will be found in Section 9. nd tht its

15 5 Using methods similr to tht given for solid cylinder, it is left s n eercise to show tht the moment of inerti of n open hollow cylinder bout n is perpendiculr to its length pssing through its centre of mss is m ( + l ), where is the rdius nd l is the length. The moment of inerti of bseless hollow cone of mss m, bse rdius, bout the is of the cone could be found by integrtion. However, those who hve n understnding of the wy in which the moment of inerti depends on the distribution of mss should redily see, without further do, tht the moment of inerti is m. (Look t the cone from bove; it looks just like disc, nd indeed it hs the sme rdil mss distribution s uniform disc.).8 Torus The rottionl inertis of solid nd hollow toruses (lrge rdius, smll rdius b) re given below for reference nd without derivtion. They cn be derived by integrl clculus, nd their derivtion is recommended s chllenge to the reder. Solid torus: m ( 4 + b ) 8 m ( 4 + 5b ) 4 Hollow torus: m ( + b ) 4 m ( + 5b ).9 Liner tritomic molecule

16 6 y m m m Here is n interesting problem. It should be strightforwrd to clculte the rottionl inerti of the bove molecule with respect to n is perpendiculr to the molecule nd pssing through the centre of mss. In prctice it is quite esy to mesure the rottionl inerti very precisely from the spcing between the lines in moleculr bnd in the infrred region of the spectrum. If you know the three msses (which you do if you know the toms tht mke up the molecule) cn you clculte the two intertomic spcings nd y? Tht would require determining two unknown quntities, nd y, from single mesurement of the rottionl inerti, I. Evidently tht cnnot be done; second mesurement is required. Cn you suggest wht might be done? We shll nswer tht shortly. In the mentime, it is n eercise to show tht the rottionl inerti is given by + hy + by + c,.9. where m ( m m )/ M.9. + h m m / M.9. ( m m ) M b m / M m + m + m.9.5 c I.9.6 For emple, suppose the molecule is the liner molecule OCS, nd the three msses re 6, nd respectively, nd, from infrred spectroscopy, it is determined tht the moment of inerti is. (For this hypotheticl illustrtive emple, I m not concerning myself with units). In tht cse, eqution.9. becomes.7& + 7.6& y + 4.9& y..9.7 We need nother eqution to solve for nd y. Wht cn be done chemiclly is to prepre n isotopiclly-substituted molecule (isotopomer) such s 8 OCS, nd mesure its moment of inerti from its spectrum, mking the probbly very justified ssumption tht the intertomic distnces re unffected by the isotopic substitution. This results in second eqution: ' + h' y + b' y + c'..9.8 Let's suppose tht the new moment of inerti is I ', nd I leve it to the reder to work out the numericl vlues of ', h' nd b' with the stern cution to retin ll the deciml plces on your clcultor. Tht is, do not round off the numbers until the very end of the clcultion.

17 7 You now hve two equtions,.9. nd.9.8, to solve for nd y. These re two simultneous qudrtic equtions, nd it my be tht some guidnce in solving them would be helpful. I hve three suggestions.. Tret eqution.9. s qudrtic eqution in nd solve it for in terms of y. Then substitute this in eqution.9.8. I epect you will very soon become bored with this method nd will wnt to try something little less tedious.. You hve two equtions of the form S(, y), S'(, y). There re stndrd wys of solving these itertively by n etension of the Newton-Rphson process. This is described, for emple, in section.9 of Chpter of my Celestil Mechnics notes, nd this generl method for two or more nonliner equtions should be known by nyone who epects to engge in much numericl clcultion. For this prticulr cse, the detiled procedure would be s follows. This is n itertive method, nd it is first necessry to mke guess t the solutions for nd y. The guesses need not be prticulrly good. Tht done, compute the following si quntities: S ( + hy ) + by + c S ' ( ' + h' y) + b' y + c' S ( + hy ) S ( y h + by ) S' ( ' + h' y) S' ( h' + b' y) Here the subscripts denote the prtil derivtives. Now if y nd (true) (guess) + ε y(true) y(guess) + η the errors ε nd η cn be found from the solution of S ε + S η + S nd S' ε + S' y η + S' If we clculte F S S' S S' The solutions for the errors re y y y

18 8 ε F( S' S S S') y η F( S S' S' S) y This will enble better guess to be mde, nd the procedure cn be repeted until the errors re s smll s desired. Generlly only very few itertions re required. If this is not the cse, progrmming mistke is indicted.. While method cn be used for ny nonliner simultneous equtions, in this prticulr cse we hve two simultneous qudrtic equtions, nd little fmilirity with conic sections provides rther nice method. Thus, if S nd S' re equtions.9. nd.9.8 respectively. Ech of these equtions represents conic section, nd they intersect t four points. We wish to find the point of intersection tht lies in the ll-positive qudrnt - i.e. with nd y both positive. Since the two conic sections re very similr, in order to clculte where they intersect it is necessry to clculte with gret ccurcy. Therefore, do not round off the numbers until the very end of the clcultion. Form the eqution c' S cs'. This is lso qudrtic eqution representing conic section pssing through the four points. Furthermore, it hs no constnt term, nd it therefore represents the two stright lines tht pss through the four points. The eqution cn be fctorized into two liner terms, αβ, where α nd β re the two stright lines. Choose the one with positive slope nd solve it with S or with S' (or with both, s check ginst rithmetic mistkes) to find nd y. In this cse, the solutions re.59, y... Pendulums In section., we discussed the physicl mening of the rottionl inerti s being the rtio of the pplied torque to the resulting ngulr ccelertion. In liner motion, we re fmilir with the eqution F m. The corresponding eqution when deling with torques nd ngulr ccelertion is τ I & θ. We re lso fmilir with the eqution of motion for mss vibrting t the end of spring of force constnt k : m & k. This is simple hrmonic motion of period π m / k. The mechnics of the torsion pendulum is similr. The torsion constnt c of wire is the torque required to twist it through unit ngle. If mss is suspended from torsion wire, nd the wire is twisted through n ngle θ, the restoring torque will be cθ, nd the eqution of motion is I & θ cθ, which is simple hrmonic motion of period π I / c. The torsion constnt of wire of circulr cross-section, by the wy, is proportionl to its sher modulus, the fourth power of its rdius, nd inversely s its length. The derivtion of this tkes little trouble, but it cn be verified by dimensionl nlysis. Thus thick wire is very much hrder to twist thn thin one. A wire of nrrow rectngulr cross-section, such s strip or ribbon hs reltively smll torsion constnt.

19 9 Now let's look not t torsion pendulum, but t pendulum swinging bout n is under grvity. O h C mg We suppose the pendulum, of mss m, is swinging bout point O, which is t distnce h from the centre of mss C. The rottionl inerti bout O is I. The line OC mkes n ngle θ with the verticl, so tht the horizontl distnce between O nd C is h sin θ. The torque bout O is mgh sin θ, so tht the eqution of motion is I & θ mghsin θ... For smll ngles, this is I & θ mghθ... This is simple hrmonic motion of period P I π... mgh

20 We'll look t two emples - uniform rod, nd n rc of circle. First, uniform rod. O h C The centre of mss is C. The rottionl inerti bout C is ml, so the rottionl inerti bout O is I ml + mh. If we substitute this in eqution.., we find for the period of smll oscilltions P l + h π...4 gh This cn be written P ( h / l), l + π...5 g h / l or, if we write P P nd h h/l : l π g P + h...6 h The figure shows grph of P versus h.

21 Eqution..6 cn be written P + h..7 h nd, by differentition of P with respect to h, we find tht the period is lest when h /. This lest period is given by P, or P.86. Eqution..7 cn lso be written h P h This qudrtic eqution shows tht there re two positions of the support O tht give rise to the sme period of smll oscilltions. The period is lest when the two solutions of eqution..8 re equl, nd by the theory of qudrtic equtions, then, the lest period is given by P, s we lso deduced by differentition of eqution..7, nd this occurs when h /. For periods longer thn this, there re two solutions for h. Let h be the smller of these, nd let h be the lrger. By the theory of qudrtic equtions, we hve h +..9 h P

22 nd h h /... Let H h h be the distnce between two points O tht give the sme period of oscilltion. Then 4 P H ( h h ) ( h + h ) 4h h... 9 If we mesure H for given period P nd recll the definition of P we see tht this provides method for determining g. Although this is common undergrdute lbortory eercise, the grph shows tht the minimum is very shllow nd consequently H nd hence g re very difficult to mesure with ny precision. For nother emple, let us look t wire bent into the rc of circle of rdius oscillting in verticl plne bout its mid-point. In the figure, C is the centre of mss. The rottionl inerti bout the centre of the circle is m. By two pplictions of the prllel es theorem, we see tht the rottionl inerti bout the point of oscilltion is I m m h + mh mh Thus, from eqution.. we find ( ). C h nd the period is independent of the length of the rc. -h P π,.. g.. Plne Lmins. Product moment. Trnsltion of Aes (Prllel Aes Theorem). We consider set of point msses distributed in plne, or plne lmin. We hve hitherto met three second moments of inerti: A m y,.. i i B m,.. i i

23 C mi ( i + yi )... These re respectively the moments of inerti bout the - nd y-es (ssumed to be in the plne of the msses or the lmin) nd the z-is (norml to the plne). Clerly, C A + B, which is the perpendiculr es theorem for plne lmin. We now introduce nother quntity, H, clled the product moment of inerti with respect to the - nd y-es, defined by H mi i yi...4 We'll need sometime to sk ourselves whether this hs ny prticulr physicl significnce, or whether it is merely something to clculte for the ske of pssing the time of dy. In the mentime, the reder should recll the prllel es theorems (Section.5) nd, using rguments similr to those given in tht section, should derive H HC + M y...5 It my lso be noted tht eqution..4 does not contin ny squred terms nd therefore the product moment of inerti, depending on the distribution of msses, is just s likely to be negtive quntity s positive one. We shll defer discussing the physicl significnce, if ny, of the product moment until section. In the mentime let us try to clculte the product moment for plne right tringulr lmin: B b y O A The re of the tringle is b nd so the mss of the element δδy is Mδδy, b where M is the mss of the complete tringle. The product moment of the element with respect to the sides My δ δy M OA, OB is nd so the product moment of the entire tringle is b yddy. We b hve to consider crefully the limits of integrtion. We'll integrte first with respect to ; tht

24 4 is to sy we integrte long the horizontl (y constnt) strip from the side OB to the side AB. y Tht is to sy we integrte δ from where to where. The product moment is b therefore y ( ). M y. b dy b We now hve to dd up ll the horizontl strips from the side OA, where y, to B, where y b. Thus H M b b y ( ) y dy, b which, fter some lgebr, comes to H Mb. The coordintes of the centre of mss with respect to the sides OA, OB re ( b),, so tht, from eqution..5, we find tht the product moment with respect to es prllel to OA, OB nd pssing through the centre of mss is 6 Mb. Eercise: Clculte the product moments of the following eight lmins, ech of mss M, with respect to horizontl nd verticl es through the origin, nd with respect to horizontl nd verticl es through the centroid of ech. (We hve just done the first of these, bove.) The horizontl bse of ech is of length, nd the height of ech is b. You re going to hve to tke gret cre with the signs, nd with the limits of integrtion. If you get n nswer right ecept for the sign, then you hve got the nswer wrong.

25 I mke the nswers s follows. 5 Mb + 6 Mb Mb Mb 6 Mb Mb Mb Mb + Mb Mb Mb Mb 6 + Mb + Mb Mb 4 6 Mb. Rottion of Aes. y y O θ We strt by reclling result from elementry geometry. Consider two sets of es Oy nd O y, the ltter being inclined t n ngle θ to the former. Any point in the plne cn be described by the coordintes (, y) or by (, y ). These coordintes re relted by rottion mtri: cosθ sinθ,.. y sin θ cos θ y y cosθ sinθ sin θ. cos θ y.. The rottion mtri is orthogonl; one of the severl properties of n orthogonl mtri is tht its reciprocl is its trnspose.

26 6 Now let us pply this to the moments of inerti of plne lmin. Let us suppose tht the es re in the plne of the lmin nd tht O is the centre of mss of the lmin. A, B nd H re the moments of inerti with respect to the es Oy, nd A, B nd H re the moments of inerti with respect to O y. Strictly speking lmin implies continuous distribution of mtter in plne, but, since mtter, we re told, is composed of discrete toms, there is little difficulty in justifying treting lmin s though it we distribution of point msses in the plne. In ny cse the results tht follow re vlid either for collection of point msses in plne or for genuine continuous lmin. We hve, by definition: A B H my.. m..4 m y..5 Now let us pply eqution.. to eqution..: A ( sin θ+ y cos θ) sin θ m sin θcos θ my + cos θ. m my Tht is to sy (writing the third term first, nd the first term lst) A Acos θ H sinθ cosθ + B sin θ...6 In similr fshion, we obtin for the other two moments B Asin θ + H sinθ cosθ + B cos θ..7 nd Asin θcos θ + H ( cos θ sin θ) B sin θcos. H..8 θ It is usully more convenient to mke use of trigonometric identities to write these s ( B + A) ( B A) cos θ sin, A H..9 θ ( B + A) + ( B A) cos θ + sin, B H.. θ ( B ) sin. H H cos θ A θ.. These equtions enble us to clculte the moments of inerti with respect to the es O y if we know the moments with respect to the es Oy. Further, mtter of importnce, we see, from eqution.., tht if

27 7 H tn θ, B A.. the product moment H with respect to the es Oy is zero. This gives some physicl mening to the product moment, nmely: If we cn find some es (which we cn, by mens of eqution..) with respect to which the product moment is zero, these es re clled the principl es of the lmin, nd the moments of inerti with respect to the principl es re clled the principl moments of inerti. I shll use the symbols A nd B for the principl moments of inerti, nd I shll dopt the convention tht A B. Emple: Consider three point msses t the coordintes given below: Mss Coordintes 5 (, ) (4, ) (, 4) The moments of inerti re A 49, B 7, C 5. The coordintes of the centre of mss re (.,.9). If we use the prllel es theorem, we cn find the moments of inerti with respect to es prllel to the originl ones but with origin t the centre of mss. With respect to these es we find A.9, B 8., H +9.. The principl es re therefore inclined t ngles θ to the -is given (eqution..) by tn θ.57669; Tht is θ 7 o ' nd 7 o '. On using eqution..9 or with these two ngles, together with the convention tht A B, we obtin for the principl moments of inerti A 5.84 nd B 5.6. Emple. Consider the right-ngled tringulr lmin of section. The moments of inerti with respect to es pssing through the centre of mss nd prllel to the orthogonl sides of the tringle re A 8 Mb, B 8 M, H 6 Mb. The ngles tht the principl es mke b with the - side re given by tn θ. The interested reder will be ble to work out b epressions, in terms of M,, b, for the principl moments.

28 8. Momentl Ellipse P θ Consider plne lmin such tht its rdius of gyrtion bout some is through the centre of mss is k. Let P be vector in the direction of tht is, originting t the centre of mss, given by P rˆ.. k Here rˆ is unit vector in the direction of interest; k is the rdius of gyrtion, nd is n rbitrry length introduced so tht the dimensions of P re those of length, nd the length of the vector P is inversely proportionl to the rdius of gyrtion. The moment of inerti is 4 Mk M / P. Tht is to sy M P 4 Acos θ H sinθcosθ + B sin θ,.. where A, H nd B re the moments with respect to the - nd y-es. Let (, y) be the coordintes of the tip of the vector P, so tht P cosθ nd y P sin θ. Then 4 M A Hy + By... Thus, no mtter wht the shpe of the lmin, however irregulr nd symmetric, the tip of the H vector P trces out n ellipse, whose es re inclined t ngles tn B A to the -is. This is the momentl ellipse, nd the es of the momentl ellipse re the principl es of the lmin. Emple. Consider regulr n-gon. By symmetry the moment of inerti is the sme bout ny two es in the plne inclined t π/n to ech other. This is possible only if the momentl ellipse

29 9 is circle. It follows tht the moment of inerti of uniform polygonl plne lmin is the sme bout ny is in its plne nd pssing through its centroid. Eercise. Show tht the moment of inerti of uniform plne n-gon of side bout ny is in its plne nd pssing through its centroid is ( m + cot ( π / n )). Wht is this for squre? For n equilterl tringle?.4. Eigenvectors nd eigenvlues. In sections -, we hve been considering some spects of the moments of inerti of plne lmins, nd we hve discussed such mtters s rottion of es, nd such concepts s product moments of inerti, principl es, principl moments of inerti nd the momentl ellipse. We net need to develop the sme concepts with respect to three-dimensionl solid bodies. In doing so, we shll need to mke use of the lgebric concepts of eigenvectors nd eigenvlues. If you re lredy fmilir with such mtters, you my wnt to skip this section nd move on to the net. If the ides of eigenvlues nd eigenvectors re new to you, or if you re bit rusty with them, this section my be helpful. I do ssume tht the reder is t lest fmilir with the elementry rules of mtri multipliction. Consider wht hppens when you multiply vector, for emple the vector, 4 mtri, for emple the mtri, We obtin: by squre 4. The result of the opertion is nother vector tht is in quite different direction from the originl one. However, now let us multiply the vector by the sme mtri. The result is. The result of the multipliction is merely to multiply the vector by without chnging its direction. The vector is very specil one, nd it is clled n eigenvector of the mtri, nd the multiplier is clled the corresponding eigenvlue. "Eigen" is Germn for "own" in the sense of "my own book". There is one other eigenvector of the mtri; it is the vector. Try it; you should find tht the corresponding eigenvlue is.

30 In short, given squre mtri A, if you cn find vector such tht A λ, where λ is merely sclr multiplier tht does not chnge the direction of the vector, then is n eigenvector nd λ is the corresponding eigenvlue. In the bove, I told you wht the two eigenvectors were, nd you were ble to verify tht they were indeed eigenvectors nd you were ble to find their eigenvlues by strightforwrd rithmetic. But, wht if I hdn't told you the eigenvectors? How would you find them? Let A A A A A nd let be n eigenvector with corresponding eigenvlue λ. Then we must hve A A A A λ λ. Tht is, ( A λ) + A nd A + ( A λ). These two equtions re consistent only if the determinnt of the coefficients is zero. Tht is, A λ A A A λ. This eqution is qudrtic eqution in λ, known s the chrcteristic eqution, nd its two roots, the chrcteristic or ltent roots re the eigenvlues of the mtri. Once the eigenvlues re found the rtio of to is esily found, nd hence the eigenvectors. Similrly, if A is mtri, the chrcteristic eqution is A λ A A A A λ A A A A λ. This is cubic eqution in λ, the three roots being the eigenvlues. For ech eigenvlue, the rtio : : cn esily be found nd hence the eigenvectors. The chrcteristic eqution is cubic eqution, nd is best solved numericlly, not by lgebric formul. The cubic eqution cn be written in the form

31 λ + λ + λ +, nd the solutions cn be checked from the following results from the theory of equtions: λ + λ + λ, λ λ + λ λ + λλ, λλ λ..5. Solid body. The moments of inerti of collection of point msses distributed in three-dimensionl spce (or of solid three-dimensionl body, which, fter ll, is collection of point msses (toms)) with respect to es Oyz re: A B C ( y + z ) F m myz ( + ) G m z mz ( + y ) m H my Suppose tht A, B, C, F, G, H, re the moments nd products of inerti with respect to es whose origin is t the centre of mss. The prllel es theorems (which the reder should prove) re s follows: Let P be some point not t the centre of mss, such tht the coordintes of the centre of mss with respect to es prllel to the es Oyz but with origin t P re (, y, z). Then the moments nd products of inerti with respect to the es through P re where M is the totl mss. A B + + M M C + M ( y + z ) F + Myz ( z + ) G + Mz ( + y ) H + My Unless stted otherwise, in wht follows we shll suppose tht the moments nd products of inerti under discussion re referred to set of es with the centre of mss s origin.

32 .6 Rottion of es - three dimensions. Let Oyz be one set of mutully orthogonl es, nd let O y z be nother set of es inclined to the first. The coordintes (, y, z ) of point with respect to the second bsis set re relted to the coordintes (, y, z ) with respect to the first by y z c c c c c c c c c y. z.6. Here the c ij re the cosines of the ngles between the es of one bsis set with respect to the es of the other. For emple, c is the cosine of the ngle between O nd Oy. c is the cosine of the ngles between Oy nd Oz. Some reders my know how to epress these cosines in terms of complicted epressions involving the Eulerin ngles. While these re importnt, they re not essentil for following the present development, so we shll not mke use of the Eulerin ngles just here. The mtri of direction cosines is orthogonl. Among the severl properties of n orthogonl mtri is the fct tht its reciprocl (inverse) is equl to its trnspose - i.e. the reciprocl of n orthogonl mtri is found merely my interchnging the rows nd columns. This enbles us esily to find (, y, z ) in terms of (, y, z ). A number of other properties of n orthogonl mtri re useful in detecting, locting nd even correcting rithmetic mistkes in computing the elements. These properties re. The sum of the squres of the elements in ny row or column is unity. This merely epresses the fct tht the mgnitude of unit vector long ny of the si es is indeed unity.. The sum of the products of corresponding elements of ny two rows or of ny two columns is zero. This merely epresses the fct tht the sclr product of ny two orthogonl vectors is zero. It will be noted tht checking for property will not detect ny mistkes in sign of the elements, wheres checking for property will do so.. Every element is equl to ± its own cofctor. This epresses the fct tht the cross product of two unit orthogonl vectors is equl to the third. 4. The determinnt of the mtri is ±. If the sign is negtive, it mens tht the chirlities (hndedness) of the two bsis sets of es re opposite; i.e. one of them is right-hnded set nd the other is left-hnded set. It is usully convenient to choose both sets s righthnded.

33 If it is possible to find set of es with respect to which the product moments F, G nd H re ll zero, these es re clled the principl es of the body, nd the moments of inerti with respect to these es re the principl moments of inerti, for which we shll use the nottion A, B, C, with the convention A B C. We shll see shortly tht it is indeed possible, nd we shll show how to do it. A vector whose length is inversely proportionl to the rdius of gyrtion trces out in spce n ellipsoid, known s the momentl ellipsoid. In the study of solid body rottion (whether by stronomers studying the rottion of steroids or by chemists studying the rottion of molecules) bodies re clssified s follows.. A B C The ellipsoid is triil ellipsoid, nd the body is n symmetric top.. A < B C The ellipsoid is prolte spheroid nd the body is prolte symmetric top.. A B < C The ellipsoid is n oblte spheroid nd the body is n oblte symmetric top. 4. A B C The ellipsoid is sphere nd the body is sphericl top. 5. One moment is zero. The ellipsoid is n infinite ellipticl cylinder, nd the body is liner top. Emple. We know from section.5 tht the moment of inerti of plne squre lmin of side bout n is through its centroid nd perpendiculr to its re is m, nd it will hence be obvious tht the moment of inerti of uniform solid cube of side bout n is pssing through the mid-points of opposite sides is lso m. It will clerly be the sme bout n is pssing through the mid-points of ny pirs of opposite sides. Therefore the cube is sphericl top nd the momentl ellipsoid is sphere. Therefore the moment of inerti of uniform solid cube bout ny is through its centre (including, for emple, digonl) is lso m. Emple. Wht is the rtio of the length to the dimeter of uniform solid cylinder such tht it is sphericl top? [Answer: I mke it /. 866.] Let us note in pssing tht ( + y + z ), A + B + C m mr.6. which is independent of the orienttion of the bsis es In other words, regrdless of how A, B nd C my depend on the orienttion of the es with respect to the body, the sum A + B + C is invrint under rottion of es. We shll del with the determintion of the principl es in section.8 - but don't skip section.7..7 Solid Body Rottion. The Inerti Tensor.

34 4 It is intended tht this chpter should be limited to the clcultion of the moments of inerti of bodies of vrious shpes, nd not with the huge subject of the rottionl dynmics of solid bodies, which requires chpter on its own. In this section I mention merely for interest two smll topics involving the principl es, nd third topic in bit more detil s necessry before proceeding to section.8. Everyone knows tht the reltion between trnsltionl kinetic energy nd liner momentum is E p /( m). Similrly rottionl kinetic energy is relted to ngulr momentum L by E L /( I ), where I is the moment of inerti. If n isolted body (such s n steroid) is rotting bout non-principl is, it will be subject to internl stresses. If the body is nonrigid this will result in distortions (strins) which my cuse the body to vibrte. If in ddition the body is inelstic the vibrtions will rpidly die out (if the dmping is greter thn criticl dmping, indeed, the body will not even vibrte). Energy tht ws originlly rottionl kinetic energy will be converted to het (which will be rdited wy.) The body loses rottionl kinetic energy. In the bsence of eternl torques, however, L remins constnt. Therefore, while E diminishes, I increses. The body djusts its rottion until it is rotting round its is of mimum moment of inerti, t which time there re no further stresses, nd the sitution remins stble. In generl the rottionl motion of solid body whose momentl ellipse is triil is quite complicted nd chotic, with the body tumbling over nd over in pprently rndom fshion. However, if the body is nonrigid nd inelstic (s ll rel bodies re in prctice), it will eventully end up rotting bout its is of mimum moment of inerti. The time tken for body, initilly tumbling choticlly over nd over, until it reches its finl blissful stte of rottion bout its is of mimum moment of inerti, depends on how fst it is rotting. For most irregulr smll steroids the time tken is comprble to or longer thn the ge of formtion of the solr system, so tht it is not surprising to find some steroids with non-principl is (NPA) rottion. However, few rpidly-rotting NPA steroids hve been discovered, nd, for rpid rottors, one would epect PA rottion to hve been reched long time go. It is thought tht something (such s collision) must hve hppened to these rpidly-rotting NPA steroids reltively recently in the history of the solr system. Another interesting topic is tht of the stbility of rigid rottor tht is rotting bout principl is, ginst smll perturbtions from its rottionl stte. Although I do not prove it here (the proof cn be done either mthemticlly, or by qulittive rgument) rottion bout either of the es of mimum or of minimum moment of inerti is stble, wheres rottion bout the intermedite is is unstble. The reder cn observe this for him- or herself. Find nything tht is triil - such s smll block of wood shped s rectngulr prllelepiped with unequl sides. Identify the es of gretest, lest nd intermedite moment of inerti. Toss the body up in the ir t the sme time setting it rotting bout one or the other of these es, nd you will be ble to see for yourself tht the rottion is stble in two cses but unstble in the third. I now del with third topic in rther more detil, nmely the reltion between ngulr momentum L nd ngulr velocity ω. The reder will be fmilir from elementry (nd twodimensionl) mechnics with the reltion L Iω. Wht we re going to find in the three-

35 5 dimensionl solid-body cse is tht the reltion is L Iω. Here L nd ω re, of course, vectors, but they re not necessrily prllel to ech other. They re prllel only if the body is rotting bout principl is of rottion. The quntity I is tensor known s the inerti tensor. Reders will be fmilir with the eqution F m. Here the two vectors re in the sme direction, nd m is sclr quntity tht does not chnge the direction of the vector tht it multiplies. A tensor usully (unless its mtri representtion is digonl) chnges the direction s well s the mgnitude of the vector tht it multiplies. The reder might like to think of other emples of tensors in physics. There re severl. One tht comes to mind is the permittivity of n nisotropic crystl; in the eqution D εe, D nd E re not prllel unless they re both directed long one of the crystllogrphic es. If there re no eternl torques cting on body, L is constnt in both mgnitude nd direction. The instntneous ngulr velocity vector, however, is not fied either in spce or with respect to the body - unless the body is rotting bout principl is nd the inerti tensor is digonl. So much for preview nd qulittive description. Now down to work. I m going to hve to ssume fmilirity with the eqution for the components of the cross product of two vectors: ˆ ˆ ˆ.7. A B ( A B A B ) + ( A B A B ) y + ( A B A B ) z. y z z y z I m lso going to ssume tht the reder knows tht the ngulr momentum of prticle of mss m t position vector r (components (, y, z) ) nd moving with velocity v (components ( &, y&, z& )) is mr v. For collection of prticles, (or n etended solid body, which, I'm told, consists of collection of prticles clled toms), the ngulr momentum is L m r v [ m ( yz& zy& ) ˆ + m( z& z& ) yˆ + m( y& y& ) zˆ ] I lso ssume tht the reltion between liner velocity v ( & y&, z& ), nd ngulr velocity ω, ω, ω is understood to be v ω r, so tht, for emple, z& ω y ω. Then ω ( ) y z L [ m( y( ω y ω ) z( ω ω z) ) ˆ + ( etc. ) yˆ ( etc. ) zˆ ] Finlly, we obtin y z + ( ω my ω my ω mz + ω mz ) + etc. y A ˆ ˆ + ˆ. ( ω Hω Gω ) + ( ) y ( )z y z z z ˆ y y y

36 6 L A H G ω L Ly H B F ω y.7. L G F C z ω z This is the eqution L Iω referred to bove. form The inerti tensor is sometimes written in the I I I I y z I I I y yy yz I I I z yz zz, so tht, for emple, I y H. It is symmetric mtri (but it is not n orthogonl mtri)..8. Determintion of the Principl Aes. We now need to ddress ourselves to the determintion of the principl es. Unlike the twodimensionl cse, we do not hve nice, simple eplicit epression similr to eqution.. to clculte the orienttions of the principl es. The determintion is best done through numericl emple. Consider four msses whose positions nd coordintes re s follows: M y z Reltive to the first prticle, the coordintes re From this, it is esily found tht the coordintes of the centre of mss reltive to the first prticle re (.7,.9,.), nd the moments of inerti with respect to es through the first prticle re

37 A 4 B 64 C 8 F 5 G H 7 From the prllel es theorems we cn find the moments of inerti with respect to es pssing through the centre of mss: The inerti tensor is therefore A 6. B 5. C 5. F G 6.. H We understnd from wht hs been written previously tht if ω, the instntneous ngulr velocity vector, is long ny of the principl es, then Iω will be in the sme direction s ω. In l, m, n re the direction cosines of principl is, then other words, if ( ) A H G H B F G l F m C n l λm, n where λ is sclr quntity. In other words, vector with components l, m, n (direction cosines of principl is) is n eigenvector of the inerti tensor, nd λ is the corresponding principl moment of inerti. There will be three eigenvectors (t right ngles to ech other) nd three corresponding eigenvlues, which we ll initilly cll λ, λ, λ, though, s soon s we know which is the lrgest nd which the smllest, we'll cll A, B, C, ccording to our convention A B C. The chrcteristic eqution is

38 8 A λ H G H B λ F G F C λ. In this cse, this results in the cubic eqution + λ + λ λ, where The three solutions for λ, which we shll cll A, B, C in order of incresing size re A B C nd these re the principl moments of inerti. From the theory of equtions, we note tht the sum of the roots is ectly equl to, nd we lso note tht it is equl to A + B + C, consistent with wht we wrote in section.6. (See eqution.6.) The sum of the digonl elements of mtri is known s the trce of the mtri. Mthemticlly we sy tht "the trce of symmetric mtri is invrint under n orthogonl trnsformtion". Two other reltions from the theory of equtions my be used s check on the correctness of the rithmetic. The product of the solutions equls, which is lso equl to the determinnt of the inerti tensor, nd the sum of the products tken two t time equls. We hve now found the mgnitudes of the principl moments of inerti; we hve yet to find the direction cosines of the three principl es. Let's strt with the is of lest moment of inerti, for which the moment of inerti is A Let the direction cosines of this is be l, m n. Since this is n eigenvector with eigenvlue we must hve ( ), l 6. m 5. n l m n These re three liner equtions in l m, n, with no constnt term. Becuse of the lck of constnt term, the theory of equtions tells us tht the third eqution, if it is consistent with the other two, must be liner combintion of the first two. We hve, in effect, only two independent equtions, nd we re going to need third, independent eqution if we re to solve

39 9 for the three direction cosines. If we let l' l / n nd m' m / n, then the first two equtions become l ' +.7m'..7l' m' 6.. The solutions re l' m' The correctness of the rithmetic cn nd should be checked by verifying tht these solutions lso stisfy the third eqution. The dditionl eqution tht we need is provided by Pythgors's theorem, which gives for the reltion between three direction cosines l + m + n, or n, l' + m' + whence n! Thus we hve, for the direction cosines of the is corresponding to the moment of inerti A, (Check tht l + m + n. ) l m n m ±. 88 ± It does not mtter which sign you choose - fter ll, the principl is goes both wys. Similr clcultions for B yield l m n ± m.9 76 ± nd for C

40 l 4 ± m ± n m For the first two es, it does not mtter whether you choose the upper or the lower sign. For the third es, however, in order to ensure tht the principl es form right-hnded set, choose the sign such tht the determinnt of the mtri of direction cosines is +. We hve just seen tht, if we know the moments nd products of inerti A, B, C, F, G, H with respect to some es (i.e. if we know the elements of the inerti tensor) we cn find the principl moments of inerti A, B, C by digonlizing the inerti tensor, or finding its eigenvlues. If, on the other hnd, we know the principl moments of inerti of system of prticles (or of solid body, which is collection of prticles), how cn we find the moment of inerti I bout n is whose direction cosines with respect to the principl es re (l, m, n)? First, some geometry. Let Oyz be coordinte system, nd let P (, y, z ) be point whose position vector is r i + y j + z k. Let L be stright line pssing through the origin, nd let the direction cosines of this line be (l, m, n ). A unit vector e directed long L is represented by e li + m j + n k. The ngle θ between r nd e is found from the sclr product r e, given by r cos θ r e. I.e. ( + y + z ) cosθ l + my + nz. The perpendiculr distnce p from P to L is p ( + y + ) sinθ. rsinθ z If we write sin ( cos θ), θ we soon obtin ( l+ my nz). p + y + z + Noting tht l m n, m n l, n l m, we find, fter further mnipultion:

41 4 ( y + z ) + m ( z + ) + n ( + y ) ( mnyz + nlz lmy). p l + Now return to our collection of prticles, nd let Oyz be the principl es of the system. The moment of inerti of the system with respect to the line L is I Mp, where I hve omitted subscript i on ech symbol. Mking use of the epression for p nd noting tht the product moments of the system with respect to Oyz re ll zero, we obtin I l A + m B + n C..8. Also, let A, B, C, F, G, H be the moments nd products of inerti with respect to set of nonprincipl orthogonl es; then the moment of inerti bout some other is with direction cosines l, m, n with respect to these nonprincipl es is I l A + m B + n C mnf nlg lmh..8. Emple. A Brick. We sw in section 6 tht the moment of inerti of uniform solid cube of mss M nd side bout body digonl is M, nd we sw how very esy this ws. At tht time the problem of finding the moment of inerti of uniform solid rectngulr prllelepiped of sides, b, c must hve seemed intrctble, but by now it is not t ll hrd.

42 Thus we hve: A B C l m n M M 4 ( + c ) ( c + ) ( + b ) M b ( + b + c ) b ( + b + c ) c ( ). + b + c M ( b c + c + b ) We obtin: I ( + b + c ). We note: i. This is dimensionlly correct; ii. It is symmetric in, b, c; iii. If b c, it reduces to M..9 Moment of Inerti with Respect to Point. By moment of inerti we hve hitherto ment the second moment of mss with respect to n is. We were esily ble to identify it with the rottionl inerti with respect to the is, nmely the rtio of n pplied torque to the resulting ngulr ccelertion. I m now going to define the (second) moment of inerti with respect to point, which I shll tke unless otherwise specified to men the origin of coordintes. If we hve collection of mss points m i t distnces r i from the origin, I define i i i i i ( + y z ) I m r m +.9. s the (second) moment of inerti with respect to the origin, lso sometimes clled the geometric moment of inerti. I cnnot relte it in n obvious wy to simple dynmicl concept in the sme wy tht I relted moment of inerti with respect to n is to rottionl inerti, but we shll see tht it is by no mens merely tedious eercise in rithmetic, nd it does hve its uses. The symbol I hs probbly been used rther lot in this chpter; so to describe the geometric moment of inerti I m going to use the symbol I. i i i