Chapter 6 Rigid Body Dynamics

Transcription

1 Chapter 6 Rgd Body Dynamcs 6. Introducton In practce, t s often not possble to dealze a system as a partcle. In ths secton, we construct a more sophstcated descrpton of the world, n whch obects rotate, n adon to translatng. Ths general branch of physcs s called Rgd Body Dynamcs. Rgd body dynamcs has many applcatons. In vehcle dynamcs, we are often more worred about controllng the orentaton of our vehcle than ts path an arcraft must eep ts shny sde up, and we don t want a spacecraft tumblng uncontrollably. Rgd body mechancs s used extensvely to desgn power generaton and transmsson systems, from et engnes, to the nternal combuston engne, to gearboxes. typcal problem s to convert rotatonal moton to lnear moton, and vce-versa. Rgd body moton s also of great nterest to people who desgn prosthetc devces, mplants, or coach athletes: here, the goal s to understand human moton, to protect athletes from nury or mprove ther performance, or to desgn devces that replcate the complcated moton of a human ont correctly. For example, Professor Crsco s orthopaedcs lab at Brown studes human moton and the forces they generate at human onts, to help understand how nures occur and how they can be prevented. The moton of a rgd body s often very counter-ntutve. That s why there are so many toys that explot the propertes of rgd bodes: the moton of a spnnng top; a boomerang; the rattlebac and a Frsbee can all be explaned usng the equatons derved n ths secton. Here s a quc outlne of how we analyze moton of rgd bodes.. rgd body s dealzed as an nfnte number of very small partcles.. We already now the equatons of moton for a system of partcles (Secton 4 of the notes): N ext dp d The force-momentum equaton F = = mv = N ext dh d The moment angular momentum equaton r F = = r mv = N ext dt d The wor-netc energy equaton F v = = mv v = 3. These equatons tell us how a rgd body moves. But to use them, we would need to eep trac trac of an nfnte number of partcles! To smplfy the problem, we set up some mathematcal methods that allow us to express the poston and velocty of every pont n a rgd body n terms of the poston r G, velocty v G and acceleraton a G of ts center of mass, and ts rotaton tensor R(quantfyng ts orentaton) and ts angular velocty ω, and angular acceleraton α. Ths allows us to wrte the lnear momentum, angular momentum, and netc energy of a rgd body n the form p= Mv G h= rg MvG + IGω T = MvG vg + ω IGω where M s the total mass of the body and I G s ts mass moment of nerta. 4. We can then derve the rgd body equatons of moton: ext ext F = Ma r F = Mr a + I α+ ω I ω [ ] G G G G G

2 6. Descrbng Moton of a Rgd Body We descrbe moton of a partcle usng ts poston, velocty and acceleraton. We can descrbe the poston of a rgd body n the same way - we could specfy the poston, velocty and acceleraton of any convenent pont n the body (we usually use the center of mass). But we also need a way to descrbe the orentaton of a rgd body, and ts rotatonal moton. In ths secton, we defne the varous mathematcal quanttes that we use to descrbe rotaton, angular velocty, and angular acceleraton. 6.. Descrbng rotatons: The Rotaton Tensor (or matrx) Rotatons are quantfed by a mathematcal obect called a rotaton tensor. It s defned as follows:. Choose some convenent ntal orentaton of the rgd body (eg for the rectangular prsm n the fgure, we chose to mae the faces perpendcular to the {,, } drectons.. When the body s rotated, every lne n the body (eg the sdes) moves to a new orentaton, wthout changng ts length. We can descrbe ths orentaton change as a mappng. Let and B be two arbtrary ponts n the body. Let p, p B be the ntal postons of these ponts, and let r, r B be ther fnal postons. We ntroduce the rotaton tensor R whch has the property that r r = Rp ( p ) B B When we solve problems, we always express vectors as components n some bass. When we do ths, R becomes a matrx. For example, f pb p = x0+ y0+ z0 rb r = x+ y+ z we would wrte x Rxx Rxy Rxz x0 y = Ryx Ryy R yz y 0 z R z yz Rzy R zz 0 Here, R, R,... are a set of nne numbers (or sometmes formulas). Followng the usual rules of matrxvector multplcaton, ths s ust a short-hand notaton for x= R x + R y + R z xx 0 xy 0 xz 0 y = R x + R y + R z yx 0 yy 0 yz 0 z = Rzxx0 + Rzy y0 + Rzz z0 The subscrpts on R are meant to you help remember what each element n the matrx does for example, R xx maps the x 0 onto x, R xy maps the y 0 onto x, and so on. B p B -p r B -r B By defnton, a second order tensor maps a vector onto another vector. In actual calculatons R s always ust a matrx, but tensor sounds better.

3 3 So when we solve a problem, how do we go about fndng R? Let me count the ways: Rotatons n two dmensons: Lfe s smple n D. In ths case our rgd body must le n the, plane, so we can only rotate t about an axs parallel to the drecton. counterclocwse rotaton through an angle θ about the axs s produced by cosθ snθ R = snθ cosθ For example, a vector L that start parallel to the axs s mapped to cosθ snθ L Lcosθ Lcosθ Lsnθ snθ cosθ 0 = = + Lsnθ p B -p r B -r B B θ Rotaton about a nown axs 3D s a bt more dffcult. ny rotaton can always be expressed as a rotaton through some angle θ about some axs parallel to a unt vector n (we always use the rght hand screw conventon). In some problems you can see what n and θ are: then you can wrte down a unt vector parallel to n n= nx+ ny+ nz and then use the Rodrguez Formula cos θ + ( cos θ) nx ( cos θ) nn x y sn θnz ( cos θ) nn x z + snθn y R = ( cos θ) nn x y + snθnz cos θ + ( cos θ) ny ( cos θ) nn y z snθn x ( cos θ) nn x z sn θny ( cos θ) nn y z + snθnx cos θ + ( cos θ) n z (Ths formula s mpossble to remember that s what Google s for). If you are gven a rotaton matrx R, and need to fnd n and θ, you can use the formulas: + cosθ = R + R + R xx yy zz ( Rzy Ryz ) ( Rxz Rzx ) ( Ryx Rxy ) n = + + snθ The second formula blows up f sn( θ ) = 0. If θ s zero or π you can smply set R = (the dentty), and n can be anythng you le. For θ = π you can use R cos cos xx θ Ryy θ Rzz cosθ n= ± ± cosθ cosθ cosθ The sgns of the square roots have to be chosen so that n (Tp: t s easy to remember ths but t s hard to remember where to put the negatve sgn. You can always fgure ths out by notng that a 90 degree counter-clocwse rotaton maps a vector parallel to the drecton onto a vector parallel to the drecton.)

4 4 nn x y = Rxy / nn x z = Rxz / nn y z = Ryz / In robotcs, game engnes, and vehcle dynamcs the axs-angle representaton of a rotaton s often stored as a quaternon. We won t use that here, but menton t n passng n case you come across t n practce. quaternon s four numbers [ q0, qx, qy, q x] that are related to n and θ through the formulas: q0 = cos( q / ) q = n sn( q / ) q = n sn( q / ) q = n sn q / Mappng the coordnate axes x x y y z z ( ) In some problems we mght now what happens to vectors that are parallel to the {,,} drectons n the ntal rgd body (eg we mght now what happens to the sdes of our rectangular prsm). For example, we mght now that {,, } map to (unt) vectors abc,,. In that case we can wrte down each of abc,, as components n {,, } a= ax+ ay+ az b= bx+ by+ bz c= cx+ cy+ cz and use the formula ax bx cx R = ay by cy az bz c z a c b sequence of rotatons Suppose we rotate an obect twce (perhaps about two dfferent axes). How do we descrbe the result of two rotatons? That s not hard. Suppose we do the frst rotaton wth one mappng rb r = R () ( pb p ) Now we rotate our body agan ths maps rb r onto some new vector ub u : () ( ub u) = R ( rb r ) We can therefore wrte () () ( ub u) = R R ( pb p ) We see that Sequental rotatons are matrx products () () R = R R Health warnng: Matrx products (and hence sequences of rotatons) do not commute () () () () R R R R

5 5 For example, the fgure below shows the change n orentaton caused by (a) a 90 degree postve rotaton about followed by a 90 degree postve rotaton about (the fgure on the left); and (b) a 90 degree postve rotaton about followed by a 90 degree postve rotaton about (the fgure on the rght). () rotaton () rotaton () rotaton () rotaton b Orthogonalty of R The rotaton tensor (matrx) has a very mportant property: If you multply R by ts transpose, the result s always the dentty matrx. nother way to say ths s that The transpose of R s equal to ts nverse Let s try ths wth the D rotaton matrx T cosθ snθ cosθ snθ cos θ + sn θ 0 0 RR = snθ cosθ snθ cosθ = = 0 sn θ cos θ 0 + T cosθ snθ cosθ snθ 0 R R = = snθ cosθ snθ cosθ 0 matrx or tensor wth ths property s sad to be orthogonal. Why s ths? It turns out that a length-preservng mappng must be an orthogonal tensor. To see ths, let s calculate the length of the rotated vector rb r = Rp ( B p ). We need to remember two vector/matrx operatons:. We can calculate the length of a vector by dottng t wth tself and tang the square root Ru Ru = u R T Ru. For a vector u and a matrx R, we now (or can show!) that ( ) ( ) ( ) Ths means

6 6 ( rb r) ( rb r) = { R( pb p) } { R( pb p) } = ( pb p) { R T R( pb p ) } But we want the length of rb r to equal the length of pb p, whch means we need R to satsfy T { } T { } { } T {( ) } ( pb p) R R( pb p) = ( pb p) ( pb p) ( pb p) R R( pb p) ( pb p) ( pb p) = 0 ( pb p) R R ( pb p) = 0 where s the dentty tensor (we normally use I for the dentty tensor, but rgd body dynamcs uses I to denote the mass moment of nerta so t s already been taen.). Ths must hold for all vectors so R R =. In fact, a rgorous mathematcal dervaton of rotatons starts wth the statement that R must preserve the length of all vectors, and then derves all the other materal n ths secton from that statement. Ths s not easy to follow the frst tme around, but wll probably be the approach used n more advanced courses. T Examples:. Wrte down the rotaton matrx for the D rotaton shown n the fgure The obect rotates 90 degree counterclocwse about the axs, so p B -p B B r B -r cosθ snθ 0 R = = snθ cosθ 0. The obect shown n the fgure s frst rotated 90 degrees about the axs, and then 80 degrees about the axs. Fnd the rotaton tensor. We can construct the two rotatons usng the Rodrguez formula. For the frst rotaton θ = π / n= n = n = n = 0 x y z 0 0 () R = = π, n= ny = nx = nz = 0 For the second rotaton θ The total rotaton s therefore 0 0 () R =

7 () () R = R R = = Fnd the axs-angle representaton for the combned rotaton n problem (). We can calculate the axs and angle of ths rotaton usng the formulas + cosθ = R + R + R cosθ = θ = π xx yy zz R cos cos xx θ Ryy θ Rzz cosθ n= ± ± cosθ cosθ cosθ ( ) 0 ( ) 0 ( ) = ± ± = ± ( ) ( ) ( ) ( ) To decde whch of these two choces to use we notce that R yz =, whch tells us that nn y z < 0. The answer s therefore θ = π, n= ( ) It s ncredbly dffcult to vsualze the effect of a rotaton about an arbtrary axs (at least for me). In fact ths formula loos wrong how can a 80 degree rotaton end up tppng the box on ts sde? But the answer s rght, as the anmaton (whch wll only show up n the html verson of the notes) shows. 6.. Descrbng rotatonal moton: The angular velocty vector and spn tensor We descrbed the locaton of a partcle n space usng ts poston vector, and ts moton usng velocty. We need to come up wth somethng smlar to velocty for rotatons. Defnton of an angular velocty vector Vsualze a spnnng obect, le the cube shown n the fgure. The box rotates about an axs n the example, the axs s the lne connectng two cube dagonals. In adon, the obect turns through some number of revolutons every mnute. We would specfy the angular velocty of the shaft as a vector ω, wth the followng propertes:. The drecton of the vector s parallel to the axs of the shaft (the axs of rotaton). Ths drecton would be specfed by a unt vector n parallel to the shaft.. There are, of course, two possble drectons for n. By conventon, we always choose a drecton such n xs of rotaton

8 8 that, when vewed n a drecton parallel to n (so the vector ponts away from you) the shaft appears to rotate clocwse. Or conversely, f n ponts towards you, the shaft appears to rotate counterclocwse. (Ths s the `rght hand screw conventon ) Vewed along n Vewed n drecton opposte to n 3. The magntude of the vector s the angular speed dθ / of the obect, n radans per second. If you now the revs per mnute n turned by the shaft, the number of radans per sec follows as dθ / = 0πn. The magntude of the angular velocty s often denoted by ω = dθ / The angular velocty vector s then dθ ω= n= ωn. Snce angular velocty s a vector, t has components ω= ωx+ ωy+ ωz n a fxed Cartesan bass. s always, n two dmensons, everythng s very smple. In ths case obects can only rotate about the axs, and we can wrte the angular velocty vector as dθ ω= where θ s the counterclocwse angle of rotaton of any lne embedded n the body. Wrtng down angular veloctes: For D problems, we always now the drecton of the angular velocty and can ust use down (of course f we now the value or a formula for ω z we can use t). ω= ωz to wrte t For 3D problems, we can often use vector adon to wrte down ω. example: We can llustrate ths wth a smple Example: The propeller on the arcraft shown n the fgure spns (about ts axs) at 000 rpm. The arcraft travels at speed 00 m/hr n a turn wth radus m. What s the angular velocty vector of () the body of the arcraft, and () the propeller? Express your answer n the normal-tangentalvertcal bass. () Note that: The crcumference of the crcle s s= πr= πm. The arplane completes a full crcle n t = s/ V = ( π / 00) 3600 = 36πsec.

9 9 full turn s π radans, so the arcraft body turns at a rate π / (36 π ) = (/8) rad/s about the axs. () The propeller turns at 000 rpm relatve to the body of the plane. The angular velocty of the prop wth respect to a statonary observer s therefore the vector sum of the 000 rpm about the t axs, plus the angular velocty of the body. Ths gves 0p ωprop = [ 000 p / 3600] t+ = t+ rad/s Relaton between the rotaton matrx and the angular velocty vector: the spn tensor We mght guess that the angular velocty vector s the dervatve of the rotaton tensor. Ths s sort of correct, but the full story s a bt more complcated. The relatonshp between R and ω s constructed as follows:. We defne the spn tensor W as d T W= R R T. The spn tensor s always sew ( W= W ), and we can read off the angular velocty vector by loong at ts components. Specfcally, f ω= ωx+ ωy+ ωz then 0 ωz ωy W = ωz 0 ωx ωy ωx 0 We can use ths formula n two ways: () Gven R, we can calculate W and then read off the angular velocty vector components. lternatvely, f we now ω, we can calculate R by frst constructng W, then ntegratng the formula dr = WR ngular velocty-rotaton relatons n D We can chec ths for the specal case of a D rotaton: dθ dθ dθ snθ cosθ 0 cosθ snθ dr T cosθ snθ R = snθ cosθ R = dθ dθ snθ cosθ = dθ cosθ snθ 0 dθ s expected, we fnd that ω z =. Ths means that n D, angular velocty and the angle of rotaton θ are related by the same formulas as dstance traveled and speed for poston. We can use all the same rules of calculus to go bac and forth between them.

10 0 ngular velocty-spn tensor formula There s an mportant formula relatng W and ω. Let rb r be a vector onng any two ponts n a rgd body. Then Wr ( r ) = ω ( r r ) B B You can see ths by ust multplyng out the defnton of W and comparng the result to the cross product: f rb r = x+ y+ z, then 0 ωz ωy x ωyz ωzx Wr ( B r ) = ωz 0 ω x y = ωzx ωxz ωy ωx 0 z ωxy ωyx Hopefully you can see that ths s the same as the cross product! 6..3 The angular acceleraton vector ngular acceleraton s the tme dervatve of angular velocty d α = ω For 3D, we can use dω dω x y dω α z x = αy = αz = For 3D, we can t express the angular acceleratons or veloctes as dervatves of rotaton angles, because these can t be defned for a general moton. For a D problem, the drecton of angular velocty and acceleraton are nown, so we have α= αz ω= ωz The components are related by dωz d θ dω α z z = = = ω z dθ For D problems, we can use all the usual rules of calculus to go from angular acceleraton to angular velocty to angle, and vce-versa (ust le dstance-speed-acceleraton formulas for straght lne moton) Relatve velocty and acceleraton of two ponts n a rgd body We now now how to descrbe rotatonal moton. Our next order of busness s to dscuss a couple of very mportant formulas that we use to analyze the moton of a system of rgd bodes, and also to derve formulas for the angular momentum and netc energy of a rgd body.. Consder a rgd body: r B -r v B,a B B v,a ω,α

11 Let ω be the (nstantaneous) angular velocty of the body, and W the correspondng spn tensor Let and B be two arbtrary ponts n a rgd body, and let r, r B and v, v B, a, a B be ther (nstantaneous) poston, velocty and acceleraton vectors. Then the relatve poston and velocty of and B are related by vb v = ω ( rb r) vb v = Wr ( B r) The relatve acceleraton of and B are related ther relatve postons and velocty by a a = α ( r r ) + ω ( v v ) = α ( r r ) + ω ω ( r r ) [ ] B B B B B For D problems only: we can smplfy these, because we now ω s always parallel to the drecton. Therefore v v = ω ( r r ) B z B B = αz B ωz B α α ( r r ) ( r r ) Proof: These fornulas are easy to prove. Remember the mappng: B = ( B ) B = d d ( B ) ( B ) = R r r R p p v v r r p p lso, T T rb r = R( pb p) R ( rb r) = R R( pb p) = ( pb p ) Hence d R T vb v = R ( rb r) = W( rb r ) Remember that Wr ( B r) = ω ( rb r ), so the acceleraton formula then follows as d d ω a a = v v = r r + ω d r r = α r r + ω v v ( ) ( ) ( ) ( ) ( ) B B B B B B v B,a B r B -r B v,a θ ω z α z 6.3 nalyzng moton n connected rgd bodes The formulas n 6..3 are used to analyze moton n machnes. typcal problem s llustrated n the fgure. n actuator moves pont B on the car ac shown n the fgure horzontally wth constant velocty V. What are the velocty and acceleraton of the platform (CF)? You could probably solve ths rather smple example wth elementary trg, but we need a more systematc method for general problems, especally to analyze 3D moton. Here s the general procedure F D L θ L B C V

12 . Defne varables to denote the unnown angular veloctes and angular acceleratons of each rgd body n the system. Wrte down all the nown veloctes n the system 3. Use the rgd body formulas vb v = ω ( rb r ) to wrte down equatons relatng veloctes of the connectons, onts, or contacts on each rgd body 4. Wrte down constrant equatons relatng veloctes of the two connected rgd bodes at each connecton, ont, or contact 5. Solve the equatons for unnown veloctes of connectons, and the angular veloctes of the rgd bodes. 6. Fnally, once the veloctes are nown, wrte down equatons for the acceleratons of pars of onts/contacts/connectons on each rgd body ab a = α ( rb r) + ω ( vb v ) 7. Wrte down constrants equatons for acceleratons at connected ponts 8. Solve the equatons n 6,7 for unnown acceleratons and angular acceleratons. Ths all sounds terrbly complcated, so let s solve a few examples to show how t wors n practce. Example : In the fgure shown the ln B rotates counterclocwse wth constant angular speed 4 rad/s. Pont C on member BC s constraned to move horzontally. Calculate the velocty and acceleraton of pont C. Calculatng the velocty: We now s statonary, and are gven the angular velocty of B. We can use the rgd body formula to fnd the velocty of B: vb v = ωzb ( rb r) = 4 v = 8 B m B m C We don t now the angular velocty of BC, so we ntroduce ω zbc as an unnown, and use the rgd body formula for member BC to wrte down an equaton for the velocty of C vc vb = ωzbc ( rc rb ) = ωzbc ( ) = ωbc+ ωbc v = 8+ ω + ω C BC BC We now that C can only move horzontally. Ths means that ts component of velocty must be zero. Ths shows that ω = 0, v = 8 Calculatng the acceleraton: BC We now s statonary, and are gven the angular velocty and angular acceleraton of B. We can use the rgd body formula to fnd the acceleraton of B: C

13 3 αb α = αzb ( rb r) ωzb ( rb r) = 3 αb = 3 We don t now the angular acceleraton of BC, so we ntroduce α zbc as an unnown and use the rgd body formula for member BC to wrte down an equaton for the acceleraton of C αc αb = αzbc ( rc rb ) ωzb ( rc rb ) = αzbc ( ) 0 αc = 3+ αbc+ αzbc Pont C can only move horzontally, so t can t have any vertcal acceleraton. Ths means that the component of acceleraton s zero: αzbc 3 = 0 αzbc = 6 α = 3 C Example : For a more complcated example, we can solve the car ac problem posed at the start of ths secton. n actuator moves pont B on the car ac shown n the fgure horzontally wth constant velocty V. What are the velocty and acceleraton of the platform (CF)? The system contans 3 rgd bodes (C, BD, CF 3 ). We don t now the angular veloctes or acceleratons of any of them, so we denote them by unnowns ω, ωzbd ω zcf, α zc, αzbd α zcf zc F D L θ L B C V Calculatng the velocty: We start at pont(s) wth nown velocty: s statonary, and the velocty of B s gven: v = 0 vb = V Pont E les on both member C and on member BD. We use the rgd body formulas to wrte down an equaton for the velocty of E on each member (notce we use the D equatons): ve v = ωzc ( re r) ve vb = ωzbd ( re rb ) The two members C and BD are pnned together at E and so must have the same velocty. We can elmnate v E and wrte out the poston vectors n, components ωzc ( Lcos30+ Lsn 30 ) V= ωzbd ( Lcos30+ Lsn 30 ) ( ωzc Lsn 30 V) + ωzc Lcos30= ωzbdlsn 30 ωzbdlcos30 The, components gve two equatons for ω zc, ω zbd 3 You may be wonderng why only a sngle pont was defned at C and E, but there are two ponts at D and F. That s because at C and E the members are pnned together, but there s a roller at D. t E, members C, BD always have the same velocty and acceleraton we can ust use a sngle varable to denote the velocty of ths pont. The same s true at C. Members CF and BD touch at F and D, but pont D on B does not have the same horzontal velocty as pont F CF, so we need to be able to dstngush between them.

14 4 ωzc Lsnθ V = ωzbdlsnθ ωzc Lcosθ = ωzbdlcosθ ωzc Lsnθcosθ Vcosθ = 0 ωzc = V / (Lsn θ) ωzbd = V / (Lsn θ) We can now use the rgd body formulas for members C and BD to fnd the veloctes of C and D V vc v = ωzc ( rc r) vc = (Lcosθ+ Lsn θ) = V Vcotθ Lsnθ V vd vb = ωzbd ( rd rb ) vd = V+ (Lcosθ+ Lsn θ) = Vcotθ Lsnθ We can use the rgd body formula for CF to relate the veloctes of C and F vf vc = ωzcf ( rf rc ) vf = V Vcotθ ωzcf Lcosθ Pont D on CD and pont F on CF must have the same vertcal velocty (the roller at D allows ther horzontal veloctes to dffer). Ths can be expressed as vf = vd Vcotθ ωzcf Lcosθ = Vcot θ ωzcf = 0 ll ponts on CF therefore have the same velocty (equal to the velocty of C) vcf = V Vcotθ Calculatng the acceleraton. We can now calculate the acceleratons. We start at a nown pont: Ponts and B have zero acceleraton. We can use the rgd body formula to calculate the acceleraton of E on each of C and BD: αe α = αzc ( re r) ωzc ( re r) αe αb = αzd ( re rb ) ωzd ( re rb ) The two members are connected at E and so must have the same acceleraton there. Ths shows that αzc ( Lcosθ+ Lsn θ) ωzc ( Lcosθ+ Lsn θ) = αzd ( Lcosθ+ Lsn θ) ωzd ( Lcosθ+ Lsn θ) αzc ( Lcosθ Lsn θ) ωzc ( Lcosθ+ Lsn θ) = αzd ( Lcosθ Lsn θ) ωzd ( Lcosθ+ Lsn θ) The, components gve two equatons for the unnown angular acceleratons: αzc Lsnθ ωzc Lcosθ = αzdlsnθ + ωzdlcosθ αzc Lcosθ ωzc Lsnθ = αzdlcosθ ωzdlsnθ αzc Lsnθcos θ = ωzdl(cos θ sn θ) + ωzc L 3 αzc = V cos θ / (4L sn θ) 3 αzd = αzc = V cos θ / (4L sn θ) We can use the rgd body acceleraton formulas to calculate the veloctes of D and C:

15 5 C = αzc ( C ) ωzc ( D ) α α r r r r V cosθ V αc = (Lcosθ Lsn θ) (Lcosθ+ Lsn θ) 3 4L sn θ 4L sn θ V cosθ V αc = L 3 sn θ L sn θ V αd αb = αzd ( rd rb ) ωzd ( rd rb ) αd = L 3 sn θ We can use the rgd body formula to relate the acceleratons of C and F αf αc = αzcf ( rf rc ) ωzcf ( rf rc ) V cosθ V αf = + α ( cos ) 3 zcf L θ L sn L sn θ θ Fnally, we now that D and F must have the same vertcal acceleraton (so they reman n contact). Ther horzontal acceleratons may dffer, because of the roller attached to D. Ths gves α = α D F V V α cos 0 3 zcf L θ = α 3 CF = L sn θ L sn θ Snce CF has zero angular velocty and angular acceleraton, all ponts on CF have the same acceleraton (whch must equal that of pont C). Therefore V cosθ V acf = L 3 sn θ L sn θ 6.3. Summary of constrant equatons at onts and contacts s the examples n the precedng secton show, the eys to analyzng moton n a system of connected rgd bodes are: () the formulas for relatve velocty and acceleraton of two ponts n a rgd body, and () constrants that relate the veloctes and acceleratons on two bodes at ponts where they touch. There are three common types of connecton between rgd bodes:. pn ont: the two connected members must have the same velocty and acceleraton at the connected pont v = v a = a B B B. slder ont: the two connected members must have the same velocty and acceleraton normal to the slder v n= v n a n= a n B B B n

16 6 3. Contact between two obects wthout relatve slp (sldng) at the contact (frcton forces must act to prevent the slp, n general): The veloctes of the touchng obects must be equal at the contact pont. The tangental components of acceleraton must also be equal (the normal components of acceleraton dffer) v = v a t = a t B B t B n 6.3. The Rollng Wheel Wheels are everywhere. They can be analyzed usng the general rgd body equatons, but t s helpful to be able to avod all the tedous cross products. In ths secton we summarze specal formulas for velocty and acceleraton of ponts on a wheel. Moton of a wheel rollng wthout slp on a statonary surface It s surprsngly dffcult to vsualze the moton of a wheel. The fgure above mght help: t shows the traectory of one pont on the crcumference of the wheel. The pont traces qute a complcated path. The mportant thng to notce s: If a wheel rolls wthout slp on a statonary surface, the pont touchng the surface s statonary Each pont s only n contact wth the ground for an nstant, and whle t touches the ground t has a large vertcal acceleraton, but t s nstantaneously statonary. We now ths from the lst of constrants n Sect 6.3., of course, but t s stll not an easy thng to vsualze. More generally, the ground need not necessarly be statonary (or the wheel could touch another surface). In ths case we now that the contactng ponts on two bodes n rollng contact have equal velocty at the contact. ngular velocty-lnear velocty formula: Wth ths nsght, we can use the rgd body formulas to calculate the nstantaneous velocty vector for any pont on the wheel. ssume that The wheel rolls wth angular velocty ω= ωz counterclocwse rotaton s postve. The center of the wheel moves wth velocty vo = vxo The rollng wheel formula gves v = ω R xo z ωz C R O v xo

17 7 To see ths, you can smply use the rgd body formula to go from the contact pont (whch s statonary) to O v v = ω ( r r ) v = ω ( R) = ω R O C O C O z z More generally, we can calculate the velocty of any pont on the wheel we mght be nterested n. In fact, we can ust wrte down the velocty of any pont n the wheel by notcng that nstantaneously all ponts are n crcular moton about the contact pont (ust magne the ds s rotatng about C). See f you can show all the followng: v = ω z R ( + ) vd = ωzr v B = ω z R ( ) Notce that the drecton of the velocty at each pont s always perpendcular to the lne connectng to the pont to C. ωz D R O C B θ ngular acceleraton-lnear acceleraton formula: ssume that The wheel rolls wth angular acceleraton α= αz counterclocwse rotaton s postve. The center of the wheel moves wth acceleraton ao = axo The rollng wheel formula gves a = a R xo z ωz α z R O a xo You can derve ths formula n two dfferent ways: () Dfferentate the velocty formula vxo = ωzr wth respect to tme () Use the rgd body formula: ( α α ) = α ( r r ) ω ( r r ) O C O C z O C αo = αc αzr ωzr We now that the component of acceleraton at pont C has to be the same as the component of acceleraton of the ground (.e. zero). (The components don t have to be equal). We also now that O has no acceleraton, because t remans at the same heght above the ground. Therefore xo= yc az ωz xo az yc ωz a a R R a = R a = R We can calculate the acceleraton of any other pont on the ds usng the rgd body formula. C Example: The bloc B has horzontal acceleraton a and horzontal speed v. Calculate the angular velocty and angular acceleraton of the rollers. Then, calculate the lnear velocty and acceleraton of O To solve problems le ths we use two deas: () the formulas relatng velocty and acceleratons of ponts on the ds; and () ωz α z B v,a R R O O C

18 8 the tangental velocty and acceleraton of contactng ponts are equal. Here, we now the tangental velocty at C s zero; the tangental velocty at s v. We can use the wheel formulas vx = ωzr ωz = v/ ( R) Smlarly, the tangental acceleraton at s a. The rollng wheel formula gves a = a R a = a/ ( R) x z z To fnd the velocty and acceleraton at O, we can use vxo = ωzr= v/ a = a R= a/ Gears Gears can be analyzed n much the same way as a rollng wheel. Gears are used to ncrease or decrease angular veloctes (they act le mechancal amplfers): for example, n the anmaton the small gear s rotatng at twce the angular rate of the large one. They also modfy the torques (or moments) appled to the gears: f a gear system ncreases angular velocty, t reduces torque by the same factor (so the torque on the small gear n the anmaton s half that on the large one). Some clever gear systems can even be used to add angular veloctes (see the dscusson of epcyclc gears below. xo z There are many dfferent gear desgns. Here, we focus only on two-dmensonal spur gears. Spur gears have a rather complcated geometry, whch we don t have tme to dscuss n detal n ths course. They are desgned to behave le two wheels whch roll aganst each other wth no slp at the contact. The wheel radus s equal to the ptch crcle radus of the gears (whch s slghtly smaller than physcal dameter of the gears, because the teeth have to overlap). Gear manufacturers often specfy the number of teeth on a gear nstead of ts radus. The number of teeth and the radus have to be related, because the teeth have to be the same crcumferental dstance apart for the gear par to mesh. We analyze moton of gears usng two deas: () Two meshed gears must have equal veloctes at the pont where they touch. () The rgd body formula, relatng the velocty of ponts on the crcumference of the gear to the velocty of ts center: ωz B vc = vo + ωz ( rc r O) In practce we don t usually bother dong the cross product, and R nstead ust wrte down the velocty on the crcumference drectly usng the fgure provded: C O v = v O + ω z R v B = v O ω z R v C = v O ω z R D v D = v O + ω z R You don t have to remember these ust vsualze every pont on the gear movng n crcular moton (counterclocwse) around O, and wrte down the vectors (be careful wth sgns!). θ

19 9 Example : The left gear n the fgure rotates wth counterclocwse angular velocty ω z. The large gear has radus R and N teeth, the small one has radus R and N teeth. Calculate the angular velocty of the smaller gear. Note: The veloctes of the two touchng gears are equal at C The gear rotaton/velocty formula gves ωz R ω zr= ωzr = ωz R Notce that we assume both gears rotate counterclocwse. The formula tells us that the second gear has a negatve angular velocty ths means that t s actually rotatng clocwse. The anmaton at the top of ths secton confrms that ths ndeed s the case. Snce the dstance between two teeth s the same on each gear, the number of teeth must be proportonal to the radus. Ths means we can also wrte the gear rato n terms of N and N ωz N ω = N z ωz O R C ωz R O Example : n epcyclc gearbox s a specal arrangement of gears that has many applcatons. The setch shows a smple example. The gearbox can be drven n three dfferent places: one drve shaft s connected to the central sun gear (); the other s attached to the planet carrer, whch s oned to the center of the pnon gears B,C and D. The outer gear (E called the rng gear ) can also be drven separately. Epcyclc gearboxes are used n all automatc vehcle transmssons. They are also very useful n splt power drves, where two motors need to be connected together to drve a sngle axle. Hybrd vehcles, whch have both an electrc motor and an nternal combuston engne drvng the same axle, are one example. You can fnd a very nce descrpton of the Toyota Prus splt power transmsson here: the webste ncludes a Flash anmaton that lets you change the speeds of the motors n the system and vsualze the moton of the gears. The fgure shows a schematc dagram llustratng the general geometry and moton of the system. We have four rgd bodes: The central sun gear, radus R S, angular velocty ω zs The planet carrer, angular velocty ω zpc N S teeth, rotatng at The rng gear, radus R R, wth N R teeth, angular velocty ω zr The planet gear, radus r = ( RR RS)/, NP = ( NR NS)/ teeth, rotatng at angular velocty ω zp ω zs Sun gear R S R R Planet Carrer ω zpc Rng gear ω zp r Planet gear ω zr

20 0 In any applcaton, we are gven the angular velocty of two of the drve shafts (any two of ω zs, ω zpc, ω zr ), and must calculate the thrd. The planet gear s not connected to any drve shaft, so we usually don t care very much about ts angular speed, but we wll need to fnd to solve for the unnown one of ω zs, ω zpc, ω zr. Ths seems a terrbly dffcult problem, but t can be solved n a very smple way wth a trc. We start by solvng a smpler verson of the problem. Suppose that the planet carrer s statonary ( ω zpc =0) and the sun gear rotates wth angular speed ω zs (see the anmaton). What s the angular velocty of the rng gear? The sun gear and the planet gear are ust a standard gear par so we now that R ω S zs RS = ωzpr ωzp = ωzs r The two touchng ponts on the planet gear and the rng gear must have the same velocty, so (usng the rotatng gear formula) r ωzpr= ωzrrr ωr = ωzp R R We can elmnate ω zp to get the answer: R ω S zr = ωzs R R ωzp Now let s try the harder problem. The anmaton shows a general stuaton, where nonzero. How can we fnd now? ωzr Ths s dffcult to analyze because the center of the planet gear s not fxed, so t s hard for us to vsualze the moton, and the standard gear formulas don t wor. But we can smplfy the problem by analyzng moton n a reference frame that rotates wth the planet carrer. For example, magne attachng a vdeocamera to the planet carrer ths camera would show the planet carrer to be statonary, wth the surroundng world rotatng n the opposte drecton. The angular velocty of the planet carrer would be subtracted from all the other angular veloctes. In ths reference frame, we can use the result we ust calculated: ( ωzr ωzpc ) R = S ( ω ω ) R zs zpc R Ths result s general, and can be re-arranged to tell you the angular veloctes for any gven combnaton of ω zs, ω zpc and ω zr. ω zs, ω zpc are both

21 6.4 Lnear momentum, angular momentum and netc energy of rgd bodes In ths secton, we determne how to calculate the angular momentum and netc energy of a rgd body, and defne two mportant quanttes: () the center of mass of a rgd body (whch you already now), and () the Inerta tensor (matrx) of a rgd body. To eep thngs smple, we won t consder a general rgd body rght away. Instead, we wll calculate the lnear momentum, angular momentum, and netc energy of a system of N partcles that are connected together by rgd, massless lns. m 3 d 3 d 4 m 4 Defntons of nertal propertes: For ths system, we wll defne N The total mass M = m = The poston of the center of mass r N G= m r M = m r The poston vector of each mass relatve to the center of mass d = r r G d The velocty of the center of mass v G G = r d r G d r m The mass moment of nerta about the center of mass (a tensor, whch can be expressed as a matrx f we choose a coordnate system and set d = dx+ dy + dz ) IGxx IGxy IGxz dy + dz dxdy dxd z N I G = IGyx IGyy IGyz = m dxdy dx dz dyd + z = IGzx IGzy IGzz dxdz dydz dx + d y The mass moment of nerta s sometmes also wrtten n a more abstract but very compact way as N IG = d d d = Here, s the dentty tensor, and d symbol s called the dadc product of two vectors). ( m m ) d s a tensor wth components dxd x, dxd y, dxd z, etc (the Formulas for lnear and angular momentum and netc energy: We wll show that: The total lnear momentum s p= Mv G The total angular momentum (about the orgn) s h= rg MvG + IGω

22 The total netc energy s T = MvG vg + ω IGω These are actually general results that hold for all rgd bodes, as long as we use a more general defnton of M and I G. Smplfed formulas for two dmensons: For planar problems, d z = 0 (snce all the masses are n the plane), and ω= ωz. In ths case, we can use The total lnear momentum s p= Mv G The total angular momentum (about the orgn) s h= rg MvG + IGzzωz The total netc energy s T = MvG v G + IGzzωz Here I Gzz s ust the bottom dagonal term of the full nerta matrx (.e. ust a sngle number) N ( ) N IGzz = m dx + dy = m d = = r r G m m 3 d d 3 d r m Example : smple 3D assembly of masses s shown n the fgure. () Fnd the mass moment of nerta. m z By symmetry, the COM s at the orgn. The nerta tensor s therefore ( ml y y + ml z z) 0 0 I G = 0 ( ml x x + ml z z) ( ml x x + ml y y) L x m x m y L z L y m z L y L z m y L x m x () ssume that the COM s statonary (.e. the assembly rotates about the orgn). Fnd formulas for the angular momentum and netc energy of the system, n terms of the angular velocty components ωx, ωy, ω z The formula gves the angular momentum

23 3 ( ml y y + ml z z) 0 0 M 0 ( ml ml) ( ml x x + ml y y) ( ml y y ml z z) ωx ( ml x x ml z z) ωy ( ml x x ml y y) ωz ωx h= rg vg + IGω = x x + z z ωy ω z = Note that h s a vector. Importantly, h s not generally parallel to the angular velocty vector, as ths example shows. The netc energy s ( ml y y + ml z z) 0 0 ωx ωx T= MvG + ω IGω= ωy 0 ( ml x x + ml z z) 0 ωy ω z ω 0 0 z ( ml x x + ml y y) T= ml y y + ml z z ωx + ml x x + ml z z ωy + ml x x + ml y y ωz ( ) ( ) ( ) These results help us understand what the formulas are predctng. Note, for example, that: The mass moment of nerta always has the form mass*length. It has unts of g-m The mass moment of nerta s a measure of how mass s dstrbuted about the center of mass. n obect has a large nerta f the mass s far from the COM, and a small one f the mass s close to the COM. The matrx-vector products n the formulas for h and T are really ust a way of calculatng the velocty of each partcle n the system n a quc way. For example, suppose we rotate our assembly of masses about the axs wth angular velocty ω (see the anmaton). z Let s calculate the netc energy of the system, but wthout usng the rgd body formulas. The two blue masses are statonary, so they have no KE. The red and green mass are both movng n a crcle about the orgn. The crcular moton formula says ther speed s V = Rωz We can calculate the total netc energy usng the usual formula T = mv = mx( Lxωz) + my( Lyωz) = ( ml x x + ml y y) ωz Ths s the same result we calculated wth the rgd body formula.

24 4 Ths explans why the formula for I Gzz contans L x and L y - the I Gzz component eeps trac of how much energy or momentum s produced by a rotaton about the z axs. The energy and momentum depend on the dstances of the masses from the z axs whch of course depends on L x and L y. Fnally, note that we can nterpret the two terms n the formulas for momentum and KE as quantfyng (separately) the effects of translaton and rotaton h = rg MvG + IGω ngular momentum Translatonal + Rotatonal T = MvG vg + ω IGω Knetc energy s Translatonal + Rotatonal Ths helps explan why we can often dealze a system as a partcle. If the rotatonal term s neglgble, the angular momentum and netc energy of a rgd body s ust the same as that of a partcle located at the COM Dervng the lnear momentum formula N By defnton p= mv. We ca re-wrte ths as follows: = N N N = d d d m = r p v m m ( M G) M G = = = r = r = = v (we used the defnton of the COM to get the last result) 6.4. Dervng the angular momentum formula Start wth the defnton: N h= r mv = Note that r = r G + d and recall the relatve velocty formula v vg = ω ( r rg) = ω d. Ths means we can re-wrte the angular momentum as Note that N h= ( rg + d) m( vg + ω d) = N N N = m G G + m G + m r v d v d ω d = = = N N N N md = m r rg = mr rg m = MrG Mr G = 0 = = = = ( ) Fnally, recall the dreaded trple cross product formula a b c= ( a c) b ( a b) c Ths means that d ω d = ( d d ) ω d ( d ω )

25 5 We can expand ths out Ths shows that ωx ( dx + dy + d ) z dx ( d d ) ω d ( d ω) = ωy ( dx + dy + dz ) dy ωxdx + ωydy + ωzd z ωz ( dx dy dz ) d + + z dy + dz dxdy dxd z = dxdy dx + dz dyd z dxdz dydz dx + d y N md ω d = IGω = ( ) Fnally collectng terms gves the requred answer N N N h= m G G + m G + m = G M G + r v d v d ω d r v IGω = = = ω ω ω x y z Dervng the netc energy formula N T = mv v = We can use v vg = ω ( r rg) = ω d N N mv v = m( vg + ω d) ( vg + ω d) = = Recall that N N N = ( vg vg) m + vg ω m + m( ) ( ) = d ω d ω d = = N m = d 0 = and expand the dot product of two cross products usng the formula ( a b) ( c d) = ( a c)( b d) ( b c)( a d ) Ths shows that

26 6 ( )( ) ( ) ( ω d) ( ω d) = ω ω d d ω d dx dy d 0 0 z dx dxdy dxd ω z x + + ωx ωx = ωy 0 dx + dy + dz 0 ωy ωy dxdy dy dyd z ω z 0 0 dx dy d ω z z ω z + + dxdz dydz d z dy dz dxdy dxd ω z x + ωx = ω y dxdy dx dz dyd + z ω y ω z dxdz dydz dx d ω y z + = ω IGω ω ω ω x y z Calculatng the center of mass and nerta of a general rgd body It s not hard to extend the results for a system of N partcles to a general rgd body. We smply regard the body to be made up of an nfnte number of vanshngly small partcles, and tae the lmt of the sums as the partcle volume goes to zero. The sums all turn nto ntegrals. 3D problems: For a body wth mass densty ρ (mass per unt volume) we have that The total mass s M = ρdv V The poston of the center of mass s ρg = M ρ ρdv The mass moment of nerta about the center of mass s dy + dz dxdy dxd z I G = ρ dxd y dx dz d yd + z dv V dxdz dydz dx + d y where d= r r G V r G r d For D problems: We now the COM must le n the, plane and we don t need to calculate the whole matrx. For a body wth mass per unt area µ we can therefore use the formulas The total mass s M = µ d r G r d

27 7 The poston of the center of mass s rg = M r µd The mass moment of nerta about the center of mass s where d= r r G IGzz = µ ( dx + d y ) d Example : To show how to use these, let s calculate the total mass, center of mass, and mass moment of nerta of a rectangular prsm wth faces perpendcular to the,, axes: Frst the total mass (sort of trval) cba M = ρdxdydz = ρabc 000 a b Now the COM cba ρg = ( x y z ) ρdxdydz a bc a b c ab c ( a b c ) ρabc + + = + + = + + abc 000 c nd fnally the mass moment of nerta ( y b/ ) + ( z c/ ) ( x a/ )( y b/ ) ( x a/ )( z c/ ) cba IG = sym ( x a / ) + ( z c / ) ( y b / )( z c / ) ρdxdydz 000 sym sym ( x a / ) + ( y b / ) 3 3 ab c + abc = ρ 0 a bc + abc a bc + ab c b + c 0 0 M = 0 a + c a + b Example : s a second example, let s calculate the mass moment of nerta of a cylnder wth mass densty ρ, length L and radus a. (We now the COM s at the center and we now the total mass so we won t bother calculatng those). We have to do the ntegral wth polar coordnates θ a z r L/ L/

28 8 dy + dz dxdy dxd z L/ π a Cylnder I G = dxd y dx dz d yd + z rrdrdθdz L/ 0 0 dxdz dydz dx + d y Now note that dx rcosθ dy rsnθ dz z = = =. We can have Mupad do the drty wor: Example 3: Let s fnsh up wth a D example. Fnd the mass, center of mass, and out of plane mass moment of nerta of the trangle shown n the fgure. The total mass s b a( yb / ) M = µ dxdy = µ ab 0 0 The poston of the COM s b a( yb / ) r G = ( x y ) µ dxdy ( a b ) µ ab + = The D mass moment of nerta s b a( yb / ) a b abµ M IGzz = ( x ) + ( y ) µ dxdy = ( a + b ) = ( a + b ) b a x=a(-y/b)

29 9 Ths s all a bg pan, and you may be contemplatng a lfe of crme nstead of an engneerng career. Fortunately, t s very rare to have to do these sorts of ntegrals n practce, because all the ntegrals for common shapes have already been done. You can google most of them. The tables below gve a short lst of all the obects we wll encounter n ths course. Table of mass moment of nerta tensors for selected 3D obects Prsm M = ρabc c a b b + c 0 0 M 0 a + c a + b Sold Cylnder M = πρa L a L/ L/ + 3 a / L 0 0 ML a / L a / L Sold Cone π M = ρ a h 3 a h/4 h + h / (4 a ) 0 0 3Ma 0 + h / (4 a ) Sold Sphere 4 3 M = πρa 3 a Ma

30 30 Sold Ellpsod 4 M = πρabc 3 a b c M 5 b + c a + c a + b Hollow Cylnder M = πρ( b a ) L a b L/ L/ L + 3( a + b ) 0 0 M 0 L + 3( a + b ) ( a + b ) Table of mass moment of nerta about perpendcular axs for selected D obects Square b IGzz = a + b a M ( ) Ds R I Gzz M = R Thn rng R IGzz = MR

31 3 Hollow ds b a ( ) M IGzz = a + b Slender rod L I Gzz M = L a/3 Trangular Plate b b/3 M ( a + b ) 8 a The Parallel xs Theorem In all the prevous calculatons we have been calculatng the mass moment of nerta about the center of mass. Ths s what always appears n the angular momentum formula. But we sometmes want to fnd the mass moment of nerta about a dfferent pont (not the COM), for reasons that wll soon be clear. The mass moment of nerta about an arbtrary pont s defned exactly the same way as the nerta about the COM, except that we use the dstances from our arbtrary pont nstead of the dstance from the COM. ry+ rz rr xy rr xz I O= r rxry rx rz ryr + z dv V rr xz rr yz rx + r y r r O -r G O It s panful to have to re-do all these ntegrals, however. If we already now I G, the parallel axs theorem lets us calculate I O drectly. Defne the vector d that ponts from G to O d= ro r G Then for a 3D obect wth mass M

32 3 dy + dz dxdy dxd z IO = I G + M dxdy dx + dz dyd z dxdz dydz dx + d y For D we have a smpler result IOzz = IGzz + M( dx + dy ) r O -r G r O Example: Let s fnd the mass moment of nerta of a cylnder about axes that pass through one end of the cylnder (O), nstead of the COM. O L L a Here, d= dx = dy = 0 dz = z The formula gves r dy + dz dxdy dxd z θ IO = I G + M dxdy dx + dz dyd z dxdz dydz dx + d y + 3 a / L 0 0 L /4 0 0 ML = a / L 0 + M 0 L / a / L a / L 0 0 ML = a / L a / L L/ L/ Calculatng moments of nerta of complex shapes by summaton The most mportant applcaton of the parallel axs theorem s n calculatng the mass moment of nerta of complcated obects (whch don t appear n our table) by addng together moments of nerta for smple shapes. We can llustrate ths wth a couple of smple examples. 3a a 3a 3a

33 33 Example : Two spheres wth radus 3a are connected by a rgd cylnder wth length 6a and radus a to create a dumbbell. ll obects have the same mass densty ρ. Calculate the total mass moment of nerta of the dumbbell. The general approach s () Fnd the COM of the entre assembly () Fnd the mass moment of nerta of each shape (the spheres and the cylnder) about ts own COM (3) Use the parallel axs theorem to fnd the moment of nerta of each shape about the combned COM (4) dd all the moments of nerta For our problem () We now the COM s at the orgn by symmetry, so we don t need to calculate t () The nerta matrces of each obect (cylnder + sphere) about ther own COM are: ( 3a) 0 0 sphere 4p 3 IG = ( 3a) r 0 ( 3a) ( 3a) (6 a) + 3a 0 0 cylnder 3 IG = ( 6 pa r ) 0 (6 a ) + 3 a a (3) We don t need to use the parallel axs theorem for the cylnder, because ts COM s already at the same place as the COM of the assembly. For the spheres, we need to move the COM a dstance 6a parallel to the drecton. Ths means that d = d = 0, d = 6a n our formula. Therefore x y z ( 3a) sphere 4p 3 4p 3 I COM = ( 3a) r 0 ( 3a) 0 + ( 3a) r ( 3a) 0 0 ( 6a) (4) We can add everythng up (note that there are two spheres). Its best to use Mupad. The answer s 99a I COM = M 0 99a 0 M = 4πa ρ a

34 34 a a a a a a a Example : Thngs are a lot smpler n D. The procedure s the same, but we only need to calculate I zz. For example, to calculate the mass moment of nerta for a square axa plate wth a hole wth an axa square cut out from the top corner we would use the followng approach. Start by calculatng the total mass and the poston of the COM. We can regard the cut-out secton as a square wth negatve densty nsde a larger axa square. The total mass s therefore ( ) M = ρ a ρa = 3ρa 3a 3a 5 The poston of the COM s ρg = 4 a ρ( a+ a) a ρ( + ) = a( + ) M 6 The mass moment of nerta of the axa square and the axa square are 8 4 Large square IGzz = 4 ra (4a + 4 a ) = ra (COM at a( + )) Small square IGzz = ra ( a + a ) = ra (COM at a( + )) 6 We now use the parallel axs theorem to fnd the moment of nerta of each square about the combned COM. For the large square: dx = a dy = a. For the small square, dx = a dy = a. The total mass moment of nerta s therefore total IGzz = ρa + 4a ρ a + a ρa ρa a + a = ρa = Ma a Rotatng the nerta tensor

35 35 ll the curous propertes of spnnng obects a gyroscope; a boomerang; the rattlebac are consequences of the fact that the mass moment of nerta of an obect changes when t s rotated. We can see ths very easly by re-vstng our assembly of masses. In the orgnal calculaton, the red, green and blue masses were located on the,, axes. We calculated the nerta tensor to be m x L z m z L y m y I ( ml y y + ml z z) 0 0 = 0 ( ml + ml) G x x z z ( ml x x + ml y y) L x m y L y m z L z L x m x Now suppose we rotate the assembly through 90 degrees about the axs. The red masses now le on the axs, and the green ones lne up wth the axs. It s not hard to see that the new mass moment of nerta s now L z m y m z L x m x L y ( ml x x + ml z z) 0 0 I G = 0 ( ml y y + ml z z) ( I, I have swtched postons) xx yy ( ml x x + ml y y) L y m x L x m z L z m y Ths seems le a huge problem f we needed to re-calculate the mass moment of nerta from scratch every tme a rgd body moves, analyzng rgd body moton would be nearly mpossble. Fortunately, we can derve a formula that tells us how the mass moment of nerta of a body changes when t s rotated. Rotaton formula for moments of nerta: Consder the rectangular prsm shown n the fgure. Let I G 0 denote the mass moment of nerta wth the prsm orented so the faces are perpendcular to,, (.e. the nerta gven n the table n Sect 6.4.5). Suppose the body s then rotated by a tensor R. The mass moment of nerta after rotaton s gven by 0 T I = RI R G G 0 B p B -p I G Example: The prsm shown n the fgure s rotated by 45 degrees about the axs. Calculate the mass moment of nerta after the rotaton I G r B -r B

36 36 Start by calculatng the rotaton (use the formulas from 6..) cos(45) sn(45) 0 R = sn(45) cos(45) We now the nerta tensor of the prsm before t s rotated s b + c M I G = 0 a + c a + b a 0 I G b c We can use Mupad to do the tedous matrx multplcatons I G Note that the nerta tensor s no longer dagonal. Rotaton formula for D moton: Fortunately, D s smple Rotatng a D obect about the axs does not change I Gzz b I Gzz I Gzz a Proof of the rotaton formula: Consder a system of N partcles. Suppose that before rotaton, the partcles are at postons d relatve to the COM. The ntal nerta tensor s 0 ( m ( ) m ) 0 N G = = 0 d = I d d d d Now rotate the system, so the partcle s move to new postons Rd. The new nerta tensor s

37 37 N N ( m ( ) m ) ( m ( 0 0 ) m 0 0 ) I = d d d d = Rd Rd Rd Rd G = = It s easy to show (ust wrte out the matrx products) whch shows that T Rd Rd = Rd dr T ( Rd ) ( Rd ) = R( d d ) R ( m ( ) m ) N T 0 T IG = R d d d d R = RIGR = Tme dervatve of the nerta tensor When we analyze moton of a rgd body, we wll need to calculate the tme dervatves of the lnear and angular momentum. Lnear momentum s no problem, but for angular momentum, we wll need to now how to dfferentate I G wth respect to tme. There s a formula for ths: dig = WIG IGW d T where W= R R s the spn tensor (see sect 6..) Proof: Start wth IG = 0 T G RI R and tae the tme dervatve d T IG d R 0 T 0 d R = IGR + RI G T dr T dr dr T dr T T Recall that RR = R + R = 0 = R R = R W Fnally note that dr / = WR and therefore dig 0 T 0 T = WRIGR + RIGR W = WIG IGW T T Tme dervatve of angular momentum To use the angular momentum conservaton equaton, we wll need to now how to calculate the tme dervatve of the angular momentum. When we do ths for a 3D problem, we need to tae nto account that the mass moment of nerta changes as the body rotates. We wll prove the followng formula: d h G M G G ( G ) = r a + I α + ω I ω For D planar problems ths can be smplfed to: d h = r G M α G + IGzzαz

38 38 Proof: We start by tang the tme dervatve of the general defnton of h dh d = ( rg MvG + IGω ) We can go ahead and do the dervatve wth the product rule: dh d G d G d G d = r M G + G M v + I ω v r ω+ I G We can smplfy ths by notng that drg / = v G and of course the cross product of v G wth tself s zero. We can also use the defnton of angular acceleraton: dω/ = α. Ths gves d h di = r G G M a G + ω + I G α Fnally we can use the formula for di G / from to see that dig ω = WIGω IGWω and recallng the spn tensor-angular velocty formula Wu = ω u for all vectors u. dig ω= ω ( IGω) IG( ω ω) = ω ( IGω ) where we recall that the cross product of a vector wth tself s always zero. Ths gves the answer stated. 6.5 Rotatonal forces revew of moments exerted by forces and torques You can fnd a detaled dscusson of forces and moments, wth lots of examples, n Secton of these notes. Moments and torques don t come up very often n partcle dynamcs, but play a very mportant role n rgd body dynamcs. We therefore revew the most mportant concepts related to torques and moments here. You need to remember, and understand, these deas: () moment s a generalzed force that causes an obect to rotate (see secton ). () force can exert a moment on a rgd body. The moment of a force (about the orgn) s defned as M = r F (3) In general, a force causes a rgd body to accelerate, and wll also nduce an angular acceleraton (so t nfluences both translatonal and rotatonal moton). (4) torque or pure moment s a specal nd of generalzed force that causes an obect to rotate, but has no effect on ts translatonal moton. s an example, a motor shaft (eg the bt on a powerdrven screwdrver!) wll exert a torque on the obect connected to t. (5) torque or pure moment s a vector quantty t has magntude and drecton. The drecton ndcates the axs assocated wth ts rotatonal force (followng the rght hand screw conventon); the magntude represents the ntensty of the rotatonal force. The magntude of a torque has unts of Newton Meters. moment s often denoted by the symbols shown n the fgure

39 39 3D moment D moment 6.5. Rate of wor done by a torque or moment: If a torque Q= Qx+ Qy+ Qz acts on an obect that rotates wth angular velocty ω, the rate of wor done on the obect by Q s P= Q ω = Qω + Q ω + Qω x x y y z z 6.5. Torsonal sprngs sold rod s a good example of a torsonal sprng. You could tae hold of the ends of the rod and twst them, causng one end to rotate relatve to the other. To do ths, you would apply a moment or a couple to each end of the rod, wth drecton parallel to the axs of the rod. The angle of twst ncreases wth the moment. Varous torson sprng desgns used n practce are shown n the pcture the mage s from ngs.html More generally, a torsonal sprng ressts rotaton, by exertng equal and opposte moments on obects connected to ts ends. For a lnear sprng the moment s proportonal to the angle of rotaton appled to the sprng. The fgure shows a formal free body dagram for two obects connected by a torsonal sprng. If obect s held fxed, and obect B s rotated through an angle θ about an axs parallel to a unt vector n, then the sprng exerts a moment Q= κθn on obect B where κ s the torsonal stffness of the sprng. Torsonal stffness has unts of Nm/radan. The potental energy of the moments exerted by the sprng can be determned by computng the wor done to twst the sprng through Q an angle θ.. The wor done by a moment Q due to twstng through a very small angle dθ about an axs parallel to a vector n s dw = Q dθn. The potental energy s the negatve of the total wor done by M,.e. θ θ θ V = Q dθn= ( κθn) dθn = κθ dθ = κθ Q Q B θ n -Q B

40 40 potental energy cannot usually be defned for most concentrated moments, because rotatonal moton s tself path dependent (the orentaton of an obect that s gven two successve rotatons depends on the order n whch the rotatons are appled). 6.6 Summary of equatons of moton for rgd bodes In ths secton, we collect together all the mportant formulas from the precedng sectons, and summarze the equatons that we use to analyze moton of a rgd body. We consder moton of a rgd body that has mass densty ρ durng some tme nterval t 0 < t < t, and defne the followng quanttes: 6.6. Forces, torques, mpulse, wor, power The total force actng on the body F The total lnear mpulse exerted by forces durng the tme nterval t () F () t t I= 0 () The total moment (ncludng torques) actng on the body () ( ) r F + Q The tot al angular mpulse exerted on the body durng the tme t () ( ) nterval = () t r F + Q t 0 The rate of wor done by forces and torques actng on the body F () Q () () ( ) P = F v + Q ω The total wor done by forces and torques on the body durng the tme nterval W = P() t t t 0 Q () F (3) v v v G r G v 3 F () 6.6. Inertal propertes The total mass s M = ρdv V The poston of the center of mass s ρg = M ρ ρdv V r G r d

41 4 The mass moment of nerta about the center of mass dy + dz dxdy dxd z I G = ρ dxd y dx dz d yd + z dv V dxdz dydz dx + d y where d= r r G For a D body wth mass per unt area µ we use The total mass s M = µ d The poston of the center of mass s rg = M r µd The mass moment of nerta about the center of mass s where d= r r G IGzz = ( dx d y ) d M µ + r G r d Descrbng moton The rotaton tensor (matrx) maps the vector connectng two ponts n a sold before t moves to ts poston after moton r r = Rp ( p ) B B The spn tensor s related to R by = dr T dr W R = WR B p B -p r B -r B Rotaton through an angle θ about an axs parallel to a unt vector n= n + n + n s x y z cos θ + ( cos θ) nx ( cos θ) nn x y sn θnz ( cos θ) nn x z + snθn y R = ( cos θ) nn x y + snθnz cos θ + ( cos θ) ny ( cos θ) nn y z snθn x ( cos θ) nn x z sn θny ( cos θ) nn y z + snθnx cos θ + ( cos θ) n z The angular velocty vector ω= ωx+ ωy+ ωz s related to W by

42 4 The angular acceleraton vector s d α = ω 0 ωz ωy W = ωz 0 ωx ωy ωx 0 The veloctes of two ponts and B n a rotatng rgd body are related by v v = ω r r ( ) B B The acceleratons of and B are related by a a = α ( r r ) + ω ( v v ) = α ( r r ) + ω ω ( r r ) [ ] B B B B B Momentum and Energy The total lnear momentum s p= Mv G The angular momentum (about the orgn) s h= rg MvG + IGω The total netc energy s T = MvG vg + ω IGω For D planar problems, we now ω= ωz. In ths case, we can use The total lnear momentum s p= Mv The total angular momentum (about the orgn) s h= rg MvG + IGzzωz The total netc energy s T = M v G + IGzzωz G Conservaton laws Lnear momentum ngular momentum () dp F = I= p p 0 () ( ) dh r F + Q = = h h 0 Wor-Power - Knetc Energy relaton () () ( ) dt F v + Q ω = W = T T0

43 43 Energy equaton for a conservatve system d ( T + V ) = 0 T 0 + V 0 = T + V Lnear and angular momentum equatons n terms of acceleratons The lnear and angular momentum conservaton equatons can also be expressed n terms of acceleratons, angular acceleratons, and angular veloctes. The results are () F = MaG () ( ) r F + Q = MrG ag + IGα+ ω IGω [ ] For D planar moton we can use the smplfed formulas () F = MαG () ( ) r F + Q = MrG αg + IGzzαz Specal equatons for analyzng bodes that rotate about a statonary pont We often want to predct the moton of a system that rotates about a fxed pvot a pendulum s a smple example. These problems can be solved usng the equatons n and 6.6.6, but can also be solved usng a useful short-cut. For an obect that rotates about a fxed pvot at the orgn: The total angular momentum (about the orgn) s h= I ω The total netc energy s T = ω IOω The equaton of rotatonal moton s () ( ) r F + Q = IOα+ ω IOω [ ] Here I O s the mass moment of nerta about O (calculated, eg, usng the parallel axs theorem) O r r O -r G O For D rotaton about a fxed pont at the orgn we can smplfy these to The total angular momentum (about the orgn) s h= I ω Ozz z r r O -r G O

44 44 The total netc energy s T = I ω Ozz z The equaton of rotatonal moton s () ( ) r F + Q = IOzzαz Proof: It s straghtforward to show these formulas. Let s show the two dmensonal verson of the netc energy formulas as an example. For fxed axs rotaton, we can use the rgd body formulas to calculate the velocty of the center of mass (O s statonary and at the orgn) vg = ω rg = ωz r G The general formula for netc energy can therefore be re-wrtten as T = MvG vg + IGzzωz = Mωz ( rg ) ( rg ) + IGzzωz = ( M rg + IGzz ) ωz = IOzzωz The other formulas can be proved wth the same method we smply express the velocty or acceleraton of the COM n the general formulas n terms of angular velocty and acceleraton, and notce that we can rearrange the result n terms of the mass moment of nerta about O. 6.7 Examples of solutons to problems nvolvng moton of rgd bodes The best way to learn how to use the equatons n secton 6.6 s ust to wor through a seres of examples Solutons to D problems Example : sold of revoluton (eg a cylnder or sphere) wth mass M and mass moment of nerta about ts COM I Gzz s released from rest at the top of a ramp. It rolls wthout slp. Calculate ts velocty at the bottom of the ramp. The system s conservatve, so we can solve the problem usng energy conservaton. The energy equaton tells us that the sum of netc and potental energy of the cylnder s constant: T0 + V0 = T+ V We can tae the datum for potental energy to be the poston of the COM at the bottom of the ramp. The ntal potental energy s therefore V0 = Mgh ; the fnal potental energy s zero. The ntal netc energy s zero, because the cylnder s statonary. The fnal netc energy s T = Mvx + IGzzωz. The energy equaton gves T0 + V0 = T+ V Mgh = Mvx + IGzzωz Fnally, snce the cylnder rolls wthout slp, we now that vx = Rωz. Hence

45 45 I Mgh = Mv Gzz x + v x R vx = gh + IGzz /( MR ) Ths formula predcts that an obect wth a smaller nerta I Gzz wll move faster than an obect wth a large nerta. sphere rolls down the ramp more qucly than a cylnder, for example, and a sold cylnder rolls more qucly than a rng. Example : For the problem treated n the precedng secton, calculate the crtcal value of frcton coeffcent necessary to prevent slp at the contact. If we want to learn about forces, we have to use the lnear and angular momentum equatons. Ths problem can be solved wth the D formulas n terms of acceleratons: () F = MαG () ( ) r F + Q = MrG αg + IGzzαz The fgure shows a free body dagram for the cylnder (or sphere) We now that the COM s always a constant heght above the ramp, so the acceleraton must be parallel to. The lnear momentum equaton gves ( Mg sn φ T ) + ( N Mg cos φ) = MaGx We can use the angular momentum equaton t s convenent to tae moments about the contact pont C. (There are no torques n ths problem). () ( ) r F + Q = MrG ag + IGzzaz RMg snφ = R agx+ IGzzaz = MRaGx+ IGzzaz a = Ra. Fnally, we can use the rollng wheel formula for acceleratons Gx z The precedng results gve: a RMg snφ = MRa Gx Gx IGzz R MgRsnφ g snφ agx = = MR + ( I / ) Gzz R + IGzz /( MR ) Fnally, substtutng bac nto the components of (): T = Mg snφ Ma Gx Mg snφ I /( ) sn Gzz MR = Mg φ = Mg sn + IGzz /( MR ) + IGzz /( MR ) The component of () gves N = Mg cosφ φ µ N Mg N R

46 46 For no slp T µ N IGzz /( MR ) µ tanφ + I /( MR ) Gzz The formula shows that obects wth large values of IGzz / MR are more lely to slp. If the nerta s very small, slp wll never occur. rng wll slp on a lower slope than a cylnder, whch wll slp on a lower slope than a sphere. Example 3: vertcal mast can be dealzed as a slender rod wth length L and mass M, whch s held n an nverted poston by a torsonal sprng wth stffness κ at ts base. Fnd the equaton of moton for the angle θ n the fgure, and hence determne the natural frequency of vbraton of the mast. M θ Ths s a conservatve system. lso, the mast rotates about a fxed pont. We can analyze the problem usng energy methods, and use the specal formulas for rotaton about a fxed pont. The netc energy formula for planar moton s T = I ω Ozz z For planar moton we now that dθ ω z = We can use the parallel axs theorem to calculate the mass moment of nerta of a rod about one end: L IOzz IGzz Md ML M = + = + = ML 3 Gravty and the torsonal sprng both contrbute to the total potental energy of the system. The total potental energy s L V = Mg cosθ + κθ Energy conservaton means that d T + V = const ( T + V ) = 0 d IOzzωz + MgLcosθ + κθ = 0 dωz dθ dθ IOzz ωz MgLsnθ + κθ = 0 dθ Recall that ω z = so d θ IOzz MgLsnθ + κθ = 0 We assume that θ s small enough that snθ θ, so IOzz d θ + θ = 0 κ MgL L κ O

47 47 Ths s a standard Case I undamped vbraton EOM, so we can ust read off the natural frequency κ MgL 3( κ MgL) ωn = = I ML Ozz Example 4: thn unform ds of radus R, mass m and mass moment of nerta mr / s placed on the ground wth a postve velocty v 0 n the horzontal drecton, and a counterclocwse rotatonal velocty (a bacspn) ω 0. The contact between the ds and the ground has frcton coeffcent µ. The ds ntally slps on the ground, and for a sutable range of values of ω 0 and v 0 ts drecton of moton may reverse. The goal of ths problem s to calculate the conons where ths reversal wll occur. ω 0 m R v 0 General dscusson of slppng contacts: Solvng problems wth sldng at a contact s always trcy, because we have to draw the frcton forces n the correct drecton. Before taclng the example, we wll summarze the general rules. We wll consder a wheel as an example, but the rules apply to contact between any obect and a statonary surface. The fgure shows a wheel that spns wth angular velocty ω= ωz whle the center moves wth speed vo = vox. The drecton of the frcton force s determned by the drecton of moton of the pont on the wheel that nstantaneously touches the ground, whch can be calculated from the formula vc = ( vox + ωzr) Frcton always acts to try to brng pont C to rest f C s movng to the rght, frcton acts to the left; f C s movng to the left, frcton acts to the rght. There are three possble cases: Forward slp: v + ω R> 0 Pont C moves n the postve drecton over the ground Ox z ωz α z C R O Slp occurs at the contact, R C O N T= µn

48 48 We have to use the frcton law T = µ N Pont C s movng to the rght, so frcton must act to the left Pure rollng v + ω R= 0. Pont C s statonary. Ox z No slp occurs at the contact. In ths case T < µ N We can draw the frcton force n ether drecton at the contact (f we choose the wrong drecton, our calculatons wll ust tell us that T s negatve). It s usually convenent to choose T to act n the postve drecton, but ths s not necessary. T C R O N Reverse slp: v + ω R< 0 Pont C moves n the negatve drecton over the ground Ox z Slp occurs at the contact, We have to use the frcton law T = µ N Pont C s movng to the left, so frcton must act to the rght R O C T=µN N Now we return to the example.

49 49 4. Draw a free body dagram showng the forces actng on the ds ust after t hts the ground. We are gven that v x0 and ω z0 are both postve so we have v + ω R> 0. Ths s forward slp, and we use the correspondng FBD. Ox z mg R µ N N 4. Hence, fnd formulas for the ntal acceleraton a and angular acceleraton α for the ds, n terms of g, R and µ. Note that the contact pont s slppng. The equatons of moton are Solvng these and usng () () ( ) F = MαG r F + Q = MrG αg + IGzzαz IGzz mn+ ( N mg) = max mnr = IGzzaz = mr / : N = mg µ N = max mr α z = µ NR ax = mg az = mg / R 4.3 Fnd formulas for the velocty and angular velocty of the ds, durng the perod whle the contact pont s stll slppng. The acceleraton and angular acceleraton are constant, so we can use the constant acceleraton formulas: vx = v0 µ gt ωz = ω 0 µ gt / R 4.4 Fnd a formula for the tme at whch the ds wll reverse ts drecton of moton. Velocty s reversed where v=0. From the prevous part, v = v 0 µ gt t = v 0 / µ g at the reversal. 4.5 Fnd a formula for the tme at whch the ds begns to roll on the ground wthout slp. Hence, show that the ds wll reverse ts drecton only f v0 < ω0 R/ Rollng wthout sldng starts when vxo = ωzr. We have that

50 50 ww 0 µ gt / R z = vx = v0 µ gt v0 µ gt = (w0 R µ gt ) when vx = w z R t= (v0 + w0 R ) / 3µ g The reversal wll only occur f rollng wthout slp occurs after the reversal of velocty. Ths means (v0 + ω0 R ) / 3µ g > v0 / µ g v0 < ω0 R / Example 5: The Sweet Spot on a softball or baseball bat, or tenns or squash racet s a pont that mnmzes the reacton forces actng on the athlete s hand when the ball s struc. In fact, any rgd body has a sweet spot the magc pont s called the center of percusson of a rgd body. For baseball and softball bats n partcular, there s a standard STM test that can be used to measure the poston of the sweet spot. The test wors le ths: the bat s suspended from the nob on handle, so t swngs le a pendulum. The perod of vbraton τ of the swngng bat s then measured. STM say that the center of percusson s then a dstance d= τ g 4π from the end of the handle. Why does ths wor? It seems that ths test has nothng whatever to do wth a ball httng the bat! We wll solve ths problem n two parts. Frst, we wll calculate a formula for the perod of vbraton n the STM test. Then we wll calculate the poston of the center of percusson. We wll see that the STM test does ndeed mae the correct predcton. We can calculate the perod usng the energy method. The fgure shows the STM pendulum test. We assume that The bat has a mass moment of nerta about ts COM I Gzz The COM s a dstance L from O The bat pvots about O, so we can use the fxed axs rotaton formula for the netc energy dθ T = I Ozz Here I= Ozz I Gzz + ML (usng the parallel axs theorem). The potental energy s V = MgL cos θ. Energy conservaton gves T= + V constant I Ozz d (T = +V ) 0 d θ dθ dθ + MgL sn θ = 0 O θ L

51 5 If θ s small then sn θ θ so the equaton of moton reduces to I Ozz d θ +θ = 0 MgL Ths s a standard Case EOM. The natural frequency s ωn = = τ π = π ωn MgL so the perod s I Ozz I Ozz MgL Next, we fnd the poston of the sweet spot. We can do ths by calculatng the reacton forces on the handle when the bat s struc, and fndng the mpact pont that mnmzes the reacton force. The fgure shows an mpact event. We assume that: The bat rotates n the horzontal plane (so gravty acts out of the plane of the fgure). The bat rotates about the pont O The ball mpacts the bat a dstance d from the handle. The ball exerts a (large) force Fmpact on the bat Ry O Rx ω L d Reacton forces Rx, R y act on the handle durng the mpact. Ths s a planar problem, so we can use the D equatons of moton. The equaton for translatonal moton gves ( Rx Fmpact ) + R y = MaG For the rotatonal equaton we can also use the short-cut for fxed axs rotaton r F ( ) + I Ozza z Q( ) = Fmpact d = IOzzaa z = z Fmpact d We can relate ag to α z usng the rgd body formula: I Ozz αg = α z rg ω zrg = α z L + ω z L We therefore see that ( ( Rx Fmpact ) + R = y M a z L + ω z L R= x Fmpact + M a z L MdL Rx Fmpact = I Ozz The sweet spot s at the poston that maes Rx = 0, whch shows that ) Fmpact

52 5 For comparson, the STM formula gves I d = Ozz ML g g τ IOzz I d = = π = Ozz 4π 4π MgL ML 6.7. Solutons to 3D problems Example : The fgure shows a wheel spnnng on a frctonless axle. The axle s supported on one sde (at ) by a pvot that allows free rotaton n any drecton. If the wheel were not spnnng, t would smply swng about le a pendulum. But f the angular speed s hgh enough, the axle remans horzontal, and the wheel turns slowly about the vertcal axs. Ths behavor s called precesson and s a bt mysterous why does spn somehow hold the wheel up? The goal of ths example s to explan ths, and to calculate a formula for the rotaton rate of the axle. We wll do ths by showng that steady precesson satsfes all the equatons of moton.. Let n be a unt vector parallel to the axle. Consder the ds at the nstant when n=, and assume that the ds spns at constant rate about the axle at ν radans per second, the ds rotates slowly at constant rate about at Ω radans per second Fnd the angular velocty and angular acceleraton at the nstant shown n the fgure The angular velocty s easy we ust add the two vectors: ω= ν+ω = ν+ω Ω d ν n B R The angular acceleraton s harder. Both ω and Ω are constant. But ths does not mean that the angular velocty vector s constant, because the axle s rotatng about the axs. The drecton of the angular velocty s changng, even though the magntude s not. We can calculate the rate of change of n by usng the rgd body formula d ( rb r) = vb v = ω ( rb r ) If we choose and B to be a unt dstance apart, then ( rb r) = n and therefore dn = ω n We can now calculate the angular acceleraton dω d dν α= = ( Ω + ν) = ν = ν( Ω + ν) ν= νω. Fnd a formula for the acceleraton of the center of mass of the ds

53 53 We can use the rgd body formula a a = α ( r r ) + ω ω ( r r ) [ ] ( ) ( ) d G G G = νω ( d) + ν+ω ν+ω ( ) = Ω d qucer way s to notce that the COM s n crcular moton around and use the crcular moton formula, wth the same result..3 Draw a free body dagram showng the forces actng on the wheel N B T Mg.3 Wrte down the equatons of translatonal and rotatonal moton for the ds () F = M a G T + ( N Mg) = MdΩ () ( ) r F + Q = MrG ag + IGα+ ω IGω d [ ] IGxx ν IGxx 0 0 ν d ( Mg) = M ( d) ( Ω d) + 0 IGyy 0 ν 0 0 IGyy 0 0 Ω I 0 Ω Gzz 0 0 I Ω Gzz Worng through the cross products and the matrx-vector products we get Mgd= I νω ( I I ) νω Gyy Gzz Gxx We see that steady precesson can ndeed satsfy all the equatons of moton. Moreover, for a ds (or any sold of revoluton) IGzz = IGyy, so we can calculate the precesson rate Mgd= I νω Gxx Mgd Ω= IGxx ν Example : The prsm shown n the fgure floats n space (no gravty). t tme t=0 ts faces are perpendcular to the,, axes as shown. It s then gven an ntal angular velocty ω= ωz0+ ωy0 wth ωy0 << ωz0 c a b

54 54 (.e. we set the body spnnng about the axs, but gve t a very small dsturbance). Investgate the nature of the subsequent moton, wth both hand calculatons and by wrtng a MTLB scrpt that wll anmate the moton of the prsm. No forces or moments act on the prsm. We can use the equatons of moton 0= Ma 0= Mr a + I α+ ω I ω [ ] G G G G G The angular momentum equaton can be wrtten out explctly IGxx 0 0 dωx / IGxx 0 0 ωx 0 0 IGyy 0 dωy / ωx, ωy, ωz 0 IGyy 0 ω y 0 + = 0 0 I / Gzz dωz I Gzz ω z (we could substtute values for IGxx, IGyy, I Gzz n terms of a,b,c and M but t s clearer to leave them) Expandng out the matrx products and cross product gves dω I x Gxx + ( IGzz IGyy ) ωω y z = 0 dωy IGyy ( IGzz IGxx ) ωω x z = 0 dω I z Gzz + ( IGyy IGxx ) ωω x y = 0 t tme t=0 ω x s zero and ω y s small. They mght ncrease, but we wll only consder behavor whle they reman small. In ths case ωω x y s extremely small so we can assume dω / 0. We can then decouple the frst two equatons le ths:. Dfferentate the second equaton wth respect to tme I ( I I ) z d ωy dωx Gyy Gzz Gxx ω z0 = 0 I. Now we can substtute for dω / usng the frst equaton, and dvde by Gyy Ths s an equaton of the form d x ωy ( IGzz IGxx )( IGzz IGyy ) + ωz 0 ωy = 0 IGyyIGxx d ωy + λω 0 y = We recognze ths as an undamped vbraton equaton (case I or case II from our table of solutons). Its soluton depends on the sgn of λ :. For λ > 0 the soluton s ωy = sn λt+ Bcos λt where, B are constants. Ths s stable moton - ω y remans small ths ndcates that the prsm wll contnue to spn about the axs. For λ < 0 the soluton s ωy = exp λt+ Bexp( λt). Ths s unstable moton - ω y wll become very large. Ths ndcates that the bloc wll tumble.

55 55 The sgn of λ s determned by the product ( IGzz IGxx )( IGzz IGyy ). Gzz. Gzz 3. Gzz. There are three possble cases: I s greater than IGxx, I Gyy (the axs has the maxmum nerta). Moton s stable I s less than IGxx, I Gyy (the axs has the mnmum nerta). Moton s stable I s between I, I. Moton s unstable. Gxx Gyy We can learn more about the moton by usng MTLB to solve the equatons of moton for us. To do ths, note that: Snce there s no moton of the center of mass, we only need to consder rotatonal moton. We now that we can descrbe the orentaton of the prsm by the rotaton tensor R and ts rate of change of orentaton by the angular velocty ω. These wll be our unnowns: we would le to solve for the nne components of R, and the three components of ω. The orentaton and angular velocty are governed by the dfferental equatons dr = WR IGα+ ω [ IGω] = 0 dω α = 0 T where IG = RIGR s the rotated nerta tensor for the bloc, and W s the spn tensor 0 ωz ωy W = ωz 0 ωx ωy ωx 0 We need to set up the MTLB ode solver to calculate R and ω as functons of tme by ntegratng these equatons. We can store the unnown rotaton matrx and the angular velocty vector n a MTLB vector: ω = Rxx, Rxy, Rxz, Ryx, Ryy, Ryz, Rzx, Rzy, Rzz, ωx, ωy, ωz We need to wrte a MTLB functon that wll calculate the tme dervatves of ths vector, gven ts current value. The calculaton nvolves the followng steps: () ssemble the vectors ω and the rotaton tensor R from the Matlab soluton vector w. Matlab has a useful functon that wll automatcally convert a matrx to a vector, and vce-versa. For example, R (a 3x3 matrx) can be converted to w (a x9 column vector) usng W(:9) = reshape(transpose(r),[9,])) To transform w (as a column vector) bac to R, you can use R = transpose(reshape(w(:9),[3,3])) () Calculate the spn tensor W 0 T (3) Calculate the rotated nerta tensor IG = RIGR (Matlab wll multply the matrces for us) (4) Solve the equatons for the angular acceleratonα : α= I ω I ω. ( [ ]) G G

56 56 Here I s the nverse of I G (Matlab wll calculate ths for us) (5) Calculate dr / = WR (as a 3x3 matrx) dw (6) ssemble the matlab vector = R xx, R xy, R xz, R yx, R yy, R yz, R zx, R zy, R zz, αx, αy, αz G Ths sounds complcated but actually MTLB s great at dong ths sort of calculaton effcently. Here s a functon: functon dw = rgd_body_eom(t,w,i0) Rvec = w(:9); % Rotaton matrx, stored as a vector omega = w(0:); % ngular velocty R = transpose(reshape(rvec,[3,3])); % R s now a 3x3 matrx, II = R*I0*transpose(R); %Current nerta tensor, n fxed coord system W = [0,-omega(3),omega();omega(3),0,-omega();-omega(),omega(),0]; alpha = -II\(cross(omega,II*omega)); % ngular accel Rdot = W*R; % Rate of change of rotaton matrx, as a 3x3 matrx Rdotvec = reshape(transpose(rdot),[9,]); dw = [Rdotvec;alpha]; end Now we ust need to set up ode45 to ntegrate (numercally) the dfferental equaton: omega0 = [0.,0.0,]; % Intal angular velocty a = [4,,]; % Dmensons (a,b,c) of the prsm tme = 0; ntal_w = [;0;0;0;;0;0;0;;transpose(omega0(:3))]; I0 = [a()^+a(3)^,0,0;0,a()^+a(3)^,0;0,0,a()^+a()^]; optons = odeset('reltol', ); sol = ode45(@(t,w) rgd_body_eom(t,w,i0),[0,tme],ntal_w,optons); anmate_rgd_body(sol,a,[0,tme]) You can download the full scrpt here. The fgures below show anmatons of the predcted behavor for the three possble types of behavor I zz s the maxmum nerta rotaton s stable

57 57 I zz s the ntermedate nerta rotaton s unstable (the bloc tumbles) I zz s the smallest nerta rotaton s stable