Physics Revision Notes ThermodynamicsPage 1 of 20 Daniel Guetta, 2008 Dynamics Rigid Body Dynamics Introduction A rigid body is a many-particle system in which the distance between particlesisfixed.Thelocationofallparticlesisdescribedby6 coordinates 3 spatial and 3 angular. Thevelocityisdeterminedbyv,thevelocityoftheCoMandw,the angular velocity. The basic two equations of angular motion are 0M = R F`` Thecentreofmassmovesasifitwereasingleparticleunderthe action of a force F0. 0= J G` The rate of change of angular momentum is equal to the total applied couple. Other basic equations: oThe velocity v of a particle at a distance r from an axis around which a rotation at speedw is happening is = v r woFor similar reasons: ddt= JJ woAngularspeedsareadditive.Toifframe1isrotatingwith 1 wrt 2wwithrespecttoframe2,whichisrotatingwith 2 wrt 3wwith respect o frame 3, then 1 wrt 3 1 wrt 2 2 wrt 3= + w w w Relating Jand w If the body is rotating atw, the total angular momentum is given by Physics Revision Notes ThermodynamicsPage 2 of 20 Daniel Guetta, 2008 22( )( )( )x y zmm rm r x y z w w w= = = = + + J r pr rr rrww-ww- In detail 2 22 22 2( )( )( )m y z mxy mxzmxy m x z myzmxz myz m x y + - - = - + - - - + = JJ IIww [The non-diagonal elements are fairly easy to derive. The diagonal ones shouldactuallyhavex2+y2+z2,becauseoneofthetermsisalways knockedoutbythesecondterminthesum].Inotherwords,Jis proportional tow, but not necessarily parallel to it. The off-axes elements are rather hard to understand they correspond to the fact that looking at a particle at a given instant, its impossible to tell exactly around which axis its moving. Also, we can find the kinetic energy [ ][ ]1212( ) ( )( )T mm= = r rr rw ww w 12T= J wThe couple is then given by = = G J J w` Note that I must be specified with its origin and with its set of axes. Properties of I Iisasymmetrictensor.Itthereforehasthreerealeigenvaluesand three perpendicular eigenvectors. With respect to the eigenvector basis: Physics Revision Notes ThermodynamicsPage 3 of 20 Daniel Guetta, 2008 1232 12[No sum][Sum]IIIIT Ia a aa aww = ==IJThe eigenvector axes are called the principal axes, and the Is are called the principal moments of inertia. An alternative way to think of this is that the principal axes are ones around which objects are happy to rotate without any torque being applied. Inw -space,surfacesofconstantTformanellipsoid,withaxesof length 1/2Ia- . Also, inw -space: gradT Ia aw = =JSo J is perpendicular to surfaces of constant T atw. We can classify the principal axes as follows: oSpherical tops all the I are equal, andI = J w , with I scalar. The body is isotropic with the same I about any axis (eg: sphere, cube). oSymmetrical tops 1 2 3I I I = . e3 axis is unique, but e1 and e2 are any two mutually perpendicular vectors perpendicular to e3 (eg: lens, cigar). oAsymmetrical tops all Is different, and axes are unique. Consider any two Is: 2 2 2 2 21 2 3 3( ) 2 I I m y z x z I mz I + = + + + = + So no I can be larger than the sum of the other two. Furthermore, ifz = 0 for every particle (ie: if we have a lamina), then 3 1 2I I I = +Consider an axis at a distance a away from a principal axis and parallel to it, and let r be the distance of each particle from the principal axis. Then: ( )2 20 00 when measuredrelative to C of M( ) ( ) 2 I m I Ma m I Ma== + + = + + = + rr a r a r a Physics Revision Notes ThermodynamicsPage 4 of 20 Daniel Guetta, 2008 This is the Parallel Axis Theorem, where each vector is considered to be a projection in a plane perpendicular to the axes. Two Basic Problems You whack it what happens? Steps for solution: oDefine principal axes with a sensible origin. oCalculate an expression for J in terms of the impulse: d d dB B Bt t t = = = = J r F r F r P tWhere B is the point at whichthe whack occurred, and rB can be taken out of the integral because the whack is assumed to be instantaneous. oWork out an expression for J in terms ofw, using the moments of inertia. oEquate the two expressions for J. oWork out the motion of the CM using standard linear mechanics. oNOTE:TheobviousorigintouseistheCM,butotherorigins can be used subject to the provisos above for using=J t`. So a pivot, for example, is fine to use. You apply a torque whats the frequency of rotation? oDefineprincipalaxeswithasensibleorigin(eg:theCMsee above). oFindanexpressionforwintheseaxes(withunknown magnitude), and find a corresponding expression for J, using the principal moments of inertia. oFindd /dt = L L w . oCalculate the torque (= r F ) and equate it withd /dt L . Free Motion Eulers Equation Free precession is a situation in which F = 0 and G = 0. In such a case, J is constant.w is constant if J is along one of the principal axes, but otherwise, it will change direction, and perhaps even magnitude. We use the Euler Equations to analyse this problem. Physics Revision Notes ThermodynamicsPage 5 of 20 Daniel Guetta, 2008 Therateofchangeofangularmomentumvectorintheprincipal-axes frame (which is rotating around with the body) and the lab frame are related by lab PAd dd d t t = + J JJ wNow,letsassumethatacoupleGisbeingappliedinthelabframe. We know that labddt = JGTherefore, using the equations above: PAddt = + JG J wFinally, we note that in the principal axes frame, 1 1 2 2 3 3( , , ) I I I w w w = J . Therefore,castingbothsidesofthisequationintotheprincipalaxes frame only 1 1 1 3 2 3 2( ) I I I t w w w = + - `And similarly with any cyclic permutation of indices. A few notes: oAll the quantities in this equation are measured with respect to thebodyframe(whichismoving).Thisistheadvantageof theseequationsallwehavetoconsideristheforcesthatthe body feels. oThetwotermsoftheRHSrefertotwotypesofwaysJcan changebecauseitcanchangeinthebodyframeandalso because the body frame is itself rotating. Free Motion Examples FREE SYMMETRIC TOP oFor a symmetrical top (I1 = I2 = I) which is free in space (ie: no torque) the Euler Equations become 1 3 3 22 3 1 33 3( ) 0( ) 00I I II I IIw w ww w ww+ - =+ - ==``` oThe last equation implies 3wis constant. Lets define Physics Revision Notes ThermodynamicsPage 6 of 20 Daniel Guetta, 2008 33I IIw-W =Then the general solution of the first two equations becomes: 12cos[ ]cos[ ]tAtw fwf W+ = W+ oInterpretation from the body frame Inthebodyframe, 1wand 2wseemtoformacirclein the xy plane, with frequencyW. How high that circle is depends on 3w . L could be abovew (if I3 > I an oblate top) or below w (if I3 < I a prolate top). oInterpretation from the fixed lab frame Inthatcase,theEulerEquationsareuseless,because they deal with the body frame, so we express things from scratch, but in terms of the body frames: ( )( )1 2 31 2 3 3 I IIw w ww w w= + += + += -W1 2 31 2 33x x xL x x xLxww WithW defined as above. Thislinearrelationshipbetweenw ,Land 3 ximplies that they are in the same plane. Furthermore,therateofchangeof 3 xis 3 x w ,because it only changes as a result of the rotation. So 33 3 3 d dt I I = -W = x L Lx x xThis is equivalent to 3 xrotating at a frequency/ L I . It turns out that we can interpretw as follows Motion of body aboutMotion of body around its own axisLI= - W3Lx w, HEAVY SYMMETRIC TOP oHere, we must define the Euler angles as follows Physics Revision Notes ThermodynamicsPage 7 of 20 Daniel Guetta, 2008 oThe total angular velocity is then given by Rotation of top Motion of top itself3 1 y q f = + + x x z w,` ` ` Which can be expressed in terms of the body-frames only: ( )( ) ( )3 1 3 23 2 1 cos sin cos siny q f q qy f q f q q= + + += + + +x x x xx x xww` ` `` ` ` ` o y`f`z3 x2 xPhysics Revision Notes ThermodynamicsPage 8 of 20 Daniel Guetta, 2008 Dynamics Normal Modes Introduction A normal mode of a system is an oscillation that has a single frequency. Allthemoregeneraloscillationsofthesystemcanbeexpressedas superpositions of these normal modes. General approach Considerasystemdefinedbygeneralisedcoordinatesqiandactedon by forces Fi,moving in a potential well U(x), and moving elastically.The kinetic energy, T, is then given by 212( )i j iT m q =x` Where( )j iiqxistheCartesiancoordinateofthejthpartofthe system, taken about an equilibrium, where all the jxare 0. Expanding about that equilibrium: ,eqeqeq( ) ( )( )jj i j i ii i ijj i ii i iq q qqq qq= + + xx xxx

`` And so: 1 12 2Tij i jT M q q = =q Mq ` ` ``Where eqeqiji jM mq q = r r Considerthepotentialenergy,aboutapointofequilibrium(ie:a minimum in U) at which all the qi are chosen to be 0. 020Eq0 since at a minimum001 d( )2 d d1( )21( )2i i ji j ixij i jU UU x U q q qq x xU x U K q qU x U= + + += + += + Tq Kq

Physics Revision Notes ThermodynamicsPage 9 of 20 Daniel Guetta, 2008 The total energy is then ( )( )01 12 2d 12 0d 2d0dij i j ij i ji ij j ij ji ij j ij jE U M q q K q qEq M q K qtEq M q K qt= + += + == + = ``` ``` `` [Non rigorous argument] the equations of motion are then: 00ij j ij jM q K q + =+ = Mq Kq```` If we seek normal modes of the form( )i tt ew= q Q , we get: ( )20 w - = K M QNon-trivial solutions only exist if ( )2det 0 w - = K MThis defines the 2wnormal mode frequencies. In practice, the steps are: oFind the K and M matrices by writing them out in terms of the variables of the system, and comparing with 12TT= q Mq ` ` 012U U = +Tq KqBoth matrices must be symmetric. oUse the determinant method above. Physics Revision Notes ThermodynamicsPage 10 of 20 Daniel Guetta, 2008 Dynamics Elasticity Introduction Hookes Law states that Strain StressF lEA lD=Where oF is the force applied to a block of material over an area A. ol Dis the extension of the block in the direction of F. ol is the original, relaxed length of the block in that direction. oE is the Youngs Modulus of the material. Furthermore, it states that w lw lsD D= -Wherew Dis the length of the block in any direction perpendicular to that of l. For an isotropic material, E ands are all we need to define the elastic properties of the material. Sincetheseequationsarealllinear,theprincipleofsuperposition applies. If we have several forces, the displacements will be the sum of the displacements with the forces acting individually. Uniform Strain the Bulk Modulus Considerarectangularblockinapressuretank,say,withidentical stress p on every face. Consideronedirectionthechangeinlengthl Dinthatdirectionis given by Due to pressure Due to pressure inin that direction other directions1 2l p p pl E E Elpl Es ssD= - + +D -= -, Theproblemissymmetrical,sothevaluewillbethesameforall directions. Physics Revision Notes ThermodynamicsPage 11 of 20 Daniel Guetta, 2008 Now, consider the change in volume V x y zV x y zD D D D= + +We therefore have 1 23VpV Es D -= -We can then define the bulk modulus 3(1 2 )EKs=- Such that the change of volume as a result of the stress p is Vp KVD= - Shear Strain the Shear Modulus Consider a cube with face area A and with shear forces acting on it If cut the cube along the diagonals A and B, we find that oThere is a stretch normal to A, of magnitudeF 2 . oThere is a compression normal to B, of magnitudeF 2 . And each of these diagonal faces has area2 A . Thelengtheningofthediagonaldwillthereforebeequaltothe lengthening of d in the following case: A B F F F F d Physics Revision Notes ThermodynamicsPage 12 of 20 Daniel Guetta, 2008 From above, this is given by: 1 2 1 22 21d F Fd E A E Ad Fd E AssD= +D += By symmetry, the other diagonal is shortened by the same amount. It is often useful to have this as a function of the twist angle: From this diagram, it is (reasonably) clear that 2 d d = D 2 d= /Therefore 2 2(1 )2d d Fl l d E Ad sqD D + = = =We therefore define the shear modulus as 2(1 )Ems=+ Such that g mq =Where g is the shear stress = F/A. F qdd D//d2 F2 F2 F2 Fd Physics Revision Notes ThermodynamicsPage 13 of 20 Daniel Guetta, 2008 Formal Definitions Stress oDefined in terms of force/unit area transmitted across planes in the medium. oRequires a tensor. We define oWe can then show that the force on any arbitrary area element is d = F S toThe tensor must be symmetric consider a small cube side dx. Because the cube must be in equilibrium, the forces on it are as follows: The net couple on the cube is ( )dxy yxS S x -But there must be no torque on the cube, or itd spin! So xy yxS S =oThe stress tensor is diagonal for suitable choices of axes. oThe stress in a solid material is therefore described by a tensor field. Strain oWhenamaterialisputunderstrain,apoint( , , ) x y zinitis moved to a point ( , , ) x X y Y z Z + + + . oThe derivatives of these X, Y and Z contain information about the strain. oAs we saw before, its worth considering two kinds of strain For the normal strains, we define: ijtForce in the i direction On a plane perpendicular to the j axis yxSyxSxySxySPhysics Revision Notes ThermodynamicsPage 14 of 20 Daniel Guetta, 2008 xxXex= yyYey= zzZez= For example, if we consider stress perpendicular to the x directioninacubeinitiallyofsidex D ,itllincreaseby xxe x D : Now, for the shear stresses, consider [Theexpressionfortheanglesaretrickytoseebut considerthatXisthechangeinx]Wethensimply define 12 xy yxY Xe ex y = = + Thisensuresthatiftheblocksimplyrotates(ie: / / Y y X x = ), these strains are 0. oSo in general, we define So, for example xx xy xzX e x e y e z = + +oThe tensor is also symmetric, due to the xy yxe e =condition. oIf the strains are non-homogenous, we sit down and cry. The relation between them willhavechangedbyeijD,assumingthat thejthcoordinateofthatpointfromthe origin is D. ( )12/ /ij j i i je X x X x = + The ith coordinate of a point in the material XyYxy DXYx Dx Dxxe x DPhysics Revision Notes ThermodynamicsPage 15 of 20 Daniel Guetta, 2008 oEachcomponentofeisrelatedtoeachcomponentoftthis gives, overall, a fourth-rank tensor of elasticity relating the two: ij ijkl klC e t =(Using the summation convention). oItlookslikethereare92=81coefficientsinC,andthat81 numbers are therefore required to define the elastic properties of a material! However, we note that since S and e are symmetry, we must be able to swap ij and kl in C without changing a thing, so there can be at most 36 different coefficients. oIf the material is isotropic, though, C must be completely frame-independent. As such, we must be able to express it in terms of thetensor ijd .Thereareonlytwowaysofdoingthisthatare also invariant under i j and l k , and so ( ) ( )ijkl ij kl ik jl il jkC ldd mdd d d = + +Soanisotropicmaterialonlyrequirestwoconstants(Eands , for example). And we have 2ij kk ij ijS e e l d m = + Examples Statics Thin tube in torsion oConsider a thin tube being twisted an anglef oWe first note that rlfq =oNext, consider a small square (dotted above) and its deformation as a result of the twist: fqrLr DPhysics Revision Notes ThermodynamicsPage 16 of 20 Daniel Guetta, 2008 From the previous result: FrrF rLmqfm=D= D// oThis force contributes a torque Dtto the rod 2rrF rLft m D = = D /oConsideringthesebitsaroundthewholerod,sothat2 r p / , we get 3r rLt pm fD= 2Wire in torsion For a wire, we simply integrate the above from r = 0 to the total radius, giving 42rLpt m f =Can under pressure oConsider a can of thickness t with closed ends with an internal pressure p. oLetthetangentialstressinthewallsbe qt ,andconsiderhalf the can The forces (= stress area) must balance, so 2 2 t p rprtqqtt = = oLettheaxialstressinthewallsbe zs ,andconsideroneofthe ends. By the same logic as above FFq/rtqtqt pPhysics Revision Notes ThermodynamicsPage 17 of 20 Daniel Guetta, 2008 222zzrt p rprtt p pt = = Bent beam oConsider a beam of length L, held in a bent position. oWe only consider longitudinal strains (valid for small deflections and thin beams). oClearly, the bits at the top of the beam will be stretched, while those at the bottom will be compressed. Somewhere in between, therell be a neutral surface neither stretched nor compressed. oConsider a small segment length/of the bent beam: oTheamountofstretchingandcompressionatanypointis proportionaltothedistancefromtheneutralsurface,y.The constant of proportionality is/R / . As such StrainyRD= =// oClearly,therellbeforcestotheleftabovetheneutralsurface, and vice versa. We therefore have FEAEF y ARD D=DD = D// oThe total torque produced about the neutral line is given by Crosssection2Crosssectionddy FEy AR==t EIBR=/qNeutral surface R+ D / /yPhysics Revision Notes ThermodynamicsPage 18 of 20 Daniel Guetta, 2008 oNow,considerabeamloadedwithweightsgivenbyW(x), where W is the forceperunitlength. Consider the statics of a small segment of the beam: Notes: Duetothebendingmoment,someverticalforcesare produced.Ignoringproductsofinfinitesimalquantities, we can write, at that point d ddS x BEIB S xR== = [Effectively,weresayingthatduetothedSneededto balance dF, the bending moment must change] Thedownwardsloadingforceneedstobebalancedbya difference in the upwards stress d d d S F W xS W= = = oNow, for small deflections 1/ y R =oAs such, we can conclude ( ) EIy W x =oBoundary conditions for various cases are as follows At a free end, S and B are clearly 0, and so0 y y = = . At a cantilevered end, yand y are given (usually 0). oFindingyisthensimplyaquestionofsolvingthatdifferential equation. However, there are a few tricky points All forces must be considered when writing down W(x), includingreactionsatcontacts.Mostoften,Wwillbea series ofd -functions. Sign conventions: Sd S S +dxdFPhysics Revision Notes ThermodynamicsPage 19 of 20 Daniel Guetta, 2008 Downwards W positive. The resulting y obtained is downwards positive, because the way the radius of curvature is specified. However,beverycarefulsometimes,the convention appears to be reversed because the bar curves downwards, and so1/R y- = . Dont worry too much about boundary conditions for y justintegrated -functionsfrom0toL(forafreeend, thisisfullyjustified).Rememberthattherelloftenbea d -functionsattheveryendoftherange,whichmight help satisfy the boundary conditions. Fromthenon,boundariesarejustprovided.Justalso remember to make the y, y and y continuous. Thecoupleprovidedbyacantilevercansimplybe worked out by evaluating = B EIyat that point. Itissometimeseasiertosimplywritedowny,the bending moment from physical considerations. oThe Euler Strut is a beam buckled between two walls: If we take y upwards, then the bending moment on any point is sinB FyFy yEIFy A xEI= - = - = Applying the boundary condition that y = 0 at x = L: 22EIFLp=Thisisindependentofdisplacement(butonlywhile1/ y R =holds). oThe Reciprocity Theorem states that The deflection at Q due to a load at P is the same as the deflection at P due to the same load at Q F FPhysics Revision Notes ThermodynamicsPage 20 of 20 Daniel Guetta, 2008 To prove, say PPQ means the deflection at P due to the load at Q. Consider loading first P and then Q. The energy stored is 2 2QQPPPQPPE F P = + + The same result must be applied the other way round, so PQ QPP P = Dynamics of Rigid Bodies Consider a small volume V of the material. It will have both external forces acting on it (eg: gravity) and internal forces (eg: elastic stresses). ext intdV r + =F F r``Everysmallparticleinthevolumeexperiencestheexternalforce, though, so Fext is given by a volume integral. Define this = int extint( )ddVVVr = - +=fF f rF f,`` Ontheotherhand,onlytheparticlesattheedgeofthevolume experience the elastic force from surrounding media, and so Fint is given by an area integral intAd dVA V = f fWe have, however, defined that the force in the x-direction, say, is ( )d dx xx xy xzF S S S = + + i j k A And so, taking only the x component of the integral above ( )Ad dxx xy xz xVS S S f V + + = i j k AUsing the Divergence Theorem on the LHS d dxyxx xzxV VSS SV f Vx y z + + = Removing the volume integrals (because this is true for any volume): /i ij jS x = f(Using the summation convention). Now, using2ij kk ij ije e t l d m = +(isotropic material), we obtain 2( ) ( ) l m m = + + f u uWhere u is the internal displacement in the solid.