Rigid Body DynamicsCSE169: Computer AnimationInstructor: Steve RotenbergUCSD, Winter 2008Cross Product[]xyyxzxxzyzzyzyxzyxbababababababbbaaa−−−=×=×bakjibaCross Product[]zyxxzzzxyxzyzyyzxxxyyxzxxzyzzybbabacbabbacbababcbabababababa⋅++−=−⋅+=+−⋅=×=−−−=×000bacbaCross Product⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⋅⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⋅++−=−⋅+=+−⋅=zyxxyxzyzzyxzyxxzzzxyxzyzyyzxxbbbaaaaaacccbbabacbabbacbababc000000Cross Product⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−=⋅=×⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⋅⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡000ˆˆ000xyxzyzzyxxyxzyzzyxaaaaaabbbaaaaaacccababaDerivative of a Rotating Vector Let’s say that vector r is rotating around the origin, maintaining a fixed distance At any instant, it has an angular velocity of ωrωr×=dtdrω×rωDerivative of Rotating Matrix If matrix A is a rigid 3x3 matrix rotating with angular velocity ω This implies that the a, b, and c axes must be rotating around ω The derivatives of each axis are ωxa, ωxb, and ωxc, and so the derivative of the entire matrix is:AωAωA⋅=×=ˆdtdProduct Rule()()dtdcabcdtdbabcdtdadtabcddtdbabdtdadtabd++=+= The product rule defines the derivative of productsProduct Rule It can be extended to vector and matrix products as well()()()dtddtddtddtddtddtddtddtddtdBABABAbabababababa⋅+⋅=⋅×+×=×⋅+⋅=⋅Dynamics of ParticlesKinematics of a Particleonaccelerati ity veloc position 22dtddtddtdxvaxvx===Mass, Momentum, and Forceforce momentum mass apfvpmdtdmm===Moment of Momentum The moment of momentum is a vector Also known as angular momentum (the two terms mean basically the same thing, but are used in slightly different situations) Angular momentum has parallel properties with linear momentum In particular, like the linear momentum, angular momentum is conserved in a mechanical systemprL ×=Moment of MomentumprL ×=p1r2r3rpp L is the same for all three of these particles•••Moment of MomentumprL ×=p1r2r3rpp L is different for all of these particles•••Moment of Force (Torque) The moment of force (or torque) about a point is the rate of change of the moment of momentum about that pointdtdLτ =Moment of Force (Torque)()frτfrvvτfrpvτprprLτprL×=×+×=×+×=×+×==×=mdtddtddtdRotational Inertia L=rxp is a general expression for the moment of momentum of a particle In a case where we have a particle rotating around the origin while keeping a fixed distance, we can re-express the moment of momentum in terms of it’s angular velocity ωRotational Inertia()() ()rrIωILωrrLωrrrωrLvrvrLprLˆˆˆˆ⋅−=⋅=⋅⋅−=××−=××=×=×=×=mmmmmmRotational Inertia⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−−−−−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−⋅⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−−−=⋅−=222222000000ˆˆyxzyzxzyzxyxzxyxzyxyxzyzxyxzyzrrrrrrrrrrrrrrrrrrmrrrrrrrrrrrrmmIIrrIRotational Inertia()()()ωILI⋅=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+−−−+−−−+=222222yxzyzxzyzxyxzxyxzyrrmrmrrmrrmrrrmrmrrmrrmrrrmRotational Inertia The rotational inertia matrix I is a 3x3 matrix that is essentially the rotational equivalent of mass It relates the angular momentum of a system to its angular velocity by the equation This is similar to how mass relates linear momentum to linear velocity, but rotation adds additional complexityωIL⋅=vp m=Systems of Particlesmomentum tal tomass ofcenter ofposition particles all of mass l tota1∑∑∑∑∑=====iiicmiiicmniitotalmmmmmvppxxVelocity of Center of Masscmtotalcmtotalcmcmiiiiiicmiiicmcmmmmmmdtdmmmdtddtdvppvvxvxxv======∑∑∑∑∑∑Force on a Particle The change in momentum of the center of mass is equal to the sum of all of the forces on the individual particles This means that the resulting change in the total momentum is independent of the location of the applied forceiiicmicmdtddtddtd∑∑∑∑====fpppppSystems of Particles()∑∑×−=×=icmicmiicmpxxLprL The total moment of momentum around the center of mass is:Torque in a System of Particles()()∑∑∑∑×=×=×==×=iicmiicmiicmcmiicmdtddtddtdfrτprτprLτprLSystems of Particles We can see that a system of particles behaves a lot like a particle itself It has a mass, position (center of mass), momentum, velocity, acceleration, and it responds to forces We can also define it’s angular momentum and relate a change in system angular momentum to a force applied to an individual particle()∑×=iicmfrτ∑=icmffInternal Forces If forces are generated within the particle system (say from gravity, or springs connecting particles) they must obey Newton’s Third Law (every action has an equal and opposite reaction) This means that internal forces will balance out and have no net effect on the total momentum of the system As those opposite forces act along the same line of action, the torques on the center of mass cancel out as wellKinematics of Rigid BodiesKinematics of a Rigid Body For the center of mass of the rigid body:22dtddtddtdcmcmcmcmcmcmxvaxvx===Kinematics of a Rigid Body For the orientation of the rigid body:onacceleratiangular locityangular ve matrixn orientatio 3x3 dtdωωωA=Offset Position Let’s say we have a point on a rigid body If r is the world space offset of the point relative to the center of mass of the rigid body, then the position x of the point in world space is:rxx+=cmOffset Velocity The velocity of the offset point is just the derivative of its positionrωvvrxxvrxx×+=+==+=cmcmcmdtddtddtdOffset Acceleration The offset acceleration is the derivative of the offset velocity()rωωrωaarωrωvvarωvv××+×+=×+×+==×+=cmcmcmdtddtddtddtdKinematics of an Offset Point The kinematic equations for an fixed point on a rigid body are:()rωωrωaarωvvrxx××+×+=×+=+=cmcmcmDynamics of Rigid BodiesRigid Bodies We treat a rigid body as a system of particles, where the distance between any two particles is fixed We will assume that internal forces are generated to hold the relative positions fixed. These internal forces are all balanced out with Newton’s third law, so that they all cancel out and have no effect on the total momentum or angular momentum The rigid body can actually have an infinite number of particles, spread out over a finite volume Instead of mass being concentrated at