Rigid Body Dynamics CSE169: Computer Animation Instructor: Steve - PowerPoint PPT Presentation

Rigid Body Dynamics CSE169: Computer Animation Instructor: Steve Rotenberg UCSD, Winter 2020

Cross Product i j k  = a b a a a x y z b b b x y z    = − − − a b a b a b a b a b a b a b y z z y z x x z x y y x

Properties of the Cross Product ◼

Cross Product    = − − − a b a b a b a b a b a b a b y z z y z x x z x y y x =  c a b =  − + c 0 b a b a b x x z y y z = +  − c a b 0 b a b y z x y x z = − + +  c a b a b 0 b z z x x y z

Cross Product =  − + c 0 b a b a b x x z y y z = +  − c a b 0 b a b y z x y x z = − + +  c a b a b 0 b z z x x y z   −     c 0 a a b x z y x       = −  c a 0 a b       y z x y       −     c a a 0 b   z y x z

Cross Product   −     c 0 a a b x z y x       = −  c a 0 a b       y z x y       −     c a a 0 b   z y x z  =  ˆ a b a b   − 0 a a z y   = − ˆ a a 0 a   z x   − a a 0   y x

Hat Operator ◼ We’ve introduced the ‘hat’ operator which converts a 𝐛 𝑈 = −ො vector into a skew-symmetric matrix ( ො 𝐛 ) ◼ This allows us to turn a cross product of two vectors into a dot product of a matrix and a vector ◼ This is mainly for algebraic convenience, as the dot product is associative (although still not commutative) ො 𝐛 ∙ 𝐜 = 𝐛 × 𝐜 𝐛 ∙ 𝐜 ≠ 𝐜 ∙ ො ො 𝐛 (non commutative) 𝐛 ∙ መ 𝐛 ∙ መ ො ො 𝐜 ∙ 𝐝 = 𝐜 ∙ 𝐝 (associative)

Derivative 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 ω ω r d r =  ω r dt ω  r

Derivative 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 ω x a , ω x b , and ω x c , and so the derivative of the entire matrix is: d A =  =  ˆ ω ω A A dt

Product Rule ◼ The product rule defines the derivative of products ( ) d ab da db = + b a dt dt dt ( ) d abc da db dc = + + bc a c ab dt dt dt dt

Product Rule ◼ It can be extended to vector and matrix products as well ( )  d a b d a d b =  +  b a dt dt dt ( )  a b a b d d d =  +  b a dt dt dt ( )  d A B d A d B =  +  B A dt dt dt

Eigenvalue Equation ◼

Symmetric Matrix ◼

Symmetric Matrix Diagonalization ◼

Dynamics of Particles

Kinematics of a Particle x position d x = v veloc ity dt 2 d v d x = = a accelerati on 2 dt dt

Mass, Momentum, and Force m mass = p m v momentum d p = = f m a force dt

Moment of Momentum ◼ The moment of momentum is a vector =  L r p ◼ 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 system ◼ It is typically represented with a capital L , which is unfortunately inconsistent with our standard of using lowercase for vectors…

Moment of Momentum ◼ L is the same for all three of these particles p • p • p r r • 3 2 =  r L r p 1

Moment of Momentum ◼ L is different for all of these particles p • p • r =  1 L r p r 2 r 3 p •

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 point d L τ = dt

Moment of Force (Torque) =  L r p L r p d d d = =  +  τ p r dt dt dt =  +  τ v p r f ( ) =  +  τ v m v r f =  τ r f

Rotational Inertia ◼ L = r x p 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 ω

Rotational Inertia =  L r p ( ) =  =  L r v r v m m ( ) ( ) =   = −   ω ω L m r r m r r = −   ˆ ˆ ω L m r r =  ω L I = −  ˆ ˆ I m r r

Rotational Inertia = −  ˆ ˆ I m r r     − − 0 r r 0 r r z y z y     = − −  − I m r 0 r r 0 r     z x z x     − − r r 0 r r 0     y x y x   − − 2 2 r r r r r r y z x y x z   = − − − 2 2 I m r r r r r r   x y x z y z   − − 2 2 r r r r r r   x z y z x y

Rotational Inertia ( )   + − − 2 2 m r r mr r mr r y z x y x z  ( )  = − + − 2 2 I mr r m r r mr r   x y x z y z ( )   − − + 2 2 mr r mr r m r r   x z y z x y =  ω L I

Rotational 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 =  ω L I ◼ This is similar to how mass relates linear momentum to linear velocity, but rotation adds additional complexity = p m v

Systems of Particles

Systems of Particles n  = m m tota l mass of all particles total i = i 1  x m = i i x position of center of mass  cm m i   = = p p m v to tal momentum cm i i i

Velocity of Center of Mass  x m x d d = = i i cm v  cm dt dt m i d x   i m v m i dt = = i i v   cm m m i i p = cm v cm m total = p m v cm total cm

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 force  = p p cm i  d p d p d p   = = = i cm i f i dt dt dt

Systems of Particles ◼ The total moment of momentum around the center of mass is:  =  L r p cm i i ( )  = −  L x x p cm i cm i

Torque in a System of Particles  =  L r p cm i i   d r p d L = = τ i i cm cm dt dt ( )   d r p = τ i i cm dt ( )  =  τ r f cm i i

Systems 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  = f f cm i ◼ We can also define it’s angular momentum and relate a change in system angular momentum to a force applied to an individual particle ( )  =  τ r f cm i i

Internal 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 well

Dynamics of Rigid Bodies

Kinematics of a Rigid Body ◼ For the position of the center of mass of the rigid body: x cm d x = cm v cm dt 2 d v d x = = cm cm a cm 2 dt dt

Kinematics of a Rigid Body ◼

Rigid 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 discrete points, we will consider the density as being variable over the volume

Rigid Body Mass ◼ With a system of particles, we defined the total mass as: n  = m m i = i 1 ◼ For a rigid body, we will define it as the integral of the density ρ over some volumetric domain Ω  =   m d 

Rigid Body Center of Mass ◼ The center of mass is:   x  d = x  cm   d