Dynamics. 1 Copyright c 2015 Roderic Grupen

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1 Dynamics The branch of physics that treats the action of force on bodies in motion or at rest; kinetics, kinematics, and statics, collectively. Websters dictionary Outline Conservation of Momentum Inertia Tensors - translation and rotation Dynamics Newton/Euler Dynamics Lagrangian Dynamics State Space Form - computed torque equation Applications: simulation; control - feedforward compensation; analysis - the acceleration ellipsoid. 1 Copyright c 2015 Roderic Grupen

2 Newton s Laws 1. a particle will remain in a state of constant rectilinear motion unless acted on by an external force; 2. the time-rate-of-change in the momentum (mv) of the particle is proportional to the externally applied forces, F = d dt (mv); and 3. any force imposed on body A by body B is reciprocated by an equal and opposite reaction force on body B by body A. Conservation of Momentum Linear: F = d [mẋ] = mẍ dt [ ] kg m [N] = sec 2 Angular: τ = d [ J dt θ ] = J θ [ kg m 2] [Nm] = sec 2 J is called the mass moment of inertia 2 Copyright c 2015 Roderic Grupen

3 Conservation of Momentum To generate an angular acceleration about O, a torque is applied around the ẑ axis τ k = r f = r kˆr d dt (m kv k ) = m k r k [ˆr d ] dt (v k) ˆt ω ˆr O f mk r k the velocity of m k due to ω O is v k = (ωẑ r kˆr) = (r k ω)ˆt, so that τ k = (m k r 2 k) ω ẑ = J k ω ẑ τ = ( ) m k rk 2 k ω = J ω. 3 Copyright c 2015 Roderic Grupen

4 Conservation of Momentum J [kg m 2 ] is the scalar moment of inertia In the rotating lamina, the counterpart of linear momentum (p = mv) is angular momentum L = Jω, so that Euler s equation can be written in the same form as Newton s second law τ = d [Jω] = J θ dt...rotating bodies conserve angular momentum (and remain in a constant state of angular velocity) unless acted upon by an external torque... O v p perigee r p O ra apogee v a r C r D r r B A D O C v B A AB = BC = CD = v t 4 Copyright c 2015 Roderic Grupen

5 Inertia Tensor z y A axis a x lamina j m k r k y axis a A r x z r a I = O t I xx I xy I xz I yx I yy I yz I zx I zy I zz MASS MOMENTS OF INERTIA I xx = (y 2 +z 2 )ρdv I yy = (x 2 +z 2 )ρdv I zz = (x 2 +y 2 )ρdv MASS PRODUCTS OF INERTIA I xy = xyρdv I xz = xzρdv I yz = yzρdv 5 Copyright c 2015 Roderic Grupen

6 EXAMPLE: Inertia Tensor of an Eccentric Rectangular Prism w x z A l y I xx = = h = = = = h l w 0 0 h l 0 0 h [( y3 0 h 0 (y 2 +z 2 )ρdxdydz (y 2 +z 2 )wρdydz ] l 3 +z2 y) wρdz 0 ( ) l z2 l wρdz ( ) l 3 z 3 + lz3 h (wρ) 3 ( 0 l 3 ) h 3 + lh3 wρ 3 or, since the mass of the rectangle m = (wlh)ρ, I xx = m 3 (l2 +h 2 ). 6 Copyright c 2015 Roderic Grupen

7 EXAMPLE: Inertia Tensor of an Eccentric Rectangular Prism...completing the other moments and products of inertia yields: A I = m 3 (l2 +h 2 ) m 4 wl m 4 hw m 4 wl m 3 (w2 +h 2 ) m 4 hl m 4 hw m 4 hl m 3 (l2 +w 2 ) 7 Copyright c 2015 Roderic Grupen

8 Parallel Axis Theorem - Translating the Inertia Tensor A z y x r cm y x z CM the moments of inertia look like: A I zz = CM I zz +m(r 2 x+r 2 y), and the products of inertia are: A I xy = CM I xy +m(r x r y ). 8 Copyright c 2015 Roderic Grupen

9 EXAMPLE: The Symmetric Rectangular Prism z w A CM y h CM I zz = A I zz m(r 2 x+r 2 y) = m 3 (l2 +w 2 ) m 4 (l2 +w 2 ) x l = m 12 (l2 +w 2 ) CM I xy = A I xy m(r x r y ) = m 4 (wl) m 4 (wl) = 0. moving the axes of rotation to the center of mass results in a diagonalized inertia tensor (l 2 +h 2 ) 0 0 CM I = m 12 0 (w 2 +h 2 ) (l 2 +w 2 ) 9 Copyright c 2015 Roderic Grupen

10 Rotating the Inertia Tensor angular momentum L 0 = I 0 ω about frame 0 in a vector quantity that is conserved. we can express it relative to frame 1 as or L 1 = 1 R 0 L 0 I 1 ω 1 = R(I 0 ω 0 ) = RI 0 R T Rω 0 and therefore, I 1 = RI 0 R T. 10 Copyright c 2015 Roderic Grupen

11 Rotating Coordinate Systems Definition (Inertial Frame) the frame where the absolute state of motion is completely known Let frame A be an inertial frame. Frame B has an absolute velocity, ω B B (written in terms of frame B coordinates). r A (t) = A R B (t)r B (t) ṙ A (t) = A R B (t) d dt [r B(t)]+ d dt [ AR B (t) ]r B (t) x ω y ω x ω y ω x ω z ω z z ω y y ω x ω z To evaluate the second term on the right, considerhowthe ˆx,ŷ,andẑ,basisvectors for frame B change by virtue of ω B. ˆx = ω z ŷ ω y ẑ ŷ = ω zˆx ω x ẑ ẑ = ω z ŷ ω y ẑ so d dt [ AR B (t) ]r B (t) = 0 ω z ω y ω z 0 ω x ω y ω x 0 r x r y r z = ω r 11 Copyright c 2015 Roderic Grupen

12 Rotating Coordinate Systems Therefore, ṙ A (t) = A R B (t) d dt [r B(t)]+ d dt [ AR B (t) ]r B (t) = A R B [ṙb +(ω B B r B ) ] and, in fact, all vector quantities expressed in local frames that are moving relative to an inertial frame are differentiated in this way d dt [ AR B (t)( ) B ] = A R B [ ] d dt ( ) B +(ω B ( ) B ) 12 Copyright c 2015 Roderic Grupen

13 Rotating Coordinate Systems: Low Pressure Systems large scale atmospheric flows converge at low pressure regions. A nonrotating planet, this would result in flow lines directed radially inward. but the earth rotates... ω consider a stationary inertial frame A and a rotating frame B attached to the earth L v L ( ω v) v A = A R B (t)v B v A = A R B [ v B +(ω v B )] so that to an observer that travels with frame B: v B = B R A [ v A ] (ω v B ) a convergent flow and a rotating system, therefore, leads to a counterclockwise flow in the northern hemisphere and a clockwise rotation in the southern hemisphere. 13 Copyright c 2015 Roderic Grupen

14 Newton/Euler Method l 3 l 1 O 2 2 l 2 3 O 3 1 O 1 0 (a) outward iteration 0 ω 0 = ω 0 = v 0 = 0 N 1 F 1 1 N 2 F 2 ω 1 2 ω 1 v 1 ω 2 ω 2 N 3 F 3 ω 3 v 2 3 ω 3 v 3 f 1 0 η 1 N 1 1 η 1 f 2 F 1 η N 2 F 2 f f 2 η2 η 3 F 3 N 3 3 η 3 f 4 = η 4 = 0 (b) inward iteration f 1 f 3 the recursive equations for these iterations are derived in Appendix B of the book 14 Copyright c 2015 Roderic Grupen

15 Recursive Newton-Euler Equations Outward Iterations Angular Velocity: ω REVOLUTE: i+1 ω i+1 = i+1 R i i ω i + θ i+1 ẑ i+1 PRISMATIC: i+1 ω i+1 = i+1 R i i ω i Angular Acceleration: ω REVOLUTE: i+1 ω i+1 = i+1 R i i ω i +( i+1 R i i ω i θ i+1 ẑ i+1 )+ θ i+1 ẑ i+1 PRISMATIC: i+1 ω i+1 = i+1 R i i ω i Linear Acceleration: v [ REVOLUTE: i+1 v i+1 = i+1 R i i v i +( i ω i i p i+1 )+( i ω i i ω i i p i+1 ) ] [ PRISMATIC: i+1 v i+1 = i+1 R i v i i + d iˆx i +2( i ω i d iˆx i )+( i ω i d iˆx i ) +( i ω i i ω i d iˆx i ) ] Linear Acceleration (center of mass): v cm i+1 v cm,(i+1) = ( i+1 ω i+1 i+1 p cm ) + ( i+1 ω i+1 i+1 ω i+1 i+1 p cm )+ i+1 v i+1 Net Force: F i+1 F i+1 = m i+1 i+1 v cm i+1 Net Moment: N i+1 N i+1 = I i+1 i+1 ω i+1 +( i+1 ω i+1 I i+1 i+1 ω i+1 ) Inward Iterations Inter-Link Forces: i f i = i F i + i R i+1 i+1 f i+1 Inter-Link Moments: i η i = i N i + i R i+1 i+1 η i+1 +( i p cm i F i ) +( i p i+1 i R i+1 i+1 f i+1 ) 15 Copyright c 2015 Roderic Grupen

16 The Computed Torque Equation State Space Form τ = M(θ) θ +V(θ θ)+g(θ)+f external forces/torques: external forces friction viscous τ = v θ coulomb τ = c(sgn( θ)) hybrid 16 Copyright c 2015 Roderic Grupen

17 EXAMPLE: Dynamic Model of Roger s Eye y θ m x l or d τ = (J θ) dt τ m +mglsin(θ) = (ml 2 ) θ τ m = M θ+g, generalized inertia M = ml 2 (a scalar); Coriolis and centripetal forces V(θ, θ) do not exist; and Gravitational loads G = mglsin(θ) 17 Copyright c 2015 Roderic Grupen

18 EXAMPLE: Dynamic Model of Roger s Arm x 3 m 2 y l y 2 x 2 m 1 l 1 x 0 y 1 x 1 y 0 M(θ) = V(θ, θ) = G(θ) = [ m2 l2 2 +2m 2 l 1 l 2 c 2 +(m 1 +m 2 )l1 2 m 2 l2 2 +m 2 l 1 l 2 c 2 m 2 l2 2 +m 2 l 1 l 2 c 2 m 2 l2 2 [ ] m2 l 1 l 2 s 2 ( θ θ 1 θ2 ) Nm m 2 l 1 l 2 s 2 θ2 1 [ ] (m1 +m 2 )l 1 s 1 g m 2 l 2 s 12 g Nm m 2 l 2 s 12 g ] 18 Copyright c 2015 Roderic Grupen

19 Simulation initial conditions: θ = M 1 (θ) θ(0) = θ 0 θ(0) = θ(0) = 0 numerical integration: θ(t) = M 1 [τ V G F] [ τ V(θ θ) G(θ) F ] θ(t+ t) = θ(t)+ θ(t) t θ(t+ t) = θ(t)+ θ(t) t+ 1 2 θ(t) t 2 19 Copyright c 2015 Roderic Grupen

20 Feedforward Dynamic Compensators + Σ q des M Σ τ ROBOT q q K B V G q ref + Σ PD controller feedforward compensator plant linearized and decoupled 20 Copyright c 2015 Roderic Grupen

21 Generalized Inertia Ellipsoid computed torque equation: τ = M θ +V(θ, θ)+g(θ) if we assume that θ 0, and we ignore gravity τ = M θ θ 1 relative inertia torque required to create a unit acceleration defined by the eigenvalues and eigenvectors of MM T 21 Copyright c 2015 Roderic Grupen

22 Acceleration Polytope gravity, actuator performance, and the current state of motion influences the ability of a manipulator to generate accelerations differentiating ṙ = J q, r = J(q) q + J(q, q) q = J [ M 1 (τ V G) ] + J q = JM 1 τ + v vel + v grav, v vel = JM 1 V+ J q, v grav = JM 1 G. and τ = L 1 τ L = diag(τ1 limit,...,τn limit ) admissible torques constitute a unit hypercube τ 1 r = JM 1 L τ + v vel + v grav = JM 1 L τ + v bias. maps the n-dimensional hypercube τ 1 to the m-dimensional acceleration polytope 22 Copyright c 2015 Roderic Grupen

23 Dynamic Manipulability Ellipsoid τ T τ = ( r v bias ) T ( [JM 1 L ] 1 ) T ( [JM 1 L ] 1 ) ( r v bias ) 1 M and L are symmetric: A T = (A 1 ) T, A 2 = A 1 A 1, and for symmetric matrices, A T = A. so that ( r v bias ) T [ J T ML 2 MJ 1] ( r v bias ) 1, dynamic manipulability ellipsoid ( r v bias )( r v bias ) T [ JM T L 2 M 1 J T] dynamic-manipulability measure κ d (q, q) = det[j(m T M) 1 J T ] 23 Copyright c 2015 Roderic Grupen

24 Conditioning Acceleration x=0.5 m x=0.3 m y x y= m y= m y x m 1 = m 2 = 0.2 kg, l 1 = l 2 = 0.25 m, τ T τ N 2 m 2. black ellipsoids - unbiased dynamic manipulability gravity biased dynamic manipulability normalized acceleration polytope with gravity bias 24 Copyright c 2015 Roderic Grupen