Rigid Body Dynamics for Movement Analysis in Biomechanics

S R S R❑ ⃗∫ ⃗ ( ).AM × V M , t dm/RSS ❑ ❑⃗ ⃗∫ ∫⃗ ⃗AB+⃗ ⃗( ) (M (BM × V , t). dm AB× V M , t). dm= = +(s A , t) /S/ R S R/S R S S❑ ⃗∫ ⃗ (BM × V M , t) . dm/RSS ❑ ❑⃗ ⃗∫ ∫⃗ ⃗ ⃗(M (AB× V , t). dm BM × V M , t) . dm= +(s A , t) / /RS R S/S R S S 11⃗ ⃗ ⃗⃗A et B : = +  ( ( ) (t )s A , t) s B ,t AB× p/ / /S R S R S R(translation theorem for angular momentum)Simpler expression of the linear momentum:❑⃗ ⃗∫(t)= ( ).p V M ,t dmBy definition: / /S R S RS|❑ ⃗d OM∫⃗ . dm if O is the origin of the reference frame R(t)=¿p dt/S R RS |( )❑ d dd⃗ | |∫ ⃗⃗ ⃗ ⃗( ) ( )(t)=p OM . dm M. OG OG= = M. = M. (GV , t)/ R RS R dt dt dt /S RRS⃗ ⃗M. (t )=¿ (Gp V , t)/ /S R S RLet us see now methods avoiding the calculation of the integral for the angularKoenig’s theoremmomentum. These methods rely on , which allows to reduce

The computation of these integrals to the particular case of motion around a fixed point.

VI. KEONIG'S THEOREM FOR THE KINETIC ENERGY AND THE ANGULAR MOMENTUM

Koening's reference frame associated to the movement of S/R: let consider- { }⃗ ⃗ ⃗O , X , Y , Z a solid S, moving /R. We note G the centre of mass of S. Let R = .0 0 0R

Definition: The Koenig's reference frame associated to the movement of S/R is K the frame { }⃗ ⃗ ⃗R G , X , Y , Z=K 0 0 0R is not fixed on S, it only has a common point G with S. As its directions remain K R movement of is a translation.

⃗ =⃗ +⃗ Using the composition of movements: (t ) (t) (t ) W W W/ / /S R S R R RK K⃗0 (translation)⃗ =⃗ (rotation speed) (t ) (t) W W/ /S R S R K 12⃗ =⃗t)+⃗ In the same way: M : (M (M (V , t) V , V M ,t) / /S R S/ R R RK K In a translation movement, all points have the same velocity at each instant of time ⃗R Then, the point M is considered rigidly

(M ¿V ,t)K /R RK⃗(GV , t)/R RK ⃗ ¿⃗Moreover, G is also fixed on S  (GV , t) (G )V ,t/R R /S RK⃗ ⃗ ¿(V M ,t) (G )V ,t/R R /S RK⃗ =⃗ ⃗Then: (M (MV , t) V , t)+¿ (G )V ,t/S R S/ R /S RKRotation about GKoenig’s theorem for the angular momentum-Refresher : A being any point, we had to calculate:❑⃗ ⃗∫ ⃗( ) ( ).s A , t = AM × V M ,t dm/ /S R S RS ⃗ ⃗And: (t)=M. (Gp V , t)/ /S R S R ⃗Using translation theorem for angular momentum: A et B : = (s A , t)/S R⃗ ⃗⃗+(B ) (G )AB×s ,t . V ,t/ /S R S R ⃗This equation is still true if B = G, t. Then, we can write: A: =  (s A , t)/S R⃗ ⃗⃗+(G (G )s , t) AG × . V ,t/ /S R S RAngular momentum in A of the particle G, on which the total mass of the solid S is assigned❑⃗ ⃗∫ ⃗(G (s , t) = GM × V M , t). dmWe have now to calculate: (by definition)/ /RS R SS❑⃗ ×(⃗ ⃗∫ ⃗(G ) (M ) (G )).s , t = GM V , t +

V, t dmWe can write: / / /S R S R S RKS❑ ❑⃗ ⃗ ⃗∫ ∫⃗ ⃗(G (M (G,s , t) = GM × V , t) . dm GM × V t) . dm+/ /RS R S S/ RKS S❑ ⃗∫⃗ ⃗ ⃗ (G )GM . dm× V ,t(G (Gs , t) = s ,t) + /S R/ /S R S R K S ⃗Rotation around a fixed point = (property of G)0⃗ ⃗ that we have to calculate in this specific movement. (G (G )s , t) = s ,t/ /S R S R K 13⃗⃗ ⃗Conclusion : A , = + (Gs , t)( AG ×s A , t) S/ R/S R K⃗(GM. V , t)/S RTheorem: A being any point, on or off the solid S, fixed or moving /R, the angularmomentum in A of the solid S is the total of:The angular momentum in A of the particle G, on which the total mass of the• solid S is assigned,The angular momentum in G of the solid S in its movement with respect to• R R, which is a rotation around the fixed point G (in ).K K❑⃗ ⃗∫ ⃗(G ) (M )s , t = GM × V , t . dmWe will have then to calculate but in a very/S/ R S RK KSspecific case.Koenig’s

1 ⃗∫ 2( )V M , t .(t)T dm

Refresher : We had to calculate: =/R S/ RS 2 S⃗ =⃗ ⃗( (With V M , t) V M , t)+¿ (G )V ,t/S R S/ R /S RK⃗ =⃗ +⃗ ⃗ ⃗2 2 2 (M (M ) (G, +2 (G )∙ (V , t) V ,t V t) V ,t V M , t)/ / / /S R S R S/ R S R S RK K❑ ❑1 1⃗∫ ∫⃗2(M )V ,t . dm2(t)T = + +dm (G )V ,t//R S RS 2 2 /S RKS S❑⃗∫⃗ (M ).V ,t dm(G )∙V ,t /S R/S R KS d |⃗ ⃗( ) RGM((1) G being the origin ofV M , t)=¿ KRdt/S R KK |( )❑ ❑d d|∫ ∫⃗ ⃗( ) ⃗GM . GM . dmdm = (1) = 0Rdt dtK RS S K1 ⃗( 2T t)(t)TConclusion : = + M (G )V ,t/RS/RS 2 /S RKTheorem : The kinetic energy associated to the movement of S / reference frame R isthe total of :The kinetic energy /R of the particle G, on which the total mass of the solid S is• assigned, RThe kinetic energy associated to the movement of S / reference frame ,• KRwhich is a rotation around the fixed point G (in ).K❑1 ⃗∫ 2( )V M ,t

(T t) dmWe will have then to calculate = but in a very/RS /S R2 KK Sspecific case. 14VII. KINETICS OF THE SPECIFIC MOVEMENT OF A SOLID S/R AROUND AFIXED POINT O (≠ G A PRIORI)Introduction : Let S being mobile /R around the point O, origin of R and belonging toS. ¿V (t)/S R|Within this context: ⃗ ⃗( )W t ≠ 0 a priori/RS¿⃗ ⃗= /RV (O, t ) 0 as O∈ S∧¿/S RAt instant t, the movement of S/R is a rotation about an axis⃗( ) parallel to and passing through O∆ t ( )W t/S R /S R⃗ (t)TSpecific expressions of and in this movement(O )- s ,t /RSS/ R⃗Calculation of : (Os , t)/S R ❑⃗ ⃗∫ ⃗(O (s , t) = OM × V M , t). dmBy definition : / /RS R SS⃗ ⃗ ⃗ ⃗Here : = + (translation theorem for(M (O ) (t )V , t) V ,t W × OM/ / /S R S R S Rvelocity on S) ❑⃗ ×(⃗∫ ⃗ ⃗(O ) (t ) )s , t = OM W × OM . dm / /RS R SS⃗ ⃗We notice that depends linearly of So, there is a linear(O (t ).s , t) W/ /S R S

of a rotation around a fixed point O: ᵣᵣ(t) = ᵇᵢ(t) (OW_s,t)/RS²/SRSR´

Then, we have to determine the inertia tensor of the solid S in O, which will give us: ᵣ´(O) = ᵣ´(O)(t) (s,t) = IW/SRSO

VIII. MATRIX OF THE INERTIA TENSOR IN ANY DIRECT ORTHONORMAL BASIS 3 3

Refresher: Let consider an operator from ℓ to ℓ, it transforms any vector ᵣ in a vector ᵣ´.

Xᵣ´ = ᵢ´ (Xᵣ)