f x f y or else = m a y

Transcription

1 Rigid Body Dynamics If the aim of kinematics is to describe the body motion, the aim of dynamics is to explain it; the history of mechanics shows that the passage from description to explanation requires the introduction of a new physical entity, that of mass or, in alternative, that of force. The fundamental law of mechanics, due to Newton, is analytically expressed as a vectorial equation between force, mass and acceleration of a translating rigid mass f = ma or else f x a x f y = m a y f z a z where the linear acceleration a = [ a x a y a z ] T is a kinematic quantity, obtained ad the time derivative of the body linear velocity. This law is true in principle only for a single point-mass particle with mass m, where the applied force f and the acceleration a can exchange the role of cause and effect: if a force f is applied to the particle, the particle accelerates with a linear acceleration equal to a, and conversely, if a particle has a linear acceleration a, the particle is subject to a force f proportional to its mass. If we jointly know the two vectors f = [ f x f y f z ] T and a = [ ax a y a z ] T, we can compute the mass from one or any of the following scalar relations as m = f x a x = f y a y = f z a z Though, in this last case, we are applying a circular argument, since no definitions of acceleration or force exist that are independent from the measurement of a mass; we are therefore compelled to use some trick, as clearly illustrated in [?]. If a single point mass is connected to others to form a rigid body, the Newton law is still valid, given that we observe some precautions. Every point-mass shall be isolated and we must consider and deduce the forces applied on it by the other masses; that is, we must introduce the constraint forces in addition to the external forces applied on the body. It is important to notice that the vector equations are dependent on the representation used to characterize its components, and change changing the reference frame 1

2 2 used to represent the force and acceleration vectors, although the masses, at least in a non-relativistic motion state, do not vary changing the reference frame. In this textbook we will use an analytical approach, as defined in [?], instead of the vectorial one due to Newton. We consider the multibody system as a system in which the dynamical equations derive from a unifying principle that implicitly includes and generates these equations. This principle is based on the fact that, in order to describe the motion of a multibody system is sufficient to consider and use in a proper way some suitable scalar quantities; these were in origin called by Leibnitz vis viva and work function, nowadays take the name of kinetic energy and potential energy. They are an example of the so-called state functions, since to each value of the state vector 1 This general principle takes the name of principle of least action; it can be roughly described in the following way. Let us consider the space Q of the generalized coordinates q Q, as sketched in Figure 8 for a two-dimensional space Q; let us assume that a particle starts its motion at time t 1 in the state Q 1 = q(t 1 ) and ends it motion at time t 2 having reached the state Q 2 = q(t 2 ). Let us further assume that the motion keeps constant the sum E = C + P of the kinetic energy C and the potential energy P that the particle has at time t 1. Given the continuity of motion, the two points Q 1 and Q 2 are connected by a continuous path (or trajectory), as the one represented by a continuous line in Figure 8; this trajectory is called the true trajectory, and at least in principle, it is unknown, since it is exactly what we want to compute as the result of the dynamical equation analysis. If we had chosen at random a different trajectory, with the only condition that the two boundary point remain fixed, and called this trajectory a perturbed trajectory, the chance to obtain exactly the true trajectory would have been minimal. Now we ask what is that characterize the true trajectory with respect to all possible other perturbed trajectories. Euler was the first mathematician to contribute to the solution of this problem, but it was Lagrange that developed a complete theory, that was later extended by Hamilton: the true trajectory is the one that minimizes the integral of the so-called vis-viva (i.e., two times the kinetic energy) of the entire motion between Q 1 and Q 2. This integral is called action and has a constant and well defined value for each perturbed trajectory at constant E. The least action principle states that the nature chooses, among the infinite number of trajectories starting in q(t 1 ) and ending in q(t 2 ), the trajectory that 1 The concept of state will be defined in Section XXX; for the moment we simply consider that the state corresponds to the two vectors q(t) and q(t).

3 3 Figure 1: True and perturbed trajectories in the configuration space Q. minimizes the definite integral S of a particular state function C (q(t), q(t)) 2 that depends on the generalized coordinates q(t) and the generalized velocities, where q(t) S = t2 t 1 C (q(t), q(t))dt (1) In order to apply this principle it is therefore necessary to compute the trajectory in the space Q that minimizes S. The integral in (1) is computed between the initial time t 1 and the final time t 2 and must obey to the boundary constraints active at these two times. The minimization of a functional 3 is based on a particular mathematical technique, called calculus of variations, whose illustration goes beyond the scope of the present text. The interested reader can find additional treatment in [?], Appendix D. FARE APPENDICE? The conditions that assure the attainment of the minimum of S provide a set of differential equations that contain the first and second time derivatives of the q i (t); this set completely describes the system motion dynamics. As a final comment, we can say that the least action principle makes use of a concept - the action - that overturns the usual representation of the physical system dynamics 2 We will see later what this state function C is. 3 A functional is a mapping between a function and a real number; the function shall be considered as a whole, i.e., not a single particular value; in this sense a functional is often the integral of the function, as in the above equation for S.

4 4 by means of a set of differential equations. With some approximation we can say that the differential equations specify the evolution of a physical quantity as the result of infinitesimal increments of time or position; summing up this infinitesimal variations we obtain the physical variables at every instant, knowing only their initial value and possibly some initial derivative: we can say that the motion has a local representation. On the contrary, the action characterizes the motion dynamics requiring only the knowledge of the states at the initial and final times; every intermediate value of the variables can be determined by the minimization of the action, that is a global, rather than local, measure. In any case the implementation of the principle of least action produces as a result a set of differential equations, different from these obtained with the local representation. In the following Sections we will illustrate how to obtain these equations..1 Point Mass System Yo introduce the Lagrangian approach and compare it with the Newtonian approach we must define some general physical entities Let us start to consider a sate of N point masses m i, with i = 1,, N), as schematically represented inn Figure 2. These systems are called multi-point systems. Figure 2: A system of N point masses m i. The position r i and the velocity v i are represented in a generic reference frame R b. Now we consider a pure rotation motion around a point O, that, for simplicity, is the origin of the reference frame R b ; all vectors will be represented in this frame. If necessary we indicate the vectors as r b i, otherwise with r i. The position of the generic mass m i is defined by the vector r i = [ x i y i z i ] T. If the rotation velocity of the system is given by the vector ω, every mass will acquire

5 .1. POINT MASS SYSTEM 5 a linear velocity v i, as the result of the rotation, whose value is v i (t) = ω(t) r i (t) = S(ω(t))r i (t) (2) We define the linear momentum 4 p i (t) as the product between the mass m i and the velocity v i (t) p i (t) = m i v i (t) In this Chapter, the symbol p indicates the linear momentum an not the cartesian pose of a body or the position of a point, as in the preceding Chapters..1.1 Moment of a Force We briefly recall that, given a point mass located in a point P represented by the vector r in the reference frame R, and a force f applied to it, the moment of the force with respect to a point O is given by the cross product r f = r m dv dt = r dp dt where the terms on the right side arise from the Newton equation. It is important to recall that the term moment indicates the cross product, while the term momentum indicates the vector p = mv. Applying the derivative rules in (??) to the right end side of (3), we obtain r f = r dp dt = d dr (r p) dt dt p = d (r p) v p dt }{{} = dh dt where the last term is always zero, since the moment p and the linear velocity of the point mass v are always collinear, and where the vector h = r p is called angular moment or moment of the momentum. Since the moment r f is often referred as the torque applied by the force with respect to the point O, we can write Note the analogy τ = d dt h f = d dt p = ṗ (3) 4 In Italian quantità di moto (lineare). τ = d dt h = ḣ

6 6 that shows the relations between the generalized force/torque and the generalized momentum. Figure 3: A point mass m with the applied force f and it linear moment p..1.2 Angular Momentum and Inertia Matrix Given a point O in space and a point mass m i with position r i, belonging to a set of point masses, we have define the angular momentum, or moment of momentum 5 h i, as h i (t) = r i (t) p i (t) = r i (t) (m i v i (t)) Replacing (2) in the above relation and omitting for simplicity the time dependency, we obtain h i = m i (r i (ω r i )) Summing up all the N point masses contributions, we have the total angular momentum N h = h i = m i (r i (ω r i )) (4) i Recalling from (??) that the triple cross product enjoys of the following property a (b c) = (a T c)b (a T b)c, where a T c and a T b are scalar products, it follows that the momentum can be written as h = ( ) m i (r T i r i )ω (r T i ω)r i i 5 In Italian momento della quantità di moto or quantità di moto angolare.

7 .1. POINT MASS SYSTEM 7 Replacing with ri 2 the square norm r i 2 = r T i r i = (x 2 i + yi 2 + zi 2 ) and writing explicitely r T i ω = (x i ω x + y i ω y + z i ω z ) we obtain or h = i h x m i r i 2 h = h y = h z i ω x ω y (x i ω x + y i ω y + z i ω z ) ω z This relation can be written in matrix form as x i y i z i m i (ri 2 x 2 i )ω x m i x i y i ω y m i x i z i ω z m i x i y i ω x + m i (ri 2 yi 2 )ω y m i y i z i ω z (5) m i x i z i ω x m i y i z i ω y + m i (r 2 i z 2 i )ω z where h = i Γ xx,i Γ xy,i Γ xz,i ω x Γ yx,i Γ yy,i Γ yz,i ω y = Γ zx,i Γ zy,i Γ zz,i ω z i Γ i ω (6) Γ xx,i = m i (r 2 i x 2 i ) = m i (y 2 i + z 2 i ) Γ yy,i = m i (r 2 i y 2 i ) = m i (x 2 i + z 2 i ) Γ zz,i = m i (r 2 i z 2 i ) = m i (x 2 i + y 2 i ) and Γ xy,i = Γ yx,i = m i x i y i Γ xz,i = Γ zx,i = m i x i z i Γ yz,i = Γ zy,i = m i y i z i If we now introduce a new matrix defined as Γ xx Γ xy Γ xz Γ = Γ yx Γ yy Γ yz = Γ xx,i Γ xy,i Γ xz,i Γ yx,i Γ yy,i Γ yz,i = i i Γ i Γ zx Γ zy Γ zz Γ zx,i Γ zy,i Γ zz,i from (6) we have or better, restating the time dependency Γ xx Γ xy Γ xz ω x h = Γ yx Γ yy Γ yz ω y = Γ ω (7) Γ zx Γ zy Γ zz ω z h(t) = Γ (t)ω(t)

8 8 The matrix Γ is called inertia matrix or inertia tensor and has on the main diagonal the inertia moments Γ xx = i m i (r 2 i x 2 i ) = i m i (y 2 i + z 2 i ) = i m i d 2 x Γ yy = i m i (r 2 i y 2 i ) = i m i (x 2 i + z 2 i ) = i m i d 2 y (8) Γ zz = i m i (r 2 i z 2 i ) = i m i (x 2 i + y 2 i ) = i m i d 2 z and outside the diagonal the inertia products Γ xy = Γ yx = i Γ xz = Γ zx = i m i x i y i m i x i z i (9) Γ yz = Γ zy = i m i y i z i Inertia moments As one can notice, the inertia moments are obtained summing the contributions of the products of each mass by the square euclidean distance d 2 x, d 2 y, d 2 z from the three axes x, y and z. Prodotti d inerzia When the inertia products are zero, the inertia matrix becomes diagonal Γ xx Γ = Γ yy Γ zz In this case the reference frame axes are aligned along what are called the principal inertia axes of the body, and the matrix is the so called principal inertia matrix. Another possible representation of h and Γ can be obtained using (4) and the properties of the skew-symmetric matrices: h = i m i (r i (ω r i )) = i m i S(r i )S(ω)r i (1) Recalling that S(ω)r i = S(r i )ω, one obtains: h = i m i S(r i )S(ω)r i = i m i S(r i )S(r i )ω (11)

9 .1. POINT MASS SYSTEM 9 and being ω a common term, at the end one has ( h = ) m i S(r i )S(r i ) ω = Γ ω (12) i with Γ = i m i S(r i )S(r i ) (13) The inertia matrix Γ depends on the r i vectors that characterize the positions of the masses m i in a given reference frame, with respect to a given point; this point can be fixed, as the origin O of the frame, or mobile (i.e., local) as the center of mass C of the body; in the first case the inertia matrix will be indicated as Γ o, in the second case as Γ c. Considering two reference frames with a common origin, but one rotated with respect to the other, the inertia matrices will be different; the rule to transform one into the other will be detailed later on. The inertia matrix Γ represents the inertial properties of a rigid body subject to a rotation, in the same way as the mass represents the inertia properties of a body subject to a translation. Equation (7) is qualitatively similar to the expression of the total linear momentum p(t) = mv(t). Since we can write it as p(t) = M v(t), where M = mi, the two momentum are p(t) = M v(t) h(t) = Γ (t)ω(t) with the difference that usually the mass matrix M does not depend on time, while Γ (t) does. Example.1.1 Let us consider a simple body system composed by two equal masses as in Figure 4; we want to compute the angular momenta and the inertia matrices with respect to the points O and C. Assume also that R a is fixed, while R b is on the barycenter (or center of mass) of the body having the axis i aligned with the segment CB. Assume r 1 = 2 ; r c = 2 ; r 2 = 2 ; c 1 = ; c 2 =

10 1 Figure 4: The two equal masses system considered in Example.1.1. and m 1 = m 2 = 1. Since R a e R b are parallel, the rotation matrix R a b = I and we can neglect it. We start computing Γ o as 8 8 Γ o = m 1 S(r 1 )S(r 1 ) m 2 S(r 2 )S(r 2 ) = and then Γ c as Γ c = m 1 S(c 1 )S(c 1 ) m 2 S(c 2 )S(c 2 ) = 2 2 Recalling the parallel axes theorem, that we will introduce in Paragraph.3.3, and assuming also (23) and (24), we obtain 8 8 Γ o Γ c = 8 8 = (m 1 + m 2 )S(r c )S(r c ) 16 Now assuming that the angular velocity of the system is ω a ab = ω b ab = 1 we can compute the momenta as follows 8 8 h o = 8 1 = ; h c = 2 =

11 .2. DISTRIBUTED MASS SYSTEMS 11.2 Distributed Mass Systems We make a further step assuming that the body whose angular moment we want to compute is composed by a continuous distribution of infinitesimal point masses dm, having point density ρ(r) = ρ(x, y, z), depending on the position (x, y, z), and where the infinitesimal mass is dm = ρ(x, y, z)dv. The coordinate r takes its values in V, where V is a finite space region that includes the body, with a total volume V = dv and total mass m V tot = dm = V ρ(x, y, z)dv. V We can apply the same arguments used above for N point masses to a system of distributed masses, replacing in (8) and (9) the summation operator with the integral operator in the volume V. We can now write in compact form the inertia matrix as: y 2 + z 2 xy xz Γ = ρ(x, y, z) xy x 2 + z 2 yz dv (14) V xz yz x 2 + y 2 obtaining the inertia moments Γ xx = Γ yy = Γ zz = and the inertia products V V V Γ xy = Γ yx = Γ xz = Γ zx = Γ yz = Γ zy = ρ(r)(y 2 + z 2 ) dv ρ(r)(x 2 + z 2 ) dv (15) ρ(r)(x 2 + y 2 ) dv V V V ρ(r)xy dv ρ(r)xz dv (16) ρ(r)yz dv Notice that ρ(r)dv is equivalent to an infinitesimal mass dm(r), therefore we can also write relation (14) as y 2 + z 2 xy xz Γ = xy x 2 + z 2 yz dm(r) (17) V xz yz x 2 + y Some Observations The inertia matrix is implicitly defined when we define the angular momentum h as in (1); this one in turn is defined introducing a rotation around a point O,

12 12 characterized by the angular velocity ω. The cartesian components of h are represented in the chosen reference frame, that often, but not always, coincides with a reference frame with its origin in O; the inertia matrix (or inertia tensor) describes the way the mass is distributed with respect to the axes of R (O; x, y, z). If the frame is rigidly attached to the body, the inertia matrix is time independent, otherwise it changes with time. This is the reason why it is common to read in textbooks inertia moment with respect to a point and/or inertia matrix with respect to the axes. This difference shall not generate confusion since it is an example of a casual use of mathematical concepts, that, once understood, does not represents a problem anymore. A reference frame attached to the body is called a body-frame and any inertia matrix with respect to this frame is always constant. If one chooses a different reference frame attached to the body, with the same origin O, but different axes, the components of h do change and therefore also the components of Γ, but they remain time-invariant. If the frame is translated, Γ changes according to the parallel axes theorem (see Section.3.3) If the body rotates with respect to the reference frame (and therefore this one is not a body-frame), the components of Γ vary with time. To conclude, we recall that what has been presented here is valid only for rigid bodies..3 Mathematical Formulation We have seen that a inertia matrix or tensor 6 of a rigid body specifies and summarize the inertial characteristics of the body with respect to the variations of the angular velocity associated to the body rotation. The matrix is always defined specifying explicitly or implicitly a point with respect to which it is computed; usually this point is the center of mass of the body, but this choice it is not strictly necessary. In order to define correctly the inertia matrix of a rigid body it is proper to introduce a number of subscripts and superscripts that make clear the meaning with respect to the reference frames defined by the user. Given a rigid body B, identified by the subscript b, and a reference system R b, called 6 The term tensor is used to indicate an abstract mathematical entity that generalizes the vectors, while we use the term matrix to indicate its representation in some defined reference frame. We prefer to use matrix instead of tensor because it is more common in modelling and control literature

13 .3. MATHEMATICAL FORMULATION 13 local frame, the position of the center of mass C of the body is defined by the vector r b c, computed as follows: dm = r b c m b tot = r b dm (18) r b c B where r b represents the position in R b of the generic point mass dm belonging to the body B; moreover m b tot = dm is the total mass of the body B. B The inertia matrix Γ b c R 3 3 around the center of mass C is defined, as seen in Section.1.2, resorting to the angular momentum definition, that we write again here, with the addition of the introduced subscripts and superscripts: B h b c = Γ b cω b (19) where h b c is the momentum of the rigid body with respect to C and ω b ω b b is the total angular velocity of the rigid body; both are 3D vectors represented in R b. Take care not to confuse the total angular velocity ω b b represented in R b and the same velocity represented in R, that we shall indicate with the symbol ω b. The use of various subscripts and superscript in Γ b c is non a useless pedantry; on the contrary it highlights the fundamental points of its definition: Γ b c is the inertia matrix of the rigid body B with respect to its center of mass C, represented in the reference frame R b, that in general can be placed everywhere one likes, but usually attached to the body. The inertia matrix changes changing the reference frame, as we will see later. If now we attach R b to the rigid body with its origin in C we have from (18) r b c = r b dm = (2) and the inertia matrix Γ b c will be defined as: [ r Γ b c = S(r b )S(r b )dm = ] b 2 I r b (r b ) T dm B B Γ xx Γ xy Γ xz = Γ yx Γ yy Γ yz Γ zx Γ zy Γ zz where we notice that r b (r b ) T is a rank one dyadic product (see Appendix??), and that S(x ) is a skew-symmetric matrix (see Appendix??). Observe the analogy of the second term in (21) with the second term in (13). B (21).3.1 Rotation of the Reference Frame If now we consider another reference frame R k, rotated around the common origin (in this case C) with respect to R b, and the rotation matrix given by R k b, then the

14 14 inertia matrix is computed according the following relation or using the usual notation R a b = (R b a) T. Γ k c = R k bγ b cr b k = R k bγ b c(r k b) T (22) Γ k cr k b = R k bγ b c. When the generic reference frame R b is attached at the body B and moving with it, i.e., it is a body-frame, the inertia matrix relative to R b is time invariant; in the other case it is time variant..3.2 Principal Inertia Matrix Often it is appropriate to refer to a privileged reference frame R, implicitly defined as that frame having its origin in the body center of mass and with respect to which the inertia matrix is diagonal, i.e., Γ x Γ c = Γ y ; Γ z This frame has the three unit vectors i, j andk, aligned along the so-called allineati principal inertia axes; the matrix Γ c is called prende il nome di principal inertia matrix..3.3 Parallel Axes Theorem Assume that we know the inertia matrix with respect to a reference frame with origin in the center of mass C, and we want to compute it with respect to another frame, with different origin O, but parallel axes. To do so we must express the relation between the infinitesimal mass ρ dv = dm with respect to the two points O e C is (vedi Fig. 6) It is straightforward to see that: r b i = r b c + c b i, i = 1,..., N where r b c = [ t x t y ] T t z represents the translation of the reference frame from O to C. As we already know, the inverse translation, from C to O, is given by r b c. This relation is valid for any vector represented in a couple of translated reference frames.

15 .3. MATHEMATICAL FORMULATION 15 Figure 5: An example of parallel axes inertia computation. The point mass element dm i that was in the position c b i, is now in position r b i. It is now possible to compute the new inertia matrix Γ b o, taking into account (21), as Γ b o = Γ b c m tot S(r b c)s(r b c) = Γ b c + m tot [ r b c 2 I r b c (r b c) T ] (23) Simplifying the notation as follows Γ xx Γ xy Γ xz Γ Γ b c Γ = Γ yx Γ yy Γ yz e Γ b o Γ xx Γ xy Γ xz = Γ yx Γ yy Γ yz Γ zx Γ zy Γ zz Γ zx Γ zy Γ zz with t r b c and with m m tot, we will obtain Γ = Γ ms(t)s(t) (24) = Γ + m [ t 2 I tt T] (t 2 y + t 2 z) t x t y t x t z = Γ + m t x t y (t 2 x + t 2 z) t y t z t x t z t y t z (t 2 x + t 2 y) that gives Γ xx = Γ xx + m ( ) t 2 y + t 2 z Γ yy = Γ yy + m (t 2 x + t 2 z) Γ zz = Γ zz + m ( (25) ) t 2 x + t 2 y.

16 16 Figure 6: Parallel axes theorem: relation between the infinitesimal mass position in the two cases. The inertia products can be written as Γ xy = Γ xy m t x t y Γ xz = Γ xz m t x t z Γ yz = Γ yz m t y t z (26) The two relations (22) and (23) allow to compute the inertia matrix with respect to any other point O of choice, knowing the inertia matrix with respect to the point C. The relation introduced above are valid both for discrete point-mass systems and for distribute mass systems..4 Angular Momentum and Euler Equation The expression of the angular momentum has been introduced in (19) without any additional explanation; here we justify it in a more formal way. We can assume that a rigid body B is composed by a continuum of point-mass particles with mass dm located at r k, with velocity v k = ṙ k and acceleration a k = v k ; the superscript k identifies the generic reference frame R k with respect to which the vectors are represented. Since B is a rigid body, there will be a number of constraint relations linking positions, velocities and accelerations, as detailed in Section??. To simplify the mathematical treatment of the rigid body dynamics, we can reduce the action of the forces and torques acting on the body B to two equivalent effects:

17 .4. ANGULAR MOMENTUM AND EULER EQUATION 17 an equivalent linear force f k c, with its line of action across the body center of mass C, and with a module equal to the module of the vector sum of the applied forces; an equivalent torque τ k c, equal to the total moment of produced by the forces acting on B, relative to C. The second Newton Law states that the resulting inertial force is equal to the time derivative of the linear momentum ( ) ( ) d d f k c = dt mv k c = m dt v k c = ma k c where v k c is the instantaneous velocity of C, m is the total mass of the body and a k c = v k c is the total acceleration of the center of mass; the relation is valid for any choice of the reference frame R k used to describe the vectors. The Euler equation states that the inertial torque is equal to the time derivative of the angular momentum ( ) d τ k c = dt h k c where in both cases we have pointed out that the total derivative must be performed with respect to an inertial 7 (or pseudo-inertial) reference frame R ; It is not necessary that R k is an inertial frame, but only that the derivative, seen as the limit of the difference quotient between vectors, is computed in the inertial frame (see also Section??). The vector h k c is the angular momentum with respect to the mass center C and is defined, in analogy to what has been introduce in Section.1.2, as follows: h k c = (c k v k )dm (27) B where c k indicates the position of an element of infinitesimal mass dm with respect to the center of mass C of the body (see also Figure 7). Since the position of an element dm with respect to the origin O of the reference frame R k can be decomposed into the sum it results r k = c k + r k c c k = r k r k c where we have indicated with r k c the position of the center of mass C and with r k the position of dm with respect to O. 7 We recall that an inertial reference frame must be fixed or in linear uniform motion, i.e.,

18 18 Figure 7: The position r k of on infinitesimal mass element dm is decomposed into the sum r k c + c k. The reference frame R k is assumed to be inertial and all vectors are represented in it. Corretto fin qua Therefore, the velocity v k = ṙ k of the generic elementary mass can be expressed in relation to the velocity of the center of mass v k c as follows v k ṙ k = ċ k + ṙ k c = v k c + ω k kb c k (28) The second term of the right hand side is obtained considering the relation (??), rewritten here for convenience comodità ṙ (t) = ω 1(t) ρ (t) + R 1ṙ 1 (t) + ḋ 1(t) (29) where ṙ 1 (t) = ḋ 1(t) =, ρ (t) c k and ω k kb ω 1(t) As usual, the angular velocity ω k kb, is the total angular velocity of B with respect to R k and is represented in R k itself. Replacing (28) in the (27), we obtain h k c = c k dm v k c + B B c k (ω k kb c k )dm = the term B ck dm goes to zero since, in accordance with the definition of the mass center in (18), one can write: c k dm = (r k r k c)dm = r k dm mr k c = B B one in which the accelerations acting on it are zero or negligible; in this last case the reference frame is pseudo-inertial. For example, a reference frame on your study table experiences only the accelerations due to the Earth rotation around its axis and around the Sun, to neglect other more complex motions of the Sun in the local galaxy, etc. In this textbook we call inertial any pseudo-inertial reference frame. B

19 .4. ANGULAR MOMENTUM AND EULER EQUATION 19 being r k dm = mr k c In conclusion, we have h k c = c k (ω k kb c k )dm = c k (c k ω k kb)dm = B B = S(c k )S(c k )ω k kbdm = S(c k )S(c k )dm ω k kb = B = definition (21) = B (3) = Γ k cω k kb If we represent the vectors and the inertia matrix in a different reference frame R l we write h l c = Γ l cω l kb. with the relation between the inertia matrices and the angular velocities therefore we can observe that Γ l c = R l kγ k cr k l (31) ω l kb = R l kω k kb h l c = Γ l cω l kb = R l kγ k c R k l R l k ω k kb = R l }{{} kγ k cω k kb = R l kh k c I To complete the Euler equation, we should define the time derivative of the angular momentum. Recalling the expression (??), we can write ḣ k c = ( ) ( ) d d (Γ k dt cω k kb) = (Γ k dt cω k kb) + ω k cb (Γ k cω k cb) k Since Γ k c is a constant in R k, it follows that ḣ k b = Γ k bα k b + ω k b (Γ k bω k b) (32) where α k b ωk b is the total angular acceleration of the body b represented in R k. In conclusion, the time derivative of the angular momentum, i.e., the inertial torques that appears in the Euler equation, can be simply written as: ḣ = Γ ω + ω (Γ ω) (33) where the simplified symbols are to be correctly interpreted.

20 2 Alternative Formulation Another way to obtain (33) is the following: we assume that the body b rotates with a total angular velocity ω with respect to an inertial reference frame R, whose origin is now located at the center-of-mass of the body. The Euler equation, that describes the dynamics of a rotating body, states that the variation of the angular momentum h is equal to the vector τ, that is the sum of all torques applied to the body: Both Γ and ω are time functions, so we have ḣ d dt (Γ ω ) = τ (34) ḣ = Γ ω + Γ ω (35) In order to easily compute Γ it is convenient to change the reference frame; if R l is the local frame attached to the body, i.e., the body-frame, having its origin in the center-of-mass, if R R l is the rotation matrix representing R l in R, and if Γ l is the inertia tensor expressed in R l, then we have seen from (22) that Γ = RΓ l R T (36) Calling ω l the total angular velocity of the body represented in the frame R l, i.e., ω l ω l b, and calling ω the same velocity represented in R, i.e., ω b ω, we can write ω = Rω l and consequently h = Γ ω = RΓ l R T Rω l = RΓ l ω l = Rh l (37) where we have introduced the new vector h l = Γ l ω l that is the angular momentum represented in R l. We can see that the angular momentum is a vector that transforms from one reference frame to another with the same rule of other vectors. Replacing h in (37) and computing the derivative, we obtain: ḣ = d dt (RΓ lω l ) = ṘΓ lω l + R Γ l ω l + RΓ l ω l (38) The inertia tensor Γ l is time invariant, since it is relative to a body-frame and the body does not change its shape, therefore Γ l = O, and (38) simplifies as follwos ḣ = S(ω )RΓ l ω l + RΓ l ω l = ω (RΓ l ω l ) + RΓ l ω l (39) This vector is represented in R but includes both ω l and ω ; if we want to represent everything in R l we should pre-multiply by R T. Recalling the properties of the skew-symmetric matrices, we obtain ḣ l = R T ḣ = R T S(ω )RΓ l ω l + R T RΓ l ω l = S(R T ω )Γ l ω l + Γ l ω l = S(ω l )Γ l ω l + Γ l ω l (4) = ω l Γ l ω l + Γ l ω l

21 .5. NEWTON-EULER EQUATIONS 21 Poiché ḣ l = τ l, segue: τ l = ω l Γ l ω l + Γ l ω l (41) where τ l = R T τ is the sum of the applied torques, represented in R l. If we prefer to represent the torque in R, we use (39), replacing Γ R instead of RΓ l, obtained from (36); in this way we obtain τ = ω Γ ω + Γ ω (42) The equations (33), (41) and (42) are identical, provided that all vectors and tensor matrices are expressed in the same reference frame..5 Newton-Euler Equations The Newton-Euler equations are vector equilibrium equations that state that all forces and torques acting on the body B, including the inertial ones due to the motion, the external ones and those due to the constraint, are in (dynamical) equilibrium. There are two vector equations, the first, due to Newton, describes the equilibrium of the linear forces, while the second, due mainly to Euler, describes the equilibrium of the linear momenta. Assuming that the rigid body is connected with other rigid bodies, we must consider also the reaction forces and torques that the other bodies transmit to the considered body. For simplicity, from now on, all vectors will be represented in a common reference frame R k, but omitting the subscript k. The motion of the rigid body B is decomposed into two distinct motions 1. A translation motion of the center-of-mass C, described by the Newton equilibrium vector equation f ac + f vc + mg C ma C = (43) where a C = v C = r C is the total acceleration, with respect to an inertial reference frame R, of the center-of-mass C, g C is the acceleration due to the gravitational field, applied to C, f ac is the resultant of all the external forces acting on the body applied in C, and f vc is the resultant of the constraint forces, with the application point brought in C. 2. A rotation motion around a generic point P, described by the Euler equilibrium vector equation τ ap + τ vp + τ f ap + τ f vp + τ gp + τ ip Γ b/p α ω Γ b/p ω = (44)

22 22 where τ ap is the resultant of the active applied external torques, τ vp is the resultant of the constraint torques, τ f ap is is the resultant of the torques due to the applied external forces, τ f vp is the resultant of torques due to the constraint forces, τ gp is the torque due to the gravitational field, and τ ip is the torque due to the inertial force ma P, ω is the total ( ) angular velocity ω ω b, and d α is the total angular acceleration α = ω b dt. Equation (44) is simplified if the point P coincides with the center-of-mass C; indeed in this case we have τ gp = τ ip = and (44)reduces to τ ac + τ vc + τ f ac + τ f vc Γ bα ω Γ b ω = (45) Notice that equations (43) and (45) are functions of a C, the linear acceleration of the center-of-mass, of the total angular acceleration α and of the total angular velocity ω. If the acceleration due to the gravitational field is negligible, then g. The Newton-Euler equations can be also written in a more compact form, introducing suitable matrices, as follows [ ] [ ] [ ] [ ] mi a C f + = totc (46) Γ b α S(ω )Γ b where τ totc f totc = f ac + f vc + mg C τ totc = τ ac + τ vc + τ f ac + τ f vc Recalling that a C = v C + ω v C, where v C is the total linear velocity of the center-of-mass, a second form of the equation is possible [ ] [ ] [ ] [ ] [ ] mi v C ms(ω ) v + C f = totc (47) Γ b α S(ω )Γ b ω τ totc If instead of the center-of-mass C we want to express the equation with respect to another point P, the resulting equation become [ ] [ ] [ ] [ ] mi ms(r pc ) T a p ms(ω )S(ω + )r pc f = totp (48) ms(r pc ) Γ b ω α S(ω )Γ b/p ω τ totp where r pc is the oriented segment from P to C, and a p = r p is the total acceleration of P. In conclusion, we want again point out that the difficulties arising with the use of these equations is due to the necessity to include the constraint forces and torques; these do not contribute to characterize the body motion, but have the only function to replace the geometrical constraints with vector quantities to be included in the equations However, if one is interested in numerical computation of the body dynamics the Newton-Euler equations, suitably organized in a recursive form, are an efficient way to do so, as we will see in Section.8.

23 .6. VIRTUAL WORK PRINCIPLE 23.6 Virtual Work Principle Virtual displacements, as seen in Section??, give the degrees-of-freedom of a rigid body and are used to define the virtual work. Given a system composed by N point masses, each located in a position defined by the vector r i, and N forces (possibly having zero magnitude) acting on them from the external world, the virtual work δw is defined as in relation (??), reported here for completeness N δw = f i δr i If the system is a continuous rigid body, on which both forces and torques act, we can write a more complete expression, as in (??) N f N τ δw = f i δr i + τ i δα i = (49) where τ i are the applied torques and δα i the respective angular virtual displacements. The virtual work principle can be applied also to dynamic equilibrium; indeed, if we introduce the inertia forces m i a i, and if we define with the name of effective forces the sum of the applied and inertia forces F i = f i m i a i acting on each mass particle, the D Alembert principle [?] states that the total virtual work of the effective forces is zero for all invertible variations that satisfy the kinematic conditions given by : F i δ r i = (f i m i a i ) δr i = (5) i i.7 Lagrange Equations in Mechanical Systems We have already seen in Section.5 how the Newton-Euler approach determines the dynamical equations of the system. We will see now how the Lagrange approach can produce an equivalent set of equations, starting from a very different viewpoint. The Lagrange approach is based on the definition of two scalar quantities, namely the total kinetic co-energy 8 and the total potential energy associated with the body. In general the Lagrange method allows to define a set of Lagrange equations, that have some advantages with respect to the vector equations provided by the Newton-Euler approach. 8 The reason of using the term co-energy instead of the term energy, will be clarified later.

24 24 The approach provides n second order scalar differential equations, directly express in the generalized coordinates q i (t) e q i (t), that provide an easier starting point for writing the state variable equations than the Newton-Euler equations. If holonomic constraint are present, in the equations the constraint force do not appear: This is very convenient when the equations are used for control design or simulation problems, but may be annoying when we want to explicitly compute constraint forces. We will discuss this topic in Section.7.5. The kinetic co-energies and the potential energies are independent of the reference frame used to represent the body motion. The kinetic co-energies and the potential energies are additive scalars: in a multi-body system the total energies/co-energies is the sum of each component energy/co-energy. We will now formally define these two quantities..7.1 Kinetic Energy and Co-energy First we will define these energies for a point mass and then for a rigid extended body. Point-mass The kinetic energy associated to a point-mass m is defined as the work necessary to increase the linear or angular momentum from to h, i.e., C(h) = h The infinitesimal work associated to the mass is given by dw = f dr dw (51) where f is the resultant of the applied forces on the mass and dr is the infinitesimal position increment. The infinitesimal increment dr differs from the virtual displacement δr since the former is consistent with the laws of dynamics, i.e., dr = vdt, while the latter if free or restricted only by the geometry of the constraints. In order to have as integrand in (51) only a quantity that depends on the integration limits, we recall that f = dh dt

25 .7. LAGRANGE EQUATIONS IN MECHANICAL SYSTEMS 25 that relates the force to the variation of the linear momentum; the resulting infinitesimal work is therefore and so we can write dw = f dr = dh dt C(h) = dr = dh dt h vdt = v dh v dh (52) The kinetic energy is a scalar state function associated to the particle, that depends only of the state variables v and h associated to the point mass. It is possible to define another state function associated to the point-mass, that we will call kinetic co-energy, defined as C (v) = v h dv (53) As shown in Figure 8, between the energy and the co-energy, a relation exists, namely C (v) = h v C(h) (54) This relation is an example of the Legendre transform. In Section?? we will present a detailed discussion of this topic. Figure 8: The relation existing between the linear momentum h and the linear velocity v; in this particular case the two quantities are scalars and the relation assumes a constant mass. In particular, if the mass particle is moving at a velocity significantly smaller that the light speed c, i.e., it is not a relativistic mass, we can write h = mv and the two

26 26 energies become C(h) = C (v) = h v 1 m h dh = 1 2m h h = 1 2m h 2 mv dv = 1 2 mv v = 1 2 m v 2 (55) As one can see, in this case the kinetic energy and co-energy are the same; this does not happens for relativistic particles. We conclude this Section pointing out that in many textbooks the kinetic energy is indicated by the symbol T, but we prefer to use this not-so-common symbol C or C, and since these are functions of the generalized coordinates and velocities, often we write C(q, q) or C (q, q) Multiple Point-masses Bodies Now we take into consideration an extended body composed by N masses m i. The kinetic co-energy will be the sum of the kinetic co-energy of each mass, C (v) = 1 2 N m i v i v i (56) In Figure 9 we have an inertial reference frame R, with origin in a fixed point O and a body-frame R b ; if we consider the velocity v i = ṙ i with respect to R, we consider first the linear velocity and then the angular velocity. The velocities in R can be computed from the general relation (??), rewritten here for convenience ṙ (t) = ω 1(t) ρ (t) + R 1ṙ 1 (t) + ḋ 1(t) where the second term R bṙ i(t) at the right hand side is zero, since the point-masses are fixed with respect to the body-frame. Now we consider a a purely translatory motion and then a purely rotational motion. Translational Motion If the motion is purely translational, (??) gives ṙ i (t) = ḋ b(t) v (t) where v is the total linear velocity with respect to R ; therefore each mass m i will have the same velocity v and the relation (56) will reduce to C = 1 2 v v N m i = 1 2 m tot v v = 1 2 m tot v 2 = 1 2 v T (m tot I )v (57)

27 .7. LAGRANGE EQUATIONS IN MECHANICAL SYSTEMS 27 Figure 9: A rigid system composed by N point masses, and the various position vectors. C is the body center-of-mass. where the mass m tot is the total body mass. So this equation states that the kinetic (co)energy of a purely translational motion is equivalent to that of one single particle with mass m tot having the total translational velocity v. We recall that the total mass m tot can be ideally concentrated in the body centerof-mass C del corpo, that obeys to the relation r c m tot = i r i m i r c = i m i m tot r i Since the velocity is equal for all points of the body, v is also the velocity of the center-of-mass C; if we use the symbol v c v for this velocity, we can write C = 1 2 m tot v c v c = 1 2 m tot v c 2 = 1 2 v T c(m tot I )v c that gives the usual rule: the kinetic energy is half the product of the total mass for total velocity squared. Rotational Motion If the motion is purely rotational, (??) reduces to ṙ i (t) = ω 1(t) r b i where ω 1 ω is the total angular velocity; therefore, using a simpler notation, the velocity of each mass follows the relation v i = ω r i

28 28 where r i is the position of the i-th mass in R. Replacing this expression in (56) we obtain C = 1 N m i (ω r i ) (ω r i ) (58) 2 Recalling that, given three generic vectors a, b, c, we can write a (b c) = b (c a) and calling a (ω r i ), b ω, c r i we obtain C = 1 2 ( N m i ω r i (ω r i ) = 1 N ) 2 ω m i r i (ω r i ) Now, recalling (4), we obtain ( C = 1 N ) 2 ω h i = 1 2 ω h where h is the total angular momentum with respect to O. Since h = Γ ω, we can conclude writing this expression C = 1 2 ωt Γ ω (59) So we interpret this equation stating that the kinetic (co)energy of purely rotating body is dependent on the total angular velocity and the inertia matrix with respect to the point O. Notice the similarity between (57) and (59): C = 1 2 v T c(m tot I )v c C = 1 2 ωt Γ ω In both cases the energy is proportional to the weighted scalar product of the velocities, where the weighing matrix is the fixed mass diagonal matrix, or the variable inertia symmetric matrix. Alternative Formulation An alternative formulation is obtained considering the center-of-mass C, whose position in R is given by r c, and representing the position of each point-mass as r i = r c + c i, where c i is the oriented segment from C to the i-th mass, as in Figure 9.

29 .7. LAGRANGE EQUATIONS IN MECHANICAL SYSTEMS 29 We compute the velocity v i = ω r i as v i = ω (r c + c i ) = ω r c + ω c i = v c + ω c i and, from (56), we have C = 1 m i v i v i = 1 m i (v c + ω c i ) (v c + ω c i ) 2 2 i i { = 1 m i v c v c + } m i (ω c i ) (ω c i ) i i (6) In this case v c is the (tangential) velocity of the center-of-mass due to the angular motion; considering (58), where now in place of r i we have c i, we can rewrite the above equation as C = 1 { } v T 2 c (m tot I )v c + ω T Γ c ω (61) where Γ c is the inertia matrix with respect to the body center-of-mass. Natural Systems In general, one can extend the arguments used above to the case in which each mass has a position that depends, in addition to the coordinates q i, also explicitly from the time t; therefore a system consisting of N point-mass particles defined by a position r i (q(t), t) and by the velocity v i (q(t), t) ṙ i (q(t), t) = n k=1 r i q k q k + r i t, have a kinetic (co)energy that is defined as [ C = 1 N m i ṙ i ṙ i = 1 N n n r i r i q j q k + 2 r i 2 2 q j q k t j=i k=1 If we introduce the following coefficients α jk = β j = γ = 1 2 N N m i r i q j r i q k m i r i t r i q j N r i m i t r i t n j=i r i q j + r i q j t r i t ] (62)

30 3 we can write (62) as where C = C 2 + C 1 + C (63) C 2 = 1 2 C 1 = C = γ n j=1 k=1 n β j q j j=1 n α jk q j q k Therefore, in general, the kinetic (co)energy is the sum of three terms: a quadratic, a linear and a constant one. If there is no direct time dependence the linear and the constant terms disappear, and the energy is a purely quadratic function. A system where the vectors r i are not explicitly depending on time, but depend on time only through the generalized coordinates q(t), the field of external forces is conservative, no constraints are present, has a kinetic (co)energy that is quadratic, as C = C 2 = 1 2 n n α jk q j q k j=1 k=1 These systems are a called natural systems; the others are called non natural systems. In a non natural system we observe that the term C behaves as a sort of potential energy and is naturally connected to the centrifugal forces 9, that are proportional to the square of the angular velocity r i t r i t, while the term C 1 produces the so called Coriolis forces, that are proportional to the product of two velocities Extended Rigid Body In analogy with the derivation used to compute the kinetic (co)energy of a multiple point-masses body, it is possible to compute the kinetic (co)energy of an extended 9 Consider, as an example, a satellite orbiting the Earth, where the gravitational (conservative) field forces are kept in equilibrium by the centrifugal force due to the angular velocity of the satellite around the Earth.

31 .7. LAGRANGE EQUATIONS IN MECHANICAL SYSTEMS 31 rigid body, as that represented in Figure 1. The relation is C (v) = 1 v v dm = 1 ρ(r) v v dv 2 B 2 B where v is the total velocity of every element with mass dm; the integral is computed considering the entire body volume V ; ρ(r) is the local density, function of the position r of the elements; the density is assumed to be uniform in time, but generally not in position. Figure 1: da definire Again we have an inertial reference frame R, with origin in a fixed point O and a body-frame R b ; with arguments that are similar to those used to derive relation (61), we have now C (v, ω) = 1 2 mv c v c ωt Γ c ω = 1 2 [ v T c (mi )v c + ω T Γ c ω ] (64) where m = B dm = B ρ(r)dv is the total body mass, Γ c is the inertia moment with respect to the center-of-mass C, v c is the total velocity of the center of mass and ω is the total angular velocity of the body. From (64) we can state the well known law for computing the kinetic (co)energy of a rigid body The kinetic (co)energy of an extended rigid body is the sum of two terms: a translational part, equal to the kinetic energy equivalent to the entire mass concentrated in the body center-of-mass, and an rotational part, equal to the kinetic energy due to the rotational inertia of the body with respect to its center-of-mass. The reference system with respect to which the inertia matrix Γ c is computed must be considered as a body-frame, with its origin in C; in this way the inertia matrix

32 32 is constant in time. However this frame may be rotate with respect to the frame where ω is represented; in this case, as given by (31) and (36), the Γ c value must be expressed in the common frame. Example.7.1 Let us consider a prismatic rigid body B (as a rod); we want to compute the kinetic energy. As shown in Figure 11 we have two possible reference frames, R a e R b. Figure 11: Computation of the inertia matrix for a rigid body and two reference frames. The first one is inertial and the total angular velocity of the body is ω a = [ θ] ; the second is a body-frame with origin in the center-of-mass C. At a given time t the two systems are related by the following rotation matrix 1 R a b = Rot(i, 9 ) = 1 1 The axes of R b are aligned with the principal inertia axes, and the inertia matrix is Γ x Γ b = Γ y Γ z Since the kinetic (co)energy is independent of the reference frame, provided that we use the total angular velocity, we can compute it in two different ways. 1. We can represent Γ in R a and use the total angular velocity ω a represented in that frame C = 1 2 (ωa ) T Γ a ω a

33 .7. LAGRANGE EQUATIONS IN MECHANICAL SYSTEMS We can represent Γ in R b and use the total angular velocity ω b represented in that frame C = 1 2 (ωb ) T Γ b ω b In the first case we have 1 Γ x 1 Γ x Γ a = R a bγ b R b a = 1 Γ y 1 = Γ z 1 Γ z 1 Γ y and consequently C = 1 2 (ωa ) T Γ a ω a = 1 [ 2 Γ x θ] Γ z = 1 2 Γ y θ y 2 In the second case we have 1 ω b = R b aω a = 1 = θ 1 θ and consequently C = 1 2 (ωb ) T Γ b ω b = 1 [ 2 θ ] Γ x Γ y θ = 1 2 Γ θ y 2 Γ z In both cases, as expected, the kinetic (co)energy is the same. Generalized Coordinates If we want to write the kinetic (co)energy as a function of the generalized velocities q i (t), we shall recall the direct kinematic velocity function, that requires to know the linear Jacobian and the geometrical angular Jacobian: v c = J l (q) q or ω = J ω (q) q (65) Here we will use the geometrical Jacobian, since we are interested in ω and not in α to express C (v, ω) in (64) as a function of q and q. Using the linear (??) and the angular geometrical jacobians we obtain C ( q, q) = 1 2 [ q T J T l (mi )J l q + q T J T ωγ c J ω q ] (66)

34 34 and finally C ( q, q) = 1 2 q T [ J T l (mi )J l + J T ωγ c J ω ] q = 1 2 q T Γ tot q (67) This relation shows that the kinetic (co)energy is a quadratic function of the overall inertial mass of the body; since the kinetic energy of a body with a non-zero mass is always positive, the matrix Γ tot is always positive definite and hence invertible..7.2 Potential Energy Gravitational Energy A force f is said to be conservative or potential when it the negative gradient of a potential function P(r); the potential function is a scalar function and depends on the position r = [ x y z ] T, defined as: [ P(r) f (r) = P(r) = x P(r) y ] T P(r) (68) z The symbol indicates the gradient operator that transforms a scalar function into a vector 1. If a potential function exists, it is conventionally called potential energy of the system; it must be observed that, if it exists, it is unique apart from an additive constant. This implies, as we will see later, that the effects on the body dynamics depend only from the potential energy variation, and not on its absolute value. The dynamical systems in which all the forces doing work are due to a potential function, are called conservative systems. In a conservative force field the potential energy P is obtained integrating the conservative forces, i.e., P(r) = r f (r) dr = r r P(r) dr r (69) where r represents a generic initial position of the mass, and the minus sign establish that the potential energy decreases if the force field perform the work on the body. A typical example of conservative force filed is the gravitational field around Earth, or in general any mass. The potential field produces the so-called weight forces. The potential energy due to a gravitational field and associated to a generic mass m is given by the following relation: P g = mg r c (7) 1 Often the gradient is interpreted as a row vector, but we prefer to follow the usual column representation rule, so we introduce the transpose symbol T.

35 .7. LAGRANGE EQUATIONS IN MECHANICAL SYSTEMS 35 where g is the local gravitational acceleration vector and r c is the body center-ofmass position vector with respect to an orthogonal plane to g that has the conventional zero value of potential energy (zero energy plane); it is the variation of energy with respect to this plane that is considered. Elastic Energy Moreover there is another force field related to potential energy; it is that due to elastic elements, as the ideal springs: these elements are the abstract model of a proportional relation between elongation and force, so that force is proportional to the deformation of the ideal spring element. If we assume a one-dimensional linear spring we can write f = k e e (71) while if we assume a one-dimensional torsional or torsion spring we write τ = k eδ (72) where e and δ are, respectively, the linear elongation, and the angular deformation from the rest position of the spring; k e and k e are the so-called elastic constants of the springs. The elongation and the deformation physically represent a difference in position or angle between a rest state of the elastic element and the deformed state of the same element. In one dimensional linear springs the force and the deformation are co-linear, while in one dimensional torsion spring the torque and the rotation axis are co-linear as well. Figure 13 depicts a one dimensional linear spring, where the force and the deformation are co-linear. It is also possible to find elastic elements where the elastic constant depends on the deformation itself, as in f = k e (e)e τ = k e(δ)δ (73) Figure 12: The scheme of a one dimensional linear spring. The potential energy is the integral of the virtual work performed by the spring deformation P(e) = e f de (74)

36 36 Figure 13: The scheme of a one dimensional torsion spring. or P(δ) = δ τ dδ (75) We can also in this case define the potential co-energies, that are or P (f ) = P (τ ) = f τ e df δ dτ The relation between P (f ) and P(e) is similar to the co-energy Legendre transformation given in (54) P (f ) = f e P(e) (76) as illustrated in Figure 14 Figure 14: da definire When the elastic elements are linear with constant k e, the potential energy and co-energy are as follows P(e) = 1 2 e T (k e I )e = 1 2 k e e 2 (77)