Advanced Dynamics with SOPHIA '03/Thylwe 1 Advanced Dynamics of Complex Systems with Lecture 1: Maple SOPHIA '03 e3 Contents 1• Maple, configuration and frames 2• Velocity, angular velocity and dyads 3• System Kvectors, tangent vectors e 4• Constraints and reduced speeds 2 • Applications e1 Introductory example 1: A particle is moving frictionless on a given surface, in this case a two-dimensional parabola. Analyse the motion and the constraint force. Dynamic law Newton's equation of motion: R - m a = 0, (1) where d 2r a = 2 , (2) dt R = Ra + Rc (applied+constraining forces), (3) R a =- mg e3 , R c = unknown?? (not completely) Advanced Dynamics with SOPHIA '03/Thylwe 2 Constraints Constraining force The constraining surface: The constraining force R c keeps the particle on the surface. Its magnitude 2 2 depends on the mass and the motion of the particle, but the direction is (r, t) = x3 - k (x1 + x2 ) = 0 , k = const. (4) where always orthogonal (or normal) to the surface. r = x e + x e + x e . 1 1 2 2 3 3 A normal to the surface is given by: Generalised coordinates N = 1 ´ 2 2 2 2 2 = - 2kq 1 cos q 2e1 + - 2kq 1 sin q 2e2 + q 1(cos q 2 + sin q 2 )e3 . Not all three coordinates are independent, only two are needed. It is also (7) possible to choose two parameters q 1 and q 2 (or generalised Another way to find this direction is to use the gradient to the surface. coordinates). For example: Consider the surface (4) to be a level surface. The neighbouring surface of ì x = q cosq higher 'value' points in the direction ï 1 1 2 r = r(q1,q2 ,t) ® í x2 = q1 sin q2 . (5) Ñ = e + e + e ï 1 2 3 2 x1 x2 x3 î x3 = kq1 Tangent vectors = - 2kx 1e1 - 2kx 2e2 + e3 . (8) Comparing the vectors, we see that N = q 1Ñ , they are indeed parallel. As for polar coordinates in a flat plane, these q 1 and q 2 also correspond The constraining force can then be written as R = N/ N , where now to local directions in space, tangents to the coordinate lines on the surface. c only the 'magnitude' is unknown. ¶r 1 = = cos q2 e1 + sin q2e 2 + 2kq1 e3 , ¶q1 ¶r Strategy 2 = = - q1 sinq2 e1 + q1 cosq2e 2 . (6) ¶q2 Since the constraining force R is partly unknown but orthogonal to the •What does 'local' mean? c •Is it necessary to normalise these tangent vectors? surface R c · k = 0, take the scalar product of the dynamical equations •Are there more tangent vectors? with two independent tangent vectors k (need not be exactly the ones we derived): R - a · = 0 . ( a m ) k These are two equations in this case, with two generalised coordinates. When they are known the complete motion of the particle is known, hence from (1), N + (R - ma) · = 0 a N provides an equation for the magnitude of the constaining force. Advanced Dynamics with SOPHIA '03/Thylwe 3 MAPLE e z Take a second look at the first problem of the introduction. Now we use e plain Maple. > restart: y with(linalg): Warning, the protected names norm and trace have been redefined and unprotected e The steps follow section 1.4.1: x Introductory example 2: The bead sliding along a rotating wire "The Parabola revisited." Consider the motion and forces in the plane z =0. Using a cylindrical basis, the acceleration and constraining force take the form: Cartesian coordinates parametrized ˙˙ 2 ˙ a = ( - )e + (2 )e , Rc = e . > Note that the gravitational force is orthogonal to the considered plane. x1:=q1*cos(q2); In this problem the bead is completely described by the coordinate . The x2:=q1*sin(q2); x3:=k*q1^2; single tangent vector corresponding to this coordinate is e . Following the strategy, we first solve the tangent projection of Newton's equation: m ˙˙ - 2 = 0 Þ (t) = c e t + c e- t . ( ) 1 2 Then the magnitude of the constraining force is found: Parameters are time dependent (t) = 2m ˙ = 2m 2 c e t- c e- t . ( 1 2 ) Observation > toTimeFunction:= The velocity of the bead: {q1=q1(t),q1t=diff(q1(t),t),q1tt=diff(q1(t),t,t),q2 v = ˙ e + e , =q2(t),q2t=diff(q2(t),t),q2tt=diff(q2(t),t,t)}: •is not orthogonal to the constraining force, whence resulting in an important energy time rate ('power'): P = Rc · v = (t) . Simplifying the resulting expressions • is not parallel to the tangent vector e . toTimeExpression:={q1(t)=q1,diff(q1(t),t)=q1t,diff( q1(t),t,t)=q1tt,q2(t)=q2,diff(q2(t),t)=q2t,diff(q2( t),t,t)=q2tt}: Advanced Dynamics with SOPHIA '03/Thylwe 4 Velocity components New example: Pendulum hanging from a rotating disc This is a similar problem. However we show how to use a simple form of vectors with Maple. > v1:=subs(toTimeExpression,diff(subs(toTimeFunction, x1),t)); > v2:=subs(toTimeExpression,diff(subs(toTimeFunction, s x2),t)); q 1 v3:=subs(toTimeExpression,diff(subs(toTimeFunction, x3),t)); q 2 > l Acceleration components > for j from 1 to 3 do a||j:=subs(toTimeExpression,diff(subs(toTimeFunctio n,v||j),t)) od: > for j from 1 to 3 do a||j od; mg The steps follow section 1.4.2: Using Lists. Pendulum on circular support: Combine components and form vectors using MAPLE lists. Position vector: > x:=(s+l*sin(q2))*cos(q1): y:=(s+l*sin(q2))*sin(q1): z:=-l*cos(q2): > r:=[x,y,z]: Advanced Dynamics with SOPHIA '03/Thylwe 5 Velocity and acceleration Projection of 'inertial force' onto the tangent plane > Pt1:=multiply(pt,tau1); > rs:=subs(toTimeFunction,r): Pt1:=simplify(multiply(pt,tau1)); vs:=map(diff,rs,t): Pt2:=simplify(multiply(pt,tau2)): v:=subs(toTimeExpression,vs): as:=map(diff,vs,t): Pt1 := - m ( - l sin( q2) q2t2 cos( q1) + l cos( q2) q2tt cos( q1) a:=subs(toTimeExpression,as): - 2 l cos(q2) q2t sin( q1) q1t - ( s + l sin( q2) ) cos( q1) q1t2 Show components - ( s + l sin( q2) ) sin( q1) q1tt ) ( s + l sin( q2) ) sin( q1) + m ( 2 > a[1]; - l sin( q2) q2t sin( q1) + l cos( q2) q2tt sin( q1) + 2 l cos( q2) q2t cos( q1) q1t - ( s + l sin( q2) ) sin( q1) q1t2 + ( s + l sin( q2) ) cos( q1) q1tt ) ( s + l sin( q2) ) cos( q1) Pt1 := - m ( - 2 l cos( q2) q2t q1t s - 2 l2 cos( q2) q2t q1t sin( q2) - q1tt s2 Two ways to form pt(=m*a), the rate of change of the - 2 q1tt s l sin( q2) - q1tt l2 + q1tt l2 cos( q2) 2) momentum vector. Projection of the gravitational force–the only applied force. First way: The gravitational force is: > pt:=evalm(m*a): > Rg:=[0, 0, -m*g]: > pt[3]; > R1:=simplify(multiply(Rg,tau1)); R2:=simplify(multiply(Rg,tau2)); Second way: > Pt:=map(x->m*x,a): > Pt[3]; Equations > Eq1:=Pt1=R1:Eq2:=Pt2=R2: Tangent vectors These equations contain second derivatives. Standard > tau1:=map(diff,r,q1); numerical routines solving differential equations use a set of first-order coupled equations. We introduce generalized > tau2:=map(diff,r,q2); speeds u1=q1t and u2=q2t, to eliminate higher derivatives. First we isolate the second-order derivatives in our equations: > Eqs:=solve({Eq1,Eq2},{q1tt,q2tt}): Advanced Dynamics with SOPHIA '03/Thylwe 6 ¶ ¶ statepEq := { q1(t) = u1(t), q2(t) = u2(t), ¶t ¶t We arrive at a system of first-order equations in standard ¶ 1 2 1 2 5 u2(t) = cos(q2(t)) u1(t) + u1(t) sin(2 q2(t)) - sin(q2(t)), form. ¶t 4 2 2 ¶ cos(q2(t)) u2(t) u1(t) > u1(t) = - 8 } Eqs:=simplify(subs({q1tt=u1t,q2tt=u2t,q1t=u1,q2t=u2 ¶t 1 + 4 sin(q2(t)) },Eqs)): > Eqs[1]; > initc:= {q1(0)=0,q2(0)=1,u1(0)=0.5,u2(0)=0}: Put into one set for MAPLE > Eqs[2]; > deqns:=statepEq union initc: 4 first-order differential equations for MAPLE. Solving and plotting > state:=Eqs union {q1t=u1,q2t=u2}; > ì ï cos( q2) u12 s + l cos( q2) u12 sin( q2) - g sin( q2) st:=dsolve(deqns,{q1(t),q2(t),u1(t),u2(t)},type=num state := íï q1t = u1, q2t = u2, u2t = , ï eric,output=procedurelist); î l st := proc(rkf45_x) ... end ( s - l sin( q2) ) u1 u2 cos( q2) l ïü u1t = - 2 ýï > ï with(plots,odeplot); s2 - l2 + l2 cos( q2) 2 þ >odeplot(st,[[t,q1(t)],[t,q2(t)]],0..20,view=[0..20 Extend ,-2...2],numpoints=100,labels=[time,q_i]); > toTimeFunction:=toTimeFunction union {u1=u1(t),u1t=diff(u1(t),t),u2=u2(t),u2t=diff(u2(t) ,t)}: Choose parameter values and initial conditions > param:={s=1,l=4,g=10}: > statep:=subs(param,state): > statepEq:=subs(toTimeFunction,statep); Advanced Dynamics with SOPHIA '03/Thylwe 7 Configurations Vector product operations Lecture 2: æn ö Orthonormal basis 1 • ´ T = n n n ´ ç n ÷ . n n ( 1 2 3) ç 2 ÷ = 0 The space can be organised using 3 orthonormal basis vectors. They are è n3 ø composed into a reference triad æn 1ö n = (n1 n2 n3). (1.1) T ´ = ç n ÷ Its transpose composition is: • n n ç 2÷ ´ (n1 n2 n3 ) è n ø æn 1ö 3 T = ç n ÷ . æn ´ n n ´ n n ´ n ö n ç 2÷ 1 1 1 2 1 3 ç è n ø = n2 ´ n1 n2 ´ n2 n2 ´ n3÷ 3 ç ÷ Here are some algebraic rules with these triads: è n3 ´ n1 n3 ´ n2 n3 ´ n3 ø Scalar product operations æ 0 n3 - n2 ö æn ö = ç - n 0 n ÷.