Rigid Body Dynamics (continued)

Transcription

1 Last time: Rigid dy Dynamics (cntinued) Discussin f pint mass, rigid bdy as useful abstractins f reality Many-particle apprach t rigid bdy mdeling: Newtn s Secnd Law, Euler s Law Cntinuus bdy apprach t rigid bdy mdeling: Newtn and Euler principles f linear mmentum and angular mmentum asserted as fundamental principles Develpment f bdy integrals fr mass, first mment f inertia, mmentum Definitin f mass center

2 Rigid dy Ntatin I made a slight change in ntatin s will review the cntinuus mdel slides frm previus lecture The ntatin change was t remve the left subscripts and superscripts

3 Cntinuus Mdel Mdel the rigid bdy as a cntinuum,, that satisfies Newtns Secnd law and Eulers Law The system linear mmentum is ~p = ~v ~v The angular mmentum abut the pint is ~ h = ~r ~v These tw vectr quantities bey the linearandangularmmentumprinciplesas expressed by Newtn and Euler î 3 ~r ~p = ~ f and ~h = ~g î 1 î 2

4 What des the integral in ~p = R ~v mean? It defines a three-dimensinal vlume integral ver the vlume f the cntinuus bdy ; in general f(~r) dv = r1max r 1min r2max (r 1 ) r3max (r 1,r 2 ) r 2min (r 1 ) Cntinuus Mdel (2) r 3min (r 1,r 2 ) f(~r) dr 3 dr 2 dr 1 î 3 ~v ~r Three types f integrals arise in the develpment f rigid bdy equatins f mtin: scalar, vectr, and tensr. Alternatively, these three mathematical bjects can be called zerth-rank, firstrank, and secnd-rank tensrs. î 1 î 2

5 Cntinuus Mdel (3) Fr example, the ttal mass, m, isthe integral ver the bdy f the density, μ(~r,t), which can in general vary frm pint t pint within the bdy, r even with time. We usually assume that the density is cnstant. m = μ dv = = a b c m = μabc μ dr 3 dr 2 dr 1 Mass is a scalar, r zerth-rank tensr, and is smetimes called the zerth mment f inertia. a ˆb 1 ˆb 3 ˆb 2 b The rectangular prism has sides f length a, b, andc. Nte that the rigin f the bdy frame shwn is nt the mass center. c

6 Cntinuus Mdel (4) The first mment f inertia abut, ~c, is the integral ver the bdy f the psitin vectr frm t each differential mass element. The vectr is dented ~r,wherethesuperscript indicates that it is the vectr frm t. ~c = ~r r c b = r b μ dv = a b c = μ 2 [a2 bc ab 2 cabc 2 ] T μ[r 1 r 2 r 3 ] T dr 3 dr 2 dr 1 a ˆb 1 ˆb 3 ˆb 2 b ~r The first mment f inertia canbeusedtfind the mass center, defined as the pint abut which c vanishes c

7 Cntinuus Mdel (5) The mass center c,isdefined as the pint abut which the firstmmentfinertia, ~c c,vanishes. ~c c = ~r c = ~ r c c b = r c b μ dv = The psitin vectr can be written as ˆb 1 ˆb 3 ˆb 2 ~r c ~r ~r c Putting it all tgether, we btain c Thus c = ~r = ~r c + ~r c r r = r c + r c (r c + r c ) = r c = r c r c = 1 2 [abc]t which is bviusly the crrect answer = mr c

8 Cntinuus Mdel (6) The linear mmentum has already been defined as ~p = ~v where ~v is the time derivative f the psitin vectr ~r f a differential mass element = μ dv This psitin vectr must be measured frm an inertial rigin, and the derivative is taken with respect t inertial space What is the velcity ~v f a differential mass element? The bdy frame mtin has tw velcities: the velcity f its rigin with respect t the inertial rigin, ~v O, and the angular velcity f the frame with respect t an inertial frame, ~ω bi Thepsitinvectrcanbewrittenas ~r O +~r,where~r O is the psitin vectr frm the inertial rigin t and ~r is the vectr frm t a pint in the bdy The velcity f a mass element is then ~v = ~v O + ~ω bi ~r

9 Thus the linear mmentum is ~p = ³~v O + ~ω bi ~r Cntinuus Mdel (7) The velcity f the rigin and the angular velcity f the frame d nt depend n the particular differential mass element, s they are cnstant with respect t the integratin, s that the mmentum becmes ~p = m~v O + ~ω bi ~r The right term is the first mment f inertia abut the rigin, s that ~p = m~v O + ~ω bi ~c If is the mass center c, thenthe linear mmentum is simply ~p = m~v Oc which we usually write as ~p = m~v

10 Cntinuus Mdel (8) Linear mmentum principle: ~p = m~v O + ~ω bi ~c ~ f = ~p Hw d we express these equatins in a rtating reference frame? Recall the frmula fr differentiating a vectr a expressed in a rtating frame F b : d dt nˆb T a where ω = ω bi T ȧ = nˆb + ω a Applying the frmula t ~p and t ~v = ~r, webtain p = m ṙ + ω r + ω c f = ṗ + ω p which we can rearrange in the usual ẋ = f(x) frmas ṙ = 1 p ω c ω r m ṗ = ω p + f The ṙ equatin is the kinematics equatin, and is frequently writteninaninertialframeinsteadf inartatingframe

11 Cntinuus Mdel (9) The translatinal equatins f mtin in vectr frm are: ~p = m~v + ~ω ~c ~ f = ~p The matrix translatinal equatins f mtin, expressed in a rtating frame and rearranged, are ṙ = 1 p ω c ω r m ṗ = ω p + f What subscripts and superscripts are mitted, and what meaning fr them shuld we assume? Nte that ~ω is generally btained frm the rtatinal equatins f mtin, and that ~ f may depend n the rtatinal mtin; i.e., r ~ f = ~ f(~r, ~r, q, ~ω) ~ f = ~ f(~r, ~r, R, ~ω) That is, the frce generally depends n the dynamic state f the bdy: its psitin, velcity, attitude, and angular velcity

12 There are tw frms f angular mmentum in cmmn use: mment f mmentum, andangular mmentum The mment f mmentum, H ~ abut is ~H = ~r ~v where ~r is the psitin vectr f a differential mass element frm pint, and~v is the velcity f the mass element with respect t inertial space. Usually chse as c r Cntinuus Mdel (1) The quantity ~v is in fact the mmentumfthe differential mass element, and by taking the crss prduct with the mment arm ~r,wearefrmingthemment f mmentum abut. F i ~r ~v ~r

13 Cntinuus Mdel (11) The angular mmentum, ~ h abut is ~ h = ~r ~r where ~r is the psitin vectr f a differential mass element frm pint, and ~r is the velcity f the mass element with respect t. The difference between the tw definitins is the velcity terms that are used The quantity ~r is nt the mmentum f the differential mass element, and by taking the crss prduct with the mment arm ~r,wearefrmingthe mment f a mmentum-like quantity abut. ~r ~r ~v ~r F i

14 We will wrk with the angular mmentum, ~ h abut : ~ h = ~r ~r The velcity term is ~r = ~ω ~r where ~ω = ~ω bi.thus ~ h = ~r [~ω ~r ] Expand the integrand using the identity: ³ ~a ~b ~c = ~ ³ b (~a ~c) ~c ~a ~b Cntinuus Mdel (12) ~ h = = [~r (~ω ~r )] [~ω (~r ~r ) ~r (~ω ~r )] Rewrite the tw terms in the integrand as ~ω (~r ~r ) = ~r ~r ~ω ~r (~ω ~r ) = ~r ~r ~ω Whenever we write three vectrs in the frm ~a ~ b ~c, parentheses are implied arund the tw vectrs with the between them. Thus, ~a ~ ³ b ~c = ~a ~b ~c, and ~a ~b~c ³ = ~a ~b ~c

15 Intrduce a new mathematical bject, the identity tensr, ~1, defined by its peratin n vectrs: ~1 ~v = ~v ~1 = ~v Use the identity tensr t rewrite the first term in the integrand as ~r ~r ~ω = ~r ~r ~1 ~ω =(r ) 2 ~1 ~ω Cmbining all these expressins, we write ~ h as h i ~ h = (r ) 2 ~1 ~ω ~r ~r ~ω nˆb T Verify that ~1 = nˆb Cntinuus Mdel (13) The angular velcity vectr is independent f the psitin vectr t each individual mass element, s h ~ h = (r ) 2 ~1 ~r ~r i ~ω The integral is the mment f inertia tensr, andiscnstant, just as the ttalmassandfirstmmentintegralsare: ~ I = h (r ) 2 ~1 ~r ~r i If the pint is clear, mit the superscript: h i ~ I = r 2 ~1 ~r~r

16 The angular mmentum is then where ~ I = ~ h = ~ I ~ω h i r 2 ~1 ~r~r Hw d we calculate ~I? Cntinuus Mdel (14) Think f ~ I as yu think f vectrs: it is an abstract mathematical bject that has numbers assciated with it when yu chse a particular reference frame. We have cmputed bdy integrals invlving ~r, andwehadtexpress ~r in a bdy-fixed frame: nˆb T ~r = r nˆb T = r Substitute these expressins int the terms in ~ I: r 2 = ~r ~r nˆb nˆb T = r T r = r T 1r = r T r T ~r~r = nˆb nˆb rr T

17 ~I = = = h i r 2 ~1 ~r~r nˆb T r T r nˆb nˆb nˆb T rr T nˆb T r T r rr T nˆb T Recall that nˆb = 1, and take the dt prduct f bth sides T with nˆb frm the left and nˆb frm the right Cntinuus Mdel (15) nˆb Taking the dt prducts gives nˆb nˆb T ~I = r T r1 rr T In the same way that taking dt prducts f a vectr ~v with the unit vectrs gives the cmpnents f the vectr in the given frame, the expressin n the left is the cmpnents f the inertia tensr in F b : v b = ~v nˆb nˆb T I b = ~I nˆb

18 Cntinuus Mdel (16) The matrix versin f the mment f inertia tensr is cmputed in similar fashin as the firstmmentfinertiavectr I b = r T r1 rr T μ dv The term in brackets is easily shwn t be r2 2 + r2 3 r 1 r 2 r 1 r 3 r 1 r 2 r1 2 + r2 3 r 2 r 3 r 1 r 3 r 2 r 3 r1 2 + r2 2 The required integratin is a b c a ˆb 1 ˆb 3 ˆb 2 b ~r Fr example, I 11 = μabc b 2 + c 2 3 = m b 2 + c 2 3 c [ ] μ dr 3 dr 2 dr 1 Each element f the matrix is integrated independently f the rest Yushuldbeabletcarry ut the required integratins.

19 Cntinuus Mdel (16) a ˆb 3 ~r c ˆb 1 ˆb 2 b Carrying ut the integratins leads t 1 I 3 b = m (b2 + c 2 ) 1 4 ab 1 4 ac 1 4 ab 1 3 (a2 + c 2 ) 1 4 bc 1 4 ac 1 4 bc 1 3 (a2 + b 2 ) Nte the pattern, especially that I T = I. Als,wehavecmputedthemmentfinertiamatrix abut pint, and with respect t the ˆb frame. We need t be able t translate the rigin, and rtate the frame.