{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 1 16 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 17 "Euler's equations" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 290 "We demon strate the free rotation of a rigid body (e.g., our textbook) by solvi ng Euler's equations for the angular velocity vector that defines the \+ motion relative to an inertial frame, but expressing it using componen ts in the body frame (such that the moments of inertia remain constant )." }}{PARA 0 "" 0 "" {TEXT -1 149 "The numerical solution demonstrate s the instability of the motion around the principal axis associated w ith the intermediate moment of inertia value." }}{PARA 0 "" 0 "" {TEXT -1 149 "Euler's equations are given in the text at the bottom of page 329, i.e., in Knudsen-Hjorth: Elements of Newtonian Mechanics, S pringer 2001 (3rd ed.)." }}{PARA 0 "" 0 "" {TEXT -1 107 "We use [w1,w2 ,w3] for the angular velocity vector, and I1,I2,I3 for the three princ ipal moments of inertia." }}{PARA 0 "" 0 "" {TEXT -1 92 "N1,N2,N3 are \+ the body-frame components of the torque which we set to zero for free \+ rotation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "E1:=I1*diff(w1 (t),t)-(I2-I3)*w2(t)*w3(t)=N1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "E2:=I2*diff(w2(t),t)-(I3-I1)*w3(t)*w1(t)=N2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "E3:=I3*diff(w3(t),t)-(I1-I2)*w1(t)*w2(t)= N3;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "Free rotation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "E1:=subs(N1=0,E1): E2:=subs(N2=0,E2 ): E3:=subs(N3=0,E3):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "For our \+ textbook we have in SI units:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "M:=1.63;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "a:=235*10^ (-3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "b:=154*10^(-3);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "c:=17*10^(-3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "We can use the result for a rectangular \+ plate spinning about the CM to find the three principal moments of ine rtia:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "I1:=M/12*(b^2+c^2) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "I2:=M/12*(a^2+c^2);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "I3:=M/12*(a^2+b^2);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 308 "The equations are ready for numer ical solution, and we can specify the initial conditions. We start the motion about the 1-axis (smallest moment of inertia, should be stable ), and admix small contributions about the other axes (it is never pos sible to hit the symmetry axis perfectly when throwing the book)." }} {PARA 0 "" 0 "" {TEXT -1 120 "We give the main initial spin axis a val ue of about one revolution per second, and the others about one percen t of that." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "E1;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "IC:=w1(0)=6.28,w2(0)=5/100,w 3(0)=5/100;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "sol:=dsolve( \{E1,E2,E3,IC\},\{w1(t),w2(t),w3(t)\},numeric,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "W1:=subs(sol,w1(t)): W2:= subs(sol,w2(t)): W3:=subs(sol,w3(t)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "plot(['W1(t)','W2(t)','W3(t)'],t=0..10,color=[red,blu e,green],axes=boxed,thickness=3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "Now we try the same thing about the 2-axis (intermediate value of moment of inertia, I1 < I2 < I3 in our case!)" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 39 "IC:=w1(0)=5/100,w2(0)=6.28,w3(0)=5/100;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "sol:=dsolve(\{E1,E2,E3,IC\}, \{w1(t),w2(t),w3(t)\},numeric,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "W1:=subs(sol,w1(t)): W2:=subs(sol,w2(t)): W3:=subs(sol,w3(t)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "pl ot(['W1(t)','W2(t)','W3(t)'],t=0..10,color=[red,blue,green],axes=boxed ,thickness=3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 297 "The motion goe s into a tumble! The blue curve indicates how we keep changing the spi n orientation between extreme values, and how the motion goes wild. It is an interesting topic for the stability theory of differential equa tions to understand from pencil-and-paper work why this behaviour occu rs." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "28" 0 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }