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by: Jamaal McGlynn


Jamaal McGlynn
GPA 3.78


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Class Notes
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This 12 page Class Notes was uploaded by Jamaal McGlynn on Tuesday October 20, 2015. The Class Notes belongs to PHYS608 at San Diego State University taught by M.Bromley in Fall. Since its upload, it has received 22 views. For similar materials see /class/225318/phys608-san-diego-state-university in Physics 2 at San Diego State University.


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Date Created: 10/20/15
Lecture 18 Outline transit 5 to 6 o Symmetric Top example Section 57 o Oscillations theory of Section 6162 Euler Equations of Motion contd 0 last time gave us Euler s equations of motion N1 2002 w3w113 1 N2 3003 w1w211 I2 N3 1001 CUQCU312 3 Continuing the dumbbell example if w 0 then 0 given up O and remembering c3 0 w sin 6 w cos 6 Then only one Torque component is non zero N96 wyw212 3 m1 m2b2w2 sin cos ie spins the dumbbell around in the x direction Archetypical Symmetrical Top 0 problem rotating rigid body xed at 9 y z O 0 choose z to be the principle axis of rotation 0 Characteristic motions of top rotation or spinning about it s own axis precession or rotation of z about z nutation or bobbing up down of z relative to z Symmetrical Top Eulerian Way For gyroscope and spinning top gtgt gtgt now 1 2 3 since symmetry along z from Euler s eqns torque with N1 N2 won t change w3 N3 13 W1WQ11 2 13 initial torque N3 N2 0 only N1 7 O and initial rotation cal wg 0 only w3 7 O constant But time evolution has cal and wg changing ie complicated motion in terms of TL Symmetrical Top Lagrangian Way 0 A somewhat easier solution from the Lagrangian T T1 T2 T3 h af 13W Now Goldstein 487 has exactly these 3 w s in terms of wa1 sin 6 sin w cos w 03 wy osin cosw sinw LuzJ cos 6 J o and in terms of Euler angle s V mgl cos 6 and T g gh2sin2 6 g ghcos 2 0 giving a Lagrangian with two cylic coords gb z ie the 19 p components are constant in time Symmetrical Top Integrals Of Motion T 12 162 d2 sin2 6 2 3 dcos62 0 so we have two rst integrals of motion d1 p 3913 COSQI3W311G dw d1 2 2 19 d Ilsm 613cos 6 13 cose11b 0 Where introduced two constants of motion a b o and a third constant of motion is mgl cos 6 0 Another rst integral is given by E const j j 2 E TV 3162 2 sin2 6 33 16 gbcos 6 mgl cos 6 c From here can nd and in terms of a b and 6 Applications 0 Given that the tag is constant in time c this means that the fourth constant of motion really is 13w i b acos 62 2 2 sin2 6 c this can be solved for 615 but it s complicated E E I 1 3162 mglcos6 i1162V 6 we won t solve it here since it just leads down into darkness ie elliptical integrals c There are limited regimes of soln see remainder Goldstein 57 o Gyroscopes pg 222223 are rapidly rotating devices mounted such that centre of mass is xed Oscillations formalism intro interested in motion about equilibrium system is in equilibrium when generalised forces 3V i 0 Q 39igt0 ie extremum at equilibrium 101 102 qon As per usual equilibrium Classi ed stable unstable in terms of the deviations of system energy from equilibrium Oscillations formalism potential o All about Taylor series fx a 00 fwm x a n0 7 0 Choose deviations of generalised coords from equilib Qi 90139 77 0 so that the potential can be expanded about each 101 av 1 82V V n V 7 7 0n 1142 71 101 102 q aqigt0n2 gqigqjgt0nm o implied summation We then set Vq01 c102 qon O o and at equilibrium we have 23 0 so 1 0 1 82V 1 VC117 Q27 7 Qn E gqaq 77139773 5in l J 0 eg 2D isotropic SHO Oscillations formalism kinetic o In the distant past 171 Cg Zj 3quotqu 39 J 0 see Goldstein equations 171 and 172 on page 25 0 our generalised coords don t explicitly depend on time 2 1 1 87 1 T Z 5W3 Z in Z 8 Z 5 Z Mijqiqj 139 i j M o The coeffs Mm are functions of qk 3771 MijQ17Q277Qn mijQO17QO277QO7L 3 13 77k Qk 0 0 only include constants since T has quadratic terms in gl 0 so we have Mm E 7711701017 C102 7 Clan Tij and 1 1 T 5mmin 537mm eg 2D isotropic SHO contd


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