RIGID BODY MOTION. Next, we rotate counterclockwise about ξ by angle. Next we rotate counterclockwise about γ by angle to get the final set (x,y z ).

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1 RGD BODY MOTON We now consder the moton of rgd bodes. The frst queston s what coordnates are needed to specf the locaton and orentaton of such an object. Clearl 6 are needed 3 to locate a partcular pont n the object, e.g., the center of mass, and 3 angles to orent t wth respect to that pont (the 3 drectons cosnes would do). We wll specf the coordnates relatve to an nertal Cartesan set fxed n space. We start b rotatng counterclockwse about z b angle ϕ to obtan a new set of axs (ξ,,γ) Next, we rotate counterclockwse about ξ b angle Next we rotate counterclockwse about γ b angle to get the fnal set (x, z ). We can now obtan the connecton between (x,,z) and (x,,z ) b matrx multplcaton. The st rotaton s represented b = -x sn ϕ + cos ϕ ξ = x cos ϕ + sn ϕ γ = z

2 The nd rotaton s gven b cos sn 0x sn cos z = cos + γ sn γ = - sn + γ cos ξ = ξ ' cos sn 0x ' 0 cos sn 0 cos sn sn cos 0 ' 0 sn cos 0 sn cos0 0 z cos sn 0 x cos sn cos cos sn sn sn sn cos cos z The 3 rd rotaton s gven b x = ξ cos + sn = - ξ sn + cos z = γ

3 x' cos sn 0 ' ' sn cos 0 ' z' 0 0 ' cos sn 0cos sn 0 x sn cos 0 cossn cos cos sn 0 0 snsn sncos cosz coscossn cossn cos sn sn cos cos sn sn x sn cos cos cossn sn sn cos cos cos cos sn sn sn sn cos cos z Ths transformaton can obvousl be vewed as rotatng a vector (from the orgn to the th partcle). Hence the new vector must have the same length as the orgnal. Such a transformaton s called orthogonal. Wrtng the transformaton equaton as x a x ' j j j we ntroduce a conventon: whenever an ndex s repeated up and down t s summed over x a x ' j j Then the condton for orthogonalt s x a x x x ' x x ' j ' j Here we have ntroduced a new knd of vector x n normal Cartesan coordnates n eucldean space we have x x However, later when we consder relatvt that wll not be so. The two forms are connected b x x j j where n our case

4 j and n relatvt j We then have x ' x a x a x j' j k j ks j j k j k s where 0 0 ks j Then x' b x s s Wth b a a a s j ks s s T j k where st a s the transpose of s a (nterchange rows and columns). We now consder the nverse transformaton A -. Then But ths s A A a a j j k k We had the orthogonalt condton

5 xx' a xa x a a xx xx ' j s T s T j j j s j s j a a s T s j j a a st s j j 0 s j s j a a a a a a j T j T j k j k j j Hence the nverse s smpl the transpose (nterchange rows and columns). We now consder a specfc sstem a smmetrc top spnnng about the smmetr axs wth the pont of contact wth the table fxed. We assume no frcton or ar resstance. We use the Euler angles to descrbe the poston of the top. To get the Legrangan we need the knetc energ and the potental energ. To fnd the knetc energ we recall that KE mv n our case v s measured relatve to the fxed pont, and thus s due entrel to rotaton. Hence v r Thus m m KE v r r v J J We evaluate KE relatve to a set of axes fxed n the bod such that ẑ s the smmetr axs and x, are then perpendcular to t. Ths wll gve xx 0 0 zz 0 0 Hence, we need x,, and z n terms of,, x,, z are relatve to axs fxed n the bod. Hence the correspond to the x, z sstem above. Snce les along the ẑ axs, we get ts x,, z components b

6 x 0 A 0 z or x cos cossn cos sn cossn sn cos cos sn sn 0 sn cos cos cossn sn sn cos cos cos cos sn 0 z sn sn sn cos cos sn sn cos sn cos Next we need the contrbuton from Ths les along axs and hence we just need the matrx cos sn 0 B sn cos x cos sn 0 cos B 0 sn cos 0 0 sn z Next we need the contrbutons from Ths les entrel along ẑ'and hence gven b 0 0 Thus the totals are x sn sn cos cossn sn cos z We can now wrte KE as

7 KE sn sn cos, cos sn sn, cos 0 0 snsncos 0 0 cos sn sn sn sn cos, cos sn sn, cos 0 0 cos snsncos cossnsn cos sn sn cos sn sn cos cossn sn cossn sn cos cos sn cos cos Next we need the potental energ. t s clearl PE mg cos Hence L sn cos mg cos L L 0 L sn cos cos sn mg sn L sn cos cos L

8 L cos Hence () sn cos cos const b () cos const a z sn cos cos sn mg sn 0 (3) We also have conservaton of energ E KEPE sn z mg cosconst (4) From () a cos (5) From () and (5) sn acos b b acos (6) sn From (6) and (5) (7) bacos a cos a b acos sn cos sn Then (6) and (7) n (4) gves bacos E z sn mg cos sn Let

9 Then E' E z bacos E' mg cos V'( ) sn where V'( ) bacos sn mg cos Let u cos cos u u / b au u u E' u u mgu u u E' u bau mg u u E' mg u bau Let E ', mg Then u u u bau / du u u u b au dt du u ubau / dt

10 u(t) t u(0) du u ubau The result depends on the constants u(0),,, b, a. Snce E sn mg cos E ' z z 0 z 0sn 0 0 mg cos 0 / and mg a 0cos0 0 we actuall need b sn cos cos ,,,, m,,,,, to completel solve the problem. Consder a partcular case 45, 0, 0, m kg,.3 m, 0, rad / sec, 0 30 rad / sec,.04 kgm,.09 kgm Then u 0.707

11 (9) ut t b a du u u u / Frst we fnd the turnng ponts where du 0 dt u u u 0 u 6.0, , Hence the turnng ponts are rad mn rad 45.0 max So the top wll nutate between 30.7 and 45. gve To fnd the perods for nutaton and precesson we use equatons (), () and (3). The sn cos mg sn sn b acos sn a b acos cos sn Wth

12 We therefore solve ths sstem numercall. u t a 0cos0 0 b sn cos cos u 0 u and The exctaton perod s easest found from equaton (9) wth mn u max. Ths wll gve half th the perod. Hence.8600 dx PN.0660sec / x x x.707 An approprate Mathematca algorthm for the general problem s as follows. Frst fnd the turnng ponts as above, and label them thetamn and thetamax. Then the algorthm s:

13 theta ; dtheta ; nvted condtons ph ; dph ; ps ; dps ; thetamn ; thetamax ; m ; ; ; ; g 9.8 a dph sntheta dphcostheta / m g Costheta ; b mg / ; a / dphcosthetadps ; b dphsntheta / Costheta / Costhetadps; t 0; t max ; dt ; Whle t t max,pr nt "theta, theta," ph,ph", ps ", ps," t ", t ; theta theta dtheta dt; f theta thetamn, theta thetamn, theta theta ; f theta thetamax, theta thetamax, theta theta ; ph ph dph dt; ps ps dps dt; dtheta dph SnthetaCostheta / / Snthetadph dps mg Sntheta / dtheta dtheta dtheta dt dph b a Costheta / Sntheta; dps a / Costhetadph; t tdt The result for the precesson perod s P 7.89sec Note that n our case has backward loops!