EULER EQUATIONS We now consider another approach to rigid body probles based on looking at the change needed in Newton s Laws if an accelerated coordinate syste is used. We start by considering how tie derivatives are effected by rotation. Consider a vector defined in the two systes by G G x,y,z xˆ G x,y,z yˆ G x,y,z zˆ x y z The origin of the prie syste (fixed in the body) is oving relative to the space (inertial) syste, but the coponents in either syste won t change since both ends of the vector are changed by the sae aount. Hence, the difference in rate of change of the coponents can be due only to rotation. Place so that one end is at the origin of the body syste. Then the other end oves in the body syste by an aount dg ' However in the space syste we have dg dg ' G' as before when we considered angular oentu. Since could be any vector we find d inertial d body x Now consider two coordinate systes one inertial and other whose origin is accelerated and whose axis are rotating with instantaneous angular velocity. Then Newton s nd Law applies in the inertial syste
R R r dr dr dr dr V r I I I E dr dr d A v r ve r E E E d F MA ae ve r r d FMA ve r r Ma Hence we can use Newton s nd Law in the accelerated syste provided we add appropriate fictitious forces. The next coon application of these is to a syste fixed on the earth. Since the earth both orbits the sun and rotates on its axis such a syste is not inertial. We consider the rotational ters 864 5 7.7 1 rad / sec For objects on the surface of the earth we have r = 6.4 1 6. We also have =. Then
v r zˆ Rsinyˆ Rsinxˆ xˆ R This is just the failiar centripetal force. Note that M v Is directed outward () as expected. Its agnitude is In coparison the force of gravity in is Hence 5 6 M 7.771 6.41 sin M sin.34 Fgx M 9.8sin and F c can norally be neglected. On the other hand the ter Fc.35 Fg car x F Mv which is known as the Coriolis force can be quite iportant. We can consider probles using this approach Consider a box with diensions a, b, c Proble 1
and ass being tossed vertically in the air. Choose body axis with origin at the center of the box and axis parallel to the sides. Let parallel to a, parallel to b, and parallel to c. Then the oent of inertial tensor will be I1 I I 3 Since the only force is gravity which acts through the origin of the body syste the torque will be zero in the internal syste ( and are parallel). Then Now dj I dj J E I1 x J I I ˆ ˆ ˆ y I1xxIyyI3zz I 3 z Hence J x y z I xi yi z ˆ ˆ ˆ x y z ˆ ˆ ˆ 1 x y 3 z zi ˆ yi ˆ zi ˆ xi ˆ yi ˆ xi ˆ x y 3 x z 1 x y 3 y z 1 z x z x 3 y z z y 1 z x 3 x z x y 1 x y xˆ I I yˆ I I zˆ I I xˆ I I yˆ I I zˆ I I y z 3 x z 1 3 x y 1 dj I1 xxˆ I ˆ ˆ yy I3 zz Thus (1) I I I 1 x y z 3 () I I I y x z 1 3 (3) I I I 3 z x y 1
Now suppose I 1 > I > I 3. Then consider trying to toss the box with x >> y, z. We have (4) I I I 1 x y z y z 3 (5) I I I y x z x z 1 3 (6) Fro () I I I 3 z x y x y 1 I1 I 3 (7) I Fro (3) y x z I I 1 (8) I3 z x y (9) I3 I I1I 3 I I 1 x z yx x const I1 I I3 Adding (1) and (8) into (5) gives I3 I I I 1 I I I y 1 3 z x y I1 I3 I1I 3 I3 I I I 1 y z xy y y cos t I I1 I3 Adding (9) and (7) into (6) gives I3I I1I 3 I3 z I I1 yz xz I1 I I I I I I I 1 3 1 3 z y x z I3 I1 I Since this also gives x y
Now suppose y x, z. Then cos t z z z z and the otion is unstable. For z x, y I I I I 3 3 x y x x I1 I3 x x et we again get stable otion (check it). In other words, rotation about the interediate axis is unstable! Proble Consider a projectile of ass launched due north fro latitude 4 N on the earth. Assue a uzzle velocity v and launch angle γ. Assue air resistance given by F v Take = 4 kg, v = 1 3 /sec, γ = 4, = 4 1 - sec/. We will solve this in two ways. First we will approxiate the earth as a flat plane (neglect curvature of the earth). Then choose axis with origin at launch point vertical (perpendicular to plane), north, and west. Then xcos ˆ zsin ˆ v xv ˆ yv ˆ zv ˆ x y z Then v zv ˆ ycos yv ˆ ˆ ˆ zcosyvxsin xvysin xˆ vysinyˆvxsin vzcoszv ˆ ycos r R cos zˆcos xˆsin
Feff gzˆ vxxˆ vyy vzzˆ xˆ vy sin yˆ v ˆ x sin vz cos zvy cos R zˆcos xˆcossin xˆ v ˆ x vysin R cos sin y vy vxsin vzcos ẑ gvz vycos Rcos dx dy dz v xˆ yˆ zˆ Hence (1) dx dv v v sin R cossin x x y dy dv vy vxsin vzcos () y (3) dz dv g v v cos R cos z z y Next we ake appropriate approxiations. Since 3 5 6 vy, v x 1 cos 4, 5.711, R 6.41, 4,.4 the approxiate sizes of the ters in (1) are vx 3, vy sin, R cos sin.4 Hence we can neglect nd and 3 rd ters on LHS. Then dv x vx vx vx e t In (3) we can neglect the 3 rd and 4 th ters on the LHS for the sae reason. Hence dv z vz g t g vz Ae d At t = we have v z = v sin γ. Hence
t g g g g vsin A A vsin A vsin e t (4) t g gt z v z z d vsin 1e The particle lands when z =. Thus the landing tie is the positive solution of g t v gt sin 1e Putting in nubers.4 4 t 3 49.8 49.8 4 1 sin 4 1e t.4.4.4 This gives Fro () we get t 18.4sec 18 t 18 1 6 1 4 x 1cos 4 e 1 cos 4e 19.31 dv y vy vxsinvzcos t t g g vy vcos e sin vsin e cos t t dvy g g vy vcos e sin vsin e cos t g f t e 1 The solution of the hoogeneous equation is
vy Ae t For the particular solution we note that t f t ae a with g a cos Try y t v cte d Then ce t cte t cte t t dae a a d ca Thus t t a vy Ae ate At t = we have a vy A t a a vy e at t a v y 1e ate t
t t at a t y vyat e a e t t t t at a a a e 1 t e t t ae t 1 e t t a 1 e t 1 e Putting in nubers 5 3 a 5.711 1 cos 4.8573 18.4 18.4 3 1 1 y.85731 1e 18.41e 83.8 Hence shell lands 9.3 1 4 N and 84 E of firing point 18 sec after firing. If we don t approxiate the earth as flat, we use spherical coordinates based at the center of earth with z along ω. Then ˆrcosˆ sin d dr drˆ v rrˆ rˆr drˆ drˆ drˆ ˆ sin ˆ d d dˆ dˆ dˆ ˆr cosˆ d d dˆ dˆ dˆ sin rˆ cos ˆ d d v rr ˆ r ˆ rsin ˆ
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ a rrˆr sin r r r rˆ cos r sinrsin r cosrsin sinrˆcos ˆ ˆr r r r sin r r r sin cos ˆ r sin r cos r sin r sin v ˆr cos ˆr sin cos ˆr sin rr ˆ sin ˆ ˆ ˆr rsin rsin cos rcos r sin r ˆ r sin r rsin ˆ cos rsin ˆ Feff grˆ rˆsin rˆ r sin cos ˆ r cos r sin ˆ ˆ ˆ ˆ ˆ r sin cos r sin rr r r sin F ff a (1) () ˆr) r g r sin r sin r r r sin ˆ r rsincosr sincosr r rsincos ˆ rsin rcos rsin r sin r cos rsin (3) Now r = R + h where R = radius of earth >> h. Thus h h R sin Rsin R R sin g (4) (5) (6) h R sin cos sin cos sin cos h h ctn ctn R R
We could use various approxiation schees to solve this syste, but it is siplest to just solve nuerically with the result t 19.7sec.858183 5.989 1 f r 5 Hence shell lands at of firing point. 6 4 d R.87665.858183 6.41 9.7 1 v Nˆ v i f d R sin 9Eˆ E F f These results are very close to those fro the approxiate ethod above because the distance traveled is sall copared to the circuference of the earth, and therefore, the earth looks nearly flat.