HW8_solution

HW8_solution - e sash Erghlem gggi The statement that i d H...

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Unformatted text preview: e sash Erghlem gggi The statement that i _ d H - ~aE H when expanded in the principal axis body frame becomes Euler's equations. If instead we use the fact that H = I dbl and expand the derivative in the inertial frame, the result is i . w + I —9— 3b1 dt 1d I obi t _ d _ H ‘ “a” H ’ { dt The inertial derivative of the angular velocity vector needs no explanation. However, the derivative of the moment of inertia matrix is most easily found by using the derivative expansion rule b i . —§E I = ~9E I + fiblx I The derivative of I in the body frame b is, of course, zero. While the cross product of a vector with, a matrix is not defined, the alternate interpretation of the cross product as a matrix multiplication (see Appendix A) leads to the correct result ebi 0 -w3 “2 w x I = w3 0 ~w1 { I } -w2 ml 0 Finally, inserting this back into the torque equation gives u ={z3b1x 1pm. 13131 While this result is correct, it is rarely if ever used. The value of the moment of inertia tensor I in the inertial frame must be used in the above equations, and this forces us to integrate the equation of i = abl motion x I as well as the moment equation. This increases the order of the differential equations by 3, and is usually unwarranted. ea are a ‘5’? ea 3;“ m. Ergblgm g—g, The pivot point on the side of the shuttle payload bay is a fixed point in inertial space. From an origin placed at this point, the center of mass of the satellite is located at r = —r b2, and the satellite revolves about the pivot point with angular velocity 3 = w b3. The usual velocity expansion then becomes .— 9 -— — = v - w x r — w b3 x ( r b2 ) w r b1 The acceleration about the center of mass can then be calculated by applying the velocity expansion rule to the velocity vector above, using the b frame: b . a. = "éf V + {Shit X V = Q r b + w2 r b 1 2 However, the satellite is not free to accelerate in the b2 direction, so we will concentrate on the vertical (bl) motion of the satellite. The horizontal motion is forbidden by another constraint force at the pivot point. Now, the center of mass of the satellite moves as if all of the external forces acted through that point, and all of the mass of the satellite were concentrated there (equation (4.5)). Thus, since the center of mass obeys F = m a, and the two external forces are PS, the spring force, and Fp, the pivot force, we have ll e O‘ l w 0‘ ma rare; saw 5§§risfif§§ $XE§ reassess i raga serrasg 41/ However, the pivot reaction force F In order to set up the rotational dynamics about the center of mass, we need to calculate the torque about the center of mass. is not really known. The spring force acts through the center of mass, and will exert no torque. The torque from r F b3. Then, p dH/dt = I «3 h3, the rotational dynamics about the center of mass takes the form the pivot force is M 2 since H = Ic'or—rF p This latter relation furnishes the pivot force directly as Fp = I But the acceleration iof the center of mass in the b1 direction is related to o by Q = a/r, so the pivot force is Fp = I a / r2. Inserting this into the equation of motion for the center of mass gives a form free of the pivot force: Q / r . I }' m + ——— a = F { r. S In a sense, the retarding effect of the rotational dynamics, responsible for the pivot force Fp, is to add an additional equivalent 2 mass I / r to the actual mass m of the satellite. There is also an equivalent rotational form of this equation of motion. If we use a = r a) to eliminate a in »the translational equation of motion, we have [ m r2 + I ] d z r FS after multiplying by r. Notice that r FS is the torque of the spring about the pivot point. This suggests an alternate way to derive the rotational form above. iIt is permissible to write rotational equations of motion about the pivot point, since this is a fixed point. Using the parallel axis theorem, the equivalent moment of inertia about the pivot point is I' 2 the pivot = I -+ m r . Since the pivot point is now the origin, reaction force now exerts no torque on the satellite, and the moment of the spring force is r FS b3. The statement that M = I & becomes as before. if? ass as - WWJM.W i éfigi g? 5; §§%M§§e 5 Problem 4—5. If the thrusters are located two meters off the rotation axis, and produce 20 N sec per pulse each, then the total angular impulse K At is H t K At [ 2 thrusters x 2 meters x 20 N sec / pulse = 80 kg mz/sec per pulse. This is also the change in the satellite angular momentum per pulse, since K At = fi At = AH. To achieve the desired final spin state, the angular momentum must be changed by 2 .AH I w — I w = 3400 kg m x 0.95 rad/sec = 3230 kg m2 /sec The total number of pulses is then the required change in the angular momentum, divided by the change per pulse: 2 N = 3230 k m sec = 40.375 80 kg mz/sec It is possible, of course, to fudge the numbers in the problem so that this result would come out as an integer. That is never true, however, in the real world. 5; ea , . W3: Eefefii'M 5%??? Prgblem 4—7. Before we can apply the Euler equations to this problem, we must calculate the torque on the rocket due to the thrust vector. For small gimbal angles 81 and 62, the thrust vector is approx1mately T = ~T 62 bl + T 91 b2 + T h3 while the vector from the center of mass to the engine is r ~ as @561: Egg w ggassewg «a? § Sséefigim s The torque is then M = r x T = Tfielbl + T392132 = TKO) 1:) + T£Kw ll 2h2 after the feedback relationships are used. It is now possible to write Euler’s equations as M1 = TWK 01 = A wl + ( C — A ) w2w3 M2 = 11K wz = A wz + ( A — C ) w1w3 M3 = 0 = C w3 + ( A — A ) wlwz The third equation immediately reduces to é3= 0, which has the consequence that w = w , a constant. If this constant is zero, the 3 30 other two Euler equations become a A w l T 2 K wl A wz = T e K wz These are first order constant coefficient equations, and furthermore are decoupled. If we solve the first equation by separation of variables, we have “1 I dwl ml 1 This can be solved for the behavior of ml as = [Twat = 10 T£K( t-t) I, o w = 0) 1 10 exp( 1%K( t — to) The exponential function will decay to zero, stable, only if GTK < 0. or e < 0, we need the feedback gain constant K < 0. and the system will be other than the absurd possibilities of T < 0 Ergblem 5—3, The angular momentum is a vector, and must be added for both rigid bodies, the satellite and the rotor, to obtain the total angular momentum of the system. The inertial angular momentum of the satellite is (ml, wz, w3}T, and its moments of inertia are A, B, and C in the principal axis b frame. If these moments of inertia include the rotor, then we only need concern ourselves with the spin of the rotor itself, wr, for the second body. The total angular momentum vector is then A o 0 ml — - - — H = 0 B o :02 + - - — - O 0 C w3 — ~ Ir wr = A ml bl + B wz b2 + ( Ir wr + C w3 ) b3 As we saw ix: the case of both the axisymmetric and general rigid inertial frame, bodies, H, although it must be constant in the In the body This magnitude is not appear to be constant seen from the body frame. frame, only the magnitude of H will be constant. 2 A w + B m + ( Irwr + C w3 ) Dividing by H2 and rearranging, this becomes 2 ( w3 + Irwr/C ) (II/<2)2 need: 5.55 aegis V gee ' ' ' ' ' ow This is the equation for a triax1al ellip501d, whose center is n .9 — I shifted to w = 0bl + 0b2 (Irwr/C)b3 be directly The total kinetic energy T is a scalar, and can calculated for each body, and then added. The result is A 0 0 w l 1 T = ~§~ {w1, w2, w3} 0 B 0 w 0 0 C w3 +—-§—-{-.-.w)—-— - - _ Ir “r mi mi wi 1 = + + ZTs/A ZTS/B ZTS/C This is the equation of an ellipsoid with its cent er at the origin of the coordinate frame. The tip of the angular velocity vector of the satellite, 3 = wlbl +‘w2b2 + w3b3 must lie on the surface of both ellipsoids at once. So, just as in the case of the triaxial rigid body, the motion of 3 can be visualized by constructing these intersection cu rves. In analogy with the single triaxial rigid body, they can be termed polhodes. Erotflgm: 5—5, It is sometimes necessary to spin a satellite, example to enable instruments to scan the sky, must still be maneuvered to point in different course of the mission. momentum, for and yet the satellite directions during the gages. :ae sees reassess ,.,, “Wu-W“lumwwwww ._ m WWW,wmeW/ngWm»meyWWMWWW On the other hand, if the angular momentum of the spinning {we}? 63‘?” was sat expensive. spacecraft is subtracted out by a counterrotating momentum wheel, the Using the results entire satellite has a zero net angular momentum. of the last problem, this will occur when where 5w is the (presumed small) additional angular rate away from 3 the required nominal value producing zero net angular momentum. The magnitude of the total angular momentum is then, using the results of the last problem _ 2 g Azwz + Bzwz + Czdw 1 2 3% The term from the rotor has been canceled by the appropriate choice of H2 = angular rate for the body of the satellite. In other words, the satellite has total angular momentum H = A wlbl + B wzbz + C 5:93:33 The angular rates wl, wz and 6w3 will probably be small, and only present during spacecraft reorientation maneuvers. Since torque equals rate of change of angular momentum M = H applying the derivative rule to the expression for H above will yield Euler’s equations in the variables wl, w and 5m . The satellite can 2’ 3 be controlled just as if it had zero net spin rate. However, its natural motion is somewhat changed. The angular momentum ellipsoid is given by mi w: 6w§ 1 = "“’””‘§’ + 2 + 2 (H/A) (H/B) (H/C) which is once again centered at the origin. the kinetic However, energy ellipsoid can be put into the form mi w: ( 5w3 — Irwr/C )2 1 =~ —————~ + m——~—— + ———~e—————__m_-— ZTS/A ZTs/B 2 TS/ c which is displaced from the origin. Again, it is possible to define pohodes for this case, as well, but they will not be the same as for a single torque free body. 3 9% ...
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HW8_solution - e sash Erghlem gggi The statement that i d H...

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