By Hanspeter Schaub, John L. Junkins
This unmarried resource offers a accomplished therapy of dynamics of aerospace structures beginning with the fundamental basics. themes variety from easy kinematics and dynamics to extra complex celestial mechanics. It publications you thru some of the derivations and proofs, yet avoids "cookbook" formulation. in its place, the reader is made to appreciate the underlying precept of the concerned equations and proven how you can follow them to varied dynamical structures. The e-book is split into components. half I covers analytical therapy of themes comparable to simple dynamic rules as much as complex power idea. specified awareness is paid to using rotating reference frames that regularly take place in aerospace structures. half II covers easy celestial mechanics treating the 2-body challenge, limited 3-body challenge, gravity box modelling, perturbation equipment, spacecraft formation flying, and orbit transfers.
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B) Find the position vector of the moon relative to the UFO. c) Find the angular velocity vectors ωE/U and ωM/U . d) What are the velocity and acceleration vectors of the moon as seen by the UFO frame? 15. Both the disk and the rod are rotating at a constant rate. Find the inertial velocity and acceleration of point P at the rim of the disk. 16. The conveyor belt speed v is constant. Find the inertial velocity and acceleration of Point P . 17. The shaft is rotating at a time varying rate φ. on the rim of the disk, while a missile is flying overhead at a fixed height h with the trajectory rm = hn ˆ 3 − tn ˆ 2.
Rc (t) = 2m¨ r1 = k(r2 − r1 ) m¨ r2 = −k(r2 − r1 ) + k(r3 − r2 ) + f m¨ r3 = −k(r3 − r2 ) + 2f This can be written in a standard ODE matrix form for a vibrating system 0 k −k 0 r1 r¨1 2m 0 0 0 m 0 r¨2 + −k 2k −k r2 = f 2f 0 −k k 0 0 m r3 r¨3 which can be solved given a set of initial conditions for ri (t0 ) and r˙i (t0 ). 2 Kinetic Energy The total kinetic energy T of the cloud of N particles can be written as the sum of the kinetic energies of each particle. 50) 42 NEWTONIAN MECHANICS CHAPTER 2 where the middle term N i=1 mi r˙ i is zero due to the definition of the center of mass in Eq.
The relative position vector σ of the point mass to O and its inertial derivative are given by σ = leˆr σ˙ = lθ˙ eˆθ The angular momentum vector HO can then be written as HO = σ × mσ˙ = −ml2 θ˙ n ˆ3 and its inertial derivative is given by ˙ O = −ml2 θ¨n H ˆ3 The torque LO about point O is written as LO = σ × F g − R n ˆ 2 × (Ff + N ) = −mgl sin (θ + α) n ˆ 3 − RFf n ˆ3 The inertial position vector rO of point O and its second inertial derivative are given by rO = d n ˆ 1 = Rθn ˆ1 ˆ1 r¨O = Rθ¨n Euler’s equation with moments about a general point in Eq.