30 H. Laos 4.2 Derivation of the EOM The EOM will make use of the Euler angles that are shown in Fig. 4.1 [11]. Figure 4.1(a) [11] shows the Euler angles θ, φ, ψ that define the motion of the system. Figure 4.1(b) shows in red and blue lines the paths of the center of gravity (spin and precession). The angular movement of axis 3 relative to z ( , nutation) is also shown. The Lagrangian (L) for the system [6] is given by L= I1 2 ˙θ 2 + ˙φ 2sin2 θ + I3 2 ˙ψ+ ˙φcosθ 2 −Mgl cosθ, (4.1) Where M is the mass of the gyroscope, I1 = I2 is the principal equatorial moment of inertia through the center of gravity (CG) of the gyroscope, and I3 is the polar moment of inertia about the symmetry axis. The dots in (4.1) refer to the differentiation with respect to time t. Because the coordinates φ and ψ are cyclic, the angular momentumpψ = I3 ˙ψ+ ˙φcosθ and pφ = I1 ˙φ sin 2 θ + pψ cosθ are conserved. The angular velocities ˙φ and ˙ψcan be eliminated by using the Routhian (R) [7]: R θ, ˙θ, t =L−pφ ˙φ pφ,pψ,θ −pψ ˙ψ pφ,pψ,θ . (4.2) The EOM is derived from d dt ∂R ∂ ˙θ − ∂R ∂θ = 0, (4.3) and then the EOM is reduced to ¨θ = 1 I1 2sin3 θ pφ −pψ cosθ pφ cosθ −pψ + Mgl I1 sin θ. (4.4) Fig. 4.1 (a) Euler angles. (b) Trajectories of the center of gravity. The author owns one of these toy gyroscopes [11]
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