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By Suzuki H.

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7 Problems with Euler angles The inverse of A is obtained by reversing the order of multiplication of Aψ, Aθ and Aφ and using the opposite angle: A−1 = A−φ A−θ A−ψ . This turns out to be exactly equal to the transpose of A, so that A−1 = AT . As with twodimensional rotations, A is an orthogonal matrix, a property which is inherited from the matrices Aψ , Aθ and Aφ . That all is not well with Euler angles can be seen when the angles ψ, θ, φ are again computed from a given rotation matrix ⎞ ⎛ A1,1 A1,2 A1,3 A = ⎝A2,1 A2,2 A2,3 ⎠.

31) which is simple if one only wants to obtain numerical values. However, when one wants to obtain scaling by the mass m of the new, hollow body, the derivations become more cumbersome. 1 as Jz = 12 m0 r02 . When the cylinder is hollowed out concentrically in the middle by a cylindrical shape so that the radius of the inner hollow is ri , the mass becomes mhc = m0 r02 − ri2 r02 = m0 − mi , where mi is the mass which has been removed from the original cylinder. So, for the moment of inertia along the symmetry axis, one obtains ✐ ✐ ✐ ✐ ✐ ✐ “Matuttis-Driv-1” — 2014/3/24 — 19:18 — page 16 — #16 ✐ 16 ✐ Understanding the Discrete Element Method J hc = 1 1 m0 r02 − mi ri2 2 2 = r2 − r2 1 1 m0 r02 − m0 − m0 0 2 i 2 2 r0 = 1 r04 1 r4 m0 2 − m0 i2 2 r0 2 r0 = mhc r02 2 r02 − ri2 = mhc 2 r0 + ri2 .

20, we visualize the flow in several ways. First, we plot with solid lines the trajectories in time, (x(t), v(t)), of the solution. 20. Finally, it has become traditional to discuss the transport of a set of initial conditions in phase space from time t0 to time t: y(t0 ) → y(t). 18 Phase portrait (attractor) for the linear oscillator (m = 1, k = 1) without damping: illustration of Liouville’s theorem on conservation of phase space volume. 1): the attractor is a whirl, where the phase space volume shrinks exponentially in spiral-shaped trajectories.