Dynamics of the Rigid Solid with General Constraints by a by Nicolae Pandrea, Nicolae-Doru Stanescu

By Nicolae Pandrea, Nicolae-Doru Stanescu

Covers either holonomic and non-holonomic constraints in a research of the mechanics of the limited inflexible body.

  • Covers all kinds of basic constraints appropriate to the forged rigid
  • Performs calculations in matrix form
  • Provides algorithms for the numerical calculations for every kind of constraint
  • Includes solved numerical examples
  • Accompanied through an internet site internet hosting programs

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Additional info for Dynamics of the Rigid Solid with General Constraints by a Multibody Approach

Example text

Ginsberg J (2007). Engineering Dynamics. Cambridge: Cambridge University Press. Goldstein H, Poole CP Jr, Safko JL (2001). Classical Mechanics. 3rd ed. Edinburg: Pearson. Greenwood DT (2006). Advanced Dynamics. Cambridge: Cambridge University Press. 44 Dynamics of the Rigid Solid Greiner W (2009). Classical Mechanics: Systems of Particles and Hamiltonian Dynamics. 2nd ed. Heidelberg: Springer. Hand LN, Finch JD (1998). Analytical Mechanics. Cambridge: Cambridge University Press. Jain A (2010). Robot and Multibody Dynamics: Analysis and Algorithms.

Observing that the matrix [Q] may be also written in the form ! 7 Composition of the Angular Velocities We consider the reference systems O0x0y0z0, O1x1y1z1 and the reference system O2x2y2z2 jointed to the rigid solid, and let [A21] be the matrix of rotation of the reference system O2x2y2z2 relative to the reference system O1x1y1z1 and [A10] be the matrix of rotation of the reference system O1x1y1z1 relative to the reference system O0x0y0z0. 122) with respect to time and keeps into account the equality r_ = ω × r and that the vector v_ and v_ O represent the accelerations a, and aO of the points P and O, respectively (Fig.

3 ð2:57Þ Determine the unitary eigenvector for the matrix 2 −cos2 φ −sin2 φcos θ 6 ½AŠ = 4 sin φ cos φð1− cos θÞ sin φ cos φð1− cos θÞ − sin2 φ −cos2 φ cos θ sin θ sin φ sin θ cos φ sin φ sin θ 3 7 cos φ sinθ 5: cos θ Solution: One may easily verify the conditions w = 0, ξ = π, ½AŠ = ½AŠT . θ θ θ T , while if θ = π, If θ 6¼ π, then one obtains fug = sin φ sin cos φ sin cos 2 2 2 then it results fug = ½ sin φ cos φ 0ŠT . 5 Symmetries: Decomposition of the Rotation into Two Symmetries In the case of a symmetry the angle of rotation is ξ = π, and the matrix of rotation reads ½AŠ = ½IŠ + 2½UŠ2 : ð2:58Þ If one considers the straight lines Δ1, Δ2 concurrent at the point O0 with the straight line Δ and perpendicular to it (Fig.

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