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MIT16_07F09_Lec30

# MIT16_07F09_Lec30 - J Peraire S Widnall 16.07 Dynamics Fall...

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J. Peraire, S. Widnall 16.07 Dynamics Fall 2008 Version 2.0 Lecture L30 - 3D Rigid Body Dynamics: Tops and Gyroscopes 3D Rigid Body Dynamics: Euler Equations in Euler Angles In lecture 29, we introduced the Euler angles as a framework for formulating and solving the equations for conservation of angular momentum. We applied this framework to the free-body motion of a symmetrical body whose angular momentum vector was not aligned with a principal axis. The angular moment was however constant. We now apply Euler angles and Euler’s equations to a slightly more general case, a top or gyroscope in the presence of gravity. We consider a top rotating about a fixed point O on a ﬂat plane in the presence of gravity. Unlike our previous example of free-body motion, the angular momentum vector is not aligned with the Z axis, but precesses about the Z axis due to the applied moment. Whether we take the origin at the center of mass G or the fixed point O , the applied moment about the x axis is M x = Mgz G sinθ , where z G is the distance to the center of mass.. Initially, we shall not assume steady motion, but will develop Euler’s equations in the Euler angle variables ψ (spin), φ (precession) and θ (nutation). 1

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Referring to the figure showing the Euler angles, and referring to our study of free-body motion, we have the following relationships between the angular velocities along the x, y, z axes and the time rate of change of the Euler angles. The angular velocity vectors for θ ˙ , φ ˙ and ψ ˙ are shown in the figure. Note that these three angular velocity vectors are not orthogonal, giving rise to some cross products when the angular velocities ω i are calculated about the three principal axes. ω x = φ ˙ sin θsinφ + θ ˙ cos ψ (1) ˙ ω y = φsinθ cos φ θ ˙ sin ψ (2) ˙ ω z = φcosθ + ψ ˙ (3) 3D Rigid Body Dynamics: Euler’s Equations We consider a symmetric body, appropriate for a top, for which the moments of inertia I xx = I yy = I 0 and I zz = I . The angular momentum is then H x = I 0 ω x (4) H y = I 0 ω y (5) H z = z (6) For the general motion of a three-dimensional body, we have Euler’s equations in body-fixed axes which rotate with the body so that the moment of inertia is constant in time. In this body-fixed coordinate system, the conservation of angular momentum is H ˙ = d ([ I ] { ω } ) = AppliedMoments (7) dt 2
d Since we have chosen to work in a rotating coordinate system so that dt I = 0, we must pay the price, applying Coriolis theorem to obtain the time derivative of the angular velocity vector in the rotating coordinate system H ˙ = d H + Ω × H , (8) dt resulting in the ”Euler” equations expressed in the x, y, z coordinate system moving with the body. In general, we must rotate with the total angular velocity ω of the body, so that the governing equation for the conservation of angular momentum become, with Ω = ω .

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