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1 € þ 1 mg ‘ cos h À 1 m‘2 €h dh þ Àk2 ‘ sin h À ‘mðÀ sin hh_ 2 þ cos hhÞ 4 2 12 ! 1 1 ¼ Àk1 ‘2 ð1 À cos hÞ sin h À k2 ‘2 sin h cos h þ mg‘ cos h À m‘2 h€ dh ¼ 0: 2 3 Since dh is arbitrary, its coefficient must be zero. 19 shows a uniform rigid disk of radius r rolling without slipping inside a circular track of radius R. Obtain the differential equation of motion for the system. Fig. 19 R φ r θ C A 24 1 Introduction to Nonlinear Vibrations and Dynamics Fig. 20 (a) R–r φ ac θ θ r C mg F N Solution Referring to the free-body diagram, we can write the force equation in the transverse direction as (Fig.

8 Using Lagrange’s equation, derive the equations of motion for small oscillations of a double pendulum, which consists of two rigid bodies, suspended at O and hinged at A, as shown in Fig. 49. The centers of gravity are C1 and C2, and the moments of inertia with respect to C1 and C2 are I1 and I2, respectively. The masses of the upper and lower bodies are m1 and m2, respectively. 50 shows a uniform cylinder rolling without slipping on a horizontal surface and constrained by a linear spring. Determine the differential equation of motion for the system.

3 Applied Examples 19 Fig. 15 È É È É Now consider body #2 (Fig. 16). Let ^i 1 ; ^j 1 ; ^i 2 ; ^j 2 be reference frames fixed in the ground, in link AC and in link CD, respectively. Use kinematic È É analysis to relate the velocity of D relative to ^i 3 ; ^j 3 and its velocity relative to È É È É È É ^i 1 ; ^j 1 , and let x be the angular velocity of ^i 3 ; ^j 3 relative to ^i 1 ; ^j 1 , vD ¼ vC þ vrel þ x Â r where vD ¼ dr D ; dt vC ¼ dr C : dt The terms are vC ¼ lh1^j2 ; vrel ¼ 0; x ¼ lh2 ^k3 ; r ¼ l2^i3 : The required unit vector transformation is ^j2 ¼ sinðh2 À h1 Þ^i3 þ cosðh2 À h1 Þ^j3 : Substituting, we get vD ¼ lh1^j2 þ 0 þ h2 ^k3 Â l2^i3 Â Ã ¼ lh1 sinðh2 À h1 Þ^i3 þ l2 h2 þ lh1 cosðh2 À h1 Þ^j3 (This also could be obtained geometrically from the law of cosines).

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