HW-H

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Unformatted text preview: .600 m + 0.270 kg 0.250 m Mld + mlr = = 0.477 m. M+m 0.500 kg + 0.270 kg (c) The period of oscillation is T = 2π I 0.205 kg ×m 2 = 2π = 1.50 s . (0.500 kg + 0.270 kg)(9.80 m/s 2 )(0.447 m) ( M + m ) gd _______________________________________________________________________ _Problem 15-49 ((My solution)) L = 1.85 m τ = I 0θ = − Mgx sin θ 1 I 0 = ML2 + Mx 2 12 For θ ≈ 0, gx sin θ θ = − ≈ −ω 2θ 12 2 L +x 12 where ω= g x 12 L + x2 12 The period T is T= 2π 2π = ω g x+ L2 12 x Note that x+ L2 L2 2 L ≥2 x = L= 12 x 12 x 12 3 where the function x + L L L2 = = 0.289 L = 0.53m has a minimum at x = 12 2 3 12 x Then T is T = 4.747 ((Mathematica)) g = 2.074 s L T1 2 1 x2 1 L 2 12 x g L2 x2 12 x 2 g rule1 x Ly x Ly T2 T1 . rule1 Simplify , L 0 1 y L 12 y 3 g f y & 12 y 1 y 1 12 y y eq1 D f y , y 1 y2 12 1 y 2 12 y eq2 Solve eq1 0, y y 1 2 T3 T2 . eq2 4.77419 y , 3 2 1 2 3 N L g rule2 L 1.85, g 9.80 L 1.85, g 9.8 T33 T3 . rule2 2.0743 Ly . eq2 0.534049 2 . rule2 N ((WileyPlus...
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This note was uploaded on 05/14/2013 for the course MATH 346 taught by Professor Professormiguelarcones during the Spring '08 term at Binghamton University.

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