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How do you find the kinetic energy of a thin rod?

A thin uniform rod (mass = 0.96 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of 1.3 s and an angular amplitude of 6.7°. (a) What is the length of the rod? (a) What is the maximum kinetic energy of the rod as it swings?

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1. (a)The formula for time period, T of suspended thin rodof length L from one end is T = 2 pi sqrt[(l+k^2/l)/g] where k is radius of gyration, l is L/2 and g is acceleration due to gravity/ k = L/[2sqrt(3)]. so 1.3 = 2x3.14xsqrt[(L/2 + L/6)/9.8] or (1.3)^2 = 4x(3.14)^2[2/(3x9.8)]L or 1.69 = 2.68 L or L = 0.629 m or 63 cm (b)The maximum kinetic energy,KE=maximum potential energy = mgh where h is teh distance throuhj which CG of thin rod rises for angular displacement of 6.7 degrees. h = (L/2)[1- cos 6.7] = 0.0021 m So KE = 0.96x9.8x0.0021 J = 0.02 J