A thin, hollow sphere of mass 2.40 kg and radius 0.620 m is rolling on a horizontal surface with a constant angular speed of 58.0 rpm. What fraction of the kinetic energy is translational kinetic energy, and what fraction of the kinetic energy is rotational kinetic energy?translational kinetic energytotal kinetic energy%3Drotational kinetic energytotal kinetic energy

m = 2.4 kg r = 0.62 m w = 58 rpm Using, total kinetic energy = translational kinetic energy + rotational kinetic energy E = (1/2)(m)(v)2 + (1/2)(2/3)(m)(r)2*(v/r)2E = (5/6)(mv2)E = (5/3) (mv2/2)E =(5/3)(translational kinetic energy)Therefore, translational kinetic energy / total kinetic energy = 3/5 = 0.6 Thus, rotational kinetic energy/total kinetic energy = 1 - (3/5) = 2/5 = 0.4

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translational kinetic energy, and what fraction of the kinetic energy is rotational kinetic energy? A thin, hollow sphere of mass 2.40 kg and radius 0.620 m is rolling on a horizontal surface with a constant angular speed of 58.0 rpm. What fraction of the kinetic energy translational kinetic energy total kinetic energy rotational kinetic energy total kinetic energy

translational kinetic energy, and what fraction of the kinetic energy is rotational kinetic energy? A thin, hollow sphere of mass 2.40 kg and radius 0.620 m is rolling on a horizontal surface with a constant angular speed of 58.0 rpm. What fraction of the kinetic energy translational kinetic energy total kinetic energy rotational kinetic energy total kinetic energy

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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