Unformatted text preview: see that each one does in fact have a translational kinetic energy: K = m 1 v 1 2 + m 2 v 2 2 + ... m n v n 2 However, we also know from our relation between linear and angular variables that v = rσ . Substituting this expression in, we see that: K = m 1 r 1 2 σ 2 + m 2 r 2 2 σ 2 + ... m n r n 2 σ 2 Since all particles are part of the same rigid body, we can factor our σ 2 : K = ( mr 2 ) σ 2 This sum, however, is simply our expression for a moment of inertia. Thus: K = Iσ 2 As we might expect, this equation is of the same form as our equation for linear kinetic energy, but with I substituted for m , and σ substituted for v . We now have rotational analogues for nearly all of our translational concepts. The last rotational equation that we need to define is power....
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 Fall '10
 DavidJudd
 Physics, Angular Momentum, Energy, Kinetic Energy, Work, Moment Of Inertia, Rigid Body, Combined Motion

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