This preview shows page 1. Sign up to view the full content.
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 = r . Substituting this expression in, we see that: K = m 1 r 1 2 2 + m 2 r 2 2 2 + ... m n r n 2 2 Since all particles are part of the same rigid body, we can factor our 2 : K = ( mr 2 ) 2 This sum, however, is simply our expression for a moment of inertia. Thus: K = I 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 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....
View
Full
Document
This note was uploaded on 02/09/2012 for the course PHY PHY2053 taught by Professor Davidjudd during the Fall '10 term at Broward College.
 Fall '10
 DavidJudd
 Physics, Energy, Work

Click to edit the document details