# 15_5_29 - = 5 6 m v 2 . The potential energy is P . E . =...

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29. By Exercise 27, the moment of inertia of a spherical shell of radius a about its diameter is I = 2 3 ma 2 . Following the argument given in Example 4(b) of Section 5.7, the kinetic energy of the sphere, rolling with speed v down a plane inclined at angle α above the horizontal (and therefore rotating with angular speed  = v/ a ) is K . E . = 1 2 m v 2 + 1 2 I  2 = 1 2 m v 2 + 1 2 2 3 ma 2 v 2 a 2
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Unformatted text preview: = 5 6 m v 2 . The potential energy is P . E . = mgh , so, by conservation of total energy, 5 6 m v 2 + mgh = constant . Differentiating with respect to time t , we get = 5 6 m 2 v d v dt + mg dh dt = 5 3 m v d v dt + mg v sin α. Thus the sphere rolls with acceleration d v dt = 3 5 g sin α....
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## This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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