problem09_p101

University Physics with Modern Physics with Mastering Physics (11th Edition)

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Unformatted text preview: 2 2 9.101: a) ds = r d = r0 d + d , so s( ) = r0 + 2 . b) Setting s = vt = r0 + 2 gives a quadratic in . The positive solution is 1 2 (t ) = = r0 + 2 vt - r0 . (The negative solution would be going backwards, to values of r smaller than r0 .) c) Differentiating, z (t ) = d = dt v r02 + 2 vt , z = d v 2 =- dt r02 = 2 vt ( ) 32 . The angular acceleration z is not constant. d) r0 = 25.0 mm; It is crucial that is measured in radians, so = (1.55 m rev ) (1 rev 2 rad ) = 0.247 m rad. The total angle turned in 74.0 min = 4440 s is = 2 2.47 10- 7 m/rad (1.25 m/s) ( 4440 s ) 1 2.47 10- 7 m/rad + 25.0 10- 3 m 2 - 25.0 10- 3 m 5 = 1.337 10 rad ( ( ) ) which is 2.13 10 4 rev. e) ...
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This document was uploaded on 02/04/2008.

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