problem10_48

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

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10.48: a) Since the gyroscope is precessing in a horizontal plane, there can be no net vertical force on the gyroscope, so the force that the pivot exerts must be equal in magnitude to the weight of the gyroscope, ( 29 ( 29 N, 617 . 1 s m 80 . 9 kg 165 . 0 2 = = = = mg ω F 1.62 N to three figures. b) Solving Eq. (10.36) for , ω ( 29 ( 29 ( 29 ( 29 , s rad 7 . 188 s 20 . 2 rad
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Unformatted text preview: 2 m kg 10 20 . 1 m 10 00 . 4 N 617 . 1 2 4 2 = ⋅ × × = Ω =--π I ωR ω which is . min rev 10 80 . 1 3 × Note that in this and similar situations, since Ω appears in the denominator of the expression for , ϖ the conversion from s rev and back to min rev must be made. c)...
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## This document was uploaded on 02/04/2008.

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