331_Physics ProblemsTechnical Physics

# 331_Physics ProblemsTechnical Physics - Chapter 11 P11.23...

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Chapter 11 333 P11.23 (a) LI M R == F H G I K J ωω 1 2 1 2 3 00 0 200 6 00 0 360 2 2 .. . . kg m rad s kg m s 2 bg a f (b) M R M R + F H G I K J L N M M O Q P P 1 22 3 4 3 00 0 200 6 00 0 540 2 2 2 . . kg m rad s kg m s 2 af P11.24 The total angular momentum about the center point is given by I hh mm =+ with I mL h hh = 2 2 3 60 0 3 146 . kg 2.70 m kg m 2 and I m mm 3 2 2 3 100 3 675 =⋅ kg 4.50 m kg m 2 In addition, ω π h = F H G I K J 21 145 10 4 rad 12 h h 3600 s rad s . while m = F H G I K J 175 10 3 rad 1 h h rad s . Thus, L × +⋅ × −− 146 1 45 10 675 1 75 10 43 kg m rad s kg m rad s ej or L 120 . k gms 2 P11.25 (a) Im L m = + = 1 12 0500 1 12 0 100 1 00 0 400 0 500 0 108 3 1 2 2 2 . . . . a f kg m 2 = 0 108 3 4 00 0 433 . kg m s 2 (b) L m R = + = 1 3 1 3 0 100 1 00 0 400 1 00 0 433 1 2 2 2 . . . a f a f = 0433400 173 . kg m s 2 *P11.26 Fm a xx = : += fm a sx We must use the center of mass as the axis in
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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