Chapter 15461P15.58ωπ==kmT2(a)kmmT2224(b)′=′=′FHGIKJmkTmTTaf2224P15.59We draw a free-body diagram of the pendulum.The force Hexerted by the hinge causes no torqueabout the axis of rotation.τα=Iandddt22θα=−τθ=+=−MgLkxhIddtsincos22For small amplitude vibrations, use theapproximations: sinθθ≈, cos≈1, and xsh≈=.mgH xkhmL L sinxkxhcosH yFIG. P15.59Therefore,ddtMgLkhI2222θω+FHGIKJωπ=+=2MgLkhMLf22fMgLkhML=+1222*P15.60(a)InxAtcosωφbg,vAt+ωωφsinwe have at t=0vAvsinmaxThis requires =°90 , sot°cos90And this is equivalent to tsinNumerically we have
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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 .