problem15_81

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

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15.81: The tension in the rope will vary with radius r . The tension at a distance r from the center must supply the force to keep the mass of the rope that is further out than r accelerating inward. The mass of this piece in , L r L m - and its center of mass moves in a circle of radius , 2 r L + and so ). ( 2 ) ( 2 2 2 2 r L L L r L ω L r L m r T - = + - = An equivalent method is to consider the net force on a piece of the rope with length dr and mass . L m dr dm = The tension must vary in such a way that . ) ( or , ) ( ) ( 2 2 dr r L dm r ω dr r T r T dr dT - = - = + - This is integrated to obtained , ) 2 ( ) ( 2 2 C r L r T + - = where C is a constant of integration. The tension must vanish at , L r = from which
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Unformatted text preview: ) 2 ( 2 L mω C = and the previous result is obtained. The speed of propagation as a function of distance is , 2 ) ( ) ( 2 2 r L ω m TL μ r T dt dr r v-= = = = where dt dr has been chosen for a wave traveling from the center to the edge. Separating variables and integrating, the time t is . 2 2 2 ∫ ∫-= = L r L dr ω dt t The integral may be found in a table, or in Appendix B. The integral is done explicitly by letting , cos , cos , sin 2 2 θ L r L dθ θ L dr θ L r =-= = so that . 2 arcsin(1) 2 and , arcsin 2 2 ω π ω t L r θ r L dr = = = =-∫...
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