459_Physics ProblemsTechnical Physics

459_Physics ProblemsTechnical Physics - Chapter 15 P15.58 =...

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Chapter 15 461 P15.58 ω π == k mT 2 (a) km m T 2 2 2 4 (b) ′= = F H G I K J m kT m T T af 2 2 2 4 P15.59 We draw a free-body diagram of the pendulum. The force H exerted by the hinge causes no torque about the axis of rotation. τα = I and d dt 2 2 θ α =− τθ =+ = MgL kxh I d dt sin cos 2 2 For small amplitude vibrations, use the approximations: sin θθ , cos 1, and xsh ≈= . m g H x k h m L L sin x k x h cos H y FIG. P15.59 Therefore, d dt MgL kh I 2 2 2 2 θω + F H G I K J ωπ = + = 2 MgL kh ML f 2 2 f MgL kh ML = + 1 2 2 2 *P15.60 (a) In xA t cos ωφ bg , vAt + ωω φ sin we have at t = 0 vA v sin max This requires 90 , so t ° cos 90 And this is equivalent to t sin Numerically 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 .

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