Chapter 10313P10.74Consider the total weight of each hand to act at the center of gravity (mid-point) of that hand. Thenthe total torque (taking CCW as positive) of these hands about the center of the clock is given byτθθ=−FHGIKJ−FHGIKJ+mgLLgmLm Lhhhmmmhhhmmm222sinsinsinsinbgIf we take t= 0 at 12 o’clock, then the angular positions of the hands at time tareθωhht=,whereωπh=6rad handmmt=,where ωπm=2 radhTherefore,τFHGIKJ+LNMOQP49060027061002...sinsinm skgmkg 4.50 m2afttor⋅FHGIKJ+LNMOQP79462782N m sin.sintt, where tis in hours.(a)(i)At 3:00, t=300. h,so⋅FHGIKJ+LNMOQP⋅79426794Nmsin.sin(ii)At 5:15, t=+=51560525h hh., and substitution gives:⋅2510 N m(iii)At 6:00,=⋅0 N m(iv)At 8:20,
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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 .