Chapter 10317P10.80Consider the free-body diagram shown. The sum of torquesabout the chosen pivot isτα∑=⇒=FHGIKJFHGIKJ=FHGIKJIFmlaml alA132322CMCM(1)(a)A==l124. m: In this case, Equation (1) becomesaFmCM2Nkg. ms=3231472063035 0..afbgFmaFHmaxx∑=⇒+=CMCMor HmaFx=−CMThus, Hx=+0630350147735.kg. N2ejorHix=.±N .(b)A120620.m: For this situation, Equation (1) yieldsaFmCM2Nkg=344063017 5...H yH xF= 14.7 NCMmgpivotAlFIG. P10.80Again, aa F∑CMCM, soHx=−175368.2or x.±N .(c)If Hx=0, then aax∑=CMCM, or aFmCM=.Thus, Equation (1) becomeslFmA=23FHGIKJFHGIKJso A==mm from the top23230827l..=.P10.81Let the ball have mass mand radius r. Then Imr=252. If the ball takes four seconds to go downtwenty-meter alley, then v= ms5. The translational speed of the ball will decrease somewhat asthe ball loses energy to sliding friction and some translational kinetic energy is converted torotational kinetic energy; but its speed will always be on the order of 500.m s , including at thestarting point.As the ball slides, the kinetic friction force exerts a torque on the ball to increase the angular speed.
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