160 Circular Motion and Other Applications of Newton’s LawsSOLUTIONS TO PROBLEMSSection 6.1Newton’s Second Law Applied to Uniform Circular MotionP6.1m=300. kg,r=0800.m. The string will break if the tension exceedsthe weight corresponding to 25.0 kg, soTMgmax..===25 0 9 80245afN.When the 3.00 kg mass rotates in a horizontal circle, the tensioncauses the centripetal acceleration,soTmvrv22...ThenvrTmTT20800 24565 3≤==.......maxafafafms22and0653≤≤v.or0808vms.FIG. P6.1P6.2In Fmvr∑=2, both mand rare unknown but remain constant. Therefore, F∑is proportional to v2and increases by a factor of 18 014 02..FHGIKJas vincreases from 14.0 m/s to 18.0 m/s. The total force at the
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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 .