158_Physics ProblemsTechnical Physics

# 158_Physics ProblemsTechnical Physics - 160 Circular Motion...

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160 Circular Motion and Other Applications of Newton’s Laws SOLUTIONS TO PROBLEMS Section 6.1 Newton’s Second Law Applied to Uniform Circular Motion P6.1 m = 300 . k g , r = 0800 . m. The string will break if the tension exceeds the weight corresponding to 25.0 kg, so TM g max .. == = 25 0 9 80 245 af N. When the 3.00 kg mass rotates in a horizontal circle, the tension causes the centripetal acceleration, so T mv r v 2 2 . . . Then v rT m TT 2 0800 245 65 3 = = . . . . . . . max a f a f a f ms 22 and 06 5 3 ≤≤ v . or 08 0 8 v m s . FIG. P6.1 P6.2 In Fm v r = 2 , both m and r are unknown but remain constant. Therefore, F is proportional to v 2 and increases by a factor of 18 0 14 0 2 . . F H G I K J as v increases 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 .

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