InterviewSolution
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InterviewSolution
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A stone of of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of `40 rev//min` in a horizontal plane What is the tension in the string ? What is the maximum speed with which the stone can be whirled around If the string can withstand a maximum tension of 200 N ? |
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Answer» Here, m = 0.25 kg , r = 1.5 m n = 40 rpm = `40/60 rps = 2/3` rps Now T = mr`omega^(2)= mr(2pin)^(2)= 4pi^(2) mrn^(2)` T = `4 xx 22/7 xx 22/7 xx 0.25 xx 1.5 xx (2/2)^(2)` = 6.6 N If `T_max` = 200 N , then from , `T_max` = m`v_max^(2)/ r impliesv_max^(2)` = `T_max xx r / m` or `v_max^(2)= 200 x 1.5 / 0.25 = 1200impliessqrt 1200` = 34.6` ms^(-1)`. |
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