Question #161091

A stone is attached to one end of a string. A child whirls the stone at a constant speed in a horizontal circle with a radius of 0.5 m. The stone makes 5 revolutions in 4 s


1
Expert's answer
2021-02-03T16:11:32-0500

Find the tension of the string. Assume that the mass of the stone is 1 kgkg . So, we have


F=mv2r=m(52πr/4)2r=m6.25πr=F=m\frac{v^2}{r}=m\cdot\frac{(5\cdot2\cdot\pi\cdot r/4)^2}{r}=m\cdot6.25 \cdot\pi\cdot r=


=16.253.140.5=9.82(N).=1\cdot6.25\cdot3.14\cdot0.5=9.82(N). Answer



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