Question

A stone of mass $ 0.\text{25 kg} $ tied to the end of a string is whirled round in a circle of radius 1.5m with speed 40 rev/min in a horizontal plane. What is the tension in the string and what is the maximum speed with which the stone can be whirled around, if the string can withstand a maximum tension of $ ~\text{2}00\text{ N} $ ?

• A. $ \text{6}.\text{6 N, 35m}{{\text{s}}^{\text{-1}}} $
• B. $ \text{6N},\text{37m}{{\text{s}}^{\text{-1}}} $
• C. $ ~\text{7}.\text{5 N, 46 m}{{\text{s}}^{\text{-1}}} $
• D. $ \text{8 N},\text{ 38 m}{{\text{s}}^{\text{-1}}} $

Answer 1

Answer: (A)

Frequency of revolution of stone, $ f=40rev/\min =\frac{40}{60}rev/s $ Angular speed of the stone, $ \omega =2\pi f $ $ \text{= 2 }\!\!\pi\!\!\text{ }\!\!\times\!\!\text{ }\frac{\text{40}}{\text{60}}\text{=}\frac{\text{4 }\!\!\pi\!\!\text{ }}{\text{3}}\text{rad }{{\text{s}}^{\text{-1}}} $ The centripetal force is provided by the tension (T) in the string i.e. $ \text{T =}\frac{m{{v}^{2}}}{r}=mr{{\omega }^{2}} $ $ =0.25\times 1.5\times {{\left( \frac{4\pi }{3} \right)}^{2}}N=6.48N\approx 6.6N $ As the string can withstand a maximum tension of 200N. $ \therefore $ $ {{T}_{\max }}=\frac{mv_{\max }^{2}}{r} $ $ \Rightarrow $ $ {{v}_{\max }}=\sqrt{\frac{r{{T}_{\max }}}{m}}=\sqrt{\frac{1.5\times 200}{0.25}} $ $ \text{= 34}\text{.64 m}{{\text{s}}^{\text{-1}}}\text{k}\approx \text{35 m}{{\text{s}}^{\text{-1}}} $ $ T=6.6N,{{v}_{\max }}=35m{{s}^{-1}} $