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, $\omega =2\pi f$ $\text{= 2 }\pi \times \frac{\text{40}}{\text{60}}\text{=}\frac{\text{4 }\pi }{\text{3}}\text{rad }{\text{s}}^{\text{-1}}$ 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. $\therefore$ $T_{\max }=\frac{m{v}_{\max }^2}{r}$ $\Rightarrow$ $v_{\max }=\sqrt{\frac{r{T}_{\max }}{m}}=\sqrt{\frac{1.5\times 200}{0.25}}$ $\text{= 34}\text{.64 m}{\text{s}}^{\text{-1}}\text{k}\approx \text{35 m}{\text{s}}^{\text{-1}}$ $T=6.6N,v_{\max }=35m{s}^{-1}$