Question

A stone of mass $0.25kg$ tied to the end of a string is whirled round in a circle of radius $1.5m$ with speed $40rev/min$ 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 $200N$?

• A. $6.6N,35{ms}^{-1}$
• B. $6N,37{ms}^{-1}$
• C. $7.5N,46{ms}^{-1}$
• D. $8N,38{ms}^{-1}$

Answer 1

Answer: (A)

$6.6N,35{ms}^{-1}$

Frequency of revolution of stone,
$f=40rev/min=\frac{40}{60}rev/s$

Angular speed of the stone, $\omega =2\pi f$
$=2\pi \times \frac{40}{60}=\frac{4\pi }{3}rad{s}^{-1}$

The centripetal force is provided by the tension $\left(T\right)$ in the string
i.e. $T=\frac{m{v}^2}{r}=mr{\omega }^2$
$=0.25\times 1.5\times {\left(\frac{4\pi }{3}\right)}^{2}N=6.48N\approx 6.6N$

As the string can withstand a maximum tension of $200N$.

$\therefore T_{max}=\frac{m{v}_{max}^2}{r}$

$\Rightarrow v_{max}=\sqrt{\frac{r{T}_{max}}{m}}=\sqrt{\frac{1.5\times 200}{0.25}}$

$=34.64{ms}^{-1}\approx 35{ms}^{-1}$

$T=6.6N,v_{max}=35{ms}^{-1}$