A stone of mass 0.25 kg tied to the end of a string is whirled
Question.
A stone 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?

solution:

Mass of the stone, $m=0.25 \mathrm{~kg}$

Radius of the circle, $r=1.5 \mathrm{~m}$

Number of revolution per second, $n=\frac{40}{60}=\frac{2}{3} \mathrm{rps}$

Angular velocity, $\omega=\frac{v}{r}=2 \pi n$ $\ldots(i)$

The centripetal force for the stone is provided by the tension T, in the string, i.e.,

$T=F_{\text {Centripctal }}$

$=\frac{m v^{2}}{r}=m r \omega^{2}=m r(2 \pi n)^{2}$

$=0.25 \times 1.5 \times\left(2 \times 3.14 \times \frac{2}{3}\right)^{2}$

$=6.57 \mathrm{~N}$

Maximum tension in the string, $T_{\max }=200 \mathrm{~N}$

$T_{\max }=\frac{m v_{\max }^{2}}{r}$

$\therefore v_{\max }=\sqrt{\frac{T_{\max } \times r}{m}}$

$=\sqrt{\frac{200 \times 1.5}{0.25}}$

$=\sqrt{1200}=34.64 \mathrm{~m} / \mathrm{s}$

Therefore, the maximum speed of the stone is $34.64 \mathrm{~m} / \mathrm{s}$.
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