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 ?

Answer 1

Mass of the stone, m = 0.25 kg
Radius of the circle, r = 1.5 m
Number of revolution per second, n = $\frac{40}{60}$= $\frac{2}{3}$ rps
Angular velocity, $\omega$ = $\frac{v}{r}$ =2$\pi$𝑛 ………….(i)
The centripetal force for the stone is provided by the tension T, in the string, i.e.,
$T$ = $F_{centripetal}$
= $\frac{mv^2}{r}$ = mr$\omega^2$ = $mr(2\pi n)^2$
= $0.25\times 1.5\times \bigg(2\times 3.14\times \frac{2}{3}\bigg)^2$
= 6.57 N
Maximum tension in the string, $T_{max}$ = 200 N
$T_{max}$ = $\frac{mv^2_{max}}{r}$

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

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

= $\sqrt{1200}$
= 34.64 m/s
Therefore, the maximum speed of the stone is 34.64 m/s.