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.