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, $\omega =\frac{v}{r}=2\pi n\dots \dots \left(i\right)$
The centripetal force for the stone is provided by the tension T, in the string, i.e.,
$T=T_{contripetal}=\frac{m{v}^2}{r}=mr{\omega }^2=mr(2\pi n{)}^2=0.25\times 1.5\times {(2\times 3.14\times \frac{2}{3})}^2=6.57N$
Maximum tension in the string, $T_{max}=200N{T}_{max}=\frac{m{v}_{max}^2}{r}\therefore v_{max}=\sqrt{\frac{T_{max}r}{m}}=\sqrt{\frac{200\times 1.5}{0.25}}=\sqrt{1200}=34.64m/s$
Therefore, the maximum speed of the stone is 34.64 m/s.