A stone of mass $0.25kg$ tied to the end of a string is whirled round in a circle of radius $1.5m$ with a speed of $40rev./\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 $200N$?
Show solutionHide solution
Ans: It is given that,
Mass of the stone, $m=0.25kg$
Radius of the circle, $r=1.5m$
Number of revolution per second, $n=\frac{40}{60}=\frac{2}{3}rps$
Angular velocity, $\omega =\frac{v}{r}=2\pi n$ ........$(1)$
The centripetal force for the stone is provided by the tension $T$ , in the string, i.e., $T={{F}_{Centripetal}}$
$\Rightarrow \frac{m{{v}^{2}}}{r}=mr{{\omega }^{2}}=mr{{\left( 2\pi n \right)}^{2}}$
$\Rightarrow {{F}_{Centripetal}}=0.25\times 1.5\times {{\left( 2\times 3.14\times \frac{2}{3} \right)}^{2}}$
$\Rightarrow {{F}_{Centripetal}}=6.57N$
Maximum tension in the string, ${{T}_{\max }}=200N$
${{T}_{\max }}=\frac{mv_{\max }^{2}}{r}$
$\Rightarrow {{v}_{\max }}=\sqrt{\frac{{{T}_{\max }}\times r}{m}}$
$\Rightarrow {{v}_{\max }}=\sqrt{\frac{200\times 1.5}{0.25}}$
\[\Rightarrow {{v}_{\max }}=\sqrt{1200}=34.64m/s\]
Thus, the maximum speed of the stone is \[34.64m/s\].