Question

A stone of mass $m$ is tied to a string and is moved in a vertical circle of radius $r$ making $n$ revolutions per minute. The total tension in the string when the stone is at its lowest point is

• A. $mg$
• B. $m(g+\pi nr^{2})$
• C. $m (g+nr)$
• D. $m\left(g+\frac{\pi^{2} n^{2} r}{900}\right)$

Answer 1

Answer: (D)

The tension at the lowest point is given by
$T_{1}=\frac{m}{r}\left(v^{2}+gr\right)$
$=m\left(\frac{v^{2}}{r}+g\right)$
$=m\left(\frac{r^{2} \omega^{2}}{r}+g\right)\left(\because v=r\omega\right) \left(i\right)$
Now $\omega=2\pi f$ where $f$ =frequency of revolution. The frequency of revolution per sec, $n =\left(f/ 60\right) $
$\therefore \omega=2\pi n/ 60=\left(\pi n /30\right) \ldots\left(ii\right)$
From equations $\left(i\right)$ and $ \left(ii\right)$, $T_{1}=m\left[\frac{r^{2}\pi^{2}n^{2}}{900 r}+g\right]$
$=m\left[g+\left(\pi^{2}n^{2}r\right)/900\right]$