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. $m(g+\pi n{r}^2)$
• 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, $n=\left(f/60\right)$

$\therefore \omega =2\pi n/60=\left(\pi n/30\right)\dots \left(ii\right)$

From equations $\left(i\right)$ and $\left(ii\right)$, $T_1=m\left[\frac{r^2{\pi }^2{n}^2}{900r}+g\right]$

$=m\left[g+\left({\pi }^2{n}^{2}r\right)/900\right]$