Question

A stone of mass $1kg$ is whirled in horizontal circle attached at the end of a $1m$ long string. If the string makes an angle of ${30}^o$ with vertical, calculate the centripetal force acting on the stone. $(g=9.8m/s^2)$

Answer 1

Given $m=1kg,r=1m,\theta ={30}^o$
Now, $\because$ $F\sin \theta =\frac{m{v}^2}{r}$
$\therefore$ Centripetal force$=\frac{1\times v^2}{r}$
and $v=\sqrt{rg\tan \theta }$
$=\sqrt{1\times 9.8\times \tan {30}^o}=2.37m/s$
Again $\because$ Centripetal force$=\frac{m{v}^2}{r}$
$=\frac{1\times {2.37}^2}{1}=5.61N$