A stone of mass $1 \mathrm{~kg}$, attached at the end of a $1 \mathrm{~m}$ long string, is whirled in a horizontal circle. If the string makes an angle of $30^{\circ}$ with the vertical, calculate the centripetal force acting on the stone.
Q 52.4
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$
\begin{aligned}
& \text { Data : } \mathrm{m}=1 \mathrm{~kg}, \mathrm{~L}=1 \mathrm{~m}, \theta=30^{\circ}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore \theta=\sin ^{-1}\left(\frac{1}{6}\right)=\sin ^{-1} 0.1667=9^{\circ} 36^{\prime} \\
& \therefore \cos \theta=\cos 9^{\circ} 36^{\prime}=0.9860 \\
\begin{aligned}
& \text { Data : } \mathrm{m}=1 \mathrm{~kg}, \mathrm{~L}=1 \mathrm{~m}, \theta=30^{\circ}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore \theta=\sin ^{-1}\left(\frac{1}{6}\right)=\sin ^{-1} 0.1667=9^{\circ} 36^{\prime} \\
& \therefore \cos \theta=\cos 9^{\circ} 36^{\prime}=0.9860 \\
\end{aligned}
$
The tension in the string,
$
\begin{aligned}
& F=\frac{m g}{\cos \theta}=\frac{0.15 \times 9.8}{0.9860}=1.491 \mathrm{~N} \\
& v=\sqrt{r g \tan \theta}
\end{aligned}
$
The centripetal force $=\frac{m v^2}{r}=\frac{m(r g \tan \theta)}{r}$
$
\begin{aligned}
& =m g \tan \theta \quad \text { MaharashtraBoardSolutions.Guru } \\
& =(1)(10)\left(\tan 30^{\circ}\right) \\
& =10 \times \frac{1}{\sqrt{3}}=\frac{10}{1.732}=\mathbf{5 . 7 7 4} \mathrm{N}
\end{aligned}
$
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