A small body of mass $0.3 \mathrm{~kg}$ oscillates in a vertical plane with the help of a string $0.5 \mathrm{~m}$ long with a constant speed of $2 \mathrm{~m} / \mathrm{s}$. It makes an angle of $60^{\circ}$ with the vertical. Calculate the tension in the string.
Q 60.3
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$
\begin{aligned}
\text { Data }: \mathrm{m} & =0.3 \mathrm{~kg}, \mathrm{r}=0.5 \mathrm{~m}, \mathrm{v}=2 \mathrm{~m} / \mathrm{s}, \theta=60^{\circ}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 \\
\frac{m v^2}{r} & =T-m g \cos \theta
\end{aligned}
$
Tension in the string, $T=\frac{m v^2}{r}+m g \cos \theta$
$
\begin{aligned}
& =\frac{(0.3)(2)^2}{0.5}+\begin{array}{c} \\
(0.3)(10) \cos 60^{\circ}
\end{array} \\
& =2.4+0.3 \times 10 \times \frac{1}{2}=2.4+0.3 \times 5=3.9 \mathrm{~N}
\end{aligned}
$
\begin{aligned}
\text { Data }: \mathrm{m} & =0.3 \mathrm{~kg}, \mathrm{r}=0.5 \mathrm{~m}, \mathrm{v}=2 \mathrm{~m} / \mathrm{s}, \theta=60^{\circ}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 \\
\frac{m v^2}{r} & =T-m g \cos \theta
\end{aligned}
$
Tension in the string, $T=\frac{m v^2}{r}+m g \cos \theta$
$
\begin{aligned}
& =\frac{(0.3)(2)^2}{0.5}+\begin{array}{c} \\
(0.3)(10) \cos 60^{\circ}
\end{array} \\
& =2.4+0.3 \times 10 \times \frac{1}{2}=2.4+0.3 \times 5=3.9 \mathrm{~N}
\end{aligned}
$
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