A small body of mass $m=0.1 \mathrm{~kg}$ at the end of a cord $1 \mathrm{~m}$ long swings in a vertical circle. Its speed is $2 \mathrm{mls}$ when the cord makes an angle $\theta=30^{\circ}$ with the vertical. Find the tension in the cord.
Q 60.15
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Data: $\mathrm{m}=0.1 \mathrm{~kg}, \mathrm{r}=1 \mathrm{~m}, \mathrm{y}=2 \mathrm{~m} / \mathrm{s}, \theta=30^{\circ}$,
$\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
When the cord makes an angle $\theta$ with the vertical, the centripetal force on the body is
$
\frac{m v^2}{r}=T-m g \cos \theta
$
The tension in the cord,
$
\begin{aligned}
T & =\frac{m v^2}{r}+m g \cos \theta \\
& =0.1\left(\frac{2^2}{1}+9.8 \times \cos 30^{\circ}\right) \\
& =0.1\left(4+9.8 \times \frac{\sqrt{3}}{2}\right)=0.1(4+4.9 \times 1.732) \\
& =0.1(4+8.486)=1.2486 \mathrm{~N}
\end{aligned}
$
$\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2$
When the cord makes an angle $\theta$ with the vertical, the centripetal force on the body is
$
\frac{m v^2}{r}=T-m g \cos \theta
$
The tension in the cord,
$
\begin{aligned}
T & =\frac{m v^2}{r}+m g \cos \theta \\
& =0.1\left(\frac{2^2}{1}+9.8 \times \cos 30^{\circ}\right) \\
& =0.1\left(4+9.8 \times \frac{\sqrt{3}}{2}\right)=0.1(4+4.9 \times 1.732) \\
& =0.1(4+8.486)=1.2486 \mathrm{~N}
\end{aligned}
$
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