A string of length $0.5 \mathrm{~m}$ carries a bob of mass $0.1 \mathrm{~kg}$ at its end. If this is to be used as a conical pendulum of period $0.4 \pi \mathrm{s}$, calculate the angle of inclination of the string with the vertical and the tension in the string.
Q 52.2
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Data : $L=0.5 \mathrm{~m}, \mathrm{~m}=0.1 \mathrm{~kg}, \mathrm{~T}=0.4 \pi \mathrm{s}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
(i) Period, $T=2 \pi \sqrt{\frac{L \cos \theta}{g}}$
$
\begin{aligned}
\therefore \cos \theta & =\frac{g T^2}{4 \pi^2 L} \\
& =\frac{10(0.4 \pi)^2}{4 \pi^2 \times 0.5} \\
& =\frac{10 \times 0.16 \pi^2}{2 \pi^2}=10 \times 0.08=0.8
\end{aligned}
$
The inclination of the string with the vertical,
$
\theta=36^{\circ} 5^{\prime}
$
(ii) The tension in the string,
$
F=\frac{m g}{\cos \theta}=\frac{0.1 \times 10}{0.8}=1.25 \mathrm{~N}
$
(i) Period, $T=2 \pi \sqrt{\frac{L \cos \theta}{g}}$
$
\begin{aligned}
\therefore \cos \theta & =\frac{g T^2}{4 \pi^2 L} \\
& =\frac{10(0.4 \pi)^2}{4 \pi^2 \times 0.5} \\
& =\frac{10 \times 0.16 \pi^2}{2 \pi^2}=10 \times 0.08=0.8
\end{aligned}
$
The inclination of the string with the vertical,
$
\theta=36^{\circ} 5^{\prime}
$
(ii) The tension in the string,
$
F=\frac{m g}{\cos \theta}=\frac{0.1 \times 10}{0.8}=1.25 \mathrm{~N}
$
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