A circular disc of 10 kg liquid is hanging from a string attached… - Vidyadip

Question

A circular disc of 10 kg liquid is hanging from a string attached to its centre. By turning the disc, it is freed by creating a twist in the wire. The period of torsional oscillation is 1.5 sec . The diameter of the disc is 15 cm . find the torsional spring constnat of the wire [The torsional spring constant is $\alpha, \tau=-\alpha \theta$ by the relation, here the elastic force pair is end is the torsion angle $\theta$ ].

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Answer

Given that
$\begin{aligned}m & =10 kg \\R & =15 cm=0.15 m \\T & =1.5 \text { second } \\\alpha & =?\end{aligned}$
Moment of inertia of circular disc
$\begin{aligned}I & =\frac{1}{2} m R^2 \\I & =\frac{1}{2} \times 10 \times(0.15)^2 kg m^2 \\I & =5 \times 225 \times 10^{-4} kg m^2 \\T & =2 \pi \sqrt{\frac{I}{\alpha}} \\\alpha & =\frac{4 \pi^2 I}{T^2} \\\text { Put value } & =\frac{4 \times\left(\frac{22}{7}\right)^2 \times 5 \times 225 \times 10^{-4}}{(1.5)^2} \\& =\frac{4 \times 22 \times 22 \times 5 \times 225 \times 10^{-4}}{49 \times 2.25} \\& =\frac{217.8}{110.25} \\\alpha & =1.976 Nm / radian\end{aligned}$

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