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A circular disc of mass $10 \mathrm{~kg}$ is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be $1.5 \mathrm{~s}$. The radius of the disc is $15 \mathrm{~cm}$. Determine the torsional spring constant of the wire.
(Torsional spring constant $\alpha$ is defined by the relation $J=-\alpha \theta$, where $J$ is the restoring couple and $\theta$ the angle of twist).
(Torsional spring constant $\alpha$ is defined by the relation $J=-\alpha \theta$, where $J$ is the restoring couple and $\theta$ the angle of twist).
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Verified Answer
Here, mass $m=10 \mathrm{~kg}, R=15 \mathrm{~cm}=0.15 \mathrm{~m}, T=1.5 \mathrm{~s}$, $\alpha=$ ?
Moment of inertia of the $\operatorname{disc}=\frac{1}{2} m R^2$
$$
\begin{aligned}
&=\frac{1}{2} \times 10 \times(0.15)^2 \\
\mathrm{~T} &=2 \pi \sqrt{\frac{\mathrm{I}}{\alpha}} \Rightarrow \alpha=\frac{4 \pi^2 \mathrm{I}}{\mathrm{T}^2} \\
&=4 \times\left(\frac{22}{7}\right)^2 \times \frac{1}{2} \times \frac{10 \times(0.15)^2}{(1.5)^2}=1.97 \mathrm{Nm} / \mathrm{rad} .
\end{aligned}
$$
Moment of inertia of the $\operatorname{disc}=\frac{1}{2} m R^2$
$$
\begin{aligned}
&=\frac{1}{2} \times 10 \times(0.15)^2 \\
\mathrm{~T} &=2 \pi \sqrt{\frac{\mathrm{I}}{\alpha}} \Rightarrow \alpha=\frac{4 \pi^2 \mathrm{I}}{\mathrm{T}^2} \\
&=4 \times\left(\frac{22}{7}\right)^2 \times \frac{1}{2} \times \frac{10 \times(0.15)^2}{(1.5)^2}=1.97 \mathrm{Nm} / \mathrm{rad} .
\end{aligned}
$$
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