A circular disc of mass $10 \;kg$ 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 $1.5 \;s$. The radius of the disc is $15\; cm .$ Determine the torsional spring constant of the wire in $N\;m\;rad^{-1}$. (Torsional spring constant $\alpha$ is defined by the relation $J=-\alpha \theta,$ where $J$ is the restoring couple and $\theta$ the angle of twist).
Medium
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Mass of the circular disc, $m=10 \,kg$
Radius of the disc, $r=15 \,cm =0.15\, m$
The torsional oscillations of the disc has a time period, $T=1.5\, s$
The moment of inertia of the disc is:
${I}=\frac{1}{2} m r^{2}$
$={2} \times(10) \times(0.15)^{2}$
$=0.1125 \,kg\, m ^{2}$
$T=2 \pi \sqrt{\frac{I}{\alpha}} a$
Time period,
is the torsional constant.
$\alpha=\frac{4 \pi^{2} I}{T^{2}}$
$=\frac{4 \times(\pi)^{2} \times 0.1125}{(1.5)^{2}}$
$=1.972\, Nm / rad$
Hence, the torsional spring constant of the wire is $1.972\, Nm$ $rad$ $^{-1}$.
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