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Q .14.23 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. (Torsional spring constant a is defined by the relationJ=-a\; \theta, where J is the restoring couple and θ the angle of twist).

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J=-a\; \theta

Moment of Inertia of the disc about the axis passing through its centre and perpendicular to it is


\\J=I\frac{\mathrm{d}^{2}\theta }{\mathrm{d} t^{2}}\\ -a\theta =\frac{MR^{2}}{2}\frac{\mathrm{d}^{2}\theta }{\mathrm{d} t^{2}}\\ \frac{\mathrm{d}^{2}\theta }{\mathrm{d} t^{2}}=-\frac{2a}{MR^{2}}\theta

The period of Torsional oscillations would be

\\T=2\pi \sqrt{\frac{MR^{2}}{2a}}\\ a=\frac{2\pi ^{2}MR^{2}}{T^{2}}\\ a=\frac{2\pi ^{2}\times 10\times (0.15)^{2}}{(1.5)^{2}}\\ a=1.97\ N\ m\ rad^{-1}

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