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  • 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.

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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best_answer

J=-a\; \theta

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

I=\frac{MR^{2}}{2}

\\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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