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 α is defined by the relation J = –α θ, where J is the restoring couple and θ the angle of twist).

Asked by Abhisek | 1 year ago | 155

Mass of the circular disc = 10 kg

Period of torsional oscillation = 1.5 s

Radius of the disc = 15 cm = 0.15 m

Restoring couple, J = –α θ

Moment of inertia, I = (\(
\dfrac{1}{2}\)) mR^{2}

= (\(
\dfrac{1}{2}\)) x 10 x (0.15)^{2}

= 0.1125 kgm^{2}

Time period is given by the relation

T = \( 2\pi \sqrt{\dfrac{I}{\alpha }}\)

So,\( \alpha =\dfrac{4\pi^{2}I }{T^{2}}α= \dfrac{4\times (3.14)^{2}\times 0.1125}{(1.5)^{2}}\)

= \( \dfrac{4.44}{2.25}\)

= 1.97 Nm/rad

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