Question

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 \(\alpha \) is defined by the relation \(J=-\alpha \theta\) , where \(J\) is the restoring couple and 𝜃 the angle of twist)

Answer 1

Given, mass of the disc, \(m=10~kg\)

Radius of the circular disc, \(r=15~cm=0.15~m\)

Moment of inertia of the disc, \(I=\dfrac{1}{2}mr^2\)

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

\(I=0.1125~kg~m^2\)

Given, the torsional oscillation of the disc has time period, \(T=1.5~s\)

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

\(\alpha =\dfrac{4\pi ^2I}{T^2}\)

\(\alpha= \dfrac{4\times 3.14\times 3.14\times 0.1125}{1.5\times 1.5}\)

\(\alpha=1.972~Nm/rad\)