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 `1.5s`. 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 `theta` the angle of twist).

Explanation

This question is of torsional pendulum. Its time period, `T= 2pi sqrt((I)/(alpha))` where I is the moment of inertia For disc `I = (1)/(2) mr^(2)` where m = mass of disc and r = radius of disc and `alpha =` torsional spring constant.
`T^(2)= (4pi^(2)I)/(alpha)`
`therefore alpha = (4pi^(2)I)/(T^2) = (4pi^(2))/(T^2)xx (1)/(2) mr^(2)`
`= (2pi^(2)mr^(2))/(T^2)`
`=(2xx (3.14)^(2)xx10xx(0.15)^(2))/((1.5)^(2)) = 1.97192`
`therefore approx 1.97 (Nm)/(rad)" or "alpha approx 2.0 (Nm)/(rad)`