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 `theta` the angle of twist).

Explanation

Mass of the circular disc,m=10kg
Radius of the disc,r=15cm=0.15m
The torsional oscillations of the disc has a time period, T=1.5s
The moment of inertia of the disc is: `I=(1)/(2)mr^(2)=(1)/(2)xx(10)xx(0.15)^(2)=0.1125kgm^(2)`
`T=2pisqrt((I)/(a))` Time period ,a is the torsional constant
`a=(4pi^(2)//)/(T^(2))=(4xx(pi)^(2)xx0.1125)/((1.5)^(2))=1.972Nm//rad` Hence, the torsional spring constant of the wire is 1.972 Nm`rad^(-1)`