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 0.15m. Determine the torsional spring constant of the wire. (Note : Torsional spring constant `'alpha'` is defined by the relation `J=-alpha theta` where J is the restoring couple is torque).

Explanation

`T=2pisqrt((I)/(alpha))or alpha=(4pi^(2)I)/(T^(2))`
`"i.e. "alpha=(4pi^(2)((MR^(2))/(2)))/(T^(2))=(2pi^(2)MR^(2))/(T^(2))" "therefore alpha=(2xx(3.142)^(2)xx10xx(0.15)^(2))/((1.5)^(2))=(4.4425)/(2.25)`
`alpha="1.974 Nm rad"^(-1)`