A circular disc of mass 10kg is suspended by a wire attahced 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 15cm. Determing the torsional spring constant of the wire. (Torsional spring constant `alpha` is definied by the relation `J=-alphatheta`, where J is the restoring coubple and `theta` the angle of twist.

Explanation

Here, `m=10kg, R=15cm=0.15m, T=1.5s, alpha=?`
Moment of inertia of disc, `I=(1)/(2)mR^(2)=(1)/(2)xx10xx(0.5)^(2)kgm^(2)`
Given, restoring couple, `J=-alphatheta or Jprop theta` and this couple tends to bring the disc back to its equilibrium position. Therefore if the disc is left free after twist, it will execute angular SHM with equilibrium position as mean position. Then time period is give by
`T=2pisqrt((I)/(alpha))` So, `alpha=(4pi^(2)I)/(T^(2))=4xx((22)/(7))^(2)xx(1)/(2)xx(10xx(0.15)^(2))/((1.5)^(2))=1.97Nm//rad`