Question

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.

Answer 1

`T=2pisqrt(I/(alpha))` or `T^(2)=(4pi^(2)I)/(alpha)`
or `alpha=(4pi^(2)I)/(T^(2))` or `alpha=(4pi^(2))/(T^(2))(1/2 MR^(2))`
or `alpha=(2pi^(2) MR^(2))/(T^(2))`
or `alpha=(2(3.14)^(2)xx10xx(0.15)^(2))/((1.5)^(2)) Nm rad^(-1)`
`=1.97 Nm rad^(-1)` .