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 alphais defined by the relation J=-alpha theta , where J is the restoring couple and theta the angle of twist).

A circular disc of mass 10 kg is suspended by a wire attached to its c

1.	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 alphais defined by the relation J=-alpha theta , where J is the restoring couple and theta the angle of twist).
Answer» Solution :Here, m = 10 kg , R=15 cm = 0.15 cm , T=1.5 s Moment of inertia of DISC , `I = 1/2 mR^2 = 1/2 xx 10 xx (0.15)^2 kgm^2` Now , `T= 2pi SQRT(T/(alpha))` So, `alpha = (4pi^2 I)/(T^2) = 4 xx ((22)/(7))^2 xx 1/2 xx (10xx (0.15)^2)/((1.5)^2) = 1.97 N` m/rad