Question

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 α is defined by the relation J = –α θ, where J is the restoring couple and θ the angle of twist).

Answer 1

Mass of the disc = m = 10 kg
radius of the disc (r) = 15 cm = 0.15 m
T = 1.5 sec

I is the moment of inertia of disc about the axis of rotation .
I = 1/2mr²
= 1/2 × 10 × (0.15)²
= 0.01125 kgm²
Now,
Time period (T) = 2π√(I/a )
a = 4π²I/T²
= 4 × (3.14)¹ × 0.1125/(1.5)2
= 1.972 Nm/rad