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).

Suppose a circular disc is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The restoring force due to wire will cause the disc to return to original position which depends on Torsional spring constant of wire α which is defined by the relation J = –α θ, where J is the restoring couple and θ the angle of twist

The disc will start rotating to and fro around mean position and will undergo simple harmonic motion

**The disc and its orientation has been shown in the figure**

We are given

Period of torsional oscillations is T = 1.5 s

The radius of the disc is R = 15 cm

Mass of the disc is m = 10 Kg

Now Time period of body undergoing such torsional oscillations is given as

Where T is the time Period of oscillations, I is the moment of inertial of the body and α is the torsional spring constant of wire

Now we are given a disc and disc id rotating about an axis passing through its centre and perpendicular to the surface, moment of inertia of a circular disc in this orientation is given by the relation

I = 1/2 mR^{2}

Where I is the moment of inertia of the disc, m is the mass of the disc of radius R

Here mass of the disc is

m = 10 Kg

radius of the disc is

R = 15 = 0.15 m

So moment of Inertia of the disc is

I = 1/2 × 10Kg × (0.15m)^{2}

= 1/2 × 10Kg × 0.0225m^{2}

= 0.1125 Kgm^{2}

So the moment of inertia of the disc is

I = 0.1125 Kgm^{2}

Now rewriting the equation

Squaring both sides we get

Or we get the torsional spring constant of the wire as

putting the values in above equation

i.e. the torsional spring constant of the wire is 1.97 Nm/rad

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