Question

A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15cm. Determine the torsional spring constant of the wire(Torsional spring constant α is defined by the relation J = -αθ, where J is the restoring couple and θ is the angle of the twist).

Answer 1

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. $T=2\pi\sqrt{\frac{I}{\alpha}}$

Where I is moment of inertia and α is torisonal constant.

we know, moment of inertia of disc about an axis normal to the plane and passing through centre of it is given by, I = 1/2 mr²

I = 1/2 × 10 × (0.15)²
= 5 × 225 × 10^-4
= 1125 × 10^-4
= 0.1125 kgm²

Now, 1.5 = 2 × 3.14 √{0.1125/α}

or, 2.25 = 4 × 3.14 × 3.14 × 0.1125/α

or, α = 4 × 3.14 × 3.14 × 0.1125/2.25

Hence, α = 1.97 Nm/rad

Answer 2

Answer:

an expression for the variation of acceleration due to gravity (a) above and (b) below the surfac