Question

Calculate the moment of inertia of uniform circular disc of mass 500 G radius 10 cm about 

1. The diameter of the disc 

2. The axis, tangent to the disc and parallel to its diameter 

3. The axis through the centre of the disc and perpendicular to its plane

Answer 1

1. Given Data: M = 500 g = 0.5 kg. R = 10 cm = 10 × 10^-2 m 

Moment of inertia of disc about diameter = Id = \(\frac{1}{4}\) MR² 

Id =\(\frac{1}{4}\) × 0.5 × 0.1 kg m² = 0.0125 kg m² 

2. Apply a parallel axes theorem, moment of inertia of the disc about a tangent to the disc and parallel to the diameter of the disc

= \(\frac{1}{4}\) MR² + MR² = \(\frac{5}{4}\) MR² = × 0.5 × 1 

= 0.0625 kgm²

3. Moment of inertia of the disc about an axis passing through the centre of disc and perpendicular to the plane of the disc

= \(\frac{1}{2}\) MR² = \(\frac{1}{2}\) × 0.5 × 0.1 = 0.025 kgm²