Question

A thin ring has mass 0.25 kg and radius 0.5 m. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is _______.

Options

• 0. 0625 kg m²

• 0.625 kg m²

• 6. 25 kg m²

• 62. 5 kg m²

Answer 1

0.0625 kg m²

 I = mr²
where m is the mass of ring and r is the minimum seperation between axis of rotation and point of observation { e.g., radius of ring }
   here, m = 0.25 Kg 
            r = 0.5 m 
now, I = 0.25 × (0.5)² = 0.25 × 0.25 = 0.0625 Kg.m²