The moment of inertia of a circular disc of radius $2m$  and mass  $1kg$ about an axis passing through the centre of mass but perpendicular to the plane of the disc is  $2kg{m}^2$. Its moment of inertia $\left(inkg{m}^2\right)$  about an axis parallel to this axis but passing through the edge of the disc is ______.

Explanation

To calculate the moment of inertia of the disc about an axis passing through the edge, we use the parallel axis theorem.

The parallel axis theorem states:

I = I_cm + Md²

Where:

• I is the moment of inertia about the new axis (through the edge of the disc),
• I_cm is the moment of inertia about the axis through the center of mass,
• M is the mass of the object, and
• d is the distance between the two axes (which in this case is the radius of the disc).

Given:

• I_cm = 2 kg·m²
• M = 1 kg
• d = 2 m (radius of the disc)

Now, applying the parallel axis theorem:

I = I_cm + Md² I = 2 kg·m² + 1 kg × (2 m)² I = 2 kg·m² + 1 kg × 4 m² I = 2 kg·m² + 4 kg·m² I = 6 kg·m²

Final Answer

 The moment of inertia about an axis passing through the edge of the disc is 6 kg·m².