Question

A circular thin disc of mass 5 kg has a diameter 1 m. The moment of inertia about an axis passing through the edge and perpendicular to the plane of the disc is

• A. $\left(8/15\right)kg-m^2$
• B. $8kg-m^2$
• C. $\left(15/8\right)kg-m^2$
• D. $1kg-m^2$

Answer 1

The correct option is C $\left(15/8\right)kg-m^2$

Moment of inertia of a disc passing through the centre perpendicular the plane of the disc is $I=M{R}^2/2$, where M and R are the mass and radius of the disc.
The moment of inertia of the disc that passes through an edge and perpendicular to the plane of the disc can be obtained using Parallel axis theorem and this moment of inertia = $M{R}^2/2+M{R}^2=3M{R}^2/2$
Substituting the values, we get, $I=3\times 5\times {0.5}^2/2=15/8kg-m^2$