Question

The moment of inertia of a uniform circular disc of radius ′R′ and mass ′M′ about an axis passing from the edge of the disc and normal to the disc is

• A. $M{R}^2$
• B. $\frac{M{R}^2}{2}$
• C. $\frac{3M{R}^2}{2}$
• D. $\frac{7M{R}^2}{2}$

Answer 1

The correct option is C

$\frac{3M{R}^2}{2}$

Step 1, Given data and diagram

Mass of the disc = M

Radius of the disc = R

Step 2, Finding the moment of inertia

We know,

The moment of inertia of the uniform circular disc of radius R and mass M about an axis passing through the center of mass of the disc is given by the formula

$I_{COM}=\frac{1}{2}M{R}^2$

Applying the parallel axis theorem for the disc

The disc's moment of inertia at an axis passing through its diameter and perpendicular to the disc's plane is given by

$I=I_{COM}+M{R}^2$

Now putting the value of moment of inertia through the center of mass

$I=\frac{M{R}^2}{2}+M{R}^2=\frac{3M{R}^2}{2}$

Hence, the moment of inertia of the disc about an axis passing from the edge of the disc and normal to it is $\frac{3}{2}M{R}^2$.

The correct option is (C) $\frac{3}{2}M{R}^2$.