Question

The moment of inertia of a thin uniform circular disc about one of its diameter is $l$. Its moment of inertia about an axis tangent to it and perpendicular to its plane is

• A. $\frac{2l}{3}$
• B. $2l$
• C. $\frac{l}{2}$
• D. $6l$

Answer 1

The correct option is D $6l$

Let mass of circular disc is $M$ and radius $R$.

Given, Moment of Inertia (MOI) about one of its diameter$=l$, $\Rightarrow \frac{M{R}^2}{4}=l$ $\Rightarrow M{R}^2=4l$ .....1

MOI about an axis passing through the centre of the disc$=\frac{M{R}^2}{2}$

Using parallel theorem of MOI, $I=I_{cm}+m{x}^2$ , Where $I_{cm}$ is MOI of centre of mass and $x$ is the distance between centre of mass and axis of rotation.

MOI about an axis tangent to disc and perpendicular to its plane$=\frac{M{R}^2}{2}+M{R}^2=\frac{3}{2}M{R}^2=6l$ (Using equation 1, $M{R}^2=4l$)