Question

A circular disc of radius $1\text{m}$ is rotating about an axis passing through its $COM$ and perpendicular to its plane. The moment of inertia about this axis is given by $I_{com}={\text{2 kg-m}}^2$. If the disc is rotated about an axis along its diameter, then the moment of inertia along that axis is $I_{dia}$. The value of $(I_{com}-I_{dia})$ is equal to

• A. ${\text{2 kg-m}}^2$
• B. ${\text{1 kg-m}}^2$
• C. ${\text{3 kg-m}}^2$
• D. ${\text{0.5 kg-m}}^2$

Answer 1

The correct option is B ${\text{1 kg-m}}^2$

Given, radius of disc $\left(r\right)$ = $1\text{m}$
Moment of inertia about axis passing through COM and perpendicular to the plane of the disc, $I_{com}={\text{2 kg-m}}^2$
From definition, $I_{com}=\frac{M{r}^2}{2}$
We get,

$M=\frac{2I_{com}}{r^2}=\frac{2\times 2}{(1{)}^2}=\text{4 kg}$
Therefore , $I_{dia}=\frac{M{r}^2}{4}=\frac{4\times (1{)}^2}{4}$
${\text{= 1 kg-m}}^2$

Hence, $I_{com}-I_{dia}=2-1=1{\text{kg-m}}^2$

Method 2:
$I_{dia}=\frac{I_{com}}{2}$ (from Perpendicular Axis Theorem)
$\therefore I_{com}-I_{dia}=\frac{I_{com}}{2}=1{\text{kg-m}}^2$