Question

Two semi-circular discs of mass density $1{\text{kg/m}}^2$ and $2{\text{kg/m}}^2$, radius $r=1\text{m}$ each are joined to form a complete disc. Find the moment of inertia ($\text{MI}$) of complete disc about an axis passing through it's centre.

• A. $\pi {\text{kg-m}}^2$
• B. $2\pi {\text{kg-m}}^2$
• C. $\frac{3\pi }{4}{\text{kg-m}}^2$
• D. $\frac{\pi }{2}{\text{kg-m}}^2$

Answer 1

The correct option is C $\frac{3\pi }{4}{\text{kg-m}}^2$

Let $\text{O}$ be the centre of complete disc, axis of rotation is as shown in figure:

Let the mass density of semi-circular discs be ${\sigma }_1$ & ${\sigma }_2$
${\sigma }_1=1{\text{kg/m}}^2,{\sigma }_2=2{\text{kg/m}}^2$.
$\because \text{mass}=\sigma \times \text{area}$
Mass of the complete disk
$m=(m_1+m_2)={\sigma }_1\left(\frac{\pi R^2}{2}\right)+{\sigma }_2\left(\frac{\pi R^2}{2}\right)$
$m=1\times \left(\frac{\pi (1{)}^2}{2}\right)+2\times \left(\frac{\pi \times (1{)}^2}{2}\right)$
$\therefore m=\frac{3\pi }{2}\text{kg}$
The $\text{MI}$ of complete disc about an axis passing through it's centre and perpendicular to the plane:
$\Rightarrow I=\frac{m{r}^2}{2}$
$I=(\frac{3\pi }{2}\times \frac{1^2}{2})=\frac{3\pi }{4}{\text{kg-m}}^2$
$\therefore I=\frac{3\pi }{4}{\text{kg-m}}^2$