Question

A thin circular plate of radius $1\text{m}$ has its density varying as $\rho \left(r\right)=8r$ where $r$ is the distance from its geometrical centre. Find the moment of inertia of the circular plate about an axis passing through its edge and perpendicular to the plate.

• A. $\frac{158\pi }{15}{\text{kg-m}}^2$
• B. $\frac{118\pi }{15}{\text{kg-m}}^2$
• C. $\frac{128\pi }{15}{\text{kg-m}}^2$
• D. $\frac{107\pi }{15}{\text{kg-m}}^2$

Answer 1

The correct option is C $\frac{128\pi }{15}{\text{kg-m}}^2$

Given,
Radius of circular plate, $R=1\text{m}$
Density of circular plate, $\rho \left(r\right)=8r$
Let us consider a thin ring of mass $dm$ and of width $dr$, at distance $r$ from the axis perpendicular to the plate and passing through its centre.


Moment of inertia of thin ring
$dI=r^{2}dm$
Integrating on both sides
${\int }_0^{I}dI={\int }_0^1{r}^2.(8r\times 2\pi rdr)$
$[\because \frac{dm}{2\pi rdr}=8r]$
$\Rightarrow I_{O{O}^{\prime }}=\frac{16\pi }{5}[r^5{]}_0^1$
Moment of Inertia about axis passing through COM
$I_{O{O}^{\prime }}=\frac{16\pi }{5}{\text{kg-m}}^2----\left(1\right)$
On applying parallel axis theorem,
$I_{A{A}^{\prime }}=I_{O{O}^{\prime }}+M{R}^2$
$\Rightarrow I_{A{A}^{\prime }}=\frac{16\pi }{5}+\left({\int }_0^{1}16\pi r^{2}dr\right)\times 1^2$
$\left[\text{from}\right(1)\text{and}M=\int dm={\int }_0^{I}16\pi r^{2}dr]\Rightarrow I_{A{A}^{\prime }}=\frac{16\pi }{5}+\frac{16\pi }{3}$
$\Rightarrow I_{A{A}^{\prime }}=\frac{128\pi }{15}{\text{kg-m}}^2$
Hence, the moment of inertia about an axis passing through the edge and perpendicular to the plane of the circular plate is $\frac{128\pi }{15}{\text{kg-m}}^2$