Question

A thin circular plate of mass $M$ and radius $R$ has its density varying as $\rho (r)={\rho }_{0}r$ with ${\rho }_0$ as constant and $r$ is the distance from its centre. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is $I=aM{R}^2$. The value of the coefficient $a$ is :

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $\frac{3}{5}$
• D. $\frac{8}{5}$

Answer 1

Answer: (D)

$M={\int }_0^R{\rho }_{0}r\left(2\pi rdr\right)=\frac{{\rho }_0\times 2\pi \times R^3}{3}$
$\underset{\text{ (MOI about COM) }}{\mathrm{I}{}_0}=\int {}_0^{\mathrm{R}}\rho {}_0\mathrm{r}(2\pi \mathrm{rdr})\times \mathrm{r}{}^2=\frac{\rho {}_0\times 2\pi \mathrm{R}{}^5}{5}$
by parallel axis theorem
$I=I_0+M{R}^2$
$=\frac{{\rho }_0\times 2\pi R^5}{5}+\frac{{\rho }_0\times 2\pi R^3}{3}\times R^2={\rho }_{0}2\pi R^5\times \frac{8}{15}$
$=M{R}^2\times \frac{8}{5}$