Question

A thin circular plate of mass $M$ and radius $R$ has its density varying as $\rho \left(r\right)={\rho }_{0}r$ 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

The correct option is D $\frac{8}{5}$

$M={\int }_0^R{\rho }_{0}r\left(2\pi rdr\right)=\frac{{\rho }_0\times 2\pi \times R^3}{3}$
Moment of inertia about CoM
$I_0={\int }_0^R{\rho }_{0}r\left(2\pi rdr\right)\times r^2=\frac{{\rho }_0\times 2\pi R^5}{5}$
By parallel axis theorem,
$I=I_0+M{R}^2$
$=\frac{{\rho }_0\times 2\pi R^5}{5}+\frac{{\rho }_0\times 2\pi R^3}{3}\times R^2$
$={\rho }_{0}2\pi R^5\times \frac{8}{15}$
$=M{R}^2\times \frac{8}{5}$.