Question

A disc of mass $M$ and radius $R$ is attached to a rectangular plate of the same mass $M$, breadth $R$ and length $2R$ as shown in figure. The moment of inertia of the system about the axis $AB$ passing through the centre of the disc and on the plane is $I=\frac{1}{\alpha }\left(\frac{31}{3}M{R}^2\right)$. Then, the value of $\alpha$ is

Answer 1

$I_{AB}=I_{\text{disc}}+I_{\text{plate}}$
Here,
$I_{AB}=$ Moment of inertia of the system about axis $AB$
$I_{\text{disc}}=$ Moment of inertia of the disc about axis $AB$
$I_{\text{plate}}=$ Moment of inertia of the plate about axis $AB$


Thus,
$I_{AB}=I_{\text{disc}}+(I_{CD}+M{\left(\frac{3R}{2}\right)}^2)$
(From Parallel Axis Theorem )
As we know,
MOI of disc about $AB=I_{\text{disc}}=\frac{M{R}^2}{4}$
MOI of plate about $CD=I_{\text{CD}}=\frac{M{R}^2}{12}$
$\therefore I_{AB}=\frac{M{R}^2}{4}+\frac{M{R}^2}{12}+M{\left(\frac{3R}{2}\right)}^2$
or $I_{AB}=\frac{31}{12}M{R}^2$
So, $\alpha =4$