Question

$I_{CM}$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. $I_{AB}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3}R$ from center. Where $R$ is the radius of the dise. The ratio of $I_{AB}$ and $I_{CM}$ is $x:9$. The value of $x$ is______

Answer 1

Answer: 17

$I_{cm}=\frac{m{R}^2}{2}$
$I_{AB}=\frac{m{R}^2}{2}+m{\left(\frac{2R}{3}\right)}^2=\frac{17}{18}m{R}^2$
$\frac{I_{AB}}{I_{cm}}=\frac{17}{9}\Rightarrow x=17$