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 $I_{A B}$ 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

Correct Answer: 17

1. Moment of inertia about the center of mass: \[ I_{\text{CM}} = \frac{1}{2} mR^2. \]
2. Using the parallel axis theorem: \[ I_{AB} = I_{\text{CM}} + m \left(\frac{2}{3} R \right)^2. \]
Substituting: \[ I_{AB} = \frac{1}{2} mR^2 + m \left(\frac{4}{9} R^2 \right). \]
Simplify: \[ I_{AB} = \frac{1}{2} mR^2 + \frac{4}{9} mR^2 = \frac{9}{18} mR^2 + \frac{8}{18} mR^2 = \frac{17}{18} mR^2. \]
3. Ratio: \[ \frac{I_{AB}}{I_{\text{CM}}} = \frac{\frac{17}{18} mR^2}{\frac{1}{2} mR^2} = \frac{17}{9}. \]
Thus: \[ x = 17. \]
Hence, the value of \(x\) is 17. The moment of inertia about an axis parallel to the center of mass axis can be calculated using the parallel axis theorem.