Rotational Motion Question 5
Question 5 - 25 January - Shift 1
$I _{CM}$ is 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 disc. The ratio of $I _{AB}$ and $I _{CM}$ is $x: 9$. The value of $x$ is
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Answer: (17)
Solution:
Formula: Moment of inertia
$I _{cm}=\frac{mR^{2}}{2}$
$I _{AB}=\frac{mR^{2}}{2}+m(\frac{2 R}{3})^{2}=\frac{17}{18} mR^{2}$
$\frac{I _{A B}}{I _{c m}}=\frac{17}{9} \Rightarrow x=17$
