Question 436: Centre of Mass and Rotational Motion

Question

$$\mathrm{I_{CM}}$$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. $$\mathrm{I_{AB}}$$ 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 $$\mathrm{I_{AB}}$$ and $$\mathrm{I_{CM}}$$ is $$x:9$$. The value of $$x$$ is _____________.

Accepted Answer

$$
I_{A B}=I_{\mathrm{cm}}+M 	imes\left(\frac{2}{3} Right)^2
$$
$$
=\frac{1}{2} M R^{2}+\frac{4}{9} M R^{2}
$$
$$
\begin{aligned}
& =\frac{(9+8) M R^{2}}{18}=\left(\frac{17}{18}ight) M R^{2}
\end{aligned}
$$
$\frac{I_{A B}}{I_{\mathrm{cm}}}=\frac{17 / 18}{1 / 2}=\left(\frac{17}{9}ight)$
Value of $x=17$

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type