Question 556: Centre of Mass and Rotational Motion

Question

$I_1$ is the moment of inertia of a circular disc about an axis passing through its centre and perpendicular to the plane of disc. $I_2$ is its moment of inertia about an axis $A B$ perpendicular to plane and parallel to axis $\mathrm{CM}$ at a distance $\frac{2 R}{3}$ from centre. The ratio of $I_1$ and $I_2$ is $x: 17$. The value of '$x$' is (R = radius of the disc)
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Accepted Answer

Using Parallel axis theorem, $\mathrm{I}_2=\mathrm{I}_1+\mathrm{Mh}^2$
For a disc, $I_1=\frac{1}{2} M R^2$ and given that, $h=\frac{2 R}{3}$
$\begin{aligned}
& 	herefore \quad \mathrm{I}_2=\frac{1}{2} \mathrm{MR}^2+\mathrm{M}\left(\frac{4 \mathrm{R}^2}{9}ight) \
& 	herefore \quad \mathrm{I}_2=\frac{1}{2} \mathrm{MR}^2\left(1+\frac{8}{9}ight)=\mathrm{I}_1 	imes \frac{17}{9} \
& 	herefore \quad \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{9}{17} \
\end{aligned}$

Answer

9

Answer

12

Answer

15

Answer

17

Centre of Mass and Rotational Motion

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