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)

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

Choose the correct answer: