Question

Three rings each of mass $m$ and radius $r$ are so placed that they touch each other. The radius of gyration of the system about the axis as shown in the figure is:

• A. $\sqrt{\frac{5}{3}}r$
• B. $\sqrt{\frac{5}{6}}r$
• C. $\sqrt{\frac{7}{3}}r$
• D. $\sqrt{\frac{7}{6}}r$

Answer 1

The correct option is D $\sqrt{\frac{7}{6}}r$

The moment of inertia of the ring about the center and perpendicular to the axis of the ring is $m{r}^2$. Let the moment of inertia about any diametrical axis be I. So using perpendicular axis theorem we get $2I=m{r}^2$ or $I=\frac{m{r}^2}{2}$.

Thus we get $I_3=\frac{m{r}^2}{2}$

Now moment of inertia of the ring about any tangential axis and in plane of the ring would be given using parallel axis theorem as $\frac{m{r}^2}{2}+m{r}^2=\frac{3}{2}m{r}^2$. Thus we get $I_1=I_2=\frac{3}{2}m{r}^2$.

Thus the total moment of inertia of the given system is given as

$I=I_1+I_2+I_3$

$I_1=I_2=\frac{3}{2}m{r}^2$

$I_3=\frac{m{r}^2}{2}$

$\therefore I=I_1+I_2+I_3=\frac{7}{2}m{r}^2$

Moment of inertia $=$ $3m{k}^2$ where k is radius of gyration.

$3m{k}^2=\frac{7}{2}m{r}^2\Rightarrow k=\sqrt{\frac{7}{6}}r$