Question

Three identical discs $(D_1,D_2,D_3)$ are arranged as shown in figure. The moment of inertia of the system about the axis $O{O}^{\prime }$ is given by $Nm{r}^2$. Then, the value of $N$ is

Answer 1

Disc $D_1$ and $D_3$ are in $YZ$ plane and $D_2$ in $XZ$ plane.
Disc $D_1$ and $D_3$ are symmetrical about axis $Y{Y}^{\prime },Z{Z}^{\prime }$ i.e diametrical axes
Applying perpendicular axis theorem for $D_1$ and $D_3$,
$I_{X{X}^{\prime }}=I_{Z{Z}^{\prime }}+I_{Y{Y}^{\prime }}$
$I_{X{X}^{\prime }}=\frac{m{r}^2}{2}$, moment of inertia about axis passing through centre and $\perp$ to its $YZ$ plane
$\Rightarrow \frac{m{r}^2}{2}=2I_{Y{Y}^{\prime }}$
[$\because I_{Y{Y}^{\prime }}=I_{Z{Z}^{\prime }}$, due to symmetry]
$\Rightarrow I_{Y{Y}^{\prime }}=\frac{m{r}^2}{4}...\left(1\right)$
Also from parallel axis theorem,
$I_{O{O}^{\prime }}=I_{Y{Y}^{\prime }}+m{d}^2$
[$\because d=r$, from figure]
Substituting from Eq.$\left(1\right)$,
$\Rightarrow I_{O{O}^{\prime }}=\frac{m{r}^2}{4}+m{r}^2$
$\Rightarrow I_{O{O}^{\prime }}=\frac{5m{r}^2}{4}...\left(2\right)$

Since $O{O}^{\prime }$ is the axis passing through centre of disc $D_2$ and $\perp$ to its $XZ$ plane
$\therefore I_{D_2}=\frac{m{r}^2}{2}$
and $I_{D_1}=I_{D_3}=I_{O{O}^{\prime }}$ due to symmetry of $D_1\&D_3$ about $O{O}^{\prime }$

Now, Moment of inertia of system of discs about axis $O{O}^{\prime }$ is sum of $\text{MOI}$ of all the discs about axis $O{O}^{\prime }$.
$\Rightarrow I=I_{D_1}+I_{D_2}+I_{D_3}...\left(3\right)$
Substituting in Eq.$\left(3\right)$, we get,
$\Rightarrow I=\frac{5m{r}^2}{4}+\frac{m{r}^2}{2}+\frac{5m{r}^2}{4}$
$\Rightarrow I=\frac{12m{r}^2}{4}=3m{r}^2$
Comparing with given ,$I=Nm{r}^2$
$\Rightarrow N=3$
The value of $N$ is $3$.