Question

Three identical circular planner disc each of mass $m$ and radius $r$ are welded as shown in figure. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point $P$ is

• A. $9m{r}^2$
• B. $\frac{1}{2}m{r}^2$
• C. $\frac{9}{2}m{r}^2$
• D. $\frac{19}{2}m{r}^2$

Answer 1

The correct option is D $\frac{19}{2}m{r}^2$

Given:
All disc are identical having mass $m$ and radius $r$


Moment of inertia of disc $1$ about the axis $I$ is $I_1$.
$I_1=\frac{m{r}^2}{2}$
moment of inertia of disc $2$ about the axis $I$ is $I_2$
Using parallel axis theorem
$I_2=I_{com}+m{d}^2$ [$\because$here, $d=2r$]
$=\frac{1}{2}m{r}^2+m(2r{)}^2$
$=\frac{9}{2}m{r}^2$

Similarly,
Moment of inertia of disc $3$ about the axis $I$ is $I_3$.
Using parallel axis theorem
$I_3=I_{com}+m{d}^2$ [$\because$here, d = 2r]
$=\frac{1}{2}m{r}^2+m(2r{)}^2$
$=\frac{1}{2}m{r}^2+m(2r{)}^2$
$=\frac{9}{2}m{r}^2$

$\because$ Moment of inertia of the system about $I$ is equal to $I_1+I_2+I_3=\frac{19m{r}^2}{2}$