Seven identical circular planar disks, each of mass $M$ and radius $R$ are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point $P$ is?

Hint: The moment of inertia of any object about an axis through its center of mass is that the minimum moment of inertia for an axis therein direction in space. So, firstly calculate the moment of inertia at the axis $O$ and then use the parallel axis theorem to calculate at point $P$.

Formula used:
${{I}_{parallel}}={{I}_{cm}}+M{{d}^{2}}$
Where:
${{I}_{cm}}$= moment of inertia about axis O
$M$= Mass of the object
$d$= distance between the main axis and the axis where MOI calculated

Complete step-by-step answer:
At first calculate the moment of inertia about point O and then apply parallel axis formula:
\[{{I}_{o}}={{I}_{cm}}+m{{d}^{2}}\]
$\begin{align}
  & =\dfrac{7}{2}M{{R}^{2}}+6\left( M\times {{\left( 2R \right)}^{2}} \right) \\
 & =\dfrac{55}{2}M{{R}^{2}} \\
 & \\
\end{align}$
\[{{I}_{p}}={{I}_{o}}+m{{d}^{2}}\]

\[\begin{align}
  & =\dfrac{55}{2}M{{R}^{2}}+7M{{\left( 3R \right)}^{2}} \\
 & =\dfrac{181}{2}M{{R}^{2}} \\
\end{align}\]



Additional Information: The expression added to the middle of mass moment of inertia are going to be recognized because the moment of inertia of some extent mass - the instant of inertia a few parallel axes is that the center of mass moment plus the instant of inertia of the whole object treated as some extent mass at the middle of mass. Moment of inertia is defined because the ratio of angular momentum of a system to its angular velocity around an optic axis, that is. If the momentum of a system is constant, then because the moment of inertia gets smaller, the angular velocity must increase.

Note: While calculating the parallel axis moment of inertia you should calculate the moment of inertia from the axis passing through the centre of gravity and then apply the parallel axis theorem, so that there will be no mistake.