Question

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 :

• A. $\frac{181}{2}M{R}^2$
• B. $\frac{19}{2}M{R}^2$
• C. $\frac{55}{2}M{R}^2$
• D. $\frac{73}{2}M{R}^2$

Answer 1

The correct option is A $\frac{181}{2}M{R}^2$


Moment of inertia of each disc of radius $R$ and about axis passing through the center and perpendicular to the plane of the disc is $\frac{M{R}^2}{2}$
From parallel axis theorem the moment of inertia of the six discs surrounding the middle disc about O is $\frac{M{R}^2}{2}+M(2R{)}^2=\frac{9M{R}^2}{2}$
Hence the total moment of inertia of the system about O is
$\frac{M{R}^2}{2}+6\times \frac{9M{R}^2}{2}=\frac{55M{R}^2}{2}$
The moment of inertia of the system about point P which is at a distance of $3R$ is
$I_p=I_0+m{d}^2$
$=\frac{55M{R}^2}{2}+7M(3R{)}^2=\frac{181}{2}M{R}^2$