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

Answer 1

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

M.O.I about the center of the system

$I_0=\frac{M{R}^2}{2}+6[I_{cm}+{\left(2R\right)}^2]$

$\therefore I_0=\frac{M{R}^2}{2}+6[\frac{M{R}^2}{2}+4M{R}^2]$

$I_0=\frac{M{R}^2}{2}+27M{R}^2=\frac{55}{2}M{R}^2$

for the system of 7M the M.O.I can be calculated by using parallel axis theorem

$I=I_0+7M{\left(3R\right)}^2$

$I=\frac{55}{2}M{R}^2+63M{R}^2=\frac{181}{2}M{R}^2$