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

Answer

Correct option: C.
$\frac{{181}}{2}M{R^2}$
c
Using parallel axes theorem, moment of inertia about $'O'$ 

$\begin{array}{l}
{I_0} = {I_{cm}} + m{d^2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{7M{R^2}}}{2} + 6\left( {M \times {{\left( {2R} \right)}^2}} \right) = \frac{{55M{R^2}}}{2}
\end{array}$

Again, moment of inertia about point P,${I_P} = {I_0} + m{d^2}$

$ = \frac{{55M{R^2}}}{2} + 7M{\left( {3R} \right)^2} = \frac{{181}}{2}M{R^2}$

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