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

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$(19 / 2) M R^{2}$

$(181 / 2) M R^{2}$

$(73 / 2) M R^{2}$

$(55 / 2) M R^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} I_{0} = I_{c m} + m d^{2} \\ I_{0} = (7 M R^{2} / 2) + 6 (M {(2 R)}^{2}) = 55 M R^{2} / 2 \\ I_{p} = I_{0} + m d^{2} \\ I_{p} = 55 M R^{2} / 2 + 7 M {(3 R)}^{2} = (181 / 2) M R^{2} \end{array}$

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