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:

$\frac{181}{2} M R^{2}$

$\frac{19}{2} M R^{2}$

$\frac{55}{2} M R^{2}$

$\frac{73}{2} M R^{2}$

Solution

$\frac{181}{2} M R^{2}$

Using parallel axes theorem, moment of inertia about 'O'

$I_{o} = I_{c m} + m d^{2} = \frac{7 M R^{2}}{2} + 6 (M x (2 R)^{2}) = \frac{55 M R^{2}}{2}$

Again, moment of inertia about point P,

I_p = I_o +md²

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

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