Seven identical circular planar disks, each of mass \(M\) and radius \(R\) are welded symmetrically as shown…

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

1. \(\frac{181}{2} M R^{2}\)
2. \(\frac{19}{2} M R^{2}\)
3. \(\frac{55}{2} M R^{2}\)
4. \(\frac{73}{2} M R^{2}\)

Solution

$\begin{aligned} & I_0=I_{cm}+md^2 \\ & =\frac{7 MR^2}{2}+6\left(M \times(2 R)^2\right)=\frac{55 MR^2}{2} \\ & I_{p}=I_{o}+md^2 \\ & =\frac{55 MR^2}{2}+7 M(3 R)^2=\frac{181}{2} MR^2 \end{aligned}$ So, option (1) is the correct option.

Asked in: JEE Mains - Rotational Motion - Chapter Test

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