Calculate the moment of inertia of a disc about an axis tangent to its inner circle and lying in the plane of the disc. The mass of the disc is $M$,the inner radius is $R_1$,and the outer radius is $R_2$.

  • A
    $\frac{M}{4}(R_1^2 + R_2^2) + MR_1^2$
  • B
    $M(R_1^2 + R_2^2) + MR_1^2$
  • C
    $\frac{M}{4}(R_1^2 - R_2^2) - MR_1^2$
  • D
    $\frac{M}{4}(R_1^2 + R_2^2) - MR_1^2$

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$I_1$ is the moment of inertia of a circular disc about an axis passing through its centre and perpendicular to the plane of the disc. $I_2$ is its moment of inertia about an axis $AB$ perpendicular to the plane and parallel to the axis $CM$ at a distance $\frac{2R}{3}$ from the centre. The ratio of $I_2$ to $I_1$ is $\frac{I_2}{I_1} = \frac{x}{9}$. The value of $x$ is ($R =$ radius of the disc).

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