From a circular disc of radius R and mass 9M, a small disc of mass M and radius R/3 is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its center is

Option 1) (40/9) MR2                                                                                                      Option 2) MR2

Option 3) 4 MR2                                                                                            Option 4) \frac{4}{9}\text{MR}^{2}

 

 

Answers (1)

 

Initial moment of inertia of the complete disc about an axis passing through its centre and perpendicular to its plane is


I = \frac{1}{2} mass \times \text{(radius)}^2=\frac{1}{2 }\times 9M\times R^2=\frac{9MR^2}{2}
moment of inertia of disc removed, about an axis passing through its centre and perpendicular to its plane is

 I' = \frac{1}{2} M\left(\frac{R}{3} \right )^2=\frac{MR^2}{18}
moment of inertia of the remaining portion of the disc = I - I' 

\frac{9MR^2}{2}- \frac{MR^2}{18}= \frac{40MR^2}{9}

Hence correct option is 1.

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