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 :

Options

A. `(73)/(2)MR^(2)`
B. `(181)/(2)MR^(2)`
C. `(19)/(2)MR^(2)`
D. `(55)/(2)MR^(2)`

Answer

B

Explanation

Let`I_(0)` be the moment of system about point O.
`I_(0)=(1)/(2)MR^(2)+(6I_(cm)^("disc")+M(2R)^(2))`
`=(1)/(2)MR^(2)+((MR^(2))/(2)+M(2R)^(2))`
If `I_(0)` is moment of inertia of system about point O, then we can use parallel axis theorem to write moment of inertia about point P as follows:
`I_(P)=I_(0)+7M(3R)^(2)`
`=[(1)/(2)MR^(2)+6((1)/(2)MR^(2)+M(2R)^(2))]+7M(3R)^(2)`
`=(181)/(2)MR^(2)`.
Hence option (b) is correct.