Two thin discs each of mass M and radius r metre are attached to form a rigid body as shown in figure. The rotational inertia of this body about an axis perpendicular to the plane of disc B and passing through its centre is

Options

A. `2Mr^(2)`
B. `3Mr^(2)`
C. `4Mr^(2)`
D. `5Mr^(2)`

Answer

D

Explanation

Moment of inertia of discs `A` and `B` about the axis through their cenre of mass and perpendicular to the plane will be
`I_("AA")=I_(BB)=1/2Mr^(2)`
Now moment of inertia of disc `S` about an axis through `B` by theorem of parallel axis will be
`I_("AA")=I_(BB) +M(2r)^(2)=1/2Mr^(2)+4Mr^(2)=9/2Mr^(2)`
So, `I=I_(BB)+I_(AB)=1/2 Mr^(2)+9/2Mr^(2)=5Mr^(2)`