Given the moment of inertia of a disc of radius R mass M about an axis along its diameter to be `(MR^(2))//4` find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Explanation

Moment of inertia of the disc about the x-axis lying along one of its diameters is `I_(x)=(1)/(4) MR^(2)` (given).
From symmetry,
`I_(y)=I_(x) =(1)/(4) MR^(2)`
From the perpendicular -axes theorem the moment of inertia about the z -axis passing through the centre of the disc and normal to it is
`I_(z)=I_(x)+I_(y)=2I_(x)=2xx(1)/(4)MR^(2)=(1)/("2)MR^(2)`
So, the moment of inertia about an axis ( denoted by z. in the Fig ) Parallel to the z-axis and passing through a point on the edge of the disc, using the parallel -axes theorem is
`I=I_(cm)+MR^(2)=I_(z)+MR^(2)=(1)/(2)MR^(2)+MR^(2) =(3)/(2)MR^(2)`