(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be `2(MR^(2))/(5)`, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters 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

(a) The moment of inertia of a sphere about it diameter = `(2)/(5)MR^(2)`. So sphere is solid sphere according to the theorem of parallel axes, the moment of inertia of about a axis to the tangent
`I=I_(C)+Md^(2)`
`=(2)/(5)MR^(2)+MR^(2)[because d=R]`
`=(7)/(5)MR^(2)`

(b) Diameter of disc are AB and CD. The moment of inertia of a disc about centre of mass and its diameter EF.
`I_(C)=(MR^(2))/(4)` hence moment of inertia of disc about AB and `CDtoI_(EF)`
`I_(EF)=I_(AB)+I_(CD)`
`=(MR^(2))/(4)+(MR^(2))/(4)=(MR^(2))/(2)`

Now according to the theorem of parallel axes moment of inertia about axis DG.
`I=I_(EF)+Md^(2)`
`=(MR^(2))/(2)+MR^(2)`
`=(3MR^(2))/(2)`