(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.

(a) Find the moment of inertia of a sphere about a tangent to the sphe

1.	(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.
Answer» Solution :(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)`