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.

#### Solution 1

`7/5 MR^2`

The moment of inertia (M.I.) of a sphere about its diameter = `2/5 MR^2`

`ML = 2/5MR^2`

According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

The M.I. about a tangent of the sphere = `2/5 MR^2 + MR^2 = 7/5 MR^2`

#### Solution 2

Moment of inertia of sphere about any diameter = 2/5 MR^{2}

Applying theorem of parallel axes, Moment of inertia of sphere about a tangent to the sphere = 2/5MR^{2} +M(R)^{2} =7/5 MR^{2}