Question

(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 2MR2/5, 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 MR2/4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

Answer 1

(a)

Given, the expression for the moment of inertia of a sphere about its diameter is 2 5 M R 2 .

The statement of parallel axes theorem states that the moment of inertia of a body about any given axis is equal to the sum of the product of its mass and the square of the distance between the two parallel axis and moment of inertia of the body about parallel axis passing through its centroid.

Let I be the moment of inertia of the sphere about the tangent, R be the radius of the sphere and I d be the moment of inertia of a sphere about its diameter.

According to parallel axis theorem,

I=M R 2 + I d

Substitute the values in the above expression.

I=M× R 2 + 2 5 M R 2 = 7 5 M R 2

Thus, the value of the moment of inertia about a tangent of the sphere is 7 5 M R 2 .

(b)

Given, the expression for the moment of inertia of a disc about its diameter is 1 4 M R 2 .

Let I c be the moment of inertia of the disc about the centre and I d ′ be the moment of inertia of a disc about its diameter.

According to perpendicular axis theorem,

I c = I d ′ + I d ′

Substitute the values in the above expression.

I c = 1 4 M R 2 + 1 4 M R 2 = 1 2 M R 2

Thus, the value of the moment of inertia about centre of the disc is 1 2 M R 2 .