Question

Derive the expression for the moment of inertia of a uniform ring about an axis passing through the centre and perpendicular to the plane.

Answer 1

Let us consider a uniform ring of mass M and radius R. To find the moment of inertia of the ring about an axis passing through its center and perpendicular to the plane, let us take an infinitesimally small mass (dm) of length (dx) of the ring. This (dm) is located at a distance R, which is the radius of the ring from the axis as shown in the figure.

The moment of inertia (dl) of this small mass (dm) is,

dI = (dm)R²

The length of the ring is its circumference (2πR). As the mass is uniformly distributed, the mass per unit length (λ) is,

λ = `"mass"/"length" = M/(2piR)`

The mass (dm) of the infinitesimally small length is,
dm = λ dx = `M/(2piR)` dx

Now, the moment of inertia (I) of the entire ring is,

I = `int dI = int (dm) R^2 = int(M/(2piR)dx)R^2`

I = `(MR)/(2pi) int dx`

To cover the entire length of the ring, the limits of integration are taken from 0 to 2πR.

I = `(MR)/(2pi) int_0^{2piR} dx`

I = `(MR)/(2pi) [x]_0^{2piR} = (MR)/(2pi) [2piR - 0]`

I = `MR^2`