Moment of Inertia of a Circular Disc
Moment of Inertia of a Circular Disc about the Axis Perpendicular to the Plane and Passing Through the Center of the Disc Diagram shows a circular disc of radius R with center O. If M is the mass of the, disc then the mass of the unit area of the surface of the disc will be represented as (σ).
Fig: Moment of Inertia of a circular disc about the axis perpendicular to the plane and passing through the center of the disc
\(\sigma = \frac{M}{\pi R^2} ......(1)\)
We can assume that the disc is made up of many same centered circular rings. Suppose that x is the radius of one of those circular ring; and its width is dx, then;
Area of the ring = circumference × width
= 2 π x dx
Mass of this ring ⇒ dM = 2 π x dxσ
All the particles of this ring are at the same distance from the center. Hence, the moment of inertia of the ring about the axis perpendicular to the plane and passing through the center is;
dl = mass of the ring × (radius)2
= 2 π x dxσ x2
= 2 π x3 dxσ …………… (2)
Now, moment of inertia of the total disc will be equal to the sum of moment of inertia of all the circular rings which are between x = 0 and x = R. Therefore;
I = \(\int d I=\int_{0}^{R} 2 \pi x^{3} \sigma d x\)
= \(2 \pi \sigma \int_{0}^{R} x^{3} d x\)