Find the moment of inertia of a uniform circular disc about an axis passing through its geometrical centre and perpendicular to its plane and radius of gyration.

Explanation

As shown in fig. consider a uniform circular disc with mass M and radius R. For the moment of inertia of this disc about the axis ZZ. passing through its geometrical centre, area of the disc `A=pir^(2)` and mass per unit area of the disc.

`therefore sigma=("Mass of the disc")/("Area of the disc")`
`(M)/(piR^(2))`
Let imagine this disc consisting of serveral concentric rings with difference radii and their centre is O.
Consider one of such rings with radius x and width dx as shown in the fig.
Area of this ring `S=2pix.dx` and
`therefore` mass `m=lambdaS`
`=(M)/(piR^(2))xx2pixdx`
`therefore m=(2Mxdx)/(R^(2))" "...(1)`
If dI is the moment of inertia of this ring about the axis ZZ. then
`dI=mr^(2)=(2Mx^(3))/(R^(2))dx" "....(2)`
The total of the moment of inertia of such concentric rings with different radius gives the moment of inertia of the disc as a whole about the ZZ. axis.
For this integrating eqn. (1) in the interval `x=0` to `x=R`,
`I=intdl=underset(0)overset(R)int(2Mx^(3))/(R^(2))dx`
`=(2M)/(R^(2))underset(0)overset(R)intx^(3)dx=(2M)/(R^(2))[(x^(4))/(4)]_(0)^(R)`
`=(2M)/(R^(2))[(R^(4))/(4)-0]`
`I=(MR^(2))/(2)`
Now comparing the eqn. (3) with `I=MK^(2)`
`k^(2)=(R^(2))/(2)`
`therefore k=(R)/(sqrt(2))` is a radius of gyration