In: Computer Science
explain how to calculate the moment of inertia of a disk, we will take the example of a uniform thin disk which is rotating about an axis through its centre.
As we have a thin disk, the mass is distributed all over the x and y plane. Then, we move on to establishing the relation for surface mass density (σ) where it is defined as or said to be the mass per unit surface area. Since the disk is uniform, therefore, the surface mass density will also be constant where;
σ= m / A
Now it is time for the simplification of the area where it can be assumed the area to be made of a collection of rings that are mostly thin in nature. The thin rings are said to be the mass increment (dm) of radius r which are at equal distance from the axis. The small area (dA) of every ring is further expressed by the length (2πr) times the small width of the rings (dr.) It is given as;
A = πr2, dA = d(πr2) = πdr2 = 2rdr
Now, we add all the rings from a radius range of 0 to R to get the full area of the disk. The radius range that is given is the value that is used in the integration of dr. If we put all these together then we get;
I = ∫r^2σ(πr)dr limit 0-R
I = 2 π σ∫r^3dr
I = 2 πσ r^4 / 4 |. Limit - 0--R
I = 2 πσ (r^4 / 4 – 0)
I = 2 π (m / 4 )(R^4 / 4)
I = 2 π (m / π r^2 )(R4 / 4)
I = ½ mR2.
Momemt of inertia of disc at centre -½ MR^2