What is the moment of inertia of a homogeneous disk with mass m and radius r for an axis passing through the center of the disk and perpendicular to the disk surface? If the angular velocity of rotation is ω, it is opposite to the direction of rotation axis. Angular momentum l = ω Mr 2/2, also known as "moment of momentum".

You can prove it by definite integral:

Take a ring, the distance from it to the center du of the disc is R to r+dr, and the mass of the ring is m * (2 * pi * r * dr)/(pi * r * r);

The moment of inertia is 2m * r 3/r 2dr.

So the moment of inertia of the disk is the definite integral of 2m * r 3/r 2 r from 0 to r.

∫2M*r^3/R^2dr = 1/2(MRR)

Extended data:

Theorem 1: If f(x) is continuous in the interval [a, b], then f(x) is integrable in [a, b].

Theorem 2: If the interval f(x) is bounded on [a, b] and there are only finite discontinuous points, then f(x) is integrable on [a, b].

Theorem 3: Let f(x) be monotone in the interval [a, b], then f(x) can be integrated in [a, b].

Definite integral and indefinite integral seem to have nothing to do, but they are closely related in essence because of the support of a mathematically important theory. It seems impossible to subdivide a graph infinitely and then accumulate it, but because of this theory, it can be transformed into calculating integral.

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