EXAMPLE 4 Find the moments of inertia Ix, Iy, and I0 of a homogeneous disk D with density rho(x, y) = rho, center the origin, an
d radius a. SOLUTION The boundary of D is the circle x2 + y2 = a2 and in polar coordinates D is described by 0 ≤ θ ≤ 2π, 0 ≤ r ≤ a. Let's compute I0 first: I0 = D (x2 + y2)rho dA = rho 2π 0 a 0 r2 r dr dθ = rho 2π 0 dθ a 0 r3 dr = 2πrho a 0 = . Instead of computing Ix and Iy directly, we use the facts that Ix + Iy = I0 and Ix = Iy (from the symmetry of the problem). Thus Ix = Iy = I0 2 = .
Moment of inertia refers to a specific rotation axis, when there’s is a moment of inertia of a point mass with respect to an axis it means the creation of the mass that is multiplied by the distance from the axis squared. The moment of inertia of whichever extended object is a build up from that fundamental definition.
Kindly check the attached image below for the step by step explanation to the question above.
It goes like this 1/2, 1/4, 1/8. But if your asking about what fraction didn’t get affected than that just zero because the original peace was cut in half.
If that’s not what you mean can you please elaborate to me!