Find the moments of inertia ix, iy, i0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the density at any point is proportional to the square of its distance from the origin. (assume that the coefficient of proportionality is k.)

I(x)  = 1444×k ×

I(y)  = 1444×k ×

I(o) = 3888×k ×  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to for θ

I(x) = y²p dA

I(x) = (a sinθ)²(k × a²) adθda

I(x) = k    da ×    (sin²θ)dθ

I(x) = k    da ×    (1-cos2θ)/2 dθ

I(x)  = k ×  

I(x)  = k × × (  

I(x)  = k ×   ×  

I(x)  = 1444×k ×    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×

I(o) = 3888×k ×   ......................2

your answer is, orbit around the nucleus

490 n

f=125n a=2.5m/s2