Question

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.)

Answer 1

Answer:

I(x)  = 1444×k ×${\pi}$

I(y)  = 1444×k ×${\pi}$

I(o) = 3888×k ×${\pi}$  

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 $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k  $\int_{0}^{6}a^(5)$  da ×  $\int_{0}^{\pi/2}$  (sin²θ)dθ

I(x) = k  $\int_{0}^{6}a^(5)$  da ×  $\int_{0}^{\pi/2}$  (1-cos2θ)/2 dθ

I(x)  = k $({r}^{6}/6)^(5)_0$ ×  ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x)  = k × $({6}^{6}/6)$ × (  ${\pi /4} - sin\pi /4)$

I(x)  = k ×  $({6}^{5})$ ×   ${\pi /4}$

I(x)  = 1444×k ×${\pi}$    .....................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 ×${\pi}$

I(o) = 3888×k ×${\pi}$   ......................2