A lamina with constant density ρ(x, y) = ρ occupies the given region. Find the moments of inertia ix and iy and the radii of gyration and . The part of the disk x2 + y2 ≤ a2 in the first quadrant
Answers
m = ∫∫ ρ dA = ρ * (1/4)πa^2 = (1/4)πρa^2.
Moments of inertia:
Ix = ∫∫ ρy^2 dA
....= ∫(θ = 0 to π/2) ∫(r = 0 to a) ρ(r sin θ)^2 * (r dr dθ)
....= ∫(θ = 0 to π/2) sin^2(θ) dθ * ∫(r = 0 to a) ρr^3 dr
....= ∫(θ = 0 to π/2) (1/2)(1 - cos(2θ)) dθ * ∫(r = 0 to a) ρr^3 dr
....= (1/2)(θ - sin(2θ)/2) {for θ = 0 to π/2} * (1/4)ρr^4 {for r = 0 to a}
....= (π/4) * (1/4)ρa^4
....= (1/16)πρa^4.
Iy = ∫∫ ρx^2 dA
....= ∫(θ = 0 to π/2) ∫(r = 0 to a) ρ(r cos θ)^2 * (r dr dθ)
....= ∫(θ = 0 to π/2) cos^2(θ) dθ * ∫(r = 0 to a) ρr^3 dr
....= ∫(θ = 0 to π/2) (1/2)(1 + cos(2θ)) dθ * ∫(r = 0 to a) ρr^3 dr
....= (1/2)(θ + sin(2θ)/2) {for θ = 0 to π/2} * (1/4)ρr^4 {for r = 0 to a}
....= (π/4) * (1/4)ρa^4
....= (1/16)πρa^4.
Hence, the radii of gyration are given by
x^ = Ix / M
= (1/4)a^2 and
y^ = Iy / M
= (1/4)a^2.
HOPE IT HELPS YOU:-))
Denote This is an EDIT of someone else response Credit to (Ruchika08)
m = ∫∫ ρ dA = ρ * (1/4)πa^2 = (1/4)πρa^2.
Moments of inertia:
Ix = ∫∫ ρy^2 dA
....= ∫(θ = 0 to π/2) ∫(r = 0 to a) ρ(r sin θ)^2 * (r dr dθ)
....= ∫(θ = 0 to π/2) sin^2(θ) dθ * ∫(r = 0 to a) ρr^3 dr
....= ∫(θ = 0 to π/2) (1/2)(1 - cos(2θ)) dθ * ∫(r = 0 to a) ρr^3 dr
....= (1/2)(θ - sin(2θ)/2) {for θ = 0 to π/2} * (1/4)ρr^4 {for r = 0 to a}
....= (π/4) * (1/4)ρa^4
....= (1/16)πρa^4.
Iy = ∫∫ ρx^2 dA
....= ∫(θ = 0 to π/2) ∫(r = 0 to a) ρ(r cos θ)^2 * (r dr dθ)
....= ∫(θ = 0 to π/2) cos^2(θ) dθ * ∫(r = 0 to a) ρr^3 dr
....= ∫(θ = 0 to π/2) (1/2)(1 + cos(2θ)) dθ * ∫(r = 0 to a) ρr^3 dr
....= (1/2)(θ + sin(2θ)/2) {for θ = 0 to π/2} * (1/4)ρr^4 {for r = 0 to a}
....= (π/4) * (1/4)ρa^4
....= (1/16)πρa^4.
--------(Revised Section)
Hence, the radii of gyration are given by
x^ = sqrt(Ix / M)
= sqrt((1/4)a^2) and
y^ = sqrt(Iy / M)
= sqrt((1/4)a^2).
--------(Revised Section)
Denote This is an EDIT of someone else response Credit to (Ruchika08)