Chapter 6: Applications of Double Integration
Section 6.6: Second Moments
Example 6.6.5
Find the moments of inertial Ix and Iy, the total mass m, and the radii of gyration Rx and Ry of the lamina whose shape is that of a semicircle with radius 1, and whose density is equal to the distance from the center of the circle. See Example 6.5.5.
Solution
Mathematical Solution
The relevant calculations are in Table 6.6.5(a).
m=∫0π∫01r⋅ρ ⅆr ⅆθ = π/3
Ix=∫0π∫01r⋅ρ⋅r sinθ2 ⅆr ⅆθ = π/10
Iy=∫0π∫01r⋅ρ⋅r cosθ2 ⅆr ⅆθ = π/10
Rx=Iy/m=π/10π/3=3/10
Ry=Ix/m=π/10π/3=3/10
Table 6.6.5(a) Moments of inertia and radii of gyration
Maple Solution - Interactive
A solution from first principles is detailed in Table 6.6.5(b).
Define the density ρ in polar coordinates
Context Panel: Assign Name
ρ=r→assign
Obtain m, the total mass in region R
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
Context Panel: Assign to a Name≻m
∫0π∫01r⋅ρ ⅆr ⅆθ = 13⁢π→assign to a namem
Obtain Ix, the moment of inertia about the x-axis
Context Panel: Assign to a Name≻Ix
∫0π∫01r⋅ρ⋅r sinθ2 ⅆr ⅆθ = 110⁢π→assign to a nameIx
Obtain Iy, the moment of inertia about the y-axis
Context Panel: Assign to a Name≻Iy
∫0π∫01r⋅ρ⋅r cosθ2 ⅆr ⅆθ = 110⁢π→assign to a nameIy
Obtain Rx
Iy/m = 110⁢30
Obtain Ry
Ix/m = 110⁢30
Table 6.6.5(b) Calculation of moments of inertia and radii of gyration from first principles
Maple Solution - Coded
Total mass
ρ≔r:
Use the Int and value commands.
q≔Intr⋅ρ,r=0..1,θ=0..π
∫0π∫01r2ⅆrⅆθ
m≔valueq
13⁢π
Second Moments (Moments of Inertia)
q≔Intr⋅ρ⋅r sinθ2,r=0..1,θ=0.. π
∫0π∫01r4⁢sin⁡θ2ⅆrⅆθ
Ix≔valueq
110⁢π
q≔Intr⋅ρ⋅r cosθ2,r=0..1,θ=0.. π
∫0π∫01r4⁢cos⁡θ2ⅆrⅆθ
Iy≔valueq
Radii of gyration
Rx=Iy/m
Rx=110⁢30
Ry=Ix/m
Ry=110⁢30
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