Chapter 6: Applications of Double Integration
Section 6.6: Second Moments
Example 6.6.3
Find the moments of inertial Ix and Iy, the total mass m, and the radii of gyration Rx and Ry of the lamina that has the shape of the triangle whose vertices are 0,0,a,0,b,c, where a,b,c are positive, and a>b. (Take ρ=1.) See Example 6.5.3.
Solution
Mathematical Solution
The labeled triangle in Figure 6.6.3(a) is representative, and the equations of the two non-horizontal sides are obtained by an elementary application of the point-slope form of the straight line.
The simplest iteration of the requisite double integrals uses the order dy dx, and requires two separate integrals to cover the whole triangle. (This is true even for iteration in the opposite order.)
The calculations for the moments of inertia and the radii of gyration are summarized in Table 6.6.3(a).
Figure 6.6.3(a) Representative triangle
m=∫0b∫0c x/b1 ⅆy ⅆx+∫ba∫0c x−a/b−a1 ⅆy ⅆx=a c/2
Ix=∫0b∫0c x/by2 ⅆy ⅆx+∫ba∫0c x−a/b−ay2 ⅆy ⅆx=a c3/12
Iy=∫0b∫0c x/bx2 ⅆy ⅆx+∫ba∫0c x−a/b−ax2 ⅆy ⅆx=a c a2+a b+b2/12
Rx=Iym=a c a2+a b+b2/12a c/2 = a2+a b+b26
Ry=Ixm=a c3/12a c/2 = c/6
Table 6.6.3(a) Moments of inertia and radii of gyration
Maple Solution - Interactive
Obtain the equations of the edges of the triangle in Figure 6.6.3(a)
Tools≻Load Package: Student Precalculus
Loading Student:-Precalculus
Context Panel: Student Precalculus≻Lines and Segments≻Line≻Equation
Context Panel: Simplify≻Simplify (if needed)
0,0,b,c→equation of liney=c⁢xb
a,0,b,c→equation of liney=−c⁢xa−b+c⁢aa−b= simplify y=c⁢a−xa−b
Calculations for the moments of inertia and the radii of gyration are given in Table 6.6.3(a).
Total mass
Calculus palette: Iterated double-integral template Press the Enter key.
Context Panel: Simplify≻Simplify
Context Panel: Assign to a Name≻m
∫0b∫0c x/b1 ⅆy ⅆx+∫ba∫0c x−a/b−a1 ⅆy ⅆx
12⁢c⁢b+12⁢c⁢a2−b2b−a−c⁢a⁢a−bb−a
= simplify
12⁢c⁢a
→assign to a name
m
Second Moments (Moments of Inertia)
Context Panel: Assign to a Name≻Ix (or Iy, as appropriate)
∫0b∫0c x/by2 ⅆy ⅆx+∫ba∫0c x−a/b−ay2 ⅆy ⅆx
112⁢c3⁢b+112⁢c3⁢a4−b4b−a3−13⁢c3⁢a⁢a3−b3b−a3+12⁢c3⁢a2⁢a2−b2b−a3−13⁢c3⁢a3⁢a−bb−a3
112⁢c3⁢a
Ix
∫0b∫0c x/bx2 ⅆy ⅆx+∫ba∫0c x−a/b−ax2 ⅆy ⅆx
14⁢c⁢b3+14⁢c⁢a4−b4b−a−13⁢c⁢a⁢a3−b3b−a
112⁢a2+a⁢b+b2⁢c⁢a
Iy
Radii of gyration
Expression palette: Square-root template Press the Enter key.
Rx=Iy/m
Rx=16⁢6⁢a2+6⁢a⁢b+6⁢b2
Context Panel: Simplify≻Assuming Positive
Ry=Ix/m
Ry=16⁢6⁢c2
→assuming positive
Ry=16⁢6⁢c
Table 6.6.3(a) Moments of inertia and radii of gyration for the triangle in Figure 6.6.3(a)
Maple Solution - Coded
Initialize
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Obtain the total area of the triangle
Use the MultiInt command in the Student MultivariateCalculus package, and the simplify command.
MultiInt1,x,y=Triangle0,0,a,0,b,c,output=integral assuming b>0,b<a,c>0;m≔simplifyMultiInt1,x,y=Triangle0,0,a,0,b,c assuming b>0,b<a,c>0
12⁢a⁢c
MultiInty2,x,y=Triangle0,0,a,0,b,c,output=integral assuming b>0,b<a,c>0;Ix≔simplifyMultiInty2,x,y=Triangle0,0,a,0,b,c assuming b>0,b<a,c>0
MultiIntx2,x,y=Triangle0,0,a,0,b,c,output=integral assuming b>0,b<a,c>0;Iy≔simplifyMultiIntx2,x,y=Triangle0,0,a,0,b,c assuming b>0,b<a,c>0
Rx=sqrtIy/m
Ry=simplifysqrtIx/m assuming c>0
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