Chapter 6: Applications of Double Integration
Section 6.6: Second Moments
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 an isosceles right triangle with legs of length 1, and has density that is twice the square of the distance from the vertex of the right angle. Hint: Put the right angle at the origin. See Example 6.5.8.
The relevant calculations are in Table 6.5.8(a).
m=∫01∫01−xρ ⅆy ⅆx = 1/3
Ix=∫01∫01−xy2⋅ρ ⅆy ⅆx = 7/90
Iy=∫01∫01−xx2⋅ρ ⅆy ⅆx = 7/90
Rx=Iy/m=7/901/3 = 21030 ≐ 0.48
Ry=Ix/m=7/901/3 = 21030 ≐ 0.48
Table 6.6.8(a) Moments of inertia and radii of gyration
Maple Solution - Interactive
Obtain the equation of the hypotenuse
Tools≻Load Package: Student Precalculus
Context Panel: Student Precalculus≻
Lines And Segments≻Line≻Equation
0,1,1,0→equation of liney=−x+1
A solution from first principles is detailed in Table 6.5.8(b).
Define the density function ρx,y
Context Panel: Assign Name
Obtain m, the total mass of the lamina
Iterated double-integral template
Context Panel: Evaluate and Display Inline
Context Panel: Assign to a Name≻m
∫01∫01−xρ ⅆy ⅆx = 13→assign to a namem
Obtain Ix, the moment of inertia about the x-axis
Context Panel: Assign to a Name≻Ix
∫01∫01−xρ⋅y2 ⅆy ⅆx = 790→assign to a nameIx
Obtain Iy, the moment of inertia about the y-axis
Context Panel: Assign to a Name≻Iy
∫01∫01−xρ⋅x2 ⅆy ⅆx = 790→assign to a nameIy
Context Panel: Approximate≻5 (digits)
Iy/m = 130⁢210→at 5 digits0.48303
Ix/m = 130⁢210→at 5 digits0.48303
Table 6.5.8(b) Moments of inertia and radii of gyration from first principles
Maple Solution - Coded
Define the density function ρx,y.
Obtain the equation of the hypotenuse of the right triangle
Obtain the total mass of the lamina
Display the unevaluated integral with the Int command, and evaluate the integral with the value command.
Obtain the moments of inertia Ix and Iy
Obtain the radii of gyration
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