   Chapter 15.4, Problem 19E

Chapter
Section
Textbook Problem

Find the moments of inertia Ix, Iy, lo for the lamina of Exercise 15.

To determine

To find: The moments of inertia Ix,Iy,I0 .

Explanation

Formula used:

The moments of inertia is,

Ix=limm,ni=1mj=1n(yij*)2ρ(xij*,yij*)ΔA=Dy2ρ(x,y)dAIy=limm,ni=1mj=1n(xij*)2ρ(xij*,yij*)ΔA=Dx2ρ(x,y)dAI0=Ix+Iy

Here, the density function is given by ρ(x,y) and D is the region that is occupied by the lamina.

Given:

The region D is the isosceles triangle with equal sides of length a.

The density function is proportional to the square of the distance from the vertex opposite to the hypotenuse, that is ρ(x,y)=k(x2+y2) .

Calculation:

From the given conditions it is observed that x varies from 0 to a and y varies from 0 to ax . Then, the moments of inertia Ix is,

Ix=Dy2ρ(x,y)dA=0a0axky2(x2+y2)dydx=k0a0ax(x2y2+y4)dydx .

Integrate with respect to y and apply it’s limit.

Ix=k0a(y3x23+y55)0axdx=k0a[((ax)3x23+(ax)55)((0)3x23+(0)55)]dx=k0a[(a333a2x3+3ax23x33)x2+(ax)55]dx=k0a(a3x233a2x33+3ax43x53+(ax)55)dx

Integrate with respect to x and apply it’s limit.

Ix=k[a3x33(3)3a2x43(4)+3ax53(5)x63(6)+(ax)65(6)]0a=k[(a3(a)393a2(a)412+3a(a)515(a)618(aa)630)(a3(0)393a2(0)412+3a(0)515(0)618(a0)630)]=k[(a69a64+a65a618030)(0904+05018a630)]=k[a69a64+a65a618+a630]

Simplify it further,

Ix=k[20a645a6+36a610a6+6a6180]=k[7a6180]=7ka6180

The moments of inertia Iy is,

Iy=Dx2ρ(x,y)dA=0a0axkx2(x2+y2)dydx=k0a0ax(x2y2+x4)dydx

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