   Chapter 15.4, Problem 23E

Chapter
Section
Textbook Problem

A lamina with constant density ρ (x, y) = ρ occupies the given region. Find the moments of inertia lx and Iy and the radii of gyration x and y.23. The part of the disk x2 + y2 ≤ a2 in the first quadrant

To determine

To find: The moments of inertia about x and y axes Ix,Iy and radii of gyration x¯¯,y¯¯ .

Explanation

Given:

The density function, ρ(x,y)=ρ .

The region D is disk x2+y2a2 in the first quadrant.

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)dA

The total mass of the lamina is, m=limk,li=1kj=1lρ(xij*,yij*)ΔA=Dρ(x,y)dA .

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

The radii of gyration about y and x axis is respectively my¯¯2=Ix,mx¯¯2=Iy .

If f is a polar rectangle R given by 0arb,αθβ, where 0βα2π , then, Rf(x,y)dA=αβabf(rcosθ,rsinθ)rdrdθ (1)

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (2)

Calculation:

Change the given problem into polar coordinates to make it easier by substituting x=rcosθ,y=sinθ . From the given conditions, it is observed that, r varies from 0 to a and θ varies from 0 to π2 .

Obtain the moments of inertia Ix by equation (1).

Ix=Ry2ρ(x,y)dA=0a0π2y2ρdydx=ρ0π20a(rsinθ)2(r)drdθ=ρ0π20ar3sin2θdrdθ

Integrate with respect to r and θ using the equation (2).

Ix=ρ0ar3dr0π2(1cos2θ)2dθ=ρ2[r44]0a[θsin2θ2]0π2=ρ2[(a)44(0)44][(π2sin(2π2)2)((0)sin(2(0))2)]=ρ2(a44)[(π2sinπ2)(00)]

On further simplification, the value of Ix becomes,

Ix=a4ρ8(π202)=116a4πρ

Obtain the moments of inertia Iy

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