   Chapter 15.4, Problem 24E

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 ¯ ¯ .24. The region under the curve y = sin x from x = 0 to x = π

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 under the curve y=sinx from x=0 to x=π .

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 .

Calculation:

From the given conditions, it is observed that, x varies from 0 to π and y varies from 0 to sinx .

Obtain the moment of inertia Ix .

Ix=Ry2ρ(x,y)dA=0π0sinxy2ρdydx=ρ0π0sinxy2dydx

Integrate with respect to y and apply the limit.

Ix=ρ0π[y33]0sinxdx=ρ0π[(sinx)33(0)33]dx=ρ30π[sin3x0]dx=ρ30π(1cos2x)sinxdx

Integrate with respect to x. For that substitute t=cosx,dt=sinxdx .

Ix=ρ311(1t2)dt=ρ3[tt33]11=ρ3[cosxcos3x3]0π=ρ3[(cosπcos3π3)(cos(0)cos3(0)3)]

On further simplification, the value of Ix becomes,

Ix=ρ3((1)+131+13)=ρ3(2+23)=ρ3(43)=4ρ9

Obtain the moment of inertia Iy .

Iy=Rx2ρ(x,y)dA=0π0sinxx2ρdydx=ρ0π0sinxx2dydx

Integrate with respect to y

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