   Chapter 15.6, Problem 43E

Chapter
Section
Textbook Problem

Assume that the solid has constant density k.43. Find the moments of inertia for a cube with side length L if one vertex is located at the origin and three edges lie along the coordinate axes.

To determine

To find: The momentum of inertia if the solid has constant density k.

Explanation

Given:

Side length of a cube is L and one of the vertex of a cube located at the origin and the three edges lie along the coordinate axes.

Calculation:

The region E={(x,y,z)|0xL,0yL,0zL}

Initially, calculate the moment of inertia over the x-axis. Therefore, the integration will be over (y2+z2)ρ(x,y,z) but the density is constant.

The moment of inertia over the x-axis is, Ix=E(y2+z2)ρ(x,y,z)dV.

Ix=k0L0L0L(y2+z2)dxdydz

Integrate the above integral with respect to x and apply the limit of it.

Ix=k0L0L[y2x+z2x]0Ldydz=k0L0L[y2(L)+z2(L)y2(0)+z2(0)]dydz=k0L0L[y2(L)+z2(L)00]dydz=kL0L0L[y2+z2]dydz

Integrate the above integral with respect to y and apply the limit of it.

Ix=kL0L0L[y33+z2Y]0Ldz=kL0L[(L33+z2(L))(033+z2(0))]dz=kL0L[(L33+z2(L)00)]dz=kL0L(L33+z2(L))dz

Integrate the above integral with respect to z and apply the limit of it.

Ix=kL[L33z+Lz33]0L=kL[L33(L)+L(L)33](L33(0)+L(0)33)=kL[L44+L4300]=kL[L44+L43]

On further simplification find the value of moment of inertia.

Ix=kL[L44+L43]=2kL53

Thus, the moment of inertia of the cube about the x-axis is 2kL53.

Calculate the moment of inertia over the y-axis. Therefore, the integration will be over (x2+z2)ρ(x,y,z) but the density is constant.

The moment of inertia over the y-axis is, Iy=E(x2+z2)ρ(x,y,z)dV.

Iy=k0L0L0L(x2+z2)dxdydz

Integrate the above integral with respect to x and apply the limit of it.

Iy=0L0L[x33+z2x]0Ldydz=k0L0L[(L33+z2L)(033+z2(0))]dydz=k0L0L[L33+z2L00]dydz=kL0L0L[L33+z2L]dydz

Integrate the above integral with respect to y and apply the limit of it

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