Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In Exercises 61–64, let R be the rectangular solid defined by R=(x,y,z)|0a1xa2,0b1yb2,0c2zc2.

Assuming that the density at each point in R is proportional to the distance of the point from the yz-plane, find the first moment of inertia about the x-axis and the radius of gyration about the x-axis.

Short Answer

Expert verified

The first moment of inertia about the x-axis is Ix=k(a2a1)(b2b1)(c2c1)3(b22+b12+b1b2+c22+c12+c1c2) and the radius of gyration about the x-axis islocalid="1650373580986" Rx=(b22+b12+b1b2+c22+c12+c1c2)3.

Step by step solution

01

Step 1. Given Information.

The given rectangular solid is defined byR=(x,y,z)|0a1xa2,0b1yb2,0c2zc2.

02

Step 2. Find the first moment of inertia about the x-axis.

It is given that the density at each point in R is proportional to the distance of the point from the yz-plane, so ρ(x,y,z)=kx.

Now, the moment of the inertia about the x-axis is,

Ix=T(y2+z2)ρ(x,y,z)dVIx=x=a1a2y=b1b2z=c1c2(y2+z2)(kx)dxdydzIx=kx22x=a1a2×z=c1c2y33b1b2+x=a1a2z2(y)b1b2dz

Integrate with respect to 'y'

Ix=x=a1a2(c2c1)y33y=b1b2dx+x=a1a213(c23c13)yy=b1b2dxIx=(c2c1)(b23b13)3(x)|x=a1a2+(c23c13)(b2b1)3(x)|x=a1a2Ix=(c2c1)(b23b13)(a2a1)3+(c23c13)(b2b1)(a2a1)3Ix=k(a2a1)(b2b1)(c2c1)3(b22+b12+b1b2+c22+c12+c1c2)

03

Step 3. Find the radius of the gyration.  

To find the radius of gyration about the x-axis, we have to find the mass of the rectangular solid:

M=Tρ(x,y,z)dxdydzM=x=a1a2y=b1b2z=c1c2(k)dxdydzM==k(a2a1)(b2b1)(c2c1)

So, the radius of gyration is:

Rx=IxmRx=k(a2a1)(b2b1)(c2c1)3(b22+b12+b1b2+c22+c12+c1c2)k(a2a1)(b2b1)(c2c1)Rx=(b22+b12+b1b2+c22+c12+c1c2)3

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free