   Chapter 14.4, Problem 9E

Chapter
Section
Textbook Problem

Finding the Center of Mass In Exercises 7-10, find the mass and center of mass of the lamina corresponding to the region R for each density.R: square with vertices (0, 0) (a, 0) (0, a) (a, a)(a) ρ = k (b) ρ = k y (c) ρ = k x

(a)

To determine

To calculate: The mass and center of mass of the lamina corresponding to region, R: Triangle with vertices (0,0), (a,0), (a,a) and density ρ=k.

Explanation

Given:

The lamina R: Triangle with vertices (0,0), (a,0), (a,a) and density ρ=k.

Formula used:

Mass of planar lamina is:

m=Rρ(x,y)dA

Moment of mass of variable density planar lamina is:

Mx=R(y)p(x,y)dAMy=R(x)p(x,y)dA

The center of mass is:

(x¯,y¯)=(Mym,Mxm)

The integral of xndx=xn+1n+1+c.

Calculation:

The mass of the lamina is:

m=0a0ykdxdy=0a[ky]0ydy=0akydy

On further simplification,

m=[ky22]0a=ka22

To find the center of the mass, find the moment of inertia about both axis

(b)

To determine

To calculate: The mass and center of mass of the lamina corresponding to region, R: Triangle with vertices (0,0), (a,0), (a,a) and density ρ=ky.

(c)

To determine

To calculate: The mass and center of mass of the lamina corresponding to region, R: Triangle with vertices (0,0), (a,0), (a,a) and density ρ=kx.

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