   Chapter 16.2, Problem 38E

Chapter
Section
Textbook Problem

If a wire with linear density ρ(x, y, z) lies along a space curve C, its moments of inertia about the x-, y-, and z-axes are defined asIx = ∫C (y2 + z2) ρ(x, y, z) dsIy = ∫C (x2 + z2) ρ(x, y, z) dsIz = ∫C (x2 + y2) ρ(x, y, z) dsFind the moments of inertia for the wire in Exercise 35.35. (a) Write the formulas similar to Equations 4 for the center of mass ( x ¯ , y ¯ , z ¯ ) of a thin wire in the shape of a space curve C if the wire has density function ρ(x, y, z).(b) Find the center of mass of a wire in the shape of the helix x = 2 sin t, y = 2 cos t, z = 3t, 0 ⩽ t ⩽ 2π, if the density is a constant k.

To determine

To find: The moment of inertia for a thin wire about the x-, y-, and z-axes.

Explanation

Given data:

The parametric equations of the helix shaped thin wire are given as follows.

x=2sint,y=2cost,z=3t,0t2π

The density function ρ(x,y,z) of the thin wire is a constant k .

Formula used:

Write the expression to find moment of inertia of a thin wire about the x-axis.

Ix=C(y2+z2)ρ(x,y,z)ds (1)

Here,

ρ(x,y,z) is the density function of a thin wire, which is k and

ds is the change in surface.

Write the expression to find moment of inertia of a thin wire about the y-axis.

Iy=C(x2+z2)ρ(x,y,z)ds (2)

Write the expression to find moment of inertia of a thin wire about the z-axis.

Iz=C(x2+y2)ρ(x,y,z)ds (3)

Write the expression for ds .

ds=(dxdt)2+(dydt)2+(dzdt)2dt (4)

Calculation of ds :

Substitute 2sint for x , 2cost for y , and 3t for z in equation (4),

ds=(ddt2sint)2+(ddt2cost)2+(ddt3t)2dt=(2cost)2+(2sint)2+(3)2dt=4cos2t+4sin2t+9dt=4(sin2t+cos2t)+9dt

Simplify the expression as follows.

ds=4(1)+9dt {cos2θ+sin2θ=1}=13dt

The limits of scalar parameter of curve is 0t2π .

To find the moment of inertial use 0 as lower limit and 2π as upper limit.

Calculation of moment of inertia about x-axis Ix :

Substitute 2cost for y , 3t for z , k for ρ(x,y,z) , 13dt for ds , 0 for lower limit, and 2π for upper limit of integral in equation (1),

Ix=02π[(2cost)2+(3t)2](k)(13dt)=13k02π(4cos2t+9t2)dt=13k[02π(4cos2t)dt+02π(9t2)dt]=13k{(2)02π(2cos2t)dt+[9(t33)]02π}

Simplify the expression as follows.

Ix=13k{(2)02π(1+cos2t)dt+[3t3]02π} {2cos2θ=1+cos2θ}=13k{(2)[t+sin2t2]02π+[3(2π)33(0)3]}=13k[(2)[(2π+sin2(2π)2)(0+sin2(0)2)]+(24π30)]=13k[(2)[(2π+0)(0)]+24π3]

Simplify the equation

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