   Chapter 16.2, Problem 33E

Chapter
Section
Textbook Problem

A thin wire is bent into the shape of a semicircle x2 + y2 = 4, x ⩾ 0. If the linear density is a constant k, find the mass and center of mass of the wire.

To determine

To find: The mass of a thin wire which bent into the shape of a semicircle x2+y2=4,x0 and center of mass of a thin wire which bent into the shape of a semicircle x2+y2=4,x0 .

Explanation

Given data:

The mass of a thin wire which bent into the shape of a semicircle x2+y2=4,x0 .

The linear density is a constant k .

Formula used:

Write the expression to find the mass of a thin wire which is in the shape of a curve C .

m=Cρ(x,y)ds (1)

Here,

ρ(x,y) is the linear density and

ds is the change in surface.

Write the expression to find ds .

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

Write the expression to find the x-coordinate of a center of mass of the thin wire.

x¯=1mCxρ(x,y)ds (4)

Here,

m is the mass of the thin wire.

Write the expression to find the y-coordinate of a center of mass of the thin wire.

y¯=1mCyρ(x,y)ds (5)

Write the equation of semi circle as follows.

x2+y2=4

Consider the parametric equations such that the considered parameters must satisfy the equation of the circle x2+y2=4 .

Consider the parametric equations of the curve as follows.

x=2cost,y=2sint,π2tπ2

The considered parametric equations and the limits are satisfied if the parameters substitute in the equation of circle x2+y2=4 .

As the linear density is a constant k and the limits of scalar parameter is π2tπ2 , rewrite the formula in equation (1) as follows.

m=π2π2kds (3)

Calculation of ds :

Substitute 2cost for x and 2sint for y in equation (2),

ds=(ddt2cost)2+(ddt2sint)2dt=(2sint)2+(2cost)2dt=4sin2t+4cos2tdt=4(sin2t+cos2t)dt

Simplify the expression as follows.

ds=4(1)dt {cos2θ+sin2θ=1}=2dt

Calculation of mass m :

Substitute 2dt for ds in equation (3),

m=π2π2k(2dt)=2kπ2π2dt=2k[t]π2π2=2k[π2(π2)]

m=2kπ

Thus, the mass of a thin wire which bent into the shape of a semicircle x2+y2=4,x0 is 2kπ_

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