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Chapter 6: Problem 5

Find the mass \(M\) and center of mass \(\bar{x}\) of the linear wire covering the given interval and having the given density \(\delta(x)\). $$\delta(x)=\left\\{\begin{array}{ll} 4, & 0 \leq x \leq 2 \\ 5, & 2

Short Answer

Expert verified
Mass \(M\) is 13 and center of mass \(\bar{x}\) is approximately 1.577.

Step by step solution

01

Set Up the Mass Integral

The mass of the wire, \(M\), is found by integrating the density function \(\delta(x)\) over the interval from 0 to 3. The wire is divided into two parts: \([0, 2]\) and \((2, 3]\). Therefore, we write the mass as the sum of two integrals: \[ M = \int_{0}^{2} 4 \, dx + \int_{2}^{3} 5 \, dx \] We will evaluate these integrals separately to find the mass.
02

Compute Mass on First Interval

Evaluate the first integral for the segment \([0, 2]\) with constant density 4: \[ \int_{0}^{2} 4 \, dx = 4x \Bigg|_{0}^{2} \] Substitute the limits: \[ 4 \times 2 - 4 \times 0 = 8 \] Thus, the mass from \([0,2]\) is 8.
03

Compute Mass on Second Interval

Evaluate the second integral for the segment \((2, 3]\) with constant density 5: \[ \int_{2}^{3} 5 \, dx = 5x \Bigg|_{2}^{3} \] Substitute the limits: \[ 5 \times 3 - 5 \times 2 = 15 - 10 = 5 \] Thus, the mass from \((2, 3]\) is 5.
04

Calculate Total Mass

Add the masses computed for the two intervals: \[ M = 8 + 5 = 13 \] The total mass of the wire is 13.
05

Set Up Center of Mass Integral

The center of mass \(\bar{x}\) is given by the formula: \[ \bar{x} = \frac{1}{M} \int_{a}^{b} x \cdot \delta(x) \, dx \] As with the mass, we need to consider the two intervals separately. Thus, \[ \int_{0}^{3} x \cdot \delta(x) \, dx = \int_{0}^{2} x \times 4 \, dx + \int_{2}^{3} x \times 5 \, dx \] We will find each integral separately.
06

Compute Center of Mass for First Interval

Evaluate the integral \(\int_{0}^{2} x \cdot 4 \, dx\): \[ \int_{0}^{2} 4x \, dx = 2x^2 \Bigg|_{0}^{2} \] Substitute the limits: \[ 2 \times 2^2 - 2 \times 0^2 = 8 \] The contribution to the center of mass from \([0,2]\) is 8.
07

Compute Center of Mass for Second Interval

Evaluate the integral \(\int_{2}^{3} x \cdot 5 \, dx\): \[ \int_{2}^{3} 5x \, dx = \frac{5}{2}x^2 \Bigg|_{2}^{3} \] Substitute the limits: \[ \frac{5}{2} \times 3^2 - \frac{5}{2} \times 2^2 = \frac{45}{2} - \frac{20}{2} = 12.5 \] The contribution to the center of mass from \((2,3]\) is 12.5.
08

Calculate Total Moment and Center of Mass

Add the contributions to the total moment of mass: \[ \int_{0}^{3} x \cdot \delta(x) \, dx = 8 + 12.5 = 20.5 \] Compute the center of mass: \[ \bar{x} = \frac{20.5}{13} \approx 1.577 \] The center of mass of the wire is approximately 1.577.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Calculation
When we calculate the mass of an object such as a wire, we are determining how much material it contains. To find the mass, we use
  • the integration of a density function over a defined interval,
  • representing different sections of the object.
For this problem, the wire is divided into two segments, each with a uniform density:
  • 4 in the interval \(0 \leq x \leq 2\),
  • and 5 in the interval \(2 < x \leq 3\).
By integrating the density function over these intervals and summing the results, we obtained the total mass of the wire as 13 units. This process shows how mass is tied to the distribution and amount of material along an object.
Density Function
The density function, \( \delta(x) \), describes how the material is distributed throughout an object. In our problem, the density is given as a piecewise function:
  • For \(0 \leq x \leq 2\), \( \delta(x) = 4 \), meaning the wire has a constant density of 4 units.
  • For \(2 < x \leq 3\), \( \delta(x) = 5 \), indicating a change to a constant density of 5 units.
This change in density means that different parts of the wire contain different amounts of material per unit length. Understanding this function is key in computing both mass and the position of the center of mass, highlighting how material properties are mathematically described.
Integration
Integration is a powerful mathematical tool used to add up small pieces, or increments, to find quantities like area, volume, and mass. In this exercise, integration was employed to compute:
  • the mass by summing the infinitesimal contributions of each segment of the wire,
  • as well as to compute the moment when finding the center of mass.
For the mass, we integrated the density function over the intervals, \( [0, 2] \) and \( (2, 3] \), separately. Similarly, we used integration to find how these masses contribute to the center of mass by analyzing their distribution along the wire. Integration, therefore, provides an aggregate measure of distributed quantities.
Piecewise Functions
A piecewise function is composed of multiple segments, each defined over a specific part of the domain. For this problem, the density of the wire is expressed as a piecewise function:
  • The section from \(0\) to \(2\) has a constant density value of \(4\).
  • From \(2\) to \(3\), the density jumps to \(5\).
Using piecewise functions allows us to accurately describe situations where properties, like density, change between different sections of a domain. Calculations involving piecewise functions often require separate integrations over each segment and can simplify complex real-world structures into manageable mathematical models. This piecewise characteristic of the wire's density function was crucial to accurately determining the mass and center of mass in this exercise.

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Most popular questions from this chapter

Find the centroid of the thin plate bounded by the graphs of the given functions. Use Equations (6) and (7) with \(\delta=1\) and \(M=\) area of the region covered by the plate. $$g(x)=x^{2}(x+1), \quad f(x)=2, \quad \text { and } \quad x=0$$

Find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region between the curve \(y=1 / x\) and the \(x\) -axis from \(x=1\) to \(x=2 .\) Give the coordinates to two decimal places.

Find the lengths of the curves . If you have graphing software, you may want to graph these curves to see what they look like. $$x=\int_{0}^{y} \sqrt{\sec ^{4} t-1} d t, \quad-\pi / 4 \leq y \leq \pi / 4$$

Find the volume of the solid generated by revolving the region bounded by \(y=\sqrt{x}\) and the lines \(y=2\) and \(x=0\) about a. the \(x\) -axis. b. the \(y\) -axis. c. the line \(y=2\) d. the line \(x=4\)

Find the lengths of the curves. If you have graphing software, you may want to graph these curves to see what they look like. $$y=(3 / 4) x^{4 / 3}-(3 / 8) x^{2 / 3}+5, \quad 1 \leq x \leq 8$$
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