/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Limiting center of mass \(A\) th... [FREE SOLUTION] | 91影视

91影视

Limiting center of mass \(A\) thin rod of length \(L\) has a linear density given by \(\rho(x)=\frac{10}{1+x^{2}}\) on the interval \(0 \leq x \leq L\). Find the mass and center of mass of the rod. How does the center of mass change as \(L \rightarrow \infty ?\)

Short Answer

Expert verified
Answer: The center of mass of the rod as a function of L is \(X_{cm} = \frac{1}{2\arctan(L)} \ln(1+L^{2})\).

Step by step solution

01

Finding the Mass of the Rod

To find the mass of the rod, we need to integrate the linear density function 蟻(x) over the length of the rod (0 to L). The mass can be found using the formula: \(M = \int_{0}^{L} \rho(x)dx\) Where \(0 \leq x \leq L\). Now, plug in the given density function and evaluate the integral: \(M = \int_{0}^{L} \frac{10}{1+x^{2}}dx\)
02

Evaluate the Mass Integral

The integral above is a standard integral, and we can evaluate it using arctangent: \(M = \left[10\arctan(x)\right]_{0}^{L}\) Now, plug in the limits of integration: \(M = 10\arctan(L) - 10\arctan(0)\) Since arctan(0) = 0, we have: \(M = 10\arctan(L)\) This is the mass of the rod as a function of L.
03

Finding the Center of Mass

To find the center of mass, we can use the formula: \(X_{cm} = \frac{1}{M} \int_{0}^{L} x \rho(x)dx\) Substitute the given density function and the mass equation into the formula: \(X_{cm} = \frac{1}{10\arctan(L)} \int_{0}^{L} x\frac{10}{1+x^{2}} dx\)
04

Evaluate the Center of Mass Integral

The integral above can be evaluated using substitution: Let \(u = 1+x^2\), then \(\frac{du}{dx} = 2x\) and \(\frac{1}{2}du = xdx\). Now rewrite the integral in terms of u: \(X_{cm} = \frac{1}{10\arctan(L)}\int_{1}^{1+L^{2}} \frac{10}{2u} \frac{1}{2}du\) Simplify and integrate: \(X_{cm} = \frac{5}{10\arctan(L)} \int_{1}^{1+L^{2}} \frac{1}{u}du\) \(X_{cm} = \frac{1}{2\arctan(L)} \left[\ln(u)\right]_{1}^{1+L^{2}}\) Now, substitute back x value: \(X_{cm} = \frac{1}{2\arctan(L)} \left[\ln(1+L^{2}) - \ln(1)\right]\) Since ln(1) = 0, we have: \(X_{cm} = \frac{1}{2\arctan(L)} \ln(1+L^{2})\) This is the center of mass of the rod as a function of L.
05

Limit of the Center of Mass as L approaches infinity

To find the limiting center of mass as L approaches infinity, we can use L'H么pital's Rule: \(\lim_{L\to\infty}X_{cm} = \lim_{L\to\infty}\frac{\ln(1+L^{2})}{2\arctan(L)}\) Differentiate both the numerator and denominator with respect to L: \(\lim_{L\to\infty}X_{cm} = \lim_{L\to\infty}\frac{\frac{2L}{1+L^{2}}}{\frac{2}{1+L^{2}}}\) Simplify: \(\lim_{L\to\infty}X_{cm} = \lim_{L\to\infty} L\) Thus, as L approaches infinity, the center of mass will also approach infinity.

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 91影视!

Key Concepts

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

Linear Density Function
Understanding the linear density function is fundamental when dealing with the distribution of mass along a one-dimensional object, like a rod or a wire. In physics, this function, usually denoted as \(\rho(x)\), models how mass is distributed along an object. The linear density function gives us mass per unit length at any point \(x\) along the object.

For example, a rod with a constant linear density is uniform, meaning its mass is distributed evenly. However, if the linear density changes with \(x\), some parts of the rod will be denser than others. In the given exercise, the linear density function is \(\rho(x) = \frac{10}{1+x^{2}}\), implying the rod becomes less dense as we travel along its length from one end to the other. By integrating this function over the length of the rod, we get the total mass of the rod.
Definite Integral
The definite integral is an essential concept in calculus, particularly when we want to find the total accumulation of a quantity, such as area under a curve, or in our case, the total mass of an object. Formally, the definite integral from \(a\) to \(b\) of a function \(f(x)\) is written as \(\int_{a}^{b} f(x)dx\) and provides the algebraic sum of areas between the function's graph and the x-axis, within the limits of \(a\) and \(b\).

The mass of our rod is computed by integrating the linear density function over its length, which is a practical application of the definite integral. This computation illustrates how integral calculus is not just abstract mathematics but a tool for solving real-world problems, embodying the principles of accumulation and net change.
Arctangent Integration
Arctangent integration is a common technique used in calculus to integrate functions that have the form \(1+x^{2}\) in the denominator. The arctangent function, denoted as \(\arctan(x)\) or \(\tan^{-1}(x)\), is the inverse of the tangent function. It arises naturally when evaluating integrals of the form \(\int \frac{1}{1+x^{2}}dx\), the result of which is \(\arctan(x) + C\), where \(C\) represents the constant of integration.

For our exercise, the mass integral \(M\) involves the arctangent because the linear density function has this specific form. Hence, the indefinite integral of \(\rho(x)\) leads to \(10\arctan(x)\), providing a clear relationship between arctangent integration and physical quantities like mass in the context of linear density functions.
Limit of a Function
In mathematics, the limit of a function describes the behavior of that function as the input approaches a particular value. Limits are fundamental to calculus and are used in defining derivatives, integrals, and continuity. For instance, when examining how a function behaves as we get infinitely close to a value鈥攚ithout necessarily reaching it鈥攚e use the concept of a limit.

In the context of our exercise, when we calculate the limit as \(L\to\infty\) of the center of mass, we examine how the center of mass behaves as the length of the rod extends towards infinity. Understanding limits allows us to grasp how a system behaves under extreme conditions or to predict trends without needing explicit calculations at those extreme values.
L'H么pital's Rule
L'H么pital's Rule is a technique in calculus used to evaluate limits that yield an indeterminate form, like 0/0 or \(\infty/\infty\). When faced with a limit of a ratio of functions that produces one of these indeterminate forms, L'H么pital's Rule states that we can take the derivatives of the numerator and the denominator separately to simplify the expression for another attempt at finding the limit.

This rule is particularly useful in our problem when we examine the limit of the center of mass as \(L\) grows without bound. It allows for a straightforward evaluation of an otherwise challenging limit, illustrating the balance between mathematical theory and practical application. By applying L'H么pital's Rule, we can conclude that the center of mass approaches infinity as the length of the rod becomes infinitely long.

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

Open and closed boxes Consider the region \(R\) bounded by three pairs of parallel planes: \(a x+b y=0, a x+b y=1\) \(c x+d z=0, c x+d z=1, e y+f z=0,\) and \(e y+f z=1\) where \(a, b, c, d, e,\) and \(f\) are real numbers. For the purposes of evaluating triple integrals, when do these six planes bound a finite region? Carry out the following steps. a. Find three vectors \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) each of which is normal to one of the three pairs of planes. b. Show that the three normal vectors lie in a plane if their triple scalar product \(\mathbf{n}_{1} \cdot\left(\mathbf{n}_{2} \times \mathbf{n}_{3}\right)\) is zero. c. Show that the three normal vectors lie in a plane if ade \(+b c f=0\) d. Assuming \(\mathbf{n}_{1}, \mathbf{n}_{2},\) and \(\mathbf{n}_{3}\) lie in a plane \(P,\) find a vector \(\mathbf{N}\) that is normal to \(P .\) Explain why a line in the direction of \(\mathbf{N}\) does not intersect any of the six planes and therefore the six planes do not form a bounded region. e. Consider the change of variables \(u=a x+b y, v=c x+d z\) \(w=e y+f z .\) Show that $$J(x, y, z)=\frac{\partial(u, v, w)}{\partial(x, y, z)}=-a d e-b c f$$ What is the value of the Jacobian if \(R\) is unbounded?

Consider the surface \(z=x^{2}-y^{2}\) a. Find the region in the \(x y\) -plane in polar coordinates for which \(z \geq 0\) b. Let \(R=\\{(r, \theta): 0 \leq r \leq a,-\pi / 4 \leq \theta \leq \pi / 4\\},\) which is a sector of a circle of radius \(a\). Find the volume of the region below the hyperbolic paraboloid and above the region \(R\)

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of a tetrahedron whose vertices are located at \((0,0,0),(a, 0,0),(0, b, 0),\) and \((0,0, c)\)

Use spherical coordinates to find the volume of the following solids. That part of the ball \(\rho \leq 4\) that lies between the planes \(z=2\) and \(z=2 \sqrt{3}\)

Let \(D\) be the solid bounded by the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1,\) where \(a>0, b>0,\) and \(c>0\) are real numbers. Let \(T\) be the transformation \(x=\)au, \(y=b v, z=c w\) Find the average square of the distance between points of \(D\) and the origin.

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.