Question

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

Answer: As the length (L) of the rod approaches infinity, the center of mass also approaches infinity.

Step-by-step solution

Find the Mass

To find the total mass of the rod, we need to integrate the linear density function, \(\rho(x)\), over the interval \([0, L]\). The mass, \(m\), can be found using the following formula: \(m=\int_{0}^{L}\rho(x)dx\) Now, integrate the given density function \(\rho(x) =\frac{10}{1+x^{2}}\): \(m=\int_{0}^{L}\frac{10}{1+x^{2}}dx\)

Compute the Center of Mass

To find the center of mass, we will use the following formula for the x-coordinate of the center of mass: \(x_{cm}=\frac{1}{m}\int_{0}^{L}x\rho(x)dx\) Now, integrate the product of x and the density function: \(x_{cm}=\frac{1}{m}\int_{0}^{L}\frac{10x}{1+x^{2}}dx\)

Evaluate the Integrals

First, we will evaluate the integral for mass: \(m=\int_{0}^{L}\frac{10}{1+x^{2}}dx = 10\int_{0}^{L}\frac{1}{1+x^{2}}dx = 10\left[\arctan{x}\right]_{0}^{L} = 10\left[\arctan{L} - \arctan{0}\right] = 10\arctan{L}\) Now, evaluate the integral for center of mass: \(x_{cm}=\frac{1}{m}\int_{0}^{L}\frac{10x}{1+x^{2}}dx = \frac{1}{10\arctan{L}}\int_{0}^{L}\frac{10x}{1+x^{2}}dx\) By noticing that the integral is the derivative of \(\frac{1}{2}\ln{(1+x^2)}\), we can evaluate it: \(\int_{0}^{L}\frac{10x}{1+x^{2}}dx = 10\left[\frac{1}{2}\ln{(1+x^2)}\right]_{0}^{L} = 5\ln{(1+L^2)}\) So, \(x_{cm}=\frac{1}{10\arctan{L}}\cdot 5\ln{(1+L^2)}\)

Find the Center of Mass as \(L \rightarrow \infty\)

To find the limit of the center of mass as \(L \rightarrow \infty\), we can use L'Hopital's rule since we have a \(\frac{0}{0}\) form as we take the limit: \(\lim_{L\to\infty}x_{cm} = \lim_{L\to\infty}\frac{\ln{(1+L^2)}}{2\arctan{L}}\) Differentiate both the numerator and denominator with respect to \(L\): \(\lim_{L\to\infty}x_{cm} = \lim_{L\to\infty}\frac{\frac{2L}{1+L^2}}{\frac{2}{1+L^2}}\) Now cancel the common factors and take the limit: \(x_{cm}(\infty) = \lim_{L\to\infty}\frac{2L}{2} = \infty\) As the length \(L\) of the rod approaches infinity, the center of mass also approaches infinity.

Key concepts

Linear Density

Linear density describes how mass is distributed along an object, like a rod. It tells us how much mass is present per unit length. For a thin rod, linear density can change along the length of the rod, denoted as \( \rho(x) \). In the given problem, the linear density is \( \rho(x) = \frac{10}{1+x^{2}} \). This function shows that the density decreases as \( x \) increases.

• Linear density helps in determining the overall mass of an object.
• In the context of a rod, it's crucial for finding the distribution of mass along its length.

Understanding linear density is essential in physics and engineering as it simplifies the calculation of mass for non-uniform objects by allowing the mass to be calculated through integration.

Center of Mass

The center of mass of an object is the point where the total mass can be considered to be concentrated. It's a sort of balancing point. Mathematically, for a continuous mass distribution like a rod, it's calculated using integration.
For this rod, we use:

• The mass, \( m \), is found through integrating the density function over the rod's length.
• The center of mass, \( x_{cm} \), involves both the density function and the distance variable \( x \) in the integration.

In this problem, the center of mass formula is \( x_{cm} = \frac{1}{m}\int_{0}^{L} x \rho(x)dx \). This requires evaluating the integral of \( \frac{10x}{1+x^{2}} \) from \( 0 \) to \( L \). The center of mass moves as \( L \) changes, hinting how mass is more concentrated towards one side when the rod length increases.

Integration

Integration is a fundamental calculus tool that accumulates values over an interval. In this problem, it's used to find both the mass and the center of mass of the rod.

• To find the mass, the integration accumulates the density \( \rho(x) \) across the rod length.
• To find the center of mass, a different integration considers each tiny piece of the rod's position \( x \) weighted by its density.

The specific integrals are solved using known functions: \( \int \frac{1}{1+x^2}dx \) results in \( \arctan{x} \), used here for the mass calculation. Similarly, logarithmic integration is used for the center of mass. Integration is powerful because it sums effects coherently along continuous variables like length.

L'Hopital's Rule

L'Hopital's Rule is a method for finding limits that can be applied when encountering indeterminate forms such as \( \frac{0}{0} \). This exercise uses it to find the behavior of the rod's center of mass as its length \( L \) approaches infinity.
When \( \lim_{L\to\infty} \frac{\ln{(1+L^2)}}{2\arctan{L}} \) needs evaluation:

• The numerator \( \ln(1+L^2) \) grows without bound, as does \( \arctan(L) \).
• First derivative of \( \ln{(1+L^2)} \) with respect to \( L \) is \( \frac{2L}{1+L^2} \).
• First derivative of \( 2\arctan(L) \) is \( \frac{2}{1+L^2} \).

After applying L'Hopital's Rule, it simplifies to \( \lim_{L\to\infty} L = \infty \), showing that the center of mass moves infinitely far as the rod's length increases indefinitely.