Question

A one-dimensional rod has a linear density that varies with position according to the relationship \(\lambda(x)=c x,\) where \(c\) is a constant and \(x=0\) is the left end of the rod. Where do you expect the center of mass to be located? a) the middle of the rod b) to the left of the middle of the rod c) to the right of the middle of the rod d) at the right end of the rod e) at the left end of the rod

Short answer

Answer: The center of mass is located to the right of the middle of the rod.

Step-by-step solution

Find the total mass of the rod

Calculate the total mass (M) of the rod using the formula \(M = \int_{0}^{L} \lambda(x)dx\). Since \(\lambda(x) = cx\), we have: $$M = \int_{0}^{L} cx dx$$ Now integrate the function with respect to x: $$M = \frac{1}{2}cL^2$$

Calculate the center of mass of the rod

Use the formula \(x_{cm} = \frac{1}{M}\int_{0}^{L} x\lambda(x)dx\) to find the center of mass. We already found that \(\lambda(x) = cx\) and \(M = \frac{1}{2}cL^2\). Plug these values into the formula: $$x_{cm} = \frac{1}{\frac{1}{2}cL^2}\int_{0}^{L} x(cx)dx$$ Now simplify the integral: $$x_{cm} = \frac{2}{cL^2}\int_{0}^{L} cx^2dx$$ Integrate the function with respect to x: $$x_{cm} = \frac{2}{cL^2}[\frac{1}{3}cx^3]_{0}^{L}$$ Solve the integral boundaries: $$x_{cm} = \frac{2}{cL^2}[\frac{1}{3}cL^3 - 0]$$ Now simplify the equation: $$x_{cm} = \frac{2}{3}L$$

Identify the location of the center of mass

The center of mass \(x_{cm}\) is located at \(\frac{2}{3}L\) which is to the right of the middle of the rod. So the answer is (c) to the right of the middle of the rod.

Key concepts

Linear Density

Linear density, typically denoted as \(\frac{\text{mass}}{\text{length}}\), is a measure of how much mass is distributed along a certain length of an object. Imagine that you have a string; if some parts of the string are thicker (contain more material) than others, then the linear density of the string is not uniform—it varies along the length of the string.

In the context of our exercise, the linear density \(\lambda(x)\) varies with the position along a rod by the relationship \(\lambda(x) = c x\), where \(c\) is a constant. This tells us that the mass per unit length increases the further from the left end of the rod we go. The linear density concept is crucial to solving for the rod's mass and its center of mass. Understanding linear density allows us to grasp how the mass of the rod is not evenly distributed and predict that the center of mass will not be in the geometric middle of the rod.

Integration in Physics

Integration is a mathematical tool that is foundational in physics, especially when dealing with varying quantities along a continuum. To understand a physical concept like the center of mass of an object with a non-uniform mass distribution, we turn to integration to sum up infinitesimally small quantities over the entire length or volume of the object.

In the provided exercise, integration is used twice: first, to calculate the total mass of the rod, and second, to find the position of the center of mass. We integrate the linear density function \(\lambda(x) = cx\) across the length of the rod to sum up all the tiny pieces of mass across its length. By doing this, we're essentially adding up all the contributions to mass—or in the second case, the contributions to the rod's moment about one end—starting from one end of the rod to the other. Statics and dynamics, fields within physics, frequently rely on integration to solve for quantities like center of mass, force distributions, and potential energy.

Mass Distribution

The concept of mass distribution refers to how mass is spread out in an object. It's not just about how much mass there is, but where that mass is located. In objects with uniform mass distribution, mass is spread evenly throughout, which usually makes calculations simpler since the center of mass coincides with the geometric center.

However, when an object has a non-uniform mass distribution, like our rod where mass increases with distance from one end, the center of mass is shifted toward the heavier side. The calculations, therefore, become a bit more involved, as in our exercise, and involve the use of integrals. Grasping the concept of mass distribution is paramount to solving many physics problems involving balance, rotation, and motion. Our exercise demonstrates a direct application by showing that the center of mass is located at \(\frac{2}{3}L\), which is a result of the mass being more heavily distributed towards the right end of the rod.