Question

Find the location of the center of mass for a onedimensional rod of length \(L\) and of linear density \(\lambda(x)=c x\), where \(c\) is a constant. (Hint: You will need to calculate the mass in terms of \(c\) and \(L\).)

Short answer

Answer: The center of mass of the rod is located at \(\frac{2}{3}L\).

Step-by-step solution

Understand the given variables

We have a rod of length \(L\) and a linear density function \(\lambda(x) = c x\), where \(c\) is a constant. The linear density represents the mass per unit length. The center of mass is the average location of mass distribution in an object.

Calculate the total mass of the rod

To calculate the total mass, integrate the linear density function over the entire length of the rod (from \(0\) to \(L\)): \(M = \int_0^L \lambda(x) dx = \int_0^L (cx) dx\)

Evaluate the integral for the total mass

Now we need to evaluate the integral to find the total mass: \(M = \int_0^L (cx) dx = \left[\frac{1}{2} cx^2\right]_0^L = \frac{1}{2} cL^2\) So the total mass is \(M = \frac{1}{2} cL^2\).

Calculate the location of the center of mass

The formula for the center of mass for a one-dimensional object is given by: \(x_{cm} = \frac{1}{M}\int_0^L x \lambda(x) dx\) Substituting the total mass and the linear density function: \(x_{cm} = \frac{1}{\frac{1}{2} cL^2}\int_0^L x (cx) dx\)

Evaluate the integral to find the location of the center of mass

Now we need to evaluate the integral to find the location of the center of mass: \(x_{cm} = \frac{1}{\frac{1}{2} cL^2}\int_0^L x (cx) dx = \frac{2}{cL^2}\int_0^L (cx^2) dx\) \(x_{cm} = \frac{2}{cL^2} \left[\frac{1}{3} cx^3\right]_0^L = \frac{2}{cL^2} \left(\frac{1}{3} cL^3\right)\) Simplifying the expression: \(x_{cm} = \frac{2}{3}L\) So the location of the center of mass for the given one-dimensional rod is at \(\frac{2}{3}L\).

Key concepts

Linear Density

Understanding linear density is critical when you're dealing with one-dimensional objects in physics. Linear density, typically represented by \( \lambda(x) \), is a measure of mass per unit length at a specific point along an object. In simpler terms, it tells you how 'heavy' a tiny segment of the rod is if you were to slice it at any given point.

For example, the linear density \( \lambda(x) = c x \) in the exercise suggests that the rod becomes heavier per unit length as you move along its length. Here, \( c \) is a constant that determines how quickly the density increases. A higher value of \( c \) would mean a steeper increase in mass as you measure further along the rod. The concept of linear density is crucial because it allows us to compute the total mass and the center of mass via integration, which brings us to our next topic - integral calculus.

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulation of quantities, such as areas under curves, volumes, and, in our case, the total mass of an object given its density function. It's like adding up an infinite number of infinitesimally small quantities to find a total.

To find the total mass \( M \) of the rod, the linear density function \( \lambda(x) \) is integrated over the rod's length. By calculating \( \int_0^L (cx) dx \) and evaluating the result, the total mass gives us a single value that accounts for the entire mass, despite its variable density along the length. Integral calculus's power is magnified when dealing with continuous mass distributions, as it provides a way to handle otherwise challenging problems with precision.

Mass Distribution

Mass distribution refers to how mass is spread out through an object. Knowing this distribution is essential for various calculations in physics, including finding the object's center of mass. In the exercise, the mass of the rod isn't uniformly distributed because the linear density isn't constant—it increases linearly along the length of the rod.

The concept indicates that certain sections of the rod contain more mass than others. Due to this variable distribution, the center of mass won’t simply be at the halfway point (\( \frac{1}{2}L \) in our case). Instead, it shifts towards the denser part of the rod. Recognizing how mass is distributed can also predict how an object will balance or rotate, which are critical aspects in statics and dynamics fields of physics.

One-Dimensional Objects

A one-dimensional object in physics is an idealized entity that has length but no width or height. Common examples include thin wires, rods, or strings. For these objects, the center of mass calculations is simplified because we're only concerned with one dimension. In our textbook problem, we're dealing with such an object: a rod.

Center of mass considerations for one-dimensional objects boil down to finding a point along the line of the object that represents the average location of mass. Through integration and understanding of mass distribution, we can accurately determine this point. For the provided rod with linearly increasing mass density, the center of mass is not at the mid-point but is calculated to be at \( \frac{2}{3}L \) from one end of the rod, showcasing the shift due to the non-uniform mass distribution.