91Ó°ÊÓ

In Section 8.5 we calculated the center of mass by considering objects composed of a finite number of point masses or objects that, by symmetry, could be represented by a finite number of point masses. For a solid object whose mass distribution does not allow for a simple determination of the center of mass by symmetry, the sums of Eqs. (8.28) must be generalized to integrals $$ x_{\mathrm{cm}}=\frac{1}{M} \int x d m \quad y_{\mathrm{cm}}=\frac{1}{M} \int y d m $$ where \(x\) and \(y\) are the coordinates of the small picce of the object that has mass \(d m\). The integration is over the whole of the object. Consider a thin rod of length \(I_{4}\), mass \(M\), and cross-sectional area \(A\). I et the origin of the coordinates be at the left end of the rod and the positive \(x\) -axis lie along the rod. (a) If the density \(\rho=M / V\) of the object is uniform, perform the integration described above to show that the \(x\) -coordinate of the center of mass of the rod is at its gcometrical center. (b) If the density of the object varies linearly with \(x\) - that is, \(\rho=\alpha x,\) where \(\alpha\) is a positive constant - calculate the \(x\) -coordinate of the rod's center of mass.

Step by step solution

01

Uniform Density: Setup

Start with the formula for the x-coordinate of the center of mass: \(x_{cm} = \frac{1}{M} \int x dm\). It can be assumed that the rod is split into small elements of length \(dx\) with mass \(dm\). Given that the density \(\rho = \frac{M}{V}\), where the volume for the small piece, considering its a rod, is \(V = A*dx\), the mass of each element is then \(dm = \rho A dx\). Substitute this into the integral and integrate over the length of the rod.

02

Uniform Density: Integration

Now integrate \(x_{cm} = \frac{1}{M} \int_0^L x \rho A dx\), given that the density and cross-sectional area are constants and can be taken out of the integral. Integrating \(x\) from 0 to L given that the mass \(M = \rho A L\), eventually get \(x_{cm} = \frac{1}{2} L\). This indicates that the center of mass is at its geometrical center.

03

Linearly Varying Density: Setup

Next, consider the scenario when the density of the object varies linearly with \(x\) that is, \(\rho = \alpha x\), where \(\alpha\) is a positive constant. Now the mass \(dm\) of each element is \(dm = \alpha x A dx\). Substitute this into the formula.

04

Linearly Varying Density: Integration

On integrating \(x_{cm} = \frac{1}{M} \int_0^L x \alpha x A dx\), the range of integral is the length of the rod. Calculate the mass \(M = \int_0^L dm = \alpha A \int_0^L x dx\), and substitute this together with the integral for \(x_{cm}\). After integrating, obtain \(x_{cm} = \frac{2}{3} L\).

05

Analysis

The center of mass for the rod is not at its geometrical center if the mass distribution along the rod is not uniform. The mass distribution changes the location of the center of mass.

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Uniform Density

When we talk about uniform density in physics, we're referring to the property of a material where mass is evenly distributed throughout its volume. In simpler terms, imagine having a bar of chocolate with an equal amount of cocoa spread throughout; that's uniform density. There are no heavier or lighter parts; each bite of that chocolate bar would essentially have the same cocoa content. Similarly, a rod with a uniform density would have the same mass per volume along its entire length.

This uniformity simplifies many physical calculations since the density value is constant and can therefore be factored out of integral expressions for calculating physical quantities like center of mass. In the context of the exercise you're working with, it means when performing the integration to find the center of mass, we can treat density as a constant multiplier, out of the integral, which greatly simplifies the math involved. This results in finding the center of mass at the geometrical center for objects with uniform density, which is intuitive once you consider the symmetry of mass distribution in such objects.

Linearly Varying Density

In contrast to uniform density, linearly varying density refers to a change in density that is directly proportional to a change in another variable - in this case, position along the length of the rod. Think of it like a gradient in art where the color transitions from light to dark from one end of the canvas to the other. Physically, this could resemble a rod made of metal that gradually transitions to plastic, becoming less dense as you move along it.

With linearly varying density, the density at any point on the object can be described mathematically as a linear function of its position. For the problem at hand, the density function is \( \rho = \( \alpha x\) \) where \alpha is a constant. When calculating the center of mass, this linear relationship means that density increases with distance along the rod, so mass is more concentrated towards one end. As a result, these density variations must be factored into the integration process. The final result signifies that the center of mass is shifted towards the heavier end of the rod, at \(\frac{2}{3}L\), away from the geometrical center.

Integration in Physics

Integration is a fundamental technique in physics essential for dealing with continuous distributions of mass, charge, or other properties. It enables us to sum up infinitely many infinitesimally small quantities, like the mass of an infinitely thin slice of a rod. In the context of finding the center of mass for different density distributions, integration allows us to account for every tiny piece of mass throughout the object’s length.

Whether the density is uniform or varying, integration provides the calculation of the overall effect of those small pieces to determine a precise value for the center of mass. When the density is uniform, the integration is relatively straightforward because the value for density is constant. But, with a linearly varying density, integration becomes more complex because the density value itself changes along with position. The calculated center of mass reflects the cumulative effect of these variations. It is the power of calculus, and specifically integration, that enables physicists and engineers to map out such intricate concepts and apply them to real-world problems.