/*! 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 40 A thin, uniform rod has length \... [FREE SOLUTION] | 91Ó°ÊÓ

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A thin, uniform rod has length \(L\) and mass \(M .\) A small uniform sphere of mass \(m\) is placed a distance \(x\) from one end of the rod, along the axis of the rod (Fig. 12.34\()\) . (a) Calculate the gravita- tional potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. Show that your answer reduces to the expected result when \(x\) is much larger than \(L\) . (Hint: Use the power series expansion for \(\ln (1+x)\) given in Appendix B. (b) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4 ). Show that your answer reduces to the expected result when \(x\) is much larger than \(L .\)

Short Answer

Expert verified
The gravitational potential energy is \(-GmM \ln((x+L)/x)/L\) and simplifies to \(-GmM/x\) when \(x >> L\). The force is approximately \(GmM/x^2\).

Step by step solution

01

Set up the problem

We want to calculate the gravitational potential energy between a thin rod of length \(L\) and mass \(M\), and a small spherical object with mass \(m\), placed at a distance \(x\) from one end of the rod. Assume gravitational potential energy is zero when the objects are far apart.
02

Express an element of the rod

Divide the rod into small elements of length \(dy\), and assume each has a mass \(dM = \frac{M}{L} dy\). The element is at a distance \(y\) from the sphere and contributes to gravitational potential energy.
03

Setup integral for gravitational potential energy

The potential energy between the element \(dM\) and the sphere is given by \(dU = -\frac{G m dM}{x + y}\). Integrating from \(y = 0\) to \(y = L\): \[ U = -Gm \frac{M}{L} \int_{0}^{L} \frac{1}{x + y} \, dy \].
04

Evaluate the integral

The integral \(\int \frac{1}{x+y} dy\) evaluates to \(\ln(x+y)\). So,\[ U = -Gm \frac{M}{L} \left[ \ln(x+L) - \ln(x) \right] = -Gm \frac{M}{L} \ln \left( \frac{x+L}{x} \right) \].
05

Simplify using a power series

Using the approximation \(\ln(1+y) \approx y - \frac{y^2}{2} + \cdots\) for small \(y\), if \(x >> L\), then \(\ln \left(1 + \frac{L}{x}\right) \approx \frac{L}{x}\), and the potential energy simplifies to \[ U \approx -\frac{GmM}{x} \].
06

Derive gravitational force from potential energy

According to \(F_x = -\frac{dU}{dx}\), differentiate the expression for \(U\):\[ F_x = Gm \frac{M}{L} \frac{1}{x(x+L)} \].
07

Simplify force expression for large x

For \(x >> L\), using the approximation \(\frac{1}{x(x+L)} \approx \frac{1}{x^2}\), the force simplifies to \[ F_x \approx \frac{GmM}{x^2} \]. This is the expected gravitational force as if the rod were a point mass.

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Key Concepts

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

Thin Rod and Sphere Interaction
When studying physics, understanding how objects interact through gravitational forces is essential. In this scenario, imagine a thin and uniform rod of length \(L\) and mass \(M\), and a small uniform sphere with mass \(m\). The sphere is positioned at a distance \(x\) from one end of the rod, along its axis. This setup leads to an intriguing gravitational interaction between the two bodies.
The gravitational potential energy of the system can be calculated by considering the contributions of infinitesimally small elements of the rod. Each element, with a small mass \(dM\), interacts with the sphere, contributing to the potential energy of the system. The potential energy is zero when the two objects are infinitely far apart, as this is our reference point for measurement.
By summing the interactions of all these small rod elements with the sphere through an integral, we ultimately determine the total potential energy, capturing the essence of their gravitational interaction. This approach highlights how even simple-looking systems can involve complex calculations to understand their energetic interactions.
Power Series Expansion for Logarithmic Functions
At the heart of simplifying the potential energy expression is the tool known as the power series expansion, which is used for logarithmic functions. Power series are a way to express complex functions as an infinite sum of simpler terms. This is extremely useful in cases where direct calculation is challenging or impossible.
For this exercise, we're interested in simplifying the expression \(\ln\left(\frac{x+L}{x}\right)\) when \(x\) is much larger than \(L\). In such a case, we can employ the power series expansion for \(\ln(1+y)\), which approximates to \(y - \frac{y^2}{2} + \ldots\) when \(y\) is small.
By recognizing \(\frac{L}{x}\) as our \(y\), we can approximate \(\ln(1 + \frac{L}{x})\) as \(\frac{L}{x}\). This approximation greatly simplifies the original problem, reducing the potential energy to a more manageable form and highlighting the utility of power series in making complex problems tractable.
Gravitational Force Calculation
Knowing the gravitational potential energy allows us to find the associated gravitational force exerted by the rod on the sphere. According to the relationship \(F_x = -\frac{dU}{dx}\), we can derive the force from potential energy by differentiation. This process essentially measures how potential energy changes with respect to the position \(x\).
In our scenario, after differentiating the potential energy expression, we find the force \(F_x\) takes on a specific mathematical form. Initially, this expression might seem complex. However, for large \(x\) values (where the distance between rod and sphere is much larger than the rod's length), the result simplifies remarkably.
This simplification shows that the expression for force approaches \(\frac{GmM}{x^2}\), which is familiar from Newton's universal law of gravitation, applicable to point masses. Such simplifications are not only elegant but also essential for understanding the fundamental physics governing gravitational interactions between distant objects.

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Most popular questions from this chapter

An experiment is performed in deep space with two uni- form spheres, one with mass 25.0 \(\mathrm{kg}\) and the other with mass 100.0 \(\mathrm{kg}\) . They have equal radii, \(r=0.20 \mathrm{m}\) . The spheres are released from rest with their centers 40.0 \(\mathrm{m}\) apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres. (a) Explain why linear momentum is conserved. (b) When their centers are 20.0 \(\mathrm{m}\) apart, find (i) the speed of each sphere and (ii) the magnitude of the relative velocity with which one sphere is approaching the other. (c) How far from the initial position of the center of the \(25.0-\mathrm{kg}\) sphere do the surfaces of the two spheres collide?

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