Consider the ringshaped body of Fig. E13.35. A particle with mass m is placed a distance x from the center of the ring, along the line through the center of the ring and perpendicular to its plane. (a) Calculate the gravitational potential energy U of this system. Take the potential energy to be zero when the two objects are far apart. (b) Show that your answer to part (a) reduces to the expected result when x is much larger than the radius a of the ring. (c) Use Fx = -dU/dx to find the magnitude and direction of the force on the particle (see Section 7.4). (d) Show that your answer to part (c) reduces to the expected result when x is much larger than a. (e) What are the values of U and Fx when x = 0? Explain why these results make sense.
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Mass Distribution with Calculus
Problem 34a
Textbook Question
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. E13.34). Calculate the gravitational 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.


1
Understand that the gravitational potential energy (U) between two masses is given by the formula: , where G is the gravitational constant, M and m are the masses, and r is the distance between the centers of the two masses.
Consider the rod as a collection of infinitesimally small mass elements, each contributing to the gravitational potential energy with the sphere. Let the linear mass density of the rod be .
Set up an integral to calculate the total gravitational potential energy between the rod and the sphere. Consider a small element of the rod of length at a distance x from the sphere. The mass of this element is .
The potential energy contribution from this small element is , where r is the distance from the sphere to the element. Integrate this expression from x to x + L to find the total potential energy.
To show that the result reduces to the expected form when x is much larger than L, consider the approximation where the rod can be treated as a point mass located at its center of mass. The distance from the sphere to the center of mass of the rod is approximately x + L/2, and use this in the potential energy formula to verify the expected result.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. For two masses, it is given by the formula U = -G(Mm)/r, where G is the gravitational constant, M and m are the masses, and r is the distance between their centers. This concept is crucial for calculating the energy of the rod-sphere system as it depends on their separation.
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Gravitational Potential Energy
Integration in Continuous Mass Distributions
When dealing with objects like rods, which have continuous mass distributions, integration is used to calculate quantities like gravitational potential energy. The rod can be divided into infinitesimally small elements, each contributing to the total potential energy. This requires setting up an integral over the length of the rod, considering the distance from each element to the sphere.
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Using Calculus to Solve Mass Distribution Problems
Limit Analysis
Limit analysis involves examining the behavior of a function as a variable approaches a particular value, often infinity. In this problem, analyzing the limit as x becomes much larger than L helps verify the solution by showing it simplifies to a known result. This concept ensures the solution is consistent with physical intuition and known cases, such as treating the rod as a point mass when far away.
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