Problem 32 - University Physics with Modern Physics

Question

A thin, uniform rod has length \(L\) and mass \(M . \mathrm{A}\) small uniform sphere of mass \(m\) is placed a distance \(x\) from one end of the rod, along the axis of the rod (Fig. E13.32). (a) 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 \(L .\) (Hint: Use the power series expansion for \(\ln (1+x)\) given in Appendix 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 \(L .\)

Step-by-Step Solution

Answer

The potential energy is \( -G \frac{mM}{L} \ln \left( \frac{x+L}{x} \right) \) and the force is \( G \frac{mM}{L} \left( \frac{1}{x} - \frac{1}{x+L} \right) \). When \( x \gg L \), results simplify to point masses.

Step 1: Understand the Gravitational Potential Energy Formula

The gravitational potential energy between two masses is given by the formula:\[ U = -G \frac{m_1 m_2}{r} \]where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of the two masses.

Step 2: Set Up the Rod as a System of Point Masses

Since the rod is uniform and thin, we can consider it as a series of infinitesimally small masses along its length. Consider an infinitesimal mass element \( dm \) of the rod at a position \( x' \). This gives:\[ dm = \frac{M}{L} dx' \]where \( x' \) is the position along the rod from one end.

Step 3: Integrate to Find Total Potential Energy

To find the potential energy of the system, integrate the small contribution from each mass element over the length of the rod:\[ dU = -G \frac{m \, dm}{x + x'} = -G \frac{m \frac{M}{L} dx'}{x + x'} \]Integrating from 0 to \( L \):\[ U = -G \frac{mM}{L} \int_{0}^{L} \frac{1}{x + x'} \, dx' \]

Step 4: Perform the Integration

Perform the integration:\[ U = -G \frac{mM}{L} \left[ \ln(x + x') \right]_{0}^{L} = -G \frac{mM}{L} \left( \ln(x+L) - \ln(x) \right) = -G \frac{mM}{L} \ln \left( \frac{x+L}{x} \right) \]

Step 5: Simplify the Result When \( x \gg L \)

When \( x \) is much greater than \( L \), use the approximation for the logarithm:\[ \ln \left( 1 + \frac{L}{x} \right) \approx \frac{L}{x} \]Thus:\[ U \approx -G \frac{mM}{x} \]which is the expected result for two point masses separated by distance \( x \).

Step 6: Apply the Force Relation \( F = -\frac{dU}{dx} \)

To find the force on the sphere, take the derivative of \( U \) with respect to \( x \):\[ \frac{dU}{dx} = -G \frac{mM}{L} \frac{d}{dx} \left[ \ln \left( \frac{x+L}{x} \right) \right] = -G \frac{mM}{L} \left( \frac{1}{x+L} - \frac{1}{x} \right) \]So the force is:\[ F = -\frac{dU}{dx} = G \frac{mM}{L} \left( \frac{1}{x} - \frac{1}{x+L} \right) \]

Step 7: Verify the Force Simplification for \( x \gg L \)

When \( x \) is much greater than \( L \), the expression for force simplifies to:\[ F \approx G \frac{mM}{x^2} \]which is the expected gravitational force between two masses separated by distance \( x \).

Key Concepts

Uniform Rod

In physics, a uniform rod is a straight object with consistent mass distribution across its entire length. Imagine cutting the rod into very small pieces; each has the same mass. This means the rod has the same density throughout.

In the exercise, we treat the uniform rod as a sequence of tiny mass elements that sum up to the rod's total mass. Knowing that the rod's mass is distributed uniformly helps us calculate gravitational effects, as each small mass element contributes equally along its length.

Infinitesimal Mass Element

An infinitesimal mass element is a very small part of an object, so small that we consider it nearly zero in length. Yet, it still has a tiny mass, denoted as \( dm \). For a uniform rod, these tiny pieces add up along the rod's length.

Think of \( dm \) as a small chunk of the rod with mass calculated by \( dm = \frac{M}{L} \, dx' \). Here, \( M \) is the total mass of the rod, and \( L \) is the total length. By integrating these mass elements over the rod, we account for all gravitational influences of the rod.

• Every infinitesimal piece contributes a tiny gravitational pull.
• Considered collectively, these small elements explain the gravitational potential energy between the rod and another object.

Gravitational Force

Gravitational force is the attractive force between two masses. Issac Newton first introduced this concept as part of his law of universal gravitation. It's determined by the equation:\[ F = G \frac{m_1 m_2}{r^2} \]where \( F \) is the force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between their centers. When studying a mass like the uniform rod, the gravitational force from all its infinitesimal mass elements combines, pulling on external masses.

Here, we calculate the force on a sphere due to several small mass parts along the rod. Integration helps us sum up these tiny contributions to predict the total gravitational interaction.

Integration in Physics

Integration plays a key role in physics, especially when dealing with continuous distributions of mass or charge. It allows scientists to calculate the total effect from many small contributions, like the gravitational pull from a rod.

By integrating the infinitesimal mass elements over the entire length of the rod, we find the net gravitational potential energy. In the exercise, the integral:\[ U = -G \frac{mM}{L} \int_{0}^{L} \frac{1}{x + x'} \, dx' \]helps determine the total energy. Integration thus connects the microcosm (small part assessments) to the macrocosm (whole object evaluations).

• Captures cumulative effects accurately.
• Makes large object analysis manageable.

Understanding integration equips you with a mathematical tool essential to solving many complex problems in physics.