Problem 108 - Physics for Scientists and Engineers with Modern Physics

Question

A particle moves where its potential energy is given by \(U(r)=U_{0}\left[\left(2 / r^{2}\right)-(1 / r)\right] .(a)\) Plot \(U(r)\) versus \(r .\) Where does the curve cross the \(U(r)=0\) axis? At what value of \(r\) does the minimum value of \(U(r)\) occur? (b) Suppose that the particle has an energy of \(E=-0.050 U_{0} .\) Sketch in the approximate turning points of the motion of the particle on your diagram. What is the maximum kinetic energy of the particle, and for what value of \(r\) does this occur?

Step-by-Step Solution

Answer

The \( U(r) \, \,= 0 \) at \( r = 2 \), min value at \( r=4 \); max KE at \( r=4 \).

Step 1: Understanding the Potential Energy Function

We are given the potential energy as a function of position, where \( U(r) = U_0\left[\frac{2}{r^2} - \frac{1}{r}\right] \). We need to understand this is a function of \( r \) that involves both \( \frac{1}{r^2} \) and \( \frac{1}{r} \) terms, indicative of a potential field influenced by inverse square and inverse linear laws.

Step 2: Identifying Points Where U(r) = 0

To find where the potential energy curve crosses the \( U(r) = 0 \) axis, set \( U(r) = 0 \):\[ U_0\left(\frac{2}{r^2} - \frac{1}{r}\right) = 0 \]Solving for \( r \), we have:\[ \frac{2}{r^2} = \frac{1}{r} \]Multiply both sides by \( r^2 \) to get:\[ 2 = r \]Thus, the curve crosses the \( U(r) = 0 \) axis at \( r = 2 \).

Step 3: Finding the Minimum Value of U(r)

To locate potential energy minimum, find the derivative of \( U(r) \) and set it to zero. The derivative is:\[ \frac{dU}{dr} = U_0\left(-\frac{4}{r^3} + \frac{1}{r^2}\right) \]Setting \( \frac{dU}{dr} = 0 \), we find:\[ -\frac{4}{r^3} + \frac{1}{r^2} = 0 \]Assuming \( U_0 eq 0 \), solving:\[ \frac{1}{r^2} = \frac{4}{r^3} \]This implies \( r = 4 \), thus the minimum value of \( U(r) \) occurs at \( r = 4 \).

Step 4: Finding the Energy of the Particle

The particle's energy is given as \( E = -0.050 U_0 \). This energy level needs to be plotted alongside \( U(r) \) to find turning points. These points occur where \( U(r) = E \).

Step 5: Engaging the Turning Point Equation

Equate \( U(r) = E \):\[ U_0\left(\frac{2}{r^2} - \frac{1}{r}\right) = -0.050 U_0 \]Simplifying gives:\[ \frac{2}{r^2} - \frac{1}{r} = -0.050 \]Solve this to find two values of \( r \) as turning points, relying on a quadratic substitution solve or numerical approximation, yielding roots \( r_1 \) and \( r_2 \).

Step 6: Calculating Maximum Kinetic Energy

Kinetic energy is maximized when potential energy is minimized since\( K = E - U(r) \). Since minimum \( U(r) \) occurs at \( r=4 \), the maximum kinetic energywill be \[ K_{max} = E - U(4) \].Calculate \( U(4) = U_0\left(\frac{2}{4^2} - \frac{1}{4}\right) \), plug this result into\( K_{max} = -0.050 U_0 - U(4) \) to get the value at maximum kinetic energy at \( r = 4 \).

Key Concepts

Inverse Square Law

The inverse square law is a fundamental principle that describes how certain physical quantities diminish with distance. In the case of potential energy, it plays a vital role in determining how forces, such as gravitational or electrostatic forces, act between two objects. Here, the potential energy function provided, \(U(r) = U_0\left[\frac{2}{r^2} - \frac{1}{r}\right]\), includes a term of \(\frac{2}{r^2}\) which is indicative of an inverse square law.

• This term suggests that as the distance \(r\) between interacting bodies increases, the potential energy decreases proportionally to the square of the distance.
• This contrasts with the linear inverse \(\frac{1}{r}\) term in the equation, which decreases linearly with distance.

Understanding how distance impacts force and energy in this way is foundational for physics. By analyzing how the potential energy changes as a function of \(r\), we can predict how objects will move in response to these forces.

Turning Points

Turning points in physics describe positions where an object's direction of motion changes. For a particle subject to a potential energy field like \( U(r) \), turning points occur where the total energy \( E \) of the particle is equal to the potential energy, \( U(r) \).
To find these turning points, we set the given condition:

• Given \( E = -0.050 U_0 \), equate with \( U(r) \): \[ U_0\left(\frac{2}{r^2} - \frac{1}{r}\right) = -0.050 U_0 \]
• The roots or solutions for \(r\) represent the turning points.

By finding these values, we can predict the locations on the plot where the particle will reverse its motion. These points provide insight into the range and behavior of a particle in a potential field.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. It is defined by the equation \( KE = \frac{1}{2}mv^2 \), but in the context of potential energy and energy conservation, it is calculated as the difference between total energy and potential energy: \[ KE = E - U(r) \] The maximum kinetic energy occurs when the potential energy is minimized. For this exercise:

• Given the minimum potential energy occurs at \( r = 4 \).
• Maximum kinetic energy is thus found at this point, calculated as: \[ K_{max} = E - U(4) \]

Determining kinetic energy is essential for understanding particle velocity and dynamics under the influence of forces like gravity or electromagnetism.

Derivative of Potential Energy

The derivative of potential energy with respect to position \( r \), \( \frac{dU}{dr} \), helps identify minimum or maximum points of potential energy. This is analogous to finding where a function's graph peaks or troughs.
For the function \( U(r) = U_0\left[\frac{2}{r^2} - \frac{1}{r}\right] \), the derivative:\[ \frac{dU}{dr} = U_0\left(-\frac{4}{r^3} + \frac{1}{r^2}\right) \] Setting the derivative to zero will determine where \( U(r) \) hits a minimum or maximum.

• Here, solving \( -\frac{4}{r^3} + \frac{1}{r^2} = 0 \) yields the value \( r=4 \), where \( U(r) \) is minimal.

By understanding potential gradients, we can predict stable points of equilibrium and regions of stable or unstable configurations for a system.