/*! 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 25 Prove that the potential energy ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Prove that the potential energy of a central force \(\mathbf{F}=-k r^{n} \hat{\mathbf{r}}(\text { with } n \neq-1)\) is \(U=k r^{n+1} /(n+1)\). In particular, if \(n=1,\) then \(\mathbf{F}=-k \mathbf{r}\) and \(U=\frac{1}{2} k r^{2}\).

Short Answer

Expert verified
Potential energy is \( U = \frac{k r^{n+1}}{n+1} \); for \( n=1 \), \( U = \frac{1}{2} kr^2 \).

Step by step solution

01

Define Potential Energy

Potential energy, denoted as \( U \), is defined for a central force \( \mathbf{F} = -abla U \). This means the force is the negative gradient of the potential energy.
02

Relate Force and Potential Energy

The force \( \mathbf{F} = -k r^n \hat{\mathbf{r}} \) is acting radially, so we can write \(-abla U = -k r^n \hat{\mathbf{r}} \). We assume \( U = U(r) \) depends solely on \( r \), leading to \(-\frac{dU}{dr} = -k r^n \).
03

Integrate to Find U(r)

Integrate \(-\frac{dU}{dr} = -k r^n \) with respect to \( r \) to find the potential energy function. This gives us \( U(r) = \int k r^n \, dr = k \frac{r^{n+1}}{n+1} + C \), where \( C \) is an integration constant.
04

Apply Specific Case (n=1)

Substitute \( n=1 \) into the derived formula \( U(r) = \frac{k r^{n+1}}{n+1} \). This yields \( U(r) = \frac{k r^2}{2} \). The specific case of \( \mathbf{F} = -k \mathbf{r} \) corresponds to this expression for potential energy.

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.

Central Force
When we talk about central forces, we are referring to forces that act along a line that passes through a central point. This point is often the origin in a coordinate system. Such forces are always directed towards or away from this central point, and their magnitude depends only on the distance from the center.
A central force can be represented mathematically as \( \mathbf{F} = -k r^n \hat{\mathbf{r}} \), where:- \( k \) is a constant that determines the strength of the force.- \( r \) is the radial distance from the origin.- \( n \) is the power through which this distance is raised, affecting how the force changes with distance.
This kind of force is quite common in physics. Examples include gravitational and electrostatic forces, where the force strength depends on the inverse square of the distance (when \( n = -2 \)). Understanding central forces is key to grasping concepts like orbiting planets and charged particle interactions.
Gradient of Potential Energy
The concept of the gradient comes from calculus and is crucial to understanding potential energy in the context of forces. The gradient essentially provides the rate and direction of change of a quantity. In the context of physics, it's used to derive the force exerted by a potential energy field.
The force associated with potential energy is given by the negative gradient of that potential energy \( U \), and is expressed as \( \mathbf{F} = -abla U \). This means:
  • The force points in the direction where potential energy decreases the fastest.
  • The magnitude of the force indicates how steeply the potential energy decreases.

This concept helps in finding the equilibrium positions and understanding stability, as areas where the gradient (or force) is zero correspond to potential minima or maxima. It provides a bridge between energy landscapes and the forces that drive physical systems.
Integration
Integration is a mathematical tool that is used to determine the whole from its parts. In the context of this exercise, it's the key to finding the potential energy function from a given force. When we say we integrate a force function, we mean we are calculating the accumulated effect of this force over a path or distance.
For the central force \( \mathbf{F} = -k r^n \hat{\mathbf{r}} \), following the expression \( -\frac{dU}{dr} = -k r^n \), the integration process yields the potential energy function \( U(r) \).
  • Integrating \( -k r^n \) with respect to \( r \), we compute: \[ U(r) = \int k r^n \, dr = \frac{k r^{n+1}}{n+1} + C \]
  • Here, \( C \) is the constant of integration, representing potential energy at a reference position.

Through integration, we transform force data into an understandable energy framework, facilitating the analysis of mechanical systems and predictions of motion dynamics.
Radial Force
Radial force is a term used to describe any force that acts along a radius from a central point, either towards or away from it. This kind of force is fundamental in systems with spherical symmetry, like planets orbiting stars or electrons circling nuclei.
Radial forces simplify many analysis and calculation processes because they maintain symmetry in relation to the center point. In our central force example \( \mathbf{F} = -k r^n \hat{\mathbf{r}} \), the term \( \hat{\mathbf{r}} \) is a unit vector pointing radially outward from the center. This notation indicates that the force vector points along the radius:
  • Radial symmetry ensures that the force's effect only depends on the distance from the center (\( r \)) and not the direction in space.
  • This makes it easier to calculate potential energies and predict the behavior of objects under such forces.

Radial forces are crucial in simplifying the mathematics of many physical systems, allowing us to focus only on how force and potential change with distance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two equal masses, \(m_{1}=m_{2}=m,\) are joined by a massless string of length \(L\) that passes through a hole in a frictionless horizontal table. The first mass slides on the table while the second hangs below the table and moves up and down in a vertical line. (a) Assuming the string remains taut, write down the Lagrangian for the system in terms of the polar coordinates \((r, \phi)\) of the mass on the table. (b) Find the two Lagrange equations and interpret the \(\phi\) equation in terms of the angular momentum \(\ell\) of the first mass. (c) Express \(\dot{\phi}\) in terms of \(\ell\) and eliminate \(\dot{\phi}\) from the \(r\) equation. Now use the \(r\) equation to find the value \(r=r_{0}\) at which the first mass can move in a circular path. Interpret your answer in Newtonian terms. (d) Suppose the first mass is moving in this circular path and is given a small radial nudge. Write \(r(t)=r_{0}+\epsilon(t)\) and rewrite the \(r\) equation in terms of \(\epsilon(t)\) dropping all powers of \(\epsilon(t)\) higher than linear. Show that the circular path is stable and that \(r(t)\) oscillates sinusoidally about \(r_{\mathrm{o}}\) What is the frequency of its oscillations?

A mass \(m_{1}\) rests on a frictionless horizontal table and is attached to a massless string. The string runs horizontally to the edge of the table, where it passes over a massless, frictionless pulley and then hangs vertically down. A second mass \(m_{2}\) is now attached to the bottom end of the string. Write down the Lagrangian for the system. Find the Lagrange equation of motion, and solve it for the acceleration of the blocks. For your generalized coordinate, use the distance \(x\) of the second mass below the tabletop.

Let \(F=F\left(q_{1}, \cdots, q_{n}\right)\) be any function of the generalized coordinates \(\left(q_{1}, \cdots, q_{n}\right)\) of a system with Lagrangian \(\mathcal{L}\left(q_{1}, \cdots, q_{n}, \dot{q}_{1}, \cdots, \dot{q}_{n}, t\right) .\) Prove that the two Lagrangians \(\mathcal{L}\) and \(\mathcal{L}^{\prime}=\mathcal{L}+d F / d t\) give exactly the same equations of motion.

A smooth wire is bent into the shape of a helix, with cylindrical polar coordinates \(\rho=R\) and \(z=\lambda \phi,\) where \(R\) and \(\lambda\) are constants and the \(z\) axis is vertically up (and gravity vertically down). Using \(z\) as your generalized coordinate, write down the Lagrangian for a bead of mass \(m\) threaded on the wire. Find the Lagrange equation and hence the bead's vertical acceleration \(\ddot{z}\). In the limit that \(R \rightarrow 0\), what is \(\ddot{z} ?\) Does this make sense?

Consider the well-known problem of a cart of mass \(m\) moving along the \(x\) axis attached to a spring (force constant \(k\) ), whose other end is held fixed (Figure 5.2 ). If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency \(\omega=\sqrt{k / m} .\) Using the Lagrangian approach, you can find the effect of the spring's mass \(M,\) as follows: (a) Assuming that the spring is uniform and stretches uniformly, show that its kinetic energy is \(\frac{1}{6} M \dot{x}^{2} .\) (As usual \(x\) is the extension of the spring from its equilibrium length.) Write down the Lagrangian for the system of cart plus spring. (Note: The potential energy is still \(\frac{1}{2} k x^{2}\).) (b) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency \(\omega=\sqrt{k /(m+M / 3)} ;\) that is, the effect of the spring's mass \(M\) is just to add \(M / 3\) to the mass of the cart.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.