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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.

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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.

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Most popular questions from this chapter

Consider a bead that is threaded on a rigid circular hoop of radius \(R\) lying in the \(x y\) plane with its center at \(O,\) and use the angle \(\phi\) of two- dimensional polar coordinates as the one generalized coordinate to describe the bead's position. Write down the equations that give the Cartesian coordinates \((x, y)\) in terms of \(\phi\) and the equation that gives the generalized coordinate \(\phi\) in terms of \((x, y)\).

A mass \(m_{1}\) rests on a frictionless horizontal table. Attached to it is a string which runs horizontally to the edge of the table, where it passes over a frictionless, small pulley and down to where it supports a mass \(m_{2} .\) Use as coordinates \(x\) and \(y\) the distances of \(m_{1}\) and \(m_{2}\) from the pulley. These satisfy the constraint equation \(f(x, y)=x+y=\) const. Write down the two modified Lagrange equations and solve them (together with the constraint equation) for \(\ddot{x}, \ddot{y},\) and the Lagrange multiplier \(\lambda\). Use (7.122) (and the corresponding equation in \(y\) ) to find the tension forces on the two masses. Verify your answers by solving the problem by the elementary Newtonian approach.

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 a double Atwood machine constructed as follows: A mass 4 \(m\) is suspended from a string that passes over a massless pulley on frictionless bearings. The other end of this string supports a second similar pulley, over which passes a second string supporting a mass of \(3 m\) at one end and \(m\) at the other. Using two suitable generalized coordinates, set up the Lagrangian and use the Lagrange equations to find the acceleration of the mass \(4 m\) when the system is released. Explain why the top pulley rotates even though it carries equal weights on each side.

Consider two particles moving unconstrained in three dimensions, with potential energy \(U\left(\mathbf{r}_{1}, \mathbf{r}_{2}\right) .\) (a) Write down the six equations of motion obtained by applying Newton's second law to each particle. (b) Write down the Lagrangian \(\mathcal{L}\left(\mathbf{r}_{1}, \mathbf{r}_{2}, \dot{\mathbf{r}}_{1}, \dot{\mathbf{r}}_{2}\right)=T-U\) and show that the six Lagrange equations are the same as the six Newtonian equations of part (a). This establishes the validity of Lagrange's equations in rectangular coordinates, which in turn establishes Hamilton's principle. since the latter is independent of coordinates, this proves Lagrange's equations in any coordinate system.

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