Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In one dimension, it is obvious that a force obeying Hooke's law is conservative (since \(F=-k x\) depends only on the position \(x,\) and this is sufficient to guarantee that \(F\) is conservative in one dimension). Consider instead a spring that obeys Hooke's law and has one end fixed at the origin, but whose other end is free to move in all three dimensions. (The spring could be fastened to a point in the ceiling and be supporting a bouncing mass \(m\) at its other end, for instance.) Write down the force \(\mathbf{F}(\mathbf{r})\) exerted by the spring in terms of its length \(r\) and its equilibrium length \(r_{\mathrm{o}} .\) Prove that this force is conservative. [Hints: Is the force central? Assume that the spring does not bend.]

Short Answer

Expert verified
The force is conservative with potential energy \( U(r) = \frac{1}{2}k (r - r_o)^2 \).

Step by step solution

01

Understand Hooke's Law in One Dimension

Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position. In one dimension, this is given by the equation \( F = -k x \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement.
02

Extend to Three Dimensions

In three dimensions, the position vector \( \mathbf{r} \) represents the displacement of the spring's end from the origin, with its magnitude being the length of the spring \( r = ||\mathbf{r}|| \). The equilibrium length of the spring is \( r_o \). The force exerted by the spring is directed along the line of displacement and can be expressed as \( \mathbf{F}(\mathbf{r}) = -k \left( \frac{r - r_o}{r} \right) \mathbf{r} \).
03

Express the Force in Vector Form

Since \( \mathbf{F}(\mathbf{r}) = -k (r - r_o) \frac{\mathbf{r}}{r} \), this representation shows that the force is equal to the spring constant \( k \) times the difference in lengths \( (r - r_o) \), scaled by the unit vector \( \frac{\mathbf{r}}{r} \) ensuring the force direction aligns with the displacement.
04

Determine if the Force is Central

A central force is a force that is directed along the line connecting two points (in this case the origin and the point \( \mathbf{r} \)) and depends only on the distance between them, \( r \). Our force \( \mathbf{F} \) satisfies these conditions, hence it is a central force.
05

Prove the Force is Conservative Using Potential Energy

For a force to be conservative, it must be derivable from a potential energy function \( U \), such that \( \mathbf{F}(\mathbf{r}) = -abla U \). Integrate \( \mathbf{F}(\mathbf{r}) = -k(r - r_o) \frac{\mathbf{r}}{r} \) with respect to \( \mathbf{r} \) to find \( U(r) = \frac{1}{2}k (r - r_o)^2 \).
06

Conclusion on Conservativeness

Since \( U(r) \) exists and \( \mathbf{F}(\mathbf{r}) = -abla U \), \( \mathbf{F} \) is indeed a conservative force.

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hooke's Law
Hooke's Law is a principle of physics that states the force needed to extend or compress a spring by some distance is proportional to that distance. In simpler terms, it can be imagined as a spring trying to pull or push back to its original length when stretched or compressed. The mathematical expression for Hooke's Law in one dimension is \( F = -k x \), where:
  • \( F \) is the force exerted by the spring, which is measured in Newtons (N).
  • \( k \) is the spring constant, indicating the stiffness of the spring, measured in N/m.
  • \( x \) is the displacement from the equilibrium position, measured in meters (m).
The negative sign in the formula signifies that the force exerted by the spring is in the opposite direction to the displacement. This opposition ensures that the force is always working to return the spring to its equilibrium state. When expanded into three dimensions, the principle remains the same. It still aims to return the system to equilibrium, while the direction and magnitude can be more complex due to vector quantities.
Central Force
A central force is defined as a force that is directed along the line joining two points. It only depends on the distance between these two points, and not on their specific positions. In the context of a spring that extends in three dimensions, the force behaves as a central force, because it focuses on the relative position and distance between the origin (the fixed point of the spring) and the end of the spring.
This can be shown mathematically by expressing the force vector, \( \mathbf{F}(\mathbf{r}) \), as \( -k (r - r_o) \frac{\mathbf{r}}{r} \). This formula shows:
  • The force acts directly along the radial direction from the origin.
  • The magnitude of the force is dependent solely on the difference in lengths \( (r - r_o) \).
Thus, the force remains central, indicating it constantly points towards or away from the center depending on whether the spring is compressed or stretched.
Potential Energy
Potential energy associated with a force is a form of energy stored within a system. For forces like those described by Hooke's Law, determining potential energy helps in confirming whether these forces are conservative. A force is considered conservative if it can be derived from a potential energy function, \( U \), such that the force can be obtained by the negative gradient of \( U \).
In our system, the potential energy function is given by: \[ U(r) = \frac{1}{2}k (r - r_o)^2 \]This function represents:
  • \( k \), the spring constant, revealing how stiff or soft the spring is.
  • \( r \), the current length of the spring.
  • \( r_o \), the natural or equilibrium length of the spring.
The squared term \((r - r_o)^2\) highlights that any displacement from equilibrium results in stored potential energy, either through compression or extension of the spring. This potential energy characterizes the system's ability to do work, i.e., returning the spring to its equilibrium position. By integrating the force with respect to \( \mathbf{r} \), we observe that it mirrors the change in potential energy, confirming the force's conservative nature. Such forces, including those governed by Hooke's Law, conserve mechanical energy within a system.

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

(a) Consider an electron (charge \(-e\) and mass \(m\) ) in a circular orbit of radius \(r\) around a fixed proton (charge \(+e\) ). Remembering that the inward Coulomb force \(k e^{2} / r^{2}\) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to \(-\frac{1}{2}\) times its \(\mathrm{PE}\); that is, \(T=-\frac{1}{2} U\) and hence \(E=\frac{1}{2} U\). (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius \(r\) around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy \(T_{2} .\) When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius \(r^{\prime} .\) (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy \(E\) long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables \(T_{2}, r,\) and \(r^{\prime}\).

A mass \(m\) is in a uniform gravitational field, which exerts the usual force \(F=m g\) vertically down, but with \(g\) varying with time, \(g=g(t) .\) Choosing axes with \(y\) measured vertically up and defining \(U=m g y\) as usual, show that \(\mathbf{F}=-\nabla U\) as usual, but, by differentiating \(E=\frac{1}{2} m v^{2}+U\) with respect to \(t,\) show that \(E\) is not conserved.

Verify that the gravitational force \(-G M m \hat{\mathbf{r}} / r^{2}\) on a point mass \(m\) at \(\mathbf{r},\) due to a fixed point mass \(M\) at the origin, is conservative and calculate the corresponding potential energy.

For a system of \(N\) particles subject to a uniform gravitational field g acting vertically down, prove that the total gravitational potential energy is the same as if all the mass were concentrated at the center of mass of the system; that is, \(U=\sum_{\alpha} U_{\alpha}=M g Y\) where \(M=\sum m_{\alpha}\) is the total mass and \(\mathbf{R}=(X, Y, Z)\) is the position of the \(\mathrm{CM},\) with the \(y\) coordinate measured vertically up. [Hint: We know from Problem 4.5 that \(\left.U_{\alpha}=m_{\alpha} g y_{\alpha} .\right]\)

An interesting one-dimensional system is the simple pendulum, consisting of a point mass \(m\), fixed to the end of a massless rod (length \(l\) ), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure \(4.26 .\) The pendulum's position can be specified by its angle \(\phi\) from the equilibrium position. (It could equally be specified by its distance \(s\) from equilibrium \(-\) indeed \(s=l \phi-\) but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is \(U(\phi)=m g l(1-\cos \phi)\). Write down the total energy \(E\) as a function of \(\phi\) and \(\dot{\phi}\). (b) Show that by differentiating your expression for \(E\) with respect to \(t\) you can get the equation of motion for \(\phi\) and that the equation of motion is just the familiar \(\Gamma=I \alpha\) (where \(\Gamma\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration \(\ddot{\phi}\) ). (c) Assuming that the angle \(\phi\) remains small throughout the motion, solve for \(\phi(t)\) and show that the motion is periodic with period \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

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.

Sign-up for free