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Chapter 4: Problem 16

If a particle's potential energy is \(U(\mathbf{r})=k\left(x^{2}+y^{2}+z^{2}\right),\) where \(k\) is a constant, what is the force on the particle?

Short Answer

Expert verified
The force on the particle is \((-2kx, -2ky, -2kz)\).

Step by step solution

01

Understanding Potential Energy Function

The potential energy of the particle is given as a function of position \(U(\mathbf{r})=k(x^2+y^2+z^2)\). This is a scalar function of the particle's position in three-dimensional space, where \(x, y, \) and \(z\) are the coordinates.
02

Recall the Force-Potential Energy Relationship

The force \(\mathbf{F}\) on a particle is related to its potential energy \(U(\mathbf{r})\) by \( \mathbf{F} = -abla U \), where \(abla U\) represents the gradient of the potential energy function. The negative sign indicates that the force is in the direction of decreasing potential energy.
03

Compute the Gradient of the Potential Energy

The gradient operation \(abla\) in Cartesian coordinates is given by \( abla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) \). Compute each partial derivative of \(U(\mathbf{r})\):\( \frac{\partial U}{\partial x} = \frac{\partial}{\partial x} [k(x^2+y^2+z^2)] = 2kx \)\( \frac{\partial U}{\partial y} = \frac{\partial}{\partial y} [k(x^2+y^2+z^2)] = 2ky \)\( \frac{\partial U}{\partial z} = \frac{\partial}{\partial z} [k(x^2+y^2+z^2)] = 2kz \)Thus, the gradient is \( abla U = (2kx, 2ky, 2kz) \).
04

Determine the Force on the Particle

Using the relation \( \mathbf{F} = -abla U \), substitute the gradient we calculated:\[ \mathbf{F} = -(2kx, 2ky, 2kz) = (-2kx, -2ky, -2kz) \]. This is the force acting on the particle.

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Key Concepts

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

Force
Imagine a particle in a potential energy field. The goal is to find out the force acting on it. Force is a vector quantity—it has both magnitude and direction.

In this context, the force (\( \mathbf{F} \)) on a particle is derived from the potential energy function, a scalar. The remarkable aspect is its opposing direction to the gradient of the potential energy. This relationship emphasizes that force seeks to reduce potential energy, guiding a particle towards a state of lower energy.

You can think of it intuitively: force tries to "pull" or "push" objects to minimize the potential energy. Understanding how this works helps us design systems ranging from rollercoasters to rockets by managing how and where a particle will move.
Gradient
To understand force in a potential energy field, we must appreciate the role of gradients. The gradient (\( abla U \)) is a vector that points in the direction of the greatest increase of a function, like potential energy, and whose magnitude is the rate of increase.

In simple terms, the gradient tells us the steepness and direction of the slope for a function. For a hill, the gradient points uphill—it's the direction you'd climb to ascend fastest. In the case of potential energy, the force acts in the opposite direction to the gradient, as forces often drive systems to lower energy states.

Computing a gradient involves understanding its components: partial derivatives with respect to each coordinate. The resultant vector reveals how the potential energy changes at each point, an essential insight for predicting the behavior of physical systems.
Potential Energy Function
The potential energy function, like the given (\( U(\mathbf{r}) = k(x^2 + y^2 + z^2) \)), embodies how energy is stored in a system based on position.

The function is scalar, meaning it gives a single energy value for any combination of coordinates (\( x, y, z \)). The constant (\( k \)) determines the strength of the potential field. For a particle, its position determines the potential energy, which can be thought of as 'stored energy' that can be converted to kinetic energy.

Understanding a potential energy function is crucial for predicting a system's dynamics—how particles will move or interact. It's like having a map of energy landscapes, enabling the design and analysis of a multitude of physical phenomena from gravitational fields to molecular interactions.
Partial Derivatives
Partial derivatives are the building blocks for understanding changes in multi-variable functions like our potential energy function (\( U(\mathbf{r}) \)).

They represent how a function changes as one particular variable shifts, keeping others constant. In Cartesian coordinates,(\( \frac{\partial U}{\partial x} \)) tells you how potential energy changes as you slide horizontally, while (\( \frac{\partial U}{\partial y} \)) and (\( \frac{\partial U}{\partial z} \)) do the same for vertical and depth changes, respectively.

Calculating these derivatives gives insight into the behavior of the potential energy landscape from every directional perspective. This analysis is vital for constructing the gradient, allowing us to understand how every minute change in position affects the potential energy of a particle, and ultimately defining the force it experiences.
  • Helps in creating the gradient vector
  • Provides detailed variation insights
  • Essential for predicting motion in fields

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Most popular questions from this chapter

A particle of mass \(m_{1}\) and speed \(v_{1}\) collides with a second particle of mass \(m_{2}\) at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that \(m_{1} \ll m_{2}\) and that \(m_{2} \ll m_{1}\).

Prove that if \(f(\mathbf{r})\) and \(g(\mathbf{r})\) are any two scalar functions of \(\mathbf{r},\) then \(\nabla(f g)=f \nabla g+g \nabla f\)

Both the Coulomb and gravitational forces lead to potential energies of the form \(U=\gamma / | \mathbf{r}_{1}-\) \(\mathbf{r}_{2} |,\) where \(\gamma\) denotes \(k q_{1} q_{2}\) in the case of the Coulomb force and \(-G m_{1} m_{2}\) for gravity, and \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) are the positions of the two particles. Show in detail that \(-\nabla_{1} U\) is the force on particle 1 and \(-\nabla_{2} U\) that on particle 2.

The proof that the condition \(\nabla \times \mathbf{F}=0\) guarantees the path independence of the work \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) done by \(\mathbf{F}\) is unfortunately too lengthy to be included here. However, the following three exercises capture the main points: \(^{16}\) (a) Show that the path independence of \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) is equivalent to the statement that the integral \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}\) around any closed path \(\Gamma\) is zero. (By tradition, the symbol \(\oint\) is used for integrals around a closed path \(-\) a path that starts and stops at the same point.) [Hint: For any two points 1 and 2 and any two paths from 1 to 2 , consider the work done by \(\mathbf{F}\) going from 1 to 2 along the first path and then back to 1 along the second in the reverse direction. \((\) b) Stokes's theorem asserts that \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=\int(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A,\) where the integral on the right is a surface integral over a surface for which the path \(\Gamma\) is the boundary, and \(\hat{\mathbf{n}}\) and \(d A\) are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if \(\nabla \times \mathbf{F}=0\) everywhere, then \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=0 .\) (c) While the general proof of Stokes's theorem is beyond our scope here, the following special case is quite easy to prove (and is an important step toward the general proof): Let \(\Gamma\) denote a rectangular closed path lying in a plane perpendicular to the \(z\) direction and bounded by the lines \(x=B, x=B+b, y=C\) and \(y=C+c .\) For this simple path (traced counterclockwise as seen from above), prove Strokes's theorem that \(\oint_{\Gamma} \mathbf{F} \cdot d \mathbf{r}=\int(\mathbf{\nabla} \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A\) where \(\hat{\mathbf{n}}=\hat{\mathbf{z}}\) and the integral on the right runs over the flat, rectangular area inside \(\Gamma\). [Hint: The integral on the left contains four terms, two of which are integrals over \(x\) and two over \(y\). If you pair them in this way, you can combine each pair into a single integral with an integrand of the form \(F_{x}(x, C+c, z)-F_{x}(x, C, z)\) (or a similar term with the roles of \(x\) and \(y\) exchanged). You can rewrite this integrand as an integral over \(y\) of \(\partial F_{x}(x, y, z) / \partial y\) (and similarly with the other term), and you're home.]

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 \(\left.U_{\alpha}=m_{\alpha} g y_{\alpha} .\right]\)
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