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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

Consider a small frictionless puck perched at the top of a fixed sphere of radius \(R\). If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck's speed as a function of its height, then use Newton's second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?]

(a) The force exerted by a one-dimensional spring, fixed at one end, is \(F=-k x,\) where \(x\) is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is \(U=\frac{1}{2} k x^{2},\) if we choose \(U\) to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass \(m\) suspended from the other end and constrained to move in the vertical direction only. Find the extension \(x_{\mathrm{o}}\) of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form \(\frac{1}{2} k y^{2}\) if we use the coordinate \(y\) equal to the displacement measured from the new equilibrium position at \(x=x_{\mathrm{o}}\) (and redefine our reference point so that \(U=0\) at \(y=0\) ).

(a) Consider a mass \(m\) in a uniform gravitational field \(\mathbf{g},\) so that the force on \(m\) is \(m \mathbf{g},\) where \(\mathbf{g}\) is a constant vector pointing vertically down. If the mass moves by an arbitrary path from point 1 to point \(2,\) show that the work done by gravity is \(W_{\mathrm{grav}}(1 \rightarrow 2)=-m g h\) where \(h\) is the vertical height gained between points 1 and 2. Use this result to prove that the force of gravity is conservative (at least in a region small enough so that \(\mathrm{g}\) can be considered constant). (b) Show that, if we choose axes with \(y\) measured vertically up, the gravitational potential energy is \(U=m g y\) (if we choose \(U=0\) at the origin).

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 \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):\) (a) \(f=x^{2}+z^{3} .\) (b) \(f=k y\), where \(k\) is a constant. (c) \(f=r \equiv \sqrt{x^{2}+y^{2}+z^{2}} .\) [Hint: Use the chain rule.] (d) \(f=1 / r\).
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