Question

The potential energy corresponding to a certain two-dimensional force is given by \(U(x, y)=\frac{1}{2} k\left(x^{2}+y^{2}\right) .(a)\) Derive \(F_{x}\) and \(F_{y}\) and describe the vector force at each point in terms of its coordinates \(x\) and \(y .(b)\) Derive \(F_{r}\) and \(F_{\theta}\) and describe the vector force at each point in terms of the polar coordinates \(r\) and \(\theta\) of the point. ( \(c\) ) Can you think of a physical model of such a force?

Short answer

The force components in Cartesian coordinates are \( F_{x} = - kx \) and \( F_{y} = - ky \). In polar coordinates, the components are \( F_{r} = -kr \) and \( F_{\theta} = 0 \). A possible physical model of this force is the force exerted by a spring attached to the origin and free to move in two dimensions.

Step-by-step solution

Derive \(F_{x}\) and \(F_{y}\)

Using the relation \( F = - \nabla U \), where \( \nabla \) is the gradient operator, it can be seen that the vector force at each point in terms of its coordinates \( x \) and \( y \) can be computed by taking the gradient of the potential energy with respect to \( x \) and \( y \) getting \(F_{x} = -\frac{\partial U}{\partial x} = - kx\) and \(F_{y} = -\frac{\partial U}{\partial y} = - ky \). The force at any point is thus a vector \( \vec{F} = -k (x \hat{i} + y \hat{j}) \)

Derive \(F_{r}\) and \(F_{\theta}\)

Potential Energy is a function of \( r = \sqrt{x^{2} + y^{2}} \). So, \( F_{r} = -\frac{\partial U}{\partial r} = - kr \). Using \( r = x \cos \theta = y \sin \theta \), \( F_{\theta} = - \frac{\partial U}{\partial \theta} =0 \), as \( U \) doesn't depend upon \( \theta \). So, force in polar coordinates is \( \vec{F} = -kr \hat{r} \).

Physical interpretation

This equation for force can be taken as the force exerted by a spring with force constant \( k \) and equilibrium position at the origin. The force is always directed towards the origin, no matter in which direction \( x \) and \( y \) are oriented.

Key concepts

Two-Dimensional Force

In mechanics, a two-dimensional force is an action that occurs within a plane, impacting an object at various points characterized by two coordinates, typically denoted as x and y. These forces can influence objects in both horizontal and vertical directions.
They create movements along these axes, and can be represented mathematically in formulas where these coordinates appear.

• Represented as vectors: Having both magnitude and direction, making visualization important.
• Equations: Often rely on principles like Newton's laws to understand impacts.

In our study, we explore a two-dimensional force expressed through a potential energy function:\[U(x, y)=\frac{1}{2} k(x^{2}+y^{2})\]
This indicates that the force acts over distances involving x and y, aiding in understanding how objects move across a plane when subjected to such forces. This expression is central to deriving the various components of force, such as the force at each dimension through its derivatives.

Vector Force

A vector force is an important concept in physics, representing a force with both a direction and magnitude. It is often depicted by an arrow symbolizing the direction of the applied force, while the length corresponds to its strength.
Understanding vector forces helps in comprehending how and where forces will move an object.

• They are additive, meaning multiple vector forces can combine to create a resultant force.
• Expressed using unit vectors, generally as \( \hat{i} \) and \( \hat{j} \) in two-dimensional planes.

The solution to our exercise involves representing the force vector as \( \vec{F} = -k (x \hat{i} + y \hat{j}) \).
This means that for every increment in x and y, the force changes predictably, always pointing back towards the origin. Examining vector forces this way reveals how each portion of the potential energy contributes directionally, helping understand real-world behaviors.

Polar Coordinates

Polar coordinates offer a distinctive way to describe locations within a plane using a radial distance and angle, often more relevant when dealing with radially symmetric systems.
This method employs two variables:\( r \) for distance from the origin, and \( \theta \) for the angle in relation to the positive x-axis.
Switching our force analysis to polar coordinates can sometimes simplify understanding radial forces, such as those we've calculated in our problem.

• Expresses relationships like \( x = r \cos \theta \) and \( y = r \sin \theta \).
• Often preferred in circular or rotational dynamics.

In this exercise, the conversion from Cartesian to polar forms exposes another view of the force field as \( \vec{F} = -kr \hat{r} \).
Here, the force is directly dependent on the radial distance r, emphasizing that the force's magnitude increases with greater distance from the origin, while direction shifts are not influenced by \( \theta \), simplifying analysis.

Gradient Operator

The gradient operator \( abla \) is a vector operator denoting the directional changes in a scalar field, such as potential energy in our case.Operating like a slope within any given multidimensional space, it helps calculate where, and how fast, a quantity increases the most.
Utilizing \( abla \), potential energy differentiation becomes straightforward, providing insight on how forces are distributed.

• In two-dimensions, \( abla U = \left(\frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}\right) \).
• Useful in deriving the vector field direction given a potential energy function.

The problem leverages the gradient operator for identifying forces at cartesian points, indicating \[ F = - abla U \].
This mathematical tool thus offers a framework for visualizing force distributions in fields that rely on spatial coordinates and is critical to forming our understanding of how and why forces act.