Question

A particle of mass \(m\) moving in the \(x y\) -plane is confined by a two- dimensional potential function, \(U(x, y)=\frac{1}{2} k\left(x^{2}+y^{2}\right)\). a) Derive an expression for the net force, \(\vec{F}=F_{x} \hat{x}+F_{y} \hat{y}\) b) Find the equilibrium point on the \(x y\) -plane. c) Describe qualitatively the effect of the net force. d) What is the magnitude of the net force on the particle at the coordinate (3.00,4.00) in centimeters if \(k=10.0 \mathrm{~N} / \mathrm{cm} ?\) e) What are the turning points if the particle has \(10.0 \mathrm{~J}\) of total mechanical energy?

Short answer

Answer: The net force acts as a restoring force towards the origin, similar to a spring acting on the particle.

Step-by-step solution

Find the net force

To find the net force \(\vec{F}\), we need to take the negative gradient of the potential function \(U(x, y)\). The gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field and has a magnitude equal to the rate of increase in that direction. The gradient is given by: \(\vec{F} = - \nabla{U} = -\left(\frac{\partial U}{\partial x} \hat{x} + \frac{\partial U}{\partial y} \hat{y}\right)\) Now, find the partial derivatives of U: \(\frac{\partial U}{\partial x} = kx\) \(\frac{\partial U}{\partial y} = ky\) The net force is: \(\vec{F} = -kx \hat{x} - ky \hat{y}\)

Find the equilibrium point

The equilibrium point is the point at which the net force acting on the particle is zero. Setting each component of the net force to zero, we get: \(-kx = 0\) \(-ky = 0\) Solving these equations: \(x = 0\) \(y = 0\) Thus, the equilibrium point is at the origin on the xy-plane, coordinate \((0, 0)\).

Describe the net force qualitatively

The net force acts towards the origin, because the force components have opposite signs compared to the corresponding position components. In other words, if the particle is in the positive x and y directions, the force components Fx and Fy are negative, thus pushing the particle back towards the origin. This behavior describes a restoring force, similar to a spring acting on the particle.

Find the force magnitude at (3.00, 4.00)

The magnitude of the net force can be found using the expression for the net force and the given values of \(x = 3.00 \, cm\), \(y = 4.00 \, cm\), and \(k = 10.0 \frac{N}{cm}\): \(F_x = -kx = -10 \cdot 3 = -30 \, N\) \(F_y = -ky = -10 \cdot 4 = -40 \, N\) To find the magnitude of the net force, we can use the Pythagorean theorem: \(F = \sqrt{F_x^2 + F_y^2} = \sqrt{(-30)^2 + (-40)^2} = \sqrt{2500} = 50 \, N\)

Find turning points for 10 J of mechanical energy

When the particle has a total mechanical energy of \(10.0 \, J\), its potential energy must be less than or equal to that value. The potential energy is given by \(U(x, y) = \frac{1}{2}k(x^2 + y^2)\). We can solve for the maximum radius from the origin that the particle could reach: \(\frac{1}{2}k(x^2 + y^2) \leq 10\) \(x^2 + y^2 \leq \frac{20}{k}\) With \(k = 10.0 \frac{N}{cm}\), we have: \(x^2 + y^2 \leq \frac{20}{10} = 2\) The turning points are the boundary of a circle with radius \(\sqrt{2} \, cm\) centered at the origin.

Key concepts

Net Force Calculation

Understanding the net force acting on a particle within a potential field is crucial in analyzing its motion. In our example exercise, a particle of mass m is subjected to a two-dimensional potential function, which depends on its coordinates (x, y). To calculate the net force, we use the concept of the gradient of the potential function, which represents how the potential varies in space. The negative gradient is important because it yields the direction in which the potential decreases most rapidly, determining the force exerted on the particle.

The calculation involves taking partial derivatives of the potential function with respect to x and y. Students often find this concept challenging, so remember that the partial derivative is simply the derivative with respect to one variable while keeping all others constant. The net force is expressed as a vector with x and y components, and its calculation is the first step in analyzing the motion of the particle under the influence of this conservative force field.

Equilibrium Point Analysis

In physics, equilibrium points are locations where the net force acting on an object is zero. The significance of finding these points lies in their ability to tell us about the stability of an object's position within a force field. For the given problem, we identified these points by setting the net force expressions equal to zero.

An equilibrium point is also considered stable if, when perturbed, a restoring force will act to bring the particle back to equilibrium. In the context of the given potential, the only equilibrium point for the particle is at the origin, (0, 0). Students should also be aware that while an equilibrium point signifies no net force, this doesn't mean the particle must be at rest; it could be moving with constant velocity due to inertia.

Restoring Force

A restoring force is any force that always acts to pull a system back toward equilibrium whenever it is displaced from it. In our case, the force derived from the potential function acts in this manner. It is important for learners to visualize the symmetry in the potential function and how any displacement in the x or y direction leads to a force in the opposite direction, aiming to restore the system to equilibrium.

This concept is analogous to the behavior of a spring; the more you stretch or compress it, the stronger the force becomes, trying to revert it to its natural length. Visual aids and real-life examples can greatly help students in understanding this abstract concept, as it is fundamental in various areas of physics, from simple harmonic motion to electric fields.

Mechanical Energy

Mechanical energy encompasses both kinetic energy (energy of motion) and potential energy (energy stored within a force field). For a particle in a conservative force field, such as the one described by the provided two-dimensional potential function, the total mechanical energy remains constant as long as there is no external force applied.

When solving physics problems, determining the turning points of a particle—where its velocity is zero and hence its kinetic energy—is a crucial step. This occurs at the maximum potential energy the particle can have while still retaining the total mechanical energy. In our exercise, where mechanical energy is known, we calculate the distance from the origin within which the particle can move. Understanding the conservation of mechanical energy is key for students, as it serves as a powerful tool for analyzing physical situations without the nitty-gritty details of forces and motion at every point.