Problem 29
Question
An object of mass \(m\), which is revolving in a circular orbit with constant angular velocity \(\omega\), is subject to the centrifugal force given by $$ \mathbf{F}(x, y, z)=m \omega^{2} \mathbf{r}=m \omega^{2}(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) $$ Show that \(f(x, y, z)=\frac{1}{2} m \omega^{2}\left(x^{2}+y^{2}+z^{2}\right)\) is a potential function for \(\mathbf{F}\).
Step-by-Step Solution
Verified Answer
The function given is a potential function for the force vector.
1Step 1: Understand the Definitions
The problem presents us with a force vector \( \mathbf{F}(x, y, z) = m \omega^2 \mathbf{r} \), where \( \mathbf{r} \) is the position vector \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \). Additionally, a potential function is a scalar function \( f(x, y, z) \) such that \( \mathbf{F} = -abla f \). Our task is to verify this relationship.
2Step 2: Compute the Gradient of the Potential Function
Calculate the gradient of \( f(x, y, z) = \frac{1}{2} m \omega^{2}(x^2 + y^2 + z^2) \). To do this, find the partial derivatives: \( \frac{\partial f}{\partial x} = m \omega^2 x \), \( \frac{\partial f}{\partial y} = m \omega^2 y \), and \( \frac{\partial f}{\partial z} = m \omega^2 z \). Therefore, \( abla f = (m \omega^2 x)\mathbf{i} + (m \omega^2 y)\mathbf{j} + (m \omega^2 z)\mathbf{k} \).
3Step 3: Compare the Force and the Gradient
We want to show \( \mathbf{F} = -abla f \). From Step 1, we have \( \mathbf{F} = m \omega^2 (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \). From Step 2, \( abla f = m \omega^2 (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \). Notice that \( \mathbf{F} = -abla f \) holds since \( \mathbf{F} \) is just the negative of \( abla f \).
4Step 4: Conclude the Verification
We've shown that the force vector \( \mathbf{F}(x, y, z) \) and the negative gradient of the potential function \( -abla f \) are equivalent, which proves that \( f(x, y, z)=\frac{1}{2} m \omega^{2}(x^{2}+y^{2}+z^{2}) \) is indeed a potential function for \( \mathbf{F} \).
Key Concepts
Centrifugal ForceGradientCircular MotionScalar Field
Centrifugal Force
Centrifugal force appears when an object moves in a circular path. It acts outwardly away from the center of rotation. Even though it is not a real force in a strict physical sense, it generates effects equivalent to pushing the object outward. Think of it as a push you feel when you take a sharp turn while driving a car.
For an object of mass \( m \) moving in a circle, the centrifugal force is given by \( \mathbf{F} = m \omega^2 \mathbf{r} \). Here, \( \omega \) represents the angular velocity, and \( \mathbf{r} \) is the radius vector from the center of rotation. Understanding this concept helps explain why planets orbit stars and why amusement park rides make you feel pressed against the seat.
For an object of mass \( m \) moving in a circle, the centrifugal force is given by \( \mathbf{F} = m \omega^2 \mathbf{r} \). Here, \( \omega \) represents the angular velocity, and \( \mathbf{r} \) is the radius vector from the center of rotation. Understanding this concept helps explain why planets orbit stars and why amusement park rides make you feel pressed against the seat.
Gradient
The gradient is a critical concept in multivariable calculus, denoting how a function changes as you move in space. For a potential function \( f(x, y, z) \), its gradient \( abla f \) provides a vector indicating the direction of the steepest increase of the function.
Computing the gradient involves finding the partial derivatives: \( \frac{\partial f}{\partial x} \), \( \frac{\partial f}{\partial y} \), and \( \frac{\partial f}{\partial z} \). For example, in our exercise, the gradient is calculated as \( abla f = (m \omega^2 x)\mathbf{i} + (m \omega^2 y)\mathbf{j} + (m \omega^2 z)\mathbf{k} \). This shows how the potential function changes spatially and is important for understanding force fields.
Computing the gradient involves finding the partial derivatives: \( \frac{\partial f}{\partial x} \), \( \frac{\partial f}{\partial y} \), and \( \frac{\partial f}{\partial z} \). For example, in our exercise, the gradient is calculated as \( abla f = (m \omega^2 x)\mathbf{i} + (m \omega^2 y)\mathbf{j} + (m \omega^2 z)\mathbf{k} \). This shows how the potential function changes spatially and is important for understanding force fields.
Circular Motion
Circular motion is when an object moves along the circumference of a circle. In this type of motion, an object keeps changing its direction as it moves but maintains a constant speed. The reason it can do this is due to the centripetal force, which keeps pulling it toward the center and the centrifugal perceived effect pushing outwards.
Angular velocity \( \omega \) describes how fast the object rotates, relating to how quickly the angle changes as it moves around the circle. For an orbiting object like a planet or a satellite, the forces involved are balanced in a way that maintains the circular path, leading to interesting physical properties and dynamics.
Angular velocity \( \omega \) describes how fast the object rotates, relating to how quickly the angle changes as it moves around the circle. For an orbiting object like a planet or a satellite, the forces involved are balanced in a way that maintains the circular path, leading to interesting physical properties and dynamics.
Scalar Field
A scalar field assigns a single value (a scalar) to every point in space. Think of it as a temperature map, where each point has a temperature value. This is simpler compared to vector fields where each point has both a magnitude and direction.
A potential function, like \( f(x, y, z)=\frac{1}{2} m \omega^{2}(x^{2}+y^{2}+z^{2}) \), is an example of a scalar field. The gradient of this potential function gives us a vector field that indicates forces. Scalar fields are foundational in physics for describing energies, electric potentials, and gravitational potentials, making them key for understanding spatial distributions and interactions.
A potential function, like \( f(x, y, z)=\frac{1}{2} m \omega^{2}(x^{2}+y^{2}+z^{2}) \), is an example of a scalar field. The gradient of this potential function gives us a vector field that indicates forces. Scalar fields are foundational in physics for describing energies, electric potentials, and gravitational potentials, making them key for understanding spatial distributions and interactions.
Other exercises in this chapter
Problem 28
Consider the velocity field \(\mathbf{v}(x, y, z)=-\omega y \mathbf{i}+\omega x \mathbf{j}\), \(\omega>0\) (see Example 2 and Figure 1). Note that \(\mathbf{v}\
View solution Problem 28
A squirrel weighing \(1.2\) pounds climbed a cylindrical tree by following the helical path \(x=\cos t, y=\sin t, z=4 t\), \(0 \leq t \leq 8 \pi\) (distance mea
View solution Problem 29
Let \(G\) be the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\). Evaluate each of the following: (a) \(\iint_{G} z d S\) (b) \(\iint_{G} \frac{x+y^{3}+\sin z}{1+z^{4}} d S\
View solution Problem 30
The scalar function \(\operatorname{div}(\operatorname{grad} f)=\nabla \cdot \nabla f\) (also written \(\left.\nabla^{2} f\right)\) is called the Laplacian, and
View solution