Chapter 14: Problem 29
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 \(\mathbf{F}\).
Short Answer
Step by step solution
Understanding the Problem
Calculate the Gradient of the Potential Function
Partial Derivative with respect to x
Partial Derivative with respect to y
Partial Derivative with respect to z
Assemble the Gradient Vector
Relate the Gradient to the Force
Conclusion
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Potential Function
Gradient
- \(\frac{\partial f}{\partial x}\): The rate of change of \(f\) with respect to the \(x\)-coordinate.
- \(\frac{\partial f}{\partial y}\): The rate of change of \(f\) with respect to the \(y\)-coordinate.
- \(\frac{\partial f}{\partial z}\): The rate of change of \(f\) with respect to the \(z\)-coordinate.
Circular Motion
- The mass \(m\) of the object.
- The square of the angular velocity \(\omega\).
- The radius of the circular path \(r\).
Angular Velocity
- Magnitude: The rate of rotation in radians per second.
- Direction: Perpendicular to the plane of motion.