Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

171 submissions , 136 unreviewed
4,258 questions , 1,606 unanswered
5,064 answers , 21,511 comments
1,470 users with positive rep
615 active unimported users
More ...

  Circular motion at $Re\ll 1$

+ 0 like - 0 dislike
22 views

Let's assume a bead of mass $m$ is moving on a circular trajectory in a fluid, and that the typical velocity, the radius of the bead $a$ and the viscoisity of the fluid $\eta$ are such that we are in a regime with $Re\ll1$. What are the driving forces necessary to sustain uniform circular motion with radius $R$ and angular velocity $\omega$?

The derivative of the velocity vector can be written as

$$\frac{d\vec{v}}{dt}=\frac{d\left(\omega R\hat{u}_{\theta}\right)}{dt}=\frac{d\omega}{dt}\hat{u}_{\theta}+\frac{d\hat{u}_{\theta}}{dt}\omega=F_{\theta}\hat{u}_{\theta}(t)+F_{R}\hat{u}_{R}(t)$$

where $\hat{u}_{\theta}$ is the tangent vector at an angle $\theta$.

In order to make the first term of the RHS vanish, I write the force balance tangentially to the circle, in which I have an unknown driving force and the hydrodynamic drag:

$$ F_{\theta}=F_{\theta}^{(driving)}+F_{\theta}^{(drag)}=0$$

In this regime I can write

$$F_{\theta}^{(drag)}=-6\pi\eta av=-6\pi\eta a\ \omega R$$

$$\Longrightarrow F_{\theta}^{(driving)}=6\pi\eta aR\ \omega$$

and this gives me the tangential component of the driving force needed.

What about the normal component? In this regime, does $F_{R}=-\omega^{2}R$ make any sense? Does the hydrodynamic drag have a component along $\hat{u}_{R}(t)$?

asked Apr 10 in Theoretical Physics by usumdelphini (0 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...