Centrifugal force and angular velocity

[Lsos]

According to the wikipedia page, when given an angular velocity, centrifugal force increases with radius. I always thought the larger the radius, the smaller the centrifugal force. I think I'm misunderstanding some term here (possibly "angular velocity"). Can someone please explain?

 

[jfizzix]

Angular velocity $\omega$ is the rate at which angle changes with respect to time (revolutions per second, degrees per second, radians per second, etc). If this rate is a constant, then the larger your distance to the center $r$ is, the larger the centrifugal force $F$ you will feel.

$F = m \;r\;\omega^{2}$

However, if you are saying your tangential velocity $v = r\;\omega$ is constant, then the farther out from the center you are, the smaller the centrifugal force you experience, but this also means your angular velocity is smaller too.

$F = m \frac{v^{2}}{r}$

 

[tiny-tim]

Hi Lsos!

… given an angular velocity, centrifugal force increases with radius.

Yes, if the angular velocity is fixed, the larger the radius, the more force you need to keep something in the circle.

Loosely speaking, changing the velocity from v to -v in the same time is obviously a larger acceleration if v is larger!

I always thought the larger the radius, the smaller the centrifugal force. I think I'm misunderstanding some term here (possibly "angular velocity"). Can someone please explain?

Angular velocity is angle per second.

It's equal to revolutions per second times 2π.

 

[quarkgazer]

Hi,
The centrifugal force is a fictitious force used in non-inertial reference frames. In general, it is given by $$ F_{\text{centrifugal}} = m\textbf{w} \times (\textbf{w}\times\textbf{r}). $$ Here, $\textbf{w}$ is the angular velocity of the rotating reference frame and $\textbf{r}$ is the position of the particle relative to the rotating frame's origin.

Usually it will simplify to jfizzix' answer (the first equation).

From a more qualitative perspective, imagine a large disc spinning. If you stand near the centre, you move in a small circle but not very fast. As you move out, you move faster and faster and you will find it more difficult to stay on the disc.

:)

PS: I used $\textbf{w}$ because for some reason the TeX code for omega isn't working... :\

 

[Andrew Mason]

PS: I used $\textbf{w}$ because for some reason the TeX code for omega isn't working... :\

I think your use of the bold face text is the problem. Latex assumes what follows is text. You can use vectors like this$$ F_{centrifugal} = m\vec{\omega} \times (\vec{\omega} \times \vec{r}) $$

AM

 

[Lsos]

Ok...yeah. I think I get it. Seems like a very simple concept, but I just need to wrap my aging mind around this. Thanks everyone!