# Radius as a function of Uniform Circular Motion

1. Feb 8, 2013

How exactly does radius affect centripetal acceleration? In one formula we have (a=v^2/r) which implies inversely proportional, while in the other we have (a=(w^2)*r) which implies directly proportional. I understand that increasing r increases velocity (v=w*r) which means the right answer is increasing r increases acceleration, however how do we justify that increasing r would decrease a in the formula (a=v^2/r)?

Also how can I intuitively makes sense of this? Does increasing r not mean that the same velocity has more time to change (as Khan Academy states at 3:29-4:32) -->

This MIT lecture at 5:40-6:21, states that the acceleration would decrease as radius decreases.

So which is it ? They seem to be saying opposite things. Am I misunderstanding something?

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Last edited by a moderator: Sep 25, 2014
2. Feb 8, 2013

### BruceW

well, it depends if you want to hold omega fixed, or velocity fixed. The idea is that the centripetal acceleration must be a specific value, so that a perfect circle is made. So to keep to that requirement, if you have a larger radius, you must have a smaller omega. And equivalently, you must have a larger velocity.

3. Feb 11, 2013

### mickybob

So you've got a rotating disk, and you're interesting in the centripetal acceleration acting on a point on the disk at distance r from the centre of rotation.

For a given angular rotational frequency, the centripetal acceleration acting on the point is proportional to r, as your second equation says.

The reason your first equation seems to disagree is it because it contains v, the velocity of the specific point you are looking at. But the velocity of that point also depends on its distance from r, from the equation v = wr. So when you make r bigger, you are also making v bigger.

Substitute v=wr into your first equation and you will see that both are saying the same thing.

Last edited by a moderator: Sep 25, 2014