# Doubt with centripetal acceleration

1. Sep 17, 2010

### pc2-brazil

Good afternoon,

For a body in a curvilinear motion moving with velocity $$\vec{v}$$, velocity can be divided into two components (see http://en.wikipedia.org/wiki/Angular_velocity#Two_dimensions" Wikipedia article): a component parallel to the position vector (the radius), which changes the magnitude of the radius and is given by dr/dt, and a component perpendicular to the radius vector, called tangential velocity (vt or $$v_{\bot}$$), which changes the direction of the radius vector.
The angular velocity is determined by the the tangential velocity:
$$\omega = \frac{v_t}{r}$$
And the centripetal acceleration is given by:
$$a_c = \frac{v^2}{r}$$ (1)
My question is: I have seen, in many places, that the centripetal acceleration can be written as:
$$a_c = r\omega^2 = \frac{v_t^2}{r}$$ (2)
But it doesn't seem right. Why is it defined like this so frequently?
EDIT: I realized that, in definition (2), "r" is the polar radius, and not necessarily the radius of curvature of the trajectory. Definition (1) is true when "r" is the radius of curvature. Am I right?

Thank you in advance.

Last edited by a moderator: Apr 25, 2017
2. Sep 17, 2010

### mathman

At any given point on the circle, the velocity is the tangential velocity.

3. Sep 18, 2010

### pc2-brazil

Yes, but I defined tangential velocity as the component of velocity perpendicular to the position vector, like the Wikipedia article to which I linked.
What I'm trying to understand is why definition (2) uses $$\omega$$ = vt/r, while it seems that it should use v, not vt, in order to be compatible with (1) (I must be doing some confusion here).

Last edited: Sep 18, 2010
4. Sep 18, 2010

### mathman

I am not sure what the confusion is, but vt=v.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook