# Circular motion (approximation of centripetal acc.)

1. May 1, 2012

### V0ODO0CH1LD

Say I have a body moving in a circle of radius r with a constant velocity v.

The time it takes the body to go around the circumference once is:

T = 2πr/v

Then the time it takes the body to go around a fourth of the circumference is T/4.

Now, imagine a diagram where when the body is at the leftmost portion of the circumference its velocity vector is pointing straight up, and when it is at the upmost portion of the circumference its velocity vector points entirely to the right. That is the body travelling over a fourth of the circumference.

Now, the magnitude of the acceleration vector required to make the body behave that way over a period of T/4 is (v√2)/(T/4). And (v√2)/(T/4) is an approximation of the centripetal acceleration for this case.

My question is: why is that an approximation? Why if I do the same thing over a smaller change in time I get closer to the centripetal acceleration? A circle is symmetric and the velocity is constant, right?

2. May 1, 2012

### Redbelly98

Staff Emeritus
Agreed, so far.
Not quite. This is the magnitude of the average acceleration, where the average acceleration is defined as
$$\vec{a_{avg}}=\frac{\Delta \vec{v}}{\Delta t}$$

Because it is an average, and the actual instantaneous acceleration changes substantially (because it changes direction) during the time interval.
Because, for a small time interval, the acceleration does not change direction very much, and is more closely approximated by its average over that time interval.
No, velocity is a vector and is not constant if its direction changes. Speed is constant, however.

3. May 1, 2012

### V0ODO0CH1LD

That helped a lot! Thanks!

One for additional question though. How does all of that get summarised into (ω^2)*r or (v^2)/r?

4. May 1, 2012

### Redbelly98

Staff Emeritus