Centripetal acceleration with constant speed

Question

If a particle is moving in a circular path at a constant speed, how can we represent its centripetal acceleration? Since the speed is constant, I assume that the centripetal acceleration describes how quickly the particle is changing direction. To obtain this, we would divide the total length of the circular path by the particle's speed, and that would show us how long it takes for the particle's velocity to reach back to its original vector. Is my thinking on the right track?

Answer 1

The easiest was is to figure out how long a period is, $T$. The velocity vector then traces out a circle in velocity space with radius $v$.

The centripetal acceleration is the length of that path divided by the period:

$$ a= \frac{2\pi v} T $$

Answer 2

Acceleration is change in velocity per unit time, which has units of length per unit time per unit time. If you divide the length of the circle by the speed, you will get units of time, which is not right.

One way to find the acceleration is to first write down the particle's motion as a function of $x$ and $y$, the horizontal and vertical axes respectively. For a particle moving at constant speed, this is:

$$x(t) = r \cdot \cos(At),$$

$$y(t) = r \cdot \sin(At),$$

where $r$ is the radius of the circle, $t$ is the time, and $A$ is a constant that we will determine. We know that these are the correct equations because we can verify that the particle will always be a distance $r$ from the origin:

$$distance = \sqrt{x^2 + y^2} = \sqrt{r^2\cos^2(At) + r^2\sin^2(At)} = \sqrt{r^2[\cos^2(At) + \sin^2(At)]} = \sqrt{r^2} = r.$$

And we can also verify that the speed, which is the magnitude of the derivative of the position with respect to time, is always constant:

$$x'(t) = -rA \cdot \sin(At),$$ $$y'(t) = rA \cdot \cos(At),$$ $$speed = \sqrt{x'(t)^2 + y'(t)^2} = rA.$$

Now, we can just take the derivative with respect to time once more, and once more look at the magnitude, and that will give us the acceleration:

$$x''(t) = -rA^2 \cdot \cos(At),$$ $$y''(t) = -rA^2 \cdot \sin(At),$$ $$acceleration = \sqrt{x''(t)^2 + y''(t)^2} = rA^2 = \frac{(rA)^2}{r} = \frac{(speed)^2}{r}.$$

Thus we see that the acceleration is equal to the square of the speed divided by the radius.