What does the value of acceleration in centripetal motion even mean?

Question

Say an object under constant speed, while undergoing centripetal motion, is accelerating at a speed of $8$ $\text{m/s}^{2}$.

Usually, the figure "$8$ $\text{m/s}^{2}$" (or 8 m/s per second) tells us that the magnitude of the velocity of the body increases $8$ $\text{m/s}$ each second, but how does that notion apply for the aforementioned object? Its velocity isn't changing at all...

Answer 1

In basic terms, the magnitude is not the only component of a vector like acceleration: its direction is, too. When the direction of a moving object changes -- even if it is moving at constant speed -- it is accelerating. In other words, the change in direction of a vector also contributes to changing its magnitude. This is better understood if you look at motion in, say, polar or spherical coordinates, where the centripetal term naturally pops out when you derive the general expression for acceleration.

Here is another way to understand this. In the diagram below, consider a particle moving with velocity $\mathbf{b}$ at time $t$ and subsequently with velocity $\mathbf{a}$ at time $t + \Delta t$. The magnitude of the two velocities are equal (i.e. the vectors are of equal length: $\mathbf{||a|| = ||b||}$.) The vector $\mathbf{\Delta v = a-b}$ denotes the change in velocity in time $\Delta t$. Clearly, $\mathbf{\Delta v}$ is also a vector in its own right, with a magnitude and a direction.

Answer 2

It's velocity is changing. If you trace out the velocity vector, it goes around a circle of radius $v=||\vec v||$ every period, $T$. The path length of the arc is $2\pi v$, which it covers in time $T$ for an acceleration of:

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

If the radius in real space is $r$, then:

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

so that:

$$ a= \frac{v^2} r$$