Suppose a particle moves counterclockwise in time \(\Delta t\) on a circular path. The particle goes from \(\vec{\textbf{r}}(t)\) to \(\vec{\textbf{r}}(t + \Delta t)\). Assume that from \(t\) to \(t + \Delta t\), the speed is constant.
The velocity changes from \(\vec{\textbf{v}}(t)\) to \(\vec{\textbf{v}}(t + \Delta t)\). The velocity vector has a constant magnitude and is tangent to the path (i.e. it's perpendicular to \(\vec{\textbf{r}}(t)\)). This means the triangles formed by \(\Delta \vec{\textbf{r}}\) and \(\Delta \vec{\textbf{v}}\) are similar:
Since \(\Vert \vec{\textbf{r}}(t) \Vert= \Vert\vec{\textbf{r}}(t + \Delta t) \Vert\) and \(\Vert \vec{\textbf{v}}(t) \Vert= \Vert\vec{\textbf{v}}(t + \Delta t) \Vert\), the two triangles are isosceles. Let \(r = \Vert \vec{\textbf{r}}(t)\Vert\) and let \(v = \Vert \vec{\textbf{v}}(t)\Vert\). This means \(r\) is the radius, and \(v\) is the speed, both of which are constant. Similarly, let \(\Delta r = \Vert \Delta \vec{\textbf{r}}(t)\Vert\) and let \(\Delta v = \Vert \Delta \vec{\textbf{v}}(t)\Vert\), where \(\Delta r\) gives the magnitude of the change in position, and \(\Delta v\) gives the magnitude of the change in velocity. Therefore:
We can find the magnitude of the acceleration as well, which is also constant:
The acceleration vector is tangent to the velocity vector, so it either points inside or outside the circle. Since velocity in both \(x\) and \(y\) directions are changing to go nearer the center, then the acceleration points inside the circle. This acceleration is called centripetal acceleration (\(a_c\)).
We derived the magnitude of centripetal acceleration as:
However at the start, we assume the speed from \(t\) to \(t + \Delta t\) was constant. What if the speed is not a constant? In this case, \(\vec{\textbf{v}}(t)\) and \(\vec{\textbf{v}}(t + \Delta t)\) is, in fact, different. Which mean the one of the triangles are no longer isosceles. However, if \(\Delta t \to 0\), then \(\vec{\textbf{v}}(t + \Delta t) \to \vec{\textbf{v}}(t)\). This means for very small \(\Delta t\), the triangle would be an isoscele, so the above equation for the magnitude of the centripetal acceleration is valid.
Let \(\omega\) be the angular speed. Let \(d \theta\) be the instantaneous change in radians:
Since the speed \(v\) can be defined as the instantaneous change in arc length over time, and since \(d(\text{arc length}) = d\theta \ r\):
Which means:
Also, if the speed is changing, and the centripetal acceleration vector is perpendicular to the velocity vector, then there must be some tangential acceleration vector, \(\vec{\textbf{a}}_t\), causing the change in velocity. It's scalar value would be: