A lot of my problems have objects moving in circular paths with tangential and normal components of acceleration.
If the tangential component is non-zero though, the speed is changing so the radius of its path would also be changing.
So how can its path be considered circular? The size of of its radius of curvature of constantly changing.
Best Answer
For any path in 3D or 2D the velocity and acceleration vector is decomposed as
$$ \vec{v} = v \vec{e} \\ \vec{a} = \dot{v} \vec{e} + \frac{v^2}{\rho} \vec{n} $$
where $\vec{e}$ is the tangent vector, $\vec{n}$ is a normal vector, and $\rho$ the radius of curvature. So when keeping the path circular ($\rho=\mbox{const.}$) the value of the radial acceleration increased with $v^2$. This is commonly referred to as centrifugal acceleration.
There is always a force of magnitude $N$ along $\vec{n}$ which keeps the object along the path, so for a particle of mass $m$ you have
$$ N = m \frac{v^2}{\rho} $$
In the case the tangential component is not zero then the applied force is decomposed as
$$\vec{F} = T \vec{e} + N \vec{n} = m \dot{v} \vec{e} + \frac{m v^2}{\rho} \vec{n} $$
$$ T = m \dot{v} \\ N = m \frac{v^2}{\rho} $$
In the end I do not see what the problem is you are asking about. Maybe you can rephrase it based on the above convention.