Relationship between tangential & centripetal forces when angular speed constant but radius varies

Question

The model of a circular motion in basic Physics textbooks and online resources (e.g., Wikipedia on circular motion) assumes that the motion is a circle with constant radius to derive relationships such as $v=R\omega$, $a_\text{centripetal}=R\omega^2=\frac{v^2}{R}$, and $a_\text{tangential} = R \alpha$ where $\alpha$ is the rate of change of the angular speed $\frac{d\omega}{dt}$.

Can I use those derived relationships to derive the following relationships among the forces for the following depicted motion of a varying radius?

First, $F_\text{tangential} \\ = m a_\text{tangential}\;\ldots\text{ definition of a force} \\ = m R \alpha\;\ldots\;a_\text{tangential} = R \alpha \\ = m R \frac{d\omega}{dt}\;\ldots\;\alpha\text{ is defined to be }\frac{d\omega}{dt}\\ = m R \frac{\omega_f-\omega_i}{dt}\;\ldots\;d\omega \text{ is the difference between a final and an initial angular speeds} \\ = \frac{m R \omega_f - m R \omega_i}{dt} \\ = \frac{m R_f \omega - m R_i \omega}{dt}\;\ldots\;\omega\text{ is now held constant but the radius }R\text{ is varied} \\ = m \omega \frac{R_f - R_i}{dt} \\ = m \omega \frac{dR}{dt}\;\ldots\text{ I can change }F_\text{tangential}\text{ by varying the radius }R\text{ only}$

Second, $F_\text{centripetal} = m a_\text{centripetal}\;\ldots\text{ definition of a force} \\ F_\text{centripetal} = m R \omega^2\;\ldots\;a_\text{centripetal} = R \omega^2 \\ \frac{dF_\text{centripetal}}{dt} = m \omega^2 \frac{dR}{dt}\;\ldots\;R\text{ varies with time}$

Third, $F_\text{tangential} = m \omega \frac{dR}{dt} \\ = m \omega \frac{dF_\text{centripetal}}{m \omega^2 dt}\;\ldots\text{ due to the second relationship above} \\ F_\text{tangential} = \frac{1}{\omega} \frac{dF_\text{centripetal}}{dt}$

Any fallacy in my reasoning?

Answer 1

About derived equations:

Keep in mind that all of mechanics effectively builds onto $F = ma$.

This means that when you overload derived formulas in unfamiliar situations without verifying that the derivation holds you may be walking into a minefield.

For example: consider the formula $s(t) = s_0 + v_0(t) + \frac{1}{2} at^2$. This was derived with a constant force in mind. If it is time- or position dependent, so $a(t,x,y,z) $, then you cannot simply plug it in. If you want to be certain, you have to repeat the derivations. Similar examples are relativity and non-inertial reference frames.

Addressing your question:

Before I start: it's not entirely clear what you want to solve, so you might want to customize the advice to your situation.

First, do away with $\omega$; use $\dot\phi$ and $\alpha = \ddot \phi$, as they more clearly convey that they're in essence the same thing. Your equations regarding force generally hold - though you will have to update them with time and/or position dependent equivalents. It's not clear which one is the case in your problem so I'll go with position. Your constraint is $R = R(\phi)$.

Your equations will derive from $F = mR(\phi)\ddot \phi $

Some more minefields:

That motion is not ballistic; it seems that at (0,-R) the motion has the same properties but can go to either orbit radius. Be careful with forces or constrains you apply there.

You probably can't just 'blow up'$\frac{\omega}{dt}$ to $ \frac{\omega_f-\omega_i}{dt}$, as the idea behind the differential operator is that it applies with an infinitesimally small limit.