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?

Circular motion with varying radius

  • $\begingroup$ After some googling, I found out that I was looking at Hohmann Transfer Orbit Problem. Since Hohmann didn't use an approach similar to mine, I can conclude that I cannot use the relationships derived from the model of a circular orbit with a constant radius to explain the relationship between radial and tangential forces in the Hohmann Transfer Orbit. That is, the fallacy in my thinking is using equations derived from a certain model to explain a different model. $\endgroup$ Commented Apr 7, 2014 at 3:24

1 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.

  • $\begingroup$ I am trying to see the relationship between $F_\text{tangential}$ and $F_\text{radial}$ when the radius is varied while $\ddot\phi$ is held constant. As you aptly point out to me, I was walking down a minefield. Thank you very much for showing me the right way to go by starting a derivation afresh from $F = m R(\phi)\ddot\phi$. Is $F = m R(\phi)\ddot\phi$ the resultant vector of the tangential $F_\text{tangential to path}$ and radial $F_\text{perpendicular to path}$ vectors? $\endgroup$ Commented Apr 7, 2014 at 5:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.