I am trying to do a derivation the results of angular momentum, but I am running into an issue.

To make things simple, lets start with the derivation of angular momentum:

$$L = mr^2\omega$$

If we differentiate, we get

$$dL/dt = 2mr*dr/dt*\omega + mr^2dw/dt$$

If we assume that L is constant, than we can solve for the relationship between r and $\omega$:

$$mr^2*d\omega/dt = -2mr*dr/dt*\omega$$

$$\frac{d\omega}{\omega} = -2\frac{dr}{r}$$

And then integrating and simplifying gives:

$$\omega = \frac{C}{r^2}$$

Since L is constant, the above is equivalent to:

$$\omega = \frac{L}{mr^2}$$

We can also substitute in the tangential velocity $\omega = v_T/r$ to find:

$$v_T = \frac{L}{mr}$$

Both of which look correct. However, if we substitute in the tangential velocity and tangential acceleration ($d\omega/dt = \alpha$ & $\alpha = a_T/r$) before integration:

$$mr^2*\frac{d\omega}{dt} = mr^2*\alpha = mr^2\frac{a_T}{r} = -2mr*\frac{dr}{dt}*\frac{v_T}{r}$$

$$mr\frac{dv_T}{dt} = -2m\frac{dr}{dt}*v_T$$

$$\frac{dv_T}{v_T} = -2\frac{dr}{r}$$

$$v_T = \frac{C}{r^2} = \frac{L}{mr^2}$$

Which is not correct.

What is going on? Why does making the transition from angular velocity and acceleration to linear velocity and acceleration before the integration change my result?

And, as a side note, in my real problem, I am not starting with the definition of angular momentum, but starting there in this post made it so I had to write less.


1 Answer 1


Your problem comes from your assuming $\alpha=\frac{ a_T}{r}$. Since you are not asuming radius is constant, we have

$$\alpha =\frac{d (v_T/r)}{dt}= \frac{dv_T}{dt}/r -\frac{1}{r^2}\frac{dr}{dt}v_T \neq \frac{a_T}{r}$$

  • $\begingroup$ Wow, that is simple. Kind of annoyed that I missed that. Thanks! $\endgroup$ Commented Aug 21, 2021 at 19:19

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.