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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.

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

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

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

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