# $\dot \omega$ when momentum is conserved

If momentum is conserved when there is no externel force, why is there an acceleration when a momentum parameter is changed (inertia)? How does it accelerate with no external force?

For example angular momentum is conserved when:

${\tau}=\frac {d(I\omega)}{dt}=0$ But in this case momentum is conserved, so both $I$ and $\omega$ are changing, so the product rule applies:

So it equals: $I{\dot \omega}+{\dot I}\omega=0$, hence $\dot \omega$ is acceleration, and we also know the object accelerates when the rotational inertia changes.

• The rotating frame is not a inertial frame so acceleration can change with no external force (see fictitious forces). Commented Aug 17, 2014 at 19:55

There is no law of physics that requires an external force to be present when some double time derivative of a parameter in a system changes. What you were taught in high school, that is

$$F = m \cdot \dot v$$

and

$$M = I \cdot \dot \omega$$

are actually the simplified version of

$$F = \frac{d}{dt}m \cdot v = m \cdot \dot v + \dot m \cdot v$$

and

$$M = \frac{d}{dt}I\cdot \omega =I \cdot \dot \omega + \dot I \cdot \omega$$

when the mass and the inertia of an object do not change in time.

The second one is the same equation that you wrote in the question (generalized for $M \neq 0$), the generalized version of the first one does not actually have any application (because mass is preserved, it's always $\dot m = 0$), at least in low energy physics.

• Ah, so only when the mass doesn't change an acceleration requires an externel force? And by the way that first statement just blew my mind, but it makes sense Commented Aug 17, 2014 at 20:51