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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.

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

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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.

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  • $\begingroup$ 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 $\endgroup$
    – dylan7
    Commented Aug 17, 2014 at 20:51

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