How to calculate the internal angular acceleration?

Question

Very well known topic, affect angular velocity.

https://www.youtube.com/watch?v=tab1VlV4R0Y

An object that changes the moment of inertia simultaneously changes the angular velocity. Because the change in angular velocity over time is angular acceleration

$$\ \vec \epsilon= \frac {d \vec \omega} {dt} $$

and there are no external influence here and the effect is the result of the object's actions, I called it the inside angular acceleration.

For those who say that angular acceleration is not there I will try to give you an answer today.

Answer 1

To calculate the internal angular acceleration we must use the law of conservation of angular momentum

$$\ L = I_1 \omega_1 = I_2 \omega_2 \tag 1 $$

So we count the derivative of the angular momentum knowing that both angular velocity and moment of inertia change over time

$$\ \frac {dL} {dt} = \frac {d(I \omega)} {dt} = \frac {dI} {dt} \omega + I \frac {d\omega} {dt}=0 \tag 2 $$

However, this pattern does not work is not yet complete. Using the (1) we count the proportions

$$\ \frac {\omega_1} {\omega_2} = \frac {mr_2^2} {mr_1^2} \tag 3 $$

now we can calculate the angular velocities

$$\ \omega_1 = \frac {r_2^2} {r_1^2} \omega_2 \tag 4 $$ $$\ \omega_2 = \frac {r_1^2} {r_2^2} \omega_1 $$

we can now count what angular acceleration is

$$\ \epsilon = \frac {\omega_1 - \omega_2}{dt} = \frac {mr_1^4 \omega_1 - mr_2^4 \omega_2} {mr_1^2 r_2^2} = \frac {L (r_1^2 - r_2^2)} {mr_1^2 r_2^2} \tag 5 $$

knowing that $\ I_1 - I_2=-(I_2-I_1) $ we can apply another record

$$\ \epsilon = \frac {L (I_1 - I_2)} {I_1 I_2} = L \frac {(\frac {-dI}{dt})}{I_1I_2} \tag 6 $$

we count the change in moment of inertia over time

$$\ \frac {dI}{dt} = - \epsilon \frac { (I_1 I_2)} {L} \tag 7 $$

now easier to record (6)

$$\ \frac {d \omega} {dt} = - \frac {dI} {dt} \frac {\omega_1} {I_2} = = - \frac {dI} {dt} \frac {\omega_2} {I_1} \tag 8 $$

we can now complete the formula (2)

$$\ \frac {d \omega} {dt} I_2= - \frac {dI} {dt} \omega_1 \tag 9 $$

or

$$\ \frac {d \omega} {dt} I_1= - \frac {dI} {dt} \omega_2 $$

We know that the angular acceleration times the moment of inertia is the moment of force $\ \epsilon I =M $

So we can save the formula (2) as follows $$\ \frac {dL} {dt} = \frac {dI} {dt} \omega_1 + I_2 \frac {d\omega} {dt}= M_I + M_\omega = 0 \tag {10} $$

so we have here two opposite inside moments of forces (their source is inside the object) which are zeroing and nonzero angular acceleration which shows the pattern (8)