# How to calculate the internal angular acceleration?

Very well known topic, affect angular velocity.

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.

• What's the question? – Andrei Aug 27 '19 at 21:07
• what are the formulas for this angular acceleration? you have the answer below. – Sylwester L Aug 28 '19 at 18:52
• Crossposted to physics.stackexchange.com/q/500625/2451 – Qmechanic Sep 10 '19 at 18:09

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)