This is a very well known topic and affects angular velocity.

How does changing the moment of inertia 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.

  • $\begingroup$ Crossposted from math.stackexchange.com/q/3326727/11127 $\endgroup$ – Qmechanic Sep 10 '19 at 18:08
  • 3
    $\begingroup$ Please don't editorialize. If you have questions or complaints about community moderation decisions you should take them to meta (as you have done). Abuse directed at other users--even at the community at large--violates the "Be nice." policy. $\endgroup$ – dmckee --- ex-moderator kitten Sep 10 '19 at 19:12
  • $\begingroup$ dmckee@ Can you explain why want to close this question here? What's wrong with this question? it is not "Be nice" $\endgroup$ – Sylwester L Sep 10 '19 at 19:17
  • 1
    $\begingroup$ Related meta post: physics.meta.stackexchange.com/q/11423/2451 $\endgroup$ – Qmechanic Sep 10 '19 at 19:26
  • 4
    $\begingroup$ You would have to ask the people who voted to close, but clicking through a few of your questions I found a lot of non-standard terminalogy (usually unexplained) and a lot of unclear diagrams. Then you go on to answer your questions (find, we generally like self-answers) at vastly unecessary length using clucnky un-vectorized notation. It looks to me like you are working through the learning process on this stuff. Which is great as far as it goes, but why would we want to record every step of your learning? That goes in your notebooks. $\endgroup$ – dmckee --- ex-moderator kitten Sep 10 '19 at 19:29

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

$$\ L = I_1 \omega_1 = I_2 \omega_2 \tag 1 $$ enter image description here

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 and is not yet complete. Using (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 it is 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):

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


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

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).

| cite | improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.