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Franklin Hu has a good experiment here.

Franklin uses the correct formula for angular momentum; and he realizes that the Linear or Newtonian velocity must double if angular momentum is to be conserved. Because he states that the rate of rotation must quad. If the linear velocity of the batteries had remained constant they would go around the half sized circle only twice as fast.

Because: The radius in the formula ($L = rmv$) has halved then the linear velocity must double; $\frac{1}{2} r m 2 v$ for angular momentum to be conserved. At twice the arc velocity they go around the half size circle four times as fast.

But isn't it possible that the pulling of the batteries (in) is actually a force that has accelerated the batteries to double their original speed.

If the pulling is not a force then you have a free increase in momentum: which is a violation of Newton's Laws of Motion. I shall assume that a pull is a force and Newton is still correct.

If the batteries had wrapped around an immovable post there would have been no linear (arc) acceleration: because there would have been no outside force. And I don't think there is any Earth motion involved; because the batteries would not lose motion as they wrap around the post.

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This is a nice question. The conservation of angular momentum says that

$$ \frac{d}{dt}\mathbf{L}=\frac{d}{dt}\sum_i \mathbf{L}_i = \frac{d}{dt}I\vec{\omega}=0$$

Note that $$\boldsymbol{\tau} = \frac{d\mathbf {L} }{dt} = \sum_i \mathbf{r_i}\times \mathbf{F_i}$$

and so we see that angular momentum is only conserved in the absence of a net torque.


Your question:

Is outside force needed to conserve angular momentum?

Short answer: No. If you consider a ballerina spinning in outer space with his or her arms out and then pulls them in, the two angular velocities will be related by

$$ I_1 \omega_1 = I_2\omega_2 $$

and so the ballerina will spin faster when they pull their arms in, and nowhere along the way did we apply an external force.


Longer answer (more fun):

Now, since the force applied at the center of the table is applied at the center of rotation $\mathbf{r}=0$, we indeed see that angular momentum should be conserved.

But it certainly seems like work was done. After all, the mass did move a distance $\Delta r= r_2 - r_1$ as the result of a force applied. So where did that energy go?

Since we are working with idealized conditions, we can find that the work done on the mass was precisely

$$W = \Delta K = \frac{1}{2}m (v_2^2 - v_1^2) = \frac{1}{2}m (\omega_2^2 r_2^2 - \omega_1^2 r_1^2) = \frac{1}{2}(I_2\omega_2^2 - I_1\omega_1^2) = \frac{1}{2}\left(\frac{L_2^2}{I_2} - \frac{L_1^2}{I_1}\right)$$

But $L_2 = L_1$ and so

$$W = \frac{L^2}{2}\left(\frac{1}{I_2} - \frac{1}{I_1}\right) $$

Indeed, work has been done since the moment of inertia has changed. But since this work was applied at the center of rotation, it does not contribute to the overall torque. Hence, even when work is done on the bead angular momentum can still be conserved, but a force doesn't have to be applied to conserve angular momentum (see short answer).

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