I have calculated the prediction for the final angular velocity of a typical classroom demonstration and it is obviously wrong. Where is my mistake?

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A typical classroom ball on a string demonstration of conservation of angular momentum starts at 2 rps and has the radius reduced to ten percent.

$$L_2=L_1$$ $$L=mvr$$ $$m_2v_2r_2=m_1v_1r_1$$ $$m_2=m_1$$ $$v_2r_2=v_1r_1$$ $$\text {divide both sides by } r_1^2r_2^2$$ $$\frac{v_2}{r_2r_1^2}=\frac{v_1}{r_1r_2^2}$$ $$\omega=\frac{v}{r}$$ $$\omega_2=\left(\frac{r_1}{r_2}\right)^2\omega_1$$ $$\omega_2=\left(\frac{10}{1}\right)^2\times2rps=200rps$$ Convert to revolutions per minute $$\omega_2=200\times60=12000rpm$$ Roughly the rotational speed of a formula one race car engine on full throttle at full speed.

This is very obviously wrong and it is unreasonable to claim that roughly a ten thousand percent loss of energy can occur within the second that it takes to pull in the string.

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You don't need to go through all those equations, you can just start with $L = I\omega$ then substitute $I = mr^2$. Since angular momentum is conserved, reducing $r$ to $0.1r$ would indeed increase $\omega$ by $100$ times.

So your result is correct, and your intuition is the one that's wrong.

Energy is not conserved* because you need to do work to pull the ball in. $E = \frac{1}{2} I \omega^2 = \frac{1}{2} m r^2\omega^2$. Since $r$ decreased to $0.1r$ but $\omega$ increased by a factor of $100$, the new fast-spinning ball has $100$ times more energy, and you will need to do work to pull the ball in. Note the new system has more energy, not less. It's part of the reason why [it's hard for a stellar system to collapse][1].


*More accurately, energy is conserved, because the amount of energy gained by the system is exactly equal to the amount of energy needed to pull the ball; in. The latter can be calculated from $dW = F \cdot dS$, where $F = -mr\omega^2$ is the centripetal force.

The integral equation you need is

$W = \int_r^{0.1r} mr * \frac{L^2}{m^2r^4} dr = 49.5 \frac{L^2}{m r^2} = 49.5 m r^2 \omega^2$

Meanwhile the energy of the system increased by $99 \times \frac{1}{2}I\omega^2 = 49.5 mr^2\omega^2$, exactly equal to the work you did against the centripetal force.