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

Question

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.

@boojum Reducing the radius by 10% would increase angular momentum by $(\frac{10}{9})^2$, not $1.1^2$. — Mark Saving, Aug 04 '21 at 21:25
What ended up happening? This man has been trying to disprove conservation of angular momentum for years now. Did you ever find success, or progress? — Fouad Saffar, Apr 01 '22 at 21:37

score 1 · Answer 1 · edited Aug 05 '21 at 14:25

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.

You are neglecting the fact that the prediction does not match reality. It is irrelevant if you can mathematically justify where the energy comes from. Do you recognise that a ball on a string does not accelerate like a Ferrari engine? — John, Aug 07 '21 at 06:51

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

1 Answers1