# Angular momentum: rotating ice skaters

Tags:
1. Nov 30, 2016

### Cepterus

1. The problem statement, all variables and given/known data
Two ice skaters of mass $m = 50\,\mathrm{kg}$ each are moving towards each other frictionless on parallel paths with a distance of $3\,\mathrm{m}$. They both have a velocity of $v_o=10\,\frac{\mathrm m}{\mathrm s}$.
Skater 1 is holding a massless rod of lenght $3\,\mathrm{m}$ at one end perpendicular to his direction of movement. Skater 2 grabs the other end of the rod when the two have minimal distance of each other (again, $3\,\mathrm{m}$). (At this point, they should start rotating.) Both skaters now pull themselves along the rod until their distance from each other is $1\,\rm m$.
Hereafter, they both let go of the rod.

Compare the kinetic energies of the system before and after the encounter of the two skaters.
Prove that the difference is equal to the work the two have to perform while shifting towards each other.

2. Relevant equations
Angular momentum: $L = rp=rmv$ (only in this case, because $r\perp v$)
Kinetic energy: $E_{\rm{kin}}=\frac12mv^2$

3. The attempt at a solution
The angular momentum when the two start rotating is $L=r(2m)v_0=1.5m\cdot(50kg\cdot2)\cdot10\frac ms=1500kg\frac m{s^2}$, and we now that $L$ is conserved. So let us call the skaters' velocities immediately before letting go of the rod $v'$ and the new (smaller) radius of their rotation $r'$. This means $$r(2m)v=r'(2m)v'\Rightarrow v'=\frac{r}{r'}v=\frac{1.5m}{0.5m}\cdot10\frac ms=30\frac ms .$$
Calculating the kinetic energies before and after gives us a difference of $\Delta E_{\rm{kin}}=40000J$, if I am not mistaken. But how should I proceed from here? I have no idea on how to calculate the work performed by the two ice skaters as I do not know the corresponding force with which I could calculate $W=Fs=F(r-r')$.

2. Nov 30, 2016

### Orodruin

Staff Emeritus
You do know the force. Just as you know the force required to move up in a gravitational field...

3. Nov 30, 2016

### Cepterus

The centripetal/centrifugal force? If so, the force is obviously not constant over the whole way (unlike the gravitational force). How would I calculate the work then?

4. Nov 30, 2016

### Orodruin

Staff Emeritus
Are you familiar with integrals?

5. Nov 30, 2016

### Cepterus

Oh, right. Would it be something like $$W=\int_{1.5m}^{0.5m} \frac{mv^2}{r}\rm dr$$ then?
So I guess I have to substitute $v$ (which depends on $r$) by $\frac{L}{rm}$, right?

6. Nov 30, 2016

### PeroK

Yes. One suggestion is to use $r_0, r_1$, $v_0, v_1$ for the start and end radii and velocities and use $v(r), F(r)$ for the velocity and force at a variable radius $r$ over which you are integrating.

7. Nov 30, 2016

### Cepterus

It worked out fine.

Thanks to both of you!

8. Nov 30, 2016

### PeroK

Your work in post #1 was very good, but it might be even better to keep things algebraic a bit longer. For example, with $r_1 = \frac{r_0}{3}$ you have for each skater:

$L = mv_0r_0 = mv_1r_1 = mv_1 \frac{r_0}{3}$

$v_1 = 3v_0$

$\Delta KE = \frac12 m(v_1^2 - v_0^2) = \frac12 m(9v_0^2 - v_0^2) = 4mv_0^2$

And, then you can put in the numbers. I think you see a lot more of the physics this way.

Note that it didn't actual matter that it was $3m$ and $1m$. Only the ratio.

9. Nov 30, 2016

### Cepterus

Okay, that makes sense and I will try to think of it in the future.
I just have one last question: When calculating the work, we get an integral of the form $$W=c\int_{r_0}^{r_1}\frac1{r^3}\rm d r$$ where $c$ is a positive constant. This would mean that the work is negative because the integral is. How does that make sense, or does the sign have no meaning at all?

10. Nov 30, 2016

### Orodruin

Staff Emeritus
It does, the force required to pull in is in the negative radial direction so $c$ is a negative constant.

11. Nov 30, 2016

### PeroK

The radial force acting on each skater is inwards, so that makes $F(r)$ negative. But, integrating $\frac{1}{r^3}$ generates another negative sign, so the work done on the skater is positive and their speed increases.

12. Nov 30, 2016

### Cepterus

By substituting $v(r)$ by $\frac L{rm}$ and $L$ by $r_0mv_0$ I get $c=mr_0^2v_0^2$, which cannot be negative.
Or do you mean $F(r)$ should be $-\frac {mv^2}r$ in the first place? Why?

13. Nov 30, 2016

### PeroK

Because $\vec{dr}$ is outwards, so outwards is positive.

14. Nov 30, 2016

### Cepterus

And that is the case because increasing $r$ means moving away from the center?

15. Nov 30, 2016

Yes.