# Kleppner Kolenkow 2.34 (mass whirling around a string)

1. Dec 2, 2014

### geoffrey159

1. The problem statement, all variables and given/known data
A mass m whirls around on a string which passes through a ring. Neglect gravity. Initially, the mass is distance $r_0$ from the center and is revolving at angular velocity $\omega_0$. The string is pulled with constant velocity $V$ starting at $t=0$ so that the radial distance to the mass decreases. Find $\omega(t)$ and string tension applied to the mass.

Hint: $Vt = \frac{r_0}{2}$ implies $\omega = 4\omega_0$

2. Relevant equations
Newton equations of movment in tangential and radial directions.

3. The attempt at a solution
Firstly, because of the junction between the mass and the string, I find that the radial speed of the mass is $\dot{r}(t) = -V$. Therefore, the radial distance should be $r(t) = r_0 - Vt$ for $0\le t \le \frac{r_0}{V}$, and 0 for any later time.
Then, using Newton's second law in tangential direction, I get that $a_{\theta}=0$ at any time, and therefore
$\frac{\dot{\omega}}{\omega} = \frac{2V}{r_0-Vt}$.
This can be rewritten
$\frac{d}{dt}(ln(\omega) ) = -2 \frac{d}{dt}(ln(r_0-Vt))$ ,
which gives
$\omega(t) = \omega_0 {(\frac{r_0}{r_0-Vt})}^2$.
Finally, computing this result into the equation of movement for radial direction gives me that the tension in the string is :
$T(t) = \frac{m\omega_0^2r_0^4}{{(r_0-Vt)}^3}$

4. Question
Even though it works for the hint, there is something very wrong with my proof because angular velocity and string tension are unbounded as $r(t) \rightarrow 0$. It doesn't make sense to me. Thanks for taking the time.

2. Dec 2, 2014

### Bystander

What's your objection to your result beyond "not making sense?"

3. Dec 2, 2014

### geoffrey159

Hi, thanks for replying.
It seems unlikely to me that angular speed and string tension tend to infinity as $r(t) \rightarrow 0$. That's why I think I'm wrong, very wrong :-)

4. Dec 2, 2014

### Bystander

Angular momentum is conserved --- if your "ice skater" is not just anorexic but achieved the physique of a stick figure, and pulls stick arms into the point that they're parallel and co-located with the stick body, r in I = mr2, the angular moment of inertia, goes to zero, I goes to zero, and angular momentum = I x omega pretty much demands an infinite rotational speed.

5. Dec 2, 2014

### geoffrey159

I don't know what is momentum yet, I'm a beginner in physics

6. Dec 2, 2014

### Bystander

Have you had to watch the Olympic figure skating to keep other people in the house happy enough to let you watch the hockey?

7. Dec 2, 2014

### geoffrey159

Huh, yes maybe. I've not been watching TV lately. What do you mean?

8. Dec 2, 2014

### Bystander

When ice skaters go into a spin for a big finish, they pull their arms in along their rotational axis, and the spin rate increases quite dramatically. Conservation of angular momentum. If you've got a young enough physics prof/instructor, he/she'll stand on a rotating turntable and do the same thing as a demonstration.

9. Dec 2, 2014

### geoffrey159

By dramatically, you mean infinitly ?

10. Dec 2, 2014

### Bystander

They never get to the zero radius for their angular moments of inertia, so, "no."

11. Dec 2, 2014

### geoffrey159

So what you say is that it'll be clear when I'll study angular momentum ? The resulat seems correct to you then ?

12. Dec 2, 2014

### Bystander

It looks great --- ya dun good.

13. Dec 2, 2014

### geoffrey159

Thank you for your explanations and for giving some of your time.
Geoffrey