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Kleppner Kolenkow 2.34 (mass whirling around a string)

  1. Dec 2, 2014 #1
    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 [itex] r(t) = r_0 - Vt [/itex] 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 [itex] a_{\theta}=0 [/itex] at any time, and therefore
    [itex] \frac{\dot{\omega}}{\omega} = \frac{2V}{r_0-Vt} [/itex].
    This can be rewritten
    [itex] \frac{d}{dt}(ln(\omega) ) = -2 \frac{d}{dt}(ln(r_0-Vt)) [/itex] ,
    which gives
    [itex] \omega(t) = \omega_0 {(\frac{r_0}{r_0-Vt})}^2 [/itex].
    Finally, computing this result into the equation of movement for radial direction gives me that the tension in the string is :
    [itex] T(t) = \frac{m\omega_0^2r_0^4}{{(r_0-Vt)}^3} [/itex]

    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. jcsd
  3. Dec 2, 2014 #2

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    What's your objection to your result beyond "not making sense?"
     
  4. Dec 2, 2014 #3
    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 :-)
     
  5. Dec 2, 2014 #4

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    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.
     
  6. Dec 2, 2014 #5
    I don't know what is momentum yet, I'm a beginner in physics
     
  7. Dec 2, 2014 #6

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    Have you had to watch the Olympic figure skating to keep other people in the house happy enough to let you watch the hockey?
     
  8. Dec 2, 2014 #7
    Huh, yes maybe. I've not been watching TV lately. What do you mean?
     
  9. Dec 2, 2014 #8

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    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.
     
  10. Dec 2, 2014 #9
    By dramatically, you mean infinitly ?
     
  11. Dec 2, 2014 #10

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    They never get to the zero radius for their angular moments of inertia, so, "no."
     
  12. Dec 2, 2014 #11
    So what you say is that it'll be clear when I'll study angular momentum ? The resulat seems correct to you then ?
     
  13. Dec 2, 2014 #12

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    It looks great --- ya dun good.
     
  14. Dec 2, 2014 #13
    Thank you for your explanations and for giving some of your time.
    Geoffrey
     
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