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A question from kleppner's book

  1. Aug 10, 2006 #1
    on page 108, question 2.34:
    A mass m whirls around on a string which passes through a ring, ngelect gravity.
    initially the mass is distance r0 from the centre and is revolving at angular velocity w0.
    the string is pulled with constant velocity V starting at t=0 so that the radial distance to the mass decreases.
    obtain a differential equation for w(t).

    here what i did:
    v'=w0r0
    r(t)=r0+Vt
    v'=w(t)*r(t)
    w(t)r(t)=w0r0

    but from the text this isnt correct, can someone help me here?
     
  2. jcsd
  3. Aug 10, 2006 #2
    1) Why the tangential velocity of the mass should be the same at any time?
    2) You should to impose the angular momentum conservation law.
     
  4. Aug 10, 2006 #3
    do you know a method which do not employ the angular momentum cl?
     
  5. Aug 10, 2006 #4
    If you cannot use angular momentum cl, thus you have to use cinematics.

    Write the expressions for radial and tangential velocity (I mean the component which is orthogonal to the radial one) in polar coordinates and do the same for acceleration.
    Take into account that it is a central motion (so acceleration only has the radial component).

    Note that by saying that tangential acceleration is zero you are implicitly imposing the angular momentum conservation law.
     
    Last edited: Aug 10, 2006
  6. Aug 11, 2006 #5
    here what i did so far:
    v=dr/dtr+rdr/dt
    a=d^2r/dt^2r+2w^2[tex]\theta[/tex]+r[tex]d\theta/dt[/tex][tex]\theta[/tex]-rw^2[tex]\theta[/tex]

    now how do i proceed?
     
    Last edited: Aug 11, 2006
  7. Aug 11, 2006 #6
    Well, begin by imposing that the radial component of the velocity is V and the tangential component of the acceleration is zero ...
     
  8. Aug 11, 2006 #7
    you mean that:
    dr/dt=V
    and dr^2/dt^2=0?
     
  9. Aug 11, 2006 #8
    I mean:

    [tex]\frac{d \rho}{d t} = - V[/tex]

    [tex]2 \frac{d \rho}{d t} \omega + \rho \frac{d \omega}{ dt} = 0[/tex].


    You should be able to go on, now ...

    EDIT: here [tex]\rho[/tex] is r.
     
    Last edited: Aug 11, 2006
  10. Aug 11, 2006 #9
    I think you didn't perform the correct derivation for the velocity and acceleration in polar coordinates ...
     
  11. Aug 11, 2006 #10
    isnt v=d(rr)/dt
    and a=dv/dt
    ?
    according to these two equations i wrote the above.
     
  12. Aug 11, 2006 #11
    Velocity and acceleration in polar coordinates are:

    [tex]v_r= \frac{d r}{d t}[/tex]

    [tex]v_\theta = r \frac{d \theta}{d t}[/tex]

    [tex]a_r = \frac{d^2 r}{d t^2} - r (\frac{d \theta}{d t})^2[/tex]

    [tex]a_\theta = 2 \frac{d r}{d t} \frac{d \theta}{d t} + r \frac{d^2 \theta}{d t^2}[/tex]

    with [tex] \omega = \frac{d \theta}{d t} [/tex].
     
  13. Aug 11, 2006 #12
    i know this is what i wrtoe in v.
     
  14. Aug 11, 2006 #13
    Ok. What's the problem then?
    You have all you need to solve the question.
     
  15. Aug 12, 2006 #14
    im having just one little problem, in the answer key it's written that:
    if Vt=r0/2 then w=4w0.

    while when i do the calculation:
    [tex]\int_{0}^{t}2Vdt=\int_{w0}^{w(t)}\frac{dw}{w}[/tex]
    which is:
    2Vt=ln(w(t)/w0)
    w(t)=(e^2Vt)*w0
    which is clearly not the same as what is written in the answer key, can you point me where did i go wrong here?

    thanks in advance.
     
  16. Aug 12, 2006 #15
    The integrand function on the left hand side of your equation is wrong ...

    From the second equation in post n. 8 you get:

    [tex]2 \int_{r_0}^r \frac{d \rho}{\rho} = - \int_{\omega_0}^{\omega} \frac{d \omega}{\omega} [/tex].
     
    Last edited: Aug 12, 2006
  17. Aug 12, 2006 #16
    but from this equation we get:
    2(ln(r-r0))=-ln(w-w0)
    ln(r-r0)^2=ln(1/(w-w0))
    (r-r0)2=1/(w-w0)
    if i put r=Vt=r0/2 then i get
    r0^2/4=1/(w-w0) and you havnet got rid from r0.
     
  18. Aug 12, 2006 #17
    We don't get that! :cry:

    I think you have some problems with the fundamental theorems of calculus or with logarithmic properties.

    [tex]\ln r - \ln r_0 = \ln \frac{r}{r_0}[/tex]
    [tex]\ln \omega - \ln \omega_0 = \ln \frac{\omega}{\omega_0}[/tex]
     
  19. Aug 12, 2006 #18
    oh, my mistake. frogot to switch them.

    i know these rules of logarithms, but i did a horribel mistake here.

    nevermind, it's good to this now, before the exams. (-;
     
  20. Aug 12, 2006 #19
    so idont need to use V here, or i shouldve used it here?
     
  21. Aug 12, 2006 #20
    good luck! :wink:
     
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