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Mass whirling on table pulled by string through center

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    f1StQsX.png


    2. Relevant equations
    Energy


    3. The attempt at a solution
    So I'm not exactly sure what this question even wants considering the statement is true by the work - energy theorem. Does it want me to show it explicitly for this case? If so, then, since there are no forces present along [itex]\hat{\theta }[/itex], [itex]W = m\int_{a}^{b}<\ddot{r} - r\dot{\theta }^{2}, 0>\cdot <dr,d\theta > = m\int_{l_{1}}^{l_{2}}\ddot{r} - r\dot{\theta }^{2}dr[/itex] so [itex]W = \frac{1}{2}m\dot{r}^{2}|_{l_1}^{l_2} - \int_{l_{1}}^{l_{2}}mr\dot{\theta }^{2}dr[/itex]. The change in kinetic energy is given by [itex]\Delta T = \frac{1}{2}m\dot{r}^{2}|_{l_1}^{l_2} + \frac{1}{2}mr^{2}\dot{\theta }^{2}|_{l_1}^{l_2}[/itex]. If we choose a coordinate system centered around the hole, then the position vector to the mass is parallel to the net force so the torque is identically zero thus angular momentum is conserved so [itex]r^{2}\dot{\theta }[/itex] is a constant of motion which implies [itex]r^{4}\dot{\theta }^{2}[/itex] is also a constant so [itex]-\int_{l_1}^{l_2}mr\dot{\theta }^{2}dr = -\int_{l_1}^{l_2}m\frac{r^{4}\dot{\theta }^{2}}{r^{3}}dr =-mr^{4}\dot{\theta }^{2}\int_{l_1}^{l_2}\frac{dr}{r^{3}} = \frac{1}{2}mr^{2}\dot{\theta }^{2}|_{l_1}^{l_2}[/itex] so [itex]W = \Delta T[/itex] but I'm not sure if my solution is valid and I'm also not satisfied with this long winded solution, if it is correct, because the chapter this problem is from was before angular momentum was introduced so there must be a solution that doesn't use the concept but I can't think of it so if anyone could guide me towards that it would be great. Thanks!
     
  2. jcsd
  3. Feb 3, 2013 #2
    "Slowly" means that you can ignore the radial velocity and the acceleration related to the change in the radial velocity. Under this assumption, what is the acceleration of the mass?
     
  4. Feb 3, 2013 #3
    Why can we assume all those things? Can you give a mathematically precise reason? Seems like we are assuming a lot just from the word slowly. Anyways if the radial acceleration is indeed small then [itex]\ddot{r} << r\dot{\theta }^{2}[/itex] so that [itex]a_{r} = -r\dot{\theta }^{2} + O(\frac{\ddot{r}}{r\dot{\theta }^{2}})[/itex]. Also, could you please tell me if my solution above was correct or not? Thanks
     
  5. Feb 3, 2013 #4
    There is no "mathematically precise" reason, it is a physical simplification.

    Your derivation does not take the Coriolis effect into account, which is precisely what the "slowly" simplification eliminates.
     
  6. Feb 3, 2013 #5
    I don't see why we care about the Coriolis forces. They show up in the angular component of the force which is zero here so it doesn't even show up in the line integral for work due to the dot product. Also I am really not convinced about the whole "slowly" approximation without being able to see how to express that mathematically. Otherwise it seems like hand waving to me.
     
  7. Feb 3, 2013 #6

    rcgldr

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    If you know that angular momentum is conserved, then you'd also know that the final energy state of the whirling mass is independent of the path. The net work related to radial acceleration and deceleration would be zero, since the whirling mass begins and ends with zero radial speed, and its angular momentum is conserved.

    So it seems that for this problem, the "slowly" pulled in means you ignore the work related to radial acceleration and deceleration, rather than relying on knowledge of conservation of angular momentum, or in effect, deriving the equivalent.
     
  8. Feb 3, 2013 #7
    But I don't get where my solution went wrong. I don't see how Coriolis forces would factor into the line integral the way I did it. The dot product takes care of that. Thanks for the response rcgldr!
     
  9. Feb 3, 2013 #8
    In your derivation of work, the acceleration vector is constant in direction. This is only possible when the vector is taken in a co-rotating frame. In a co-rotating frame, you have to take the Coriolis effect into account - unless you can assume that the velocity is "slow".

    From a purely mathematical point of view, a lot of physics may seem like hand waving. Which is OK, because physics is not mathematics. Learn the difference.
     
  10. Feb 3, 2013 #9
    If you can't quantify a result mathematically, approximation or not, then why should I accept it as fact? You're literally translating a simple phrase in English to a powerful physical assumption. There's a difference between being given an explicit approximation and taking something out of that and loosely interpreting an english word to mathematical results without justifying it mathematically. Anyways, can you point out where in the work integral it was assumed the direction of the radial acceleration was constant? Do you mean to say the direction of the radial acceleration is not the same as the direction of the radial displacement? Thanks.
     
  11. Feb 3, 2013 #10

    ehild

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    The problem asks the work of the person who pulls the string W=∫Fdl. You calculated the work of the string on the mass, which is the same now, I did not find any error in your derivation. You only did not take the condition "pulling slowly" into account.

    "Pulling slowly" means that you pull with a force which is almost equal to the centripetal force, it can be taken equal to it. So F=mrω2. That means a constant radial speed, so the radial speed does not contribute to the change of KE.
    Conservation of angular momentum is necessary to take into account.

    ehild
     
    Last edited: Feb 3, 2013
  12. Feb 3, 2013 #11
    The acceleration vector you had was ## (\ddot{r} - r\theta^2, 0) ##. Its direction is constant. Which is only possible in a co-rotating frame. In which case there must be a Coriolis term.

    You are free to debate whether "slow" means "absence of radial velocity", but then you must take it into account.
     
    Last edited: Feb 3, 2013
  13. Feb 3, 2013 #12
    Hi ehild. Thanks for the response. So are you saying if I take into account the radial acceleration not contributing any work then I will get the same answer I have now except the change in .5m(dr/dt)^2 from l1 to l2 terms will go away in the work expression? Why will the term go away in the kinetic energy expression where radial velocity shows up without reference to the radial acceleration? Also, if I don't take the approximation into account, is my final expression/ solution for work and for kinetic energy still correct? I just don't get where any reference to Coriolis forces come up. Thanks!
     
  14. Feb 3, 2013 #13

    ehild

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    Your derivation is correct proving that the work done on the mass is equal to the change of kinetic energy in general, but the problem asks to derive it when the radial velocity is "slow" ; that excludes the term connected to the change of the radial velocity from the KE. And I think the problem wants you to present the result in the usual form of the KE: 1/2 mv2,so expand the last expression and give the work in terms of the initial and final speeds.

    The Coriolis force is irrelevant here.
    Centrifugal force and Coriolis force are inertial forces, fictitious forces, experienced in a rotating frame of reference. You observe the motion from a rest frame.

    ehild
     
  15. Feb 3, 2013 #14
    He may be observing it from a rest frame, but his derivation is an co-rotating frame.
     
  16. Feb 4, 2013 #15

    ehild

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    No. Why do you think so?

    ehild
     
  17. Feb 4, 2013 #16
    See #11.
     
  18. Feb 4, 2013 #17

    ehild

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    Last edited: Feb 4, 2013
  19. Feb 4, 2013 #18
    ## (\ddot{r} - r\theta^2, 0) ## is a vector. Vectors are invariant. You can't have one and the same vector have constant and non-constant direction just because you change coordinates.
     
  20. Feb 4, 2013 #19
  21. Feb 4, 2013 #20

    ehild

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    A coordinate system is not the same as the frame of reference.

    Describing a motion with a polar coordinate system, there are both radial and azimuthal components of acceleration. The azimuthal component includes both the "Coriolis term" and the angular acceleration. As the string can exert only radial force, the azimuthal component of acceleration is zero. That does not mean that the acceleration or the force has constant direction as you said in #11. It means that the force is "central", acts along the radius, and the acceleration is radial.

    ehild
     
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