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Integrating centrifugal force

  1. May 2, 2014 #1

    bobie

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    1. The problem statement, all variables and given/known data

    This is not homework. Suppose I am rotating a ball on a string (m=1, r = .2, v = 10 m/s)
    2. Relevant equations

    Fc = (m) v2/r = 500 N*m
    If I reduce the length of the string to .1, v becomes 20, so
    Fc = 4000 N*m

    what is the work I have done, what kind of integration do I need,
    does the energy spent depend on the speed/time I use to pull the string?

    Thanks for your help
     
  2. jcsd
  3. May 2, 2014 #2

    BvU

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    Work is force along the direction of movement times path length along the direction of movement.
    Once the motion is circular and end of the string you are holding is stationary, you don't do any work any more: the string is perpendicular to v.

    To bring about this rotation, you make little circles to accelerate the ball. That's work. The sum of the work delivered during these initial small movements is the ultimate kinetic energy: 1/2 m v2
     
  4. May 2, 2014 #3

    bobie

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    Probably there is a misunderstanding,BvU.
    I am not referring to setting the ball spinning.
    Somebody/thing set the ball spinning at 10 m/s and gave it KE 50J.
    I am just shortening the string/spoke from .2 to .1 m.
    I work .1m against the centrifugal force v^2/2.
    Is it now clear?
    Thanks
     
    Last edited: May 2, 2014
  5. May 2, 2014 #4

    haruspex

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    Simplest is to use a conservation law to figure out how fast the ball will be moving after you've shortened the string. Then you can look at the change in KE.
    You can do it by integration. Suppose the string is shortened by some small amount dr, so that the force is more-or-less constant. What work has been done? Where has that energy gone?
     
  6. May 2, 2014 #5

    bobie

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    The problem is complicated by the fact that F = v2/r is changing all the time along the distance. otherwise 500 N *.1m would simply make 50J wouldn't it?, if we consider only the final value of F it would be 8 times greater. The real increase is 150J, 3 times or 0.888 times
    I want to derive that. Should I rather start a thread in a math forum?
     
    Last edited: May 2, 2014
  7. May 2, 2014 #6

    BvU

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    Apologize: didn't read the problem statement that was slightly hidden under the relevant equation heading.

    I remember a similar thread where I claim to have done the energy thing -- but can't find it back. (by the way, K&K turns out to be Kleppner & Kolenkov (MIT))

    So, same reasoning: wrt to the origin, the wire can not exercise a torque on the mass because force and r are aligned. With torque ##= 0 = \tau = {\bf d\over dt}(I\omega)##

    and ## I = mr^2## I get ## m\omega r^2 = ## constant. So ## \omega r^2 = \omega_0 r_0^2## or ## v r = v_0 r_0 ## as you already used.

    Now you can write down something for Fc and for the work, that is part constant and part r dependent. Integrate from 0.2 to 0.1 to find the work done. Compare with the kinetic energy at 0.2 and at 0.1 and sure enough: the increase in KE is equal to work done !
     
  8. May 2, 2014 #7

    bobie

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    But does it exercise any external torque on the system? How does that agree with:
    Since the work his done perpendicular to rotation i.e radially, how can that increase the tangential vector?
    In case the string is a problem you can imagin a telescopic spoke.
    Can you write a single formula for that integral?
    I really appreciate your help, BvU!
     
  9. May 2, 2014 #8

    haruspex

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    Quite so, which is why I suggested thinking about what has happened to the work done. If you think about that you can figure out how the force changes.
    While the radius is constant, the acceleration is perpendicular to the velocity, so the tangential speed does not change. As soon as you start to pull the string in it acquires some radial velocity, so now its total speed increases, giving it a velocity heading towards a smaller radius. As it continues to move around the axis, that same velocity heading, as a straight line, becomes tangential once more, but it retains its additional speed.
     
  10. May 2, 2014 #9

    BvU

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    Note that angular momentum hasn't changed: ##v r = v_0 r_0 ## !

    As for F, well, you wrote it yourself already:

    Fc = m v2/r .

    You know that mvr is a constant. So F is what (constant) times what (only dependent on r) ?

    The constant part is m(vr)2 and that leaves a ## {1\over r^3}## factor. So ##dW = - |F| \, dr = - {m v_0^2 r_0^2 \over r^3} \, dr ## (a minus sign because F and dr point in opposite directions) and the integral is $$ W = - m v_0^2 r_0^2\, \int_{0.2}^{0.1}\, {1\over r^3} \, dr $$
     
  11. May 2, 2014 #10

    bobie

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    Thanks, just a couple of more questions:

    - does absolutely all the work done end up in tangential KE?
    - does anything change if we change the time of the process/speed of the radial motion?
     
  12. May 2, 2014 #11

    BvU

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    Did you find W = 150 J ? And check that the original KE of 50 J goes to 200 J if the velocity doubles ? So: yes (friction effects excluded).

    Changing the time shouldn't have an influence according to the formulas. They are very patient. In practice there probably will be problems when trying to do it faster. Then there is also a small matter about the ball having a size and a moment of inertia around its axis.
     
  13. May 2, 2014 #12

    haruspex

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    Yes. There's nowhere else for it to go.
    No.
     
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