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Frictionless mass sliding in rotating pipe

  1. Jun 14, 2017 #1
    1. The problem statement, all variables and given/known data

    A pipe is spinning around with angular velocity ω = 2 rad/s about its fixed end. Inside the pipe is a frictionless block that is free to slide. The block is located 1 m from the fixed end of the pipe when the angle from the vertical, θ = π/2 (horizontal). What is the radial acceleration of the block at this instant?

    2. Relevant equations


    3. The attempt at a solution

    My immediate intuition is that there is no acceleration along the pipe, but how do I show that?

    I tried to draw a FBD but I can't even come up with any forces to write on it. There is no centripetal force, there is no friction... there is nothing.

    It's sort of like 2.33 on here
    https://www.slideshare.net/anshukg/an-introduction-to-mechanics-daniel-kleppner-solutions-part1

    But I'm not really able to follow the solution there.
     
    Last edited: Jun 14, 2017
  2. jcsd
  3. Jun 14, 2017 #2

    haruspex

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    Mine too.
    This seems like a reasonable way:
    ... except that you should also show there is no tangential acceleration.
     
  4. Jun 14, 2017 #3
    So how do I show that there is no tangential acceleration along the pipe?
     
  5. Jun 14, 2017 #4

    haruspex

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    Tangential is not along the pipe. What do you know about the pipe's movements?
     
  6. Jun 14, 2017 #5
    Ops I misspoke. I only know that it's rotating at a constant speed.
     
  7. Jun 14, 2017 #6

    haruspex

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    So what is happening to the vertical velocity?
     
  8. Jun 14, 2017 #7
    Slowing down. I just can't connect the dots with how that does anything to show me that there is no acceleration radially for the mass.
     
  9. Jun 14, 2017 #8

    haruspex

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    It is? Why?
    The questions just says acceleration. It does not restrict to radial acceleration.
     
  10. Jun 14, 2017 #9
    Because at the instant when it is perpendicular to the vertical all the velocity is in the Y direction, as it passes that position some velocity will be in x and some in y and since the angular speed is constant that means that it must accelerate in x and decelerate in y.

    Ops that was a mistake from my side. I translated the problem from German and missed that part. It specifically asks to solve for the acceleration along the pipe i.e. radial.
     
  11. Jun 14, 2017 #10
    I suppose what I descrived above is some dt after the case that I am studying, so in this case there is no tangential acceleration. I just fail to see how I'm supposed to "show" any of this mathematically. It ends up being an empty free body diagram.
     
  12. Jun 14, 2017 #11

    haruspex

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    That is a bit ambiguous. It could mean the component of acceleration in the radial direction, or it could mean ##\ddot r##. Is it clearer in the German?
    Yes.
    As you have - there are no radial forces, so no radial acceleration (unless the other interpretation is intended).
     
  13. Jun 14, 2017 #12
    Yes, it literatelly says ##\ddot r## I just don't know how to use LATEX.

    So just draw up an FBD, write something like "no forces in radial direction, therefore radial acceleration = 0."? Almost seems too easy :D.
     
  14. Jun 14, 2017 #13

    haruspex

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    ##\ddot r## is not the same as radial component of acceleration, so the answer is not zero.
    E.g. if ##\ddot r## were continuously zero then r would not change, so there would be centripetal acceleration.
     
  15. Jun 14, 2017 #14
    Ah yeah, you're right of course. Well then I'm completely lost.
     
  16. Jun 14, 2017 #15

    haruspex

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    No need to be lost. My example in the previous post was a bit of a hint.
    If ##\ddot r## were zero, what would radial acceleration be?
    So if the radial acceleration is zero....?
     
  17. Jun 14, 2017 #16
    Hmm.. To radial is zero, angular acceleration is zero since we have a constant omega. I just can't see it. With a pendulum we'd have centripetal acceleration but in this case I don't know. Since omega is constant I assume tangential acceleration is zero too.
     
  18. Jun 14, 2017 #17

    haruspex

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    Take a look at the expressions for radial and tangential acceleration in polar coordinates at https://en.m.wikipedia.org/wiki/Polar_coordinate_system#Centrifugal_and_Coriolis_terms.
     
  19. Jun 15, 2017 #18
    Ok so from an inertial reference frame Newton's second law in polar coordinates becomes

    F=m[ (r''- rθ'^2)r + (rθ'' + 2r'θ')θ]

    therefore

    a = (r''- rθ'^2)r + (rθ'' + 2r'θ')θ

    We know that the radial component of the acceleration is zero from the FBD thus:

    r'' - rθ'^2 = 0

    then r'' = rθ'^2 = (1m)(2rad/s)^2 = 4m/s^2

    is this the final answer though, since the question was asking for r'', I assume I do not need to solve for the θ term of a?

    What exactly does r'' mean in physical terms? Isn't r'' the radial acceleration, and we just looked at the FBD and determined that it's zero!? What is the difference between the r component of a and r''?
     
    Last edited: Jun 15, 2017
  20. Jun 15, 2017 #19

    haruspex

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    Yes.
    ##\dot r## is the rate of change of r, and ##\ddot r## is the rate of change of ##\dot r##. Not sure that I can give any better physical interpretation.
    If you take a non-inertial frame of reference, Cartesian coordinates centred on the mass and oriented with x along the pipe, the mass appears to be thrown in the x direction by a centrifugal force. The resulting acceleration is ##\ddot r ##.

    You could also consider throwing away the pipe. A mass travels at constant speed in a straight line past you, missing you by some distance. If r(t) is the distance from you to the mass it clearly reaches a minimum before increasing again. ##\ddot r## is always positive.
     
  21. Jun 15, 2017 #20
    Thank you! I'm still not sure that I understand the physical meaning of it though - I understand the derivative notation. What I don't understand is that when we set up the problem we said that the radial acceleration is zero at the point in time when we're looking at the problem - that far I am with you.

    When we started setting up the problem we got the equation a = (r''- rθ'^2)r + (rθ'' + 2r'θ')θ, where r is the radial component in polar coordinates. So we know that radial acceleration is zero and thus we were able to set up the problem and solve it. But what is the difference between radial acceleration and r'', it seems to me that r'' IS the radial acceleration?
     
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