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Bead on a Spinning Hoop

  1. Feb 9, 2012 #1
    A bead of mass m can slide on a frictionless circular hoop in the vertical plane with radius R. When the hoop spins around the vertical axis with a rate of ω, the bead moves up the hoop by an angle θ which depends on the angular velocity of the hoop. (Pic attached)

    a) Find θ in terms of ω, R, m and g.

    b) If there is friction between the bead and hoop given by μ, within what range can ω change without changing θ?

    For part a what I have so far is that arad=R*ω2. The centripetal force (F) then is m*arad and the normal force is F/sinθ.

    The weight needs to cancel out the normal force then so the weight in the direction of the normal force is mg*cosθ = F/sinθ

    Am I anywhere close to being right?
     

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  3. Feb 9, 2012 #2

    tiny-tim

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    hi zaper! :smile:

    no

    keep it simple

    do F = ma vertically to find N

    then do F= ma horizontally to find ω

    (and remember that the centripetal acceleration is horizontal, so use the correct r :wink:)
     
  4. Feb 9, 2012 #3
    Hey there tiny tim!

    Sorry, but I'm really struggling to keep up with this problem so forgive me for anything dumb that I do here.

    So you said to find N as F=ma vertically so that means that N=mg*cosθ

    By you're second part are you saying m*arad=m*ω2*r (r being R*sinθ)
     
  5. Feb 9, 2012 #4

    tiny-tim

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    hey there zaper! :smile:
    that's not in the vertical direction, is it? :redface:
    no, that's ma = ma, you need F = ma :wink:
     
  6. Feb 9, 2012 #5
    Ok so then N=mg*sinθ I assume. Could you please explain why this is and what exactly you want for the second equation. I'm not really following you here.
     
  7. Feb 9, 2012 #6

    tiny-tim

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    no

    start again …

    Fx = max

    Fy = may

    what are ax and ay ?​
     
  8. Feb 9, 2012 #7
    I'm really really sorry. I guess I'm way overthinking this or something. Is ax the radial acceleration?
     
  9. Feb 9, 2012 #8

    tiny-tim

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    a is the acceleration

    ax and ay are its x and y components

    (i'm assuming you're using y for up and x for radial)
     
  10. Feb 9, 2012 #9
    Ok so ax is the radial acceleration and the total acceleration is pointing up and to the left. The triangle formed by the accelerations has top angle θ and ay=ax/tanθ. Is this correct? Also does ay actually describe something here or is it just a component of the total a?
     
  11. Feb 9, 2012 #10

    tiny-tim

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    what??

    why? :confused:
     
  12. Feb 9, 2012 #11
    Wait, I mean to the right
     
  13. Feb 10, 2012 #12

    tiny-tim

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    do you mean up and to the right, or only to the right? :confused:
     
  14. Feb 10, 2012 #13
    hi zapper i have an easily understandable solution, if u have not come to the answer i may help.
     
  15. Feb 10, 2012 #14
    Up and to the right tiny tim and what's your idea abhinav?
     
  16. Feb 10, 2012 #15

    tiny-tim

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    (along the normal?)

    nooo :redface:

    i think you're confusing physics and maths

    force is physics

    acceleration is maths (geometry)

    if you know how the object is moving, then the acceleration follows automatically as a matter of maths

    it doesn't matter what forces are producing that motion, the acceleration depends only on the motion, not on the forces

    in this case, in equilibrium, the bead is moving uniformly in a horizontal circle …

    so what is its acceleration? :smile:
     
  17. Feb 10, 2012 #16
    Since there's no angular acceleration then the only acceleration is radial right?
     
  18. Feb 10, 2012 #17

    tiny-tim

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    yes, but you're looking at the wrong radius …

    the circle to look at is the circle of the bead's motion, which is horizontal, isn't it? :smile:
     
  19. Feb 10, 2012 #18
    Yes. That last post was supposed to be in terms of the horizontal circle. Sorry that I forgot to explain that
     
  20. Feb 10, 2012 #19

    tiny-tim

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    ok, so now do Ftotal = ma in both the horizontal and vertical directions :smile:
     
  21. Feb 10, 2012 #20
    Ok I think we're getting back to my point of confusion from before. We've established that the only acceleration in the horizontal circle is the radial acceleration so I'm not quite seeing where this vertical force can come from unless you mean the weight
     
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