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Homework Help: Beads sliding down a ring

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data

    A single large circular Olympic ring hangs freely at the lower end of a strong flexible rope firmly supported at the other end. Two identical small beads each of mass 30 kg are free to slide symmetrically without friction around the ring (the ring passes through the holes in the beads). The two beads are released from rest at the very top as shown and they slide symmetrically down around opposite sides of the ring. At least once during their fall to the bottom of the tension in the rope is observed to be zero. Calculate the maximum value for the mass of the ring. Answer in kg.

    a) 15
    b) 20
    c) 25
    d) 30
    e) This could never happen

    The question setup

    2. Relevant equations

    [itex]F_c = \frac {mv^2}{r}[/itex]

    3. The attempt at a solution

    My final answer was 30 kg (d), but I am unsure if my steps and logic are correct.

    The beads slide down a circular ring, so they will exert a centrifugal force to the ring that will counter gravity.

    Since centrifugal force acts away from the center, the tension can only be zero when the beads are at the top half of the ring.

    The free body diagram of the beads will look like this.

    [itex]F_c = \frac {mv^2}{r}[/itex]
    X components cancel since the 2 beads are in opposite directions.
    Y component: [itex]F_cy = \frac {mv^2}{r} * sin \theta[/itex]

    Solve for v2 using the work-energy theorem (setting the zero point at the diameter of the ring).
    [itex]mgr = \frac {mv^2}{2} + mgrsin\theta[/itex]
    [itex]v^2 = 2gr - 2grsin\theta [/itex]

    Sub v2 into the y component of the centrifugal force...
    [itex]F_cy = m\frac {2gr - 2grsin\theta}{r} * sin \theta[/itex]

    [itex]F_cy = 2mgsin\theta - 2mg sin^2 \theta[/itex]

    Differentiate the y component with respect to θ and set to zero.
    [itex]0 = 2mgcos\theta -2mg (2sin \theta)(cos\theta)[/itex]

    [itex]0 = 1 -2sin \theta[/itex]

    [itex]sin\theta = 0.5 [/itex]

    Sub sinθ = 0.5 into the y component
    [itex]F_cy = 2mgsin\theta - 2mg sin^2 \theta[/itex]

    [itex]F_cy = 2(30)g(0.5) - 2(30)g(0.25)[/itex]

    [itex]F_cy = 15g[/itex]

    There are 2 beads, so the total force upwards is 30g.
    30g = Mg (M is the mass of the ring)
    M = 30

    Again I am not sure if I did the question correctly. I used centrifugal force instead of centripetal force, and in all the physics questions I've done I've never used centrifugal force in a free body diagram. Also I am not sure if the force of gravity from the beads would act on the ring.
  2. jcsd
  3. Feb 27, 2013 #2


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    No, I would say you used centripetal force. You calculated the inward force needed on the beads to make them move in a circle. The force the beads exert on the ring is reaction to this, so is equal and opposite.
    No, but it does act on the beads, and you neglected that (thus arriving at the wrong answer). Add it to your free body diagram and see what you get for Fc now.
  4. Feb 27, 2013 #3
    So if gravity contributes to the centripetal acceleration, the component that does it would be mg cosθ. Then, the vertical component of that component would be mg (cosθ)2.

    So the new centripetal force y component is....

    [itex]15g + mg cos^2\theta[/itex]

    [itex]15g + 30g cos^2(30)[/itex]

    [itex]15g + 30g (3/4)[/itex]

    [itex]15g + 22.5g[/itex]


    The new centripetal force would be 37.5 g, but now the mass of the ring could be larger than 30 kg.
  5. Feb 27, 2013 #4


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    No, you went the wrong way there somewhere.
    Gravity is providing mg cosθ towards the centripetal force. So the contribution from the ring is mv2/r - mg cosθ.
  6. Feb 27, 2013 #5
    Why -mg cosθ? If gravity does provide the centripetal force, then the centrifugal force that the beads exert on the ring should be greater.

    Also shouldn't gravity already be counted in the first place by using the work energy theorem? :confused:
  7. Feb 27, 2013 #6
    Alright, I probably don't understand centripetal and centrifugal force.
    The way I was taught centripetal force was that it was the net force.
    Since only mg cosθ can contribute to the centripetal force, does this mean mg cosθ = mv2/r?
    Is this how you get mv2/r - mg cosθ ?

    In that case:

    [itex]mg cos\theta = m \frac{v^2}{r}[/itex]
    [itex]mg cos\theta = 2mgsinθ−2mgsin^2θ[/itex]
    [itex]cos\theta = 2sinθ−2sin^2θ[/itex]
    [itex]1 - (sin\theta)^2 = 2sinθ−2sin^2θ[/itex]
    [itex]0 = −sin^2θ + 2sinθ - 1[/itex]
    [itex]0 = sin^2θ - 2sinθ + 1[/itex]

    sin θ = -1 ?

    Edit: Forgot the square root, will fix

    Edit 2: The equation becomes worse after the square root is implemented. I get a cubic equation with a real solution of sin x = -0.44.
    Last edited: Feb 27, 2013
  8. Feb 27, 2013 #7


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    That's right.
    No. There is a normal force acting between bead and ring. Say it's N, radially inwards on the bead, radially outwards on the ring. In the radial direction, the net force on the bead is N+mg cos θ inwards. Since the resultant needs to be mv2/r, we have N = mv2/r - mg cos θ. So as far as the ring is concerned, each bead exerts a force mv2/r - mg cos θ in the radially outward direction. The vertical component of that is mv2/r cos θ - mg cos2 θ.
  9. Feb 28, 2013 #8
    So the force that the bead exerts on the ring is a normal force, not a centrifugal force?

    In that case the y component would be Fn cos θ

    [itex]N cos\theta = m\frac{v^2}{r}cos\theta - mg cos^2\theta[/itex]

    [itex]N cos\theta = 15g - 22.5g[/itex]

    [itex]N cos\theta = -7.5g[/itex]

    Negative because down was set to be positive?
    Then the mass of the ring would be:

    2(7.5)g = Mg
    M = 15

    The answer is a.

    Edit: Looking back at this post, it seems like the y component should be N sin θ instead.
    Last edited: Feb 28, 2013
  10. Feb 28, 2013 #9


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    I don't understand that step. We don't know what theta is yet.
    Plug in what you already worked out for v2 from the KE.
  11. Feb 28, 2013 #10
    I thought theta was solved in the first post to be 30.

    However, now that a normal force is introduced, does this mean I have to differentiate the y component normal force with respect to theta to optimize the upward force?

    If so:

    [itex]N = m\frac {v^2}{r} - mg cos\theta[/itex]

    [itex]N = 2mg -2mg sin\theta - mg cos\theta[/itex]

    [itex]N sin\theta = 2mg sin\theta -2mg sin^2\theta - mg cos\theta sin \theta[/itex]

    I tried differentiating this equation to find theta, but I got stuck in a trig equation. Am I doing something wrong?
  12. Feb 28, 2013 #11


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    The error I pointed out in my first post (neglecting mg on the beads) occurred before you derived theta, so that value is almost surely wrong.
    Yes. But I just noticed that I'd misremembered which way you were defining theta. Most would take it as the angle to the vertical, but I forgot you have it as the angle to the horizontal. So N + gm sin(θ) = mv2/r = 2mg(1 - sin(θ))
    Go forwards from there.
  13. Feb 28, 2013 #12
    I see, that equation becomes easier to differentiate now, if all goes well this should be the last step.

    [itex]N = 2mg - 3mgsin\theta[/itex]

    [itex]N sin\theta = 2mg sin\theta - 3mgsin^2\theta[/itex]


    [itex]0 = 2mg cos\theta - 3mg(2)(sin\theta)(cos\theta)[/itex]

    [itex]0 = 1 - 3(sin\theta)[/itex]

    [itex]sin\theta = \frac {1}{3}[/itex]

    Substitute in the original equation...

    [itex]N_{y} = 2mg \frac {1}{3} - 3mg\frac {1}{9}[/itex]

    [itex]N_{y} = mg \frac {1}{3} [/itex]


    [itex]mg \frac {2}{3} = Mg[/itex]

    M = 20. So the answer is b.

    Thank you for all help! :)

    Though one last question, the normal force acts inwards on the beads and outwards on the ring. Mathematically this makes sense to keep the centripetal acceleration pointing towards the center, but could you determine the directions of the normal force before any calculations?

    When I did my physics problems the normal force always pointed in the opposite direction from gravity, why did it point in the same direction this time?
    Last edited: Feb 28, 2013
  14. Feb 28, 2013 #13


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    Well done.
    No, the direction of the normal force can change as the bead moves around the ring. To begin with it would be outwards, so the beads push down on the ring. When the beads are at the bottom of the ring it's inwards, but again that means the beads push down. It's only over a short range of angles that the normal is force is outwards but the beads, still being above the midpoint, pull up on the ring.
    That's because in most problems the mass is sitting on some surface which cannot hold it down, so the normal force cannot be into the surface.
  15. Feb 28, 2013 #14
    I assume the normal force you're talking about is the force the beads exert on the ring instead of the force the ring exerts on the beads. However, if the normal force is outwards, why would the beads still push down at the beginning? Plus if the normal force at the bottom is inwards, shouldn't the beads should also pull up the ring?

    Also I am trying to figure out why the normal force would suddenly change from outwards to inwards at the midpoint.
  16. Feb 28, 2013 #15


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    Sorry, should have been clearer. I was referring to the force the ring exerts on the beads.
    At start, beads sit on top the ring. Normal force is outwards on the beads, downward on the ring. At half way down and beyond, the normal force must be inward on the beads, outward on the ring. (At half way it's all centripetal; after that it's centripetal plus a component for gravity on the beads.) Past half way, outward on the ring means downward on it, so the only section where the force on the ring can be upwards is in some range above the halfway mark.
    It's not a sudden change. Somewhere before the angle you found (maximising the upward force on the ring), the normal force passes from outwards on the beads, through zero, to inwards on the beads. If you put N=0 in instead you'll find sin(θ) = 2/3.
  17. Feb 28, 2013 #16
    I am afraid I am still struggling with this whole concept. The beads get pushed outwards, but the ring gets pushed inwards. However the calculation showed that the beads had to push the ring outwards.

    Perhaps a more specific question would be: why use centrifugal force for this question instead of centripetal force? I only got to using centrifugal force and making the beads above the midpoint only because doing this problem using centripetal force didn't give a solution in the choices.
  18. Feb 28, 2013 #17


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    The ring is pushing upwards on the beads when they're at the top, yes? It's just the usual normal force to balance gravity on the beads. At this point, upwards=outwards. As the beads slide around, the normal force required to balance gravity reduces, because of the angle. At the same time, a centripetal force is required to keep the beads moving in a circle. When sin(theta) = 2/3, the radial component of gravity acting on the beads is only just enough to provide the centripetal acceleration, so at this point the normal force has fallen to 0. Beyond that, the beads are moving faster, and gravity acting on the beads is becoming less and less able to provide a radially inward force, so, to provide sufficient centripetal force, the normal force is now going inward on the beads, outward (and therefore upward) on the ring. Ever after, the normal force will be inward on the beads, outward (and thus, when past halfway) downward on the ring.
    At no point did I appeal to centrifugal force, and I don't believe you did either. Centrifugal force is only an apparent force that arises when you use a non-inertial frame of reference.
  19. Feb 28, 2013 #18
    Okay, I think I finally understand everything now.

    The normal force changes direction to maintain a centripetal force, it doesn't spontaneously change direction halfway.

    Centrifugal force was never responsible for lifting the ring, it was the reaction force of the normal force.

    Thank you for all your help! =)
    I should probably review Newton's third law and circular motion again.
  20. Feb 28, 2013 #19


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    That's all correct, and you're welcome. Btw, if you search these forums you'll find this exact question has been dealt with at least twice before, but each person who brings it here has a different journey to follow.
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