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Circular mass

  1. Dec 30, 2012 #1
    1. The problem statement, all variables and given/known data
    This is the diagram http://s1302.beta.photobucket.com/user/Rameel17/media/dd_zps43d04d44.png.html

    A single large circular Olympic ring hangs freely at the lower end of a strong flexible ring which is firmly supported at the other end. Two identical beads each has a mass of 30 kg are free to slide without friction around the ring. The ring passes through the hole in the beads. The two beads are released from rest at the very top as shown of the ring. At least once during their fall to the bottom the tension in the rope is zero. Calculate the maximum mass of the king in kg.



    3. The attempt at a solution

    I am not sure where to being for this one.
     
  2. jcsd
  3. Dec 30, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi SuperHero! Welcome to PF! :smile:

    You could use conservation of energy to find the relationship between angular speed and height.

    And you could consider what happens to the centre of mass.
     
  4. Dec 30, 2012 #3
    Re: Welcome to PF!

    Okay so do you mean like this

    mgh = 1/2mvo^2
    (30kg)(9.8m/s^2)h=1/2(30kg)(v^2)
    (30kg)(9.8m/s^2)h/1/2(30kg) = v^2

    Fc = ma
    Fc = m (vc^2/R)

    right? :D
     
  5. Dec 30, 2012 #4

    tiny-tim

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    Hi SuperHero! :smile:

    (try using the X2 and X2 buttons just above the Reply box :wink:)
    there are two velocities, that of the beads, and that of the large ring :wink:
    that's not correct … ma equals the total force (including the weight)

    anyway, more useful would be:
    total change in momentum = sum of all external forces​

    (and i'm off to bed :zzz:)
     
  6. Dec 30, 2012 #5
    total change in momentum = sum of all external forces​


    What do you mean by this?
    I am really confused..
    So this is what is given to us.
    V1 of bead 1 = 0m/s
    V1 of bead 2 = 0m/s
    bead 1 mass = 30 kg
    bead 2 mass = 30 kg
    T= 0
    Mass of ring = ?

    Now what do i do?
    Are you trying to tell me that i should use total conservation of momentum? or?
     
  7. Dec 30, 2012 #6

    haruspex

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    You need to find the position of the masses at which the tension is minimised. Suppose this is when the lines from the ring's centre to the masses are at angle theta to the vertical. Use conservation of energy to find the speeds at that point. Then you can work out the normal force from ring to masses, and from that the tension in the string.
     
  8. Dec 31, 2012 #7

    tiny-tim

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    Hi SuperHero! :smile:

    I think I've misunderstood the set-up …

    my computer wouldn't show me your picture, and i thought the large ring was moving …

    please ignore what i previously wrote :redface:
    yes :smile:

    (except you could have cancelled m on both sides, and it might be better to use θ as your variable, rather than v)

    find the equation for the normal force, as haruspex :smile: suggests ​
     
  9. Jan 2, 2013 #8
    Okay so like this

    mgh = 1/2mvo^2
    (30kg)(9.8m/s^2)h=1/2(30kg)(v^2)
    (30kg)(9.8m/s^2)h/1/2(30kg) = v^2

    but i am so lost. I do not understand how i can use the above equation to find the v^2 if we don't even know the height.
     
  10. Jan 2, 2013 #9

    tiny-tim

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    Hi SuperHero! Happy new year! :smile:
    That's an equation relating h and v.

    So you can find the centripetal acceleration at a general height h.

    What is the relation between the tension in the string and the centripetal acceleration (at height h)?​
     
  11. Jan 2, 2013 #10
    will they be point down?
     
  12. Jan 2, 2013 #11

    TSny

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    In order to motivate why you should even bother with writing the conservation of energy equation or why you need to worry about centripetal acceleration, see if you can conceptualize what causes the big ring to be “lifted” upward so that the tension in the string can go slack. It must be some force acting on the ring. Can you describe all the forces acting on the big ring as the beads are sliding? Can you identify which of those forces is (are) responsible for lifting up the ring? The answer is not immediately obvious. But if you can see what it is, it will provide a motivation for what equations to write down and what to solve for.

    (I hope I'm not side-tracking the thread. If so, just ignore this.)
     
  13. Jan 2, 2013 #12
    Well some of the forces acting on the ring are force of gravity, the tension, and the normal force. The normal force is responsible for lifting up the ring. right?
     
  14. Jan 2, 2013 #13
    OKay so i was doing a little research on this type of question, apparently there is a force known as the normal force that acts against the beads and the force of gravity changes into mgcosθ.
    Furthermore they come up with this equation mgcosθ - N = mv^2/R then they have plugged this equation into the conservation of momentum equation which becomes
    mg(2R) = 1/2mv^2 + mgR(1+cosθ)
     
  15. Jan 2, 2013 #14

    TSny

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    Those aren't just some of the forces, those are all of the forces if you include the normal force from each bead. Good. And, yes, the normal forces from the beads are responsible for lifting up the ring. Or, more precisely, the vertical components of the normal forces on the ring are what lift the ring.

    So, you're going to need to find an expression for the normal force that each bead exerts on the ring. The expression will depend on the position of the beads. The position of the beads are determined by θ, where θ is the angle between the vertical and a line drawn from the center of the ring to one of the beads
     
  16. Jan 2, 2013 #15
    so like this?

    mgcosθ - Fn = m(v^2/R)
     
  17. Jan 2, 2013 #16

    TSny

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    This looks like the equation describing the forces on one of the beads, right? If so, then Fn here is the normal force on one of the beads. Just to make sure we're on track, how is that normal force related to the normal force acting on the ring?
     
  18. Jan 2, 2013 #17
    well there are two beads.
    if we add them together, then that would be the normal force pulling the ring right?
     
  19. Jan 2, 2013 #18

    TSny

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    Yes, each bead exerts a normal force on the ring. So, there are two normal forces on the ring.

    I just wanted to make sure that you see the relationship between the normal force that the ring exerts on a bead and the normal force that a bead exerts on the ring. These are "action-reaction" forces so they have the same magnitude but opposite directions. So, if you can find Fn acting on a bead, then the normal force acting on the ring will have the same magnitude but opposite direction.

    For the total centripetal force acting on a bead, you wrote mgcosθ - Fn, with a minus sign for Fn. Doesn't that mean that you are considering the force Fn on the bead as acting outward (away from the center)? If so, then the force on the ring would be in the opposite direction, which would be inward toward the center of the ring. That is not going to lift the ring. (We assume θ<90o)

    As a bead begins to slide on the ring, the normal force on the bead is indeed outward. But, as the bead picks up more and more speed, it needs more and more centripetal force in order to travel in a circle. At some point (θ<90o ), the normal force on the bead actually changes direction and points inward to help provide sufficient centripetal force on the bead. Then, the normal force on the ring is outward and tends to lift the ring. So, you want to be considering the case where the normal force on the bead is inward in the same direction as mgcosθ. So, your formula would be mgcosθ + Fn = mv2/R
     
  20. Jan 2, 2013 #19
    ok so mgcosθ + Fn = mv2/R is my equation

    so i would plug this into the conservation of energy formula

    right?
     
  21. Jan 2, 2013 #20

    haruspex

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    Yes, in the sense that you use the expression for v2 that you obtained from conservation of energy.
     
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