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Homework Help: A chain falling off a rotating disc

  1. Jun 29, 2015 #1
    1. The problem statement, all variables and given/known data
    A uniform chain of length L = πR and mass M is placed on the upper half of a
    uniform thin disc of radius R and mass M. The disc is placed vertically and can rotate freely
    about its center that is fixed in space. With a small disturbance the chain starts to fall. Find out
    the angular velocity of the disc when the chain leaves it.
    (Assume the chain does not slide off the disc, just falls off the side)

    2. Relevant equations
    L = I w = F_ext R
    I_disc = (MR^2)/2
    F_ext = (d/dt) P = v (d/dt m) + m (d/dt) v

    3. The attempt at a solution
    So, the chain is the cause of any change in rotation. Only the length of chain that is not counterbalanced across the disc causes any net acceleration, so mass (related to acceleration) is a function of time:
    My guess: m(t) = M Sin[wt/2] = M Sin[theta/2]

    When the disc has rotated 180 degrees, the chain is completely off and therefore there is no net acceleration on the disc (so that's why I picked the mass function above).

    My naive feeling is that
    w = (F_ext R)/ I

    But before I even get that far, I need to account for I, which is more complicated than just the I of the disc, because the chain is there initially:
    I = I_disc+ I_chain = (M^2 R/2) + (I_chain)

    And I_chain is a function of rotation angle?
    I_chain(theta) = ((MR^2)/Pi) * theta?

    I got that by using I = Integral[r^2 dm] where dm= lambda dL, and dL = r dTheta, and lambda= M/(Pi * R)

    So if that's true, how do I properly solve for this? I figure I can't just plug in theta=Pi and call it a day...

  2. jcsd
  3. Jun 29, 2015 #2


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    Why not use conservation of energy? If the chain doesn't slide off, then no energy should be dissipated.
  4. Jun 29, 2015 #3


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    This seems like a troublesome way of looking at it. The point of this is to find the torque as a function of the angle turned, right? (Or am I missing your idea?)

    If you want to find the torque as a function of the angle turned I suggest looking at it in two parts: The torque from the hanging part of the chain and the torque from the unbalanced part of the chain. (The second part will be an integral.) This will be a key equation if you want to solve it with the differential-equation-method.

    But again, this convoluted differential-equation-method can be bypassed by using conservation of energy.

    I don't believe it is.

    I think (but am not 100% sure) that we can treat the rotational inertia of the chain as if it were just constantly wrapped around the disc ([itex]I_{chain}=MR^2[/itex])

    The reason I think this is because the kinetic energy of the chain will depend on ω in the same way whether it is wrapped around or hanging off.

    I'd like to know what others think of this.
    Last edited: Jun 29, 2015
  5. Jun 29, 2015 #4


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    I agree with this. Nice observation. :smile:
  6. Jun 29, 2015 #5
    Ok, I thought about conservation of energy, but it's not so simple either:

    If the chain were sliding off of a table, this would be easier, but it's turning a wheel. So, I am not sure how to calculate the potential energy before the chain falls...
  7. Jun 29, 2015 #6


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    The problem becomes finding the center of mass of a (uniformly dense) semi-circle.
  8. Jun 29, 2015 #7
    Well, it's only a semi-circle for a brief moment of time, and then it's something else. Finding the center of mass of a semicircle is simple...
  9. Jun 29, 2015 #8


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    Yes, but that's the usefulness of conservation of energy: You only need to consider the initial and final states.
  10. Jun 29, 2015 #9
    Ok, true. So how would I apply the center of mass of a semicircle in an energy conservation model?

    PE_initial = KE_disc + KE_chain
    PE_initial = (Iw^2)/2 + (mv^2)/2 = (Iw^2)/2 + (m(rw)^2)/2

    (I substituted v=rw in the chain's KE because the chain and disc are moving at the same speed up until the chain leaves the disc)

    I'm not sure what to choose for PE_initial. Would I use PE= Mgh, where h is the center of mass of the chain when it starts to move?
  11. Jun 29, 2015 #10


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    Only changes in potential energy are meaningful. So this equation doesn't make much sense.
    Perhaps you mean something more like [itex]|PE_{final}-PE_{initial}|=|KE_{final}-KE_{initial}|[/itex] or more concisely [itex]|\Delta [PE]|=|\Delta [KE]|[/itex]
    (I put absolute value signs because when the PE decreases, the KE increases, so there should be a negative sign in there, but it's not important.)

    This is a great example why only changes in potential energy matter: Where do you take h to be measured from? It could be measured from anywhere, so the initial PE could be anything. Thus only the Δ[PE] is meaningful.
    (The Δ[PE] does not depend on where you measure h from, "the constant cancels out" so to speak.)

    But yes, you have the right idea: Δ[PE]=mgΔh where Δh is the change in height of the center of mass (between the initial and final states).
  12. Jun 30, 2015 #11
    Yeah I had arbitrarily set h=0 to where the CM of the chain is once it's off the disc, but you are right, it's really just Δh that matters.

    So, I've created this horrible drawing to try an illustrate what I'm thinking. The red dots symbolize the CM of the chain, and the constants a and b represent the vertical distance from the center of the disc to the CM of the chain:

    So in this case, Δh = a + b = R(π/2 + 1/√2)

    With that, I can use the equation we discussed:
    mgΔh = (I ω^2)/2 + (m ω^2 R^2)/2

    And solve for ω

    My attempt:
    ω^2 = 4g(π/2 + 1/√2)/(3R)

    (I just left the square to keep it clean)
  13. Jun 30, 2015 #12


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    Looks good except I disagree with "a"
  14. Jun 30, 2015 #13
    Where would you put the center of mass of the chain, as it was resting on the wheel? (if not at a)
  15. Jun 30, 2015 #14


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    I agree with your idea, I just disagree with your value of a.

    Can you show the calculation that led to the value of a?
  16. Jun 30, 2015 #15
    Half of the mass of the chain corresponds to half the length. Half of the length is from Pi/4 to 3Pi/4, measured from 0 being on the horizontal axis. Therefore, the height that splits the mass of the chain vertically would be R Sin [Pi/4]. If you drew a horizontal line at this height, half of the chain's mass is above, and half is below.
  17. Jun 30, 2015 #16


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    That's an interesting idea, but that's not how the center of mass works. It's not about having half the mass on one side and half on the other side, it's about "weighing" the mass by it's distance from the axis. (More specifically, [itex]y_{CoM}=\int ydm[/itex])

    What if we were to deform the top half of the semi circle so that half of the mass is still above 1/√2 like in this picture:
    Wouldn't you expect that the center of mass would change? But according to your method the center of mass does not change.
  18. Jun 30, 2015 #17
    Ah, good point...

    Ok, so Rcm = (1/M) ∫(vector r) dM

    M= λ L
    dM = λ dL = λ R dφ

    Vector r = R rhat = R Cosφ x_hat + R Sinφ y_hat

    Plugging this into the integral and solving in terms of known constants yields:
    Rcm = (2R/π) y_hat

    So you are absolutely right, it's less high than I had originally thought.
  19. Jun 30, 2015 #18


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    With that fixed, I agree with your answer for ω :smile:
  20. Jun 30, 2015 #19
    Thank you (and everyone else) so much for your help! It is REALLY appreciated!!!
  21. Jul 4, 2015 #20


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    Just to note that in practice there would be a complication. I don't think the chain would stay in contact with the disc until it is vertical. As the speed increases, it will detach at a progressively higher point. But I very much doubt the question setter intended that to be considered. (On the other hand, that would justify its being posted in the Advanced Physics forum.)
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