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Homework Help: Chain and spinning disk problem

  1. Mar 4, 2014 #1
    1. The problem statement, all variables and given/known data
    A chain is wrapped around a disk of radius R. The tension of chain is T. What is the coefficient of friction, if when the disk is spinning at angular velocity ω, the chain slips down?

    See image attached.

    2. Relevant equations
    II Newton law
    [itex] a_{centripetal} = \frac{v^2}{R}[/itex]
    k = F \ N

    3. The attempt at a solution
    I'm not even sure where to start - I don't really understand, why does the chain slip? Is it because friction can't 'hold' it anymore? Does chain tension depend on the rotation?

    Can we say that the chain slips when F(friction) < mg ?

    Attached Files:

  2. jcsd
  3. Mar 4, 2014 #2
    Yes, it's from friction. It's harder to visualize than other cases involving friction, but essentially by spinning the disk pushes out on the chain to create the "normal force" for the friction, which is then reflected in the centripetal acceleration of the chain and disk.
  4. Mar 4, 2014 #3
    So then the normal force of the chain is
    [itex] N = ma_{centripetal} = m\frac{v^2}{R} = m\omega^2R [/itex]?
    But then chain tension has no impact here, because [itex] k = \frac{F_f}{N} = \frac{mg}{m\omega^2R} = \frac{g}{\omega^2R}[/itex]? What am I missing?
  5. Mar 4, 2014 #4
    I'm not sure that the tension matters for this case since it's not pulling anything... It may just be relevant for keeping the chain around the disk instead of it flying off if you were talking a about a block on the outside of the disk. I'm not entirely sure though. Do you happen to have the final solution?
  6. Mar 4, 2014 #5
    If the chain has a hoop force T and the disk is not rotating, what normal force (per unit rim length) does the chain exert on the disk (like a rubber band would exert on the disk)? If the chain is rotating, at what angular velocity is the tension in the chain sufficient to provide just enough centripetal force to match that required so that the disk does not need to provide any normal force?

  7. Mar 5, 2014 #6
    No, I don't...

    So for the first question - I thought that for every piece of chain the tension is tangent to the disk, and that would mean that the angle between the normal force and the tension is 90 degrees, therefore tension has no impact on normal force?

    Then tension does depend on velocity?
  8. Mar 5, 2014 #7
    Try this. Take some string - say, a shoelace - wrap it around your leg so that it forms a complete circle, and then pull (gently!) the loose ends to tighten it. Does your leg agree that the tension has no impact on the normal force?
  9. Mar 5, 2014 #8
    Yes, I guess tension does have impact! But how do we express that mathematically? Is tension not right-angled with normal force?
  10. Mar 5, 2014 #9


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    Yes it is. Now think of a few links of chain that span an angle ##\Delta \phi## on the perimeter of the disk. The tensions T at the ends don't align any more and presto! there is your normal force !
  11. Mar 5, 2014 #10
    Wow! Okay, so I drew a little sketch of how I imagine it right now.


    [itex] N = Tsin\frac{\Delta\phi}{2} [/itex] and also [itex] \Delta\phi = \frac{x}{R} [/itex] where x is the rim corresponding to [itex]\Delta\phi[/itex]. I guess now integrating, to find the whole normal force?

    Attached Files:

  12. Mar 5, 2014 #11


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    Funny you should draw these T inward. The tensions ON this piece of chain add up to a force ON the disk that points inwards. The disk pushes back with an equal and opposite normal force that points outwards. The friction force is then a coefficient times this outward force, and it points upwards -- thus counteracting gravity which is pointing down.
  13. Mar 5, 2014 #12
    See if you can show that, if you have a differential section of chain extending from θ to θ+dθ, the net force that the adjacent portions of the chain exert on this differential section of chain is Tdθ = (T/r)rdθ, and this net force is directed radially inward toward the axis. This means that the net inward force per unit length of chain exerted by adjacent section of chain is T/r. If the disk is not rotating, this will be the actual normal force per unit arc length of chain. (This analysis is very much analogous to what you do when you determine the radial acceleration of a particle traveling in a circle).

    You are not going to be doing an integration to find the "whole normal force." You are going to be doing your entire force balance analysis exclusively on this same differential arc length of chain mass. As we progress, you'll see how this plays out.

    Last edited: Mar 5, 2014
  14. Mar 6, 2014 #13
    so, let's say each link of the chain has mass m. Then, mg = μ(T-mω2R), right? Here T is the tension acting inward in each link of the chain. Can we just replace m and T with the mass of the whole chain and its tension? I feel like you might be able to, but I'm not really sure.
  15. Mar 6, 2014 #14
    So my sketch is what you had in mind, just T should be inwards? And the T - N relationship I wrote - is it right?

    So does this apply:

    [itex] m\vec{a_c} = \vec{T} + \vec{N} => ma_c = \frac{T}{R}\phi - N[/itex] ?
    Thus [itex] N = \frac{T}{R}\phi - m\omega^2R[/itex]?
  16. Mar 6, 2014 #15
    ##T/R## is the force due to tension per unit length of chain. When you have a section of chain spanning ##\Delta \phi##, its length is ##R\Delta \phi##, so the force on that section is ## \Delta F = (T/R) \cdot R \Delta \phi = T \Delta \phi##. Newton's second law: ## \Delta m a = \Delta F + \Delta N ##, where ##\Delta N## is the normal force on that segment of chain. What can be said about the normal force when the chain begins to slip? What is ##\Delta m##?
  17. Mar 6, 2014 #16
    [itex]\Delta m[/itex] is the mass of the segment of the chain, and when chain begins to slip normal force is [itex] \Delta N = \frac{\Delta mg}{k}[/itex] (so the friction is equal to mg) ?
  18. Mar 6, 2014 #17


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    T should be outwards. T1 to the left, T2 to the right. Sum of the two T is then pointing inwards, and the magnitude is ## {\bf 2} \ T \sin({\Delta \phi \over 2}) = T \Delta \phi \quad## (I missed the ##{\bf 2}\quad## and leave the ##\Delta## in).

    This T resultant provides the centripetal force ## m \omega^2 R## for this little piece of chain, for which you now must also express the mass in therms of ##\Delta \phi##. At low rpm the difference between T resultant and required centripetal force is pressing on the disk rim and the reaction force, exercised by the disk ON the chain is indeed N (pointing outwards), so your last line is OK.

    At the rpm where the chain falls off, kN = mg . ##\Delta \phi## falls out and your expression for k is ready. That's all.

    [edit]voko came in while I was typing ever so slowly. Yet another crossing reply.
  19. Mar 6, 2014 #18
    Correct on the normal force. ##\Delta m##, though - what is it in terms of ##\Delta \phi##?
  20. Mar 6, 2014 #19
    [itex] \Delta m = \frac{m \Delta \phi}{2 \pi}[/itex] ?
  21. Mar 6, 2014 #20
    OK. The force acting on a differential section of chain between θ and θ+dθ (imposed on the section of chain by the adjacent regions of chain) is [itex]-T\vec{i}_rdθ[/itex]. Do a free body diagram of this small section of chain. The mass contained in this small section of chain is [itex]\frac{mdθ}{2π}[/itex], where m is the total mass of the chain. The normal force exerted on this small section of chain by the disk is [itex]nrdθ\vec{i}_r[/itex], where n is the normal force per unit length. The acceleration is [itex]-ω^2r\vec{i}_r[/itex]. So the force balance reads:

    The force balance in the vertical direction on this small section of chain reads:
    where f is the frictional force per unit length of chain. This equation reduces to:
    If the chain is about to slip, how is the frictional force per unit length f related to the normal force per unit length n?
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