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Mechanics velocity problem

  1. Oct 1, 2005 #1
    Hiya, I'm not entirely sure where I should put this question, but as it isn't homework I thought this might be the best place for it.

    Anyways, I've been banging my head on this question for almost an hour now, and I've gotten absolutely nowhere.

    The question is that there's a chain hanging off the edge of a table with dynamic coefficient of friction of 0.1, where 1/5 is hanging off the edge and 4/5 is still on the table, and I need to find the velocity of the chain once it has completely left the table.

    What on earth do I do to solve this? I keep getting 3.92 ms^-1, but the answer should be 2.63ms^-1...

    Just some guy.
  2. jcsd
  3. Oct 1, 2005 #2


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    What DE's (differential equations) did you come up with?

    I just did a quick calc and got 2.963 m/s assumming a one meter total chain length (you didn't specify the chain length, I dont think it's invarient of this).

    BTW, with this type of problem you also need to make certain assumptions about the chains initial arrangment on the table. In my above calculation I've assumed that the chain is initially fully out-streched on the table, perpendicular to the edge and with no loose links. The opposite extreme of this is the case where all the links are bunched up on the edge of the table and can fall without any appeciable motion of the remaining links required, that would actually make it a very different problem.
  4. Oct 1, 2005 #3
    ok, well I haven't properly done differential equations in my pure maths classes, nor modelled anything like this in my mechanics classes (only string/pulley systems where the mass over the edge isn't changing), so this may look a little daft!

    I started from the equation used where the system consists of two masses (one on the table, one over the edge) which don't change, which gives g(m1) - 0.1g(m2) = (m1+m2)a, where 0.1 = mu, m1 = mass over the edge and m2 = mass on the table. If the masses are changing with time and m1+m2 = 1 (as they're the same object) this (might?) give

    g(dm/dt) - 0.1g(d(1-m)/dt) = dv/dt, where m is just the mass over the edge of the table.

    Have I already gotten way off track or is this vaguely correct?
  5. Oct 1, 2005 #4


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    Your first equation, g(m1) - 0.1g(m2) = (m1+m2)a, is correct, but you shouldn't just let the total mass be 1.
    You 2nd equation though, g(dm/dt) - 0.1g(d(1-m)/dt) = dv/dt is off. You have let m1 = dm/dt, and m2 = d(1-m)/dt, which are wrong.

    As, uart mentioned, you have to take the length of the chain into account. I did the calculation also and got the exact same value for acceleration using a length of 1 m.

    Let the total mass be M, say.

    You know that the masses on the table and over the table edge change with time, yes ?
    You might find it easier to treat the mass as a linear value, i.e. m = ρl.

    It might also be easier if you treated the lhs of your equation as forces acting on the masses, rather than the accelerations produced in them.
    Last edited: Oct 1, 2005
  6. Oct 1, 2005 #5
    um, what's pl? And since the masses on the table and over the table are proprtions of the same mass what's wrong with what I did? And what's lhs?

    Just some guy
  7. Oct 1, 2005 #6


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    I also get 2.963 m/s by using energy conservation. This requires working out how much work is done dragging the chain across the table and assuming the total length of the chain is 1 m.
  8. Oct 2, 2005 #7


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    I always forget to try doing it with energy consideration :mad:
    And it's usually much easier!
    But I found it a wee bit difficult in this case because of the changing masses/lengths, so I'll explain it the way I did it originally.

    Perhaps the other guys can show you how to do it easier the energy conservatrion way.

    The original eqn you had, "g(m1) - 0.1g(m2) = (m1 + m2)a" doesn't take into account that the lengths, and hence massses, will be changing.
    That is [tex]\rho l[/tex] where [tex]\rho[/tex], rho, is the linear density. So you could have [tex]M = \rho L[/tex], where [tex]M[/tex] is the total mass and [tex]L[/tex] is the total length of the cable.
    As the cable slips over the edge of the table, it moves a distance, x say, and its vertical length (and hence its "vertical" mass) will increase by an amount x. Similarly, the horizontal length ( and hence the "horizontal" mass) will decrease by an amount x.

    Let [tex]m_1 = \rho l_1[/tex] and [tex]m_2 = \rho l_2[/tex] with [tex] l_1 + l_2 = L[/tex] and [tex]M = \rho L[/tex]

    Now substitute for [tex]m_1[/tex] and [tex]m_2[/tex] into your original eqn giving,

    [tex]g(m_1) - 0.1g(m_2) = (m_1 + m_2)a[/tex]
    [tex]g(\rho l_1) - 0.1g(\rho l_2) = (\rho L)a[/tex]

    Now when the lengths of the cable change as it slips over the table edge, you can take that into account by adding, and subtracting an amount x to/from both lengths,
    [tex]g(\rho l_1 - x) - 0.1g(\rho l_2 + x) = (\rho L)a[/tex]

    You can turn this into a DE asnd solve for v.
  9. Oct 2, 2005 #8


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    I used the following notation,

    m := mass per unit length
    L := total length of chain
    x := length of chain overhanging the edge
    u := coefficient of friction
    F := force
    x'' := acceleration (second time derivative of x).

    The equations I got for the system are as follows,

    F = m g x - u m g(L-x)

    m L x'' = m g (1 + u)x - m g u L

    x'' = g/L * (1 + u) x - g u
    Last edited: Oct 2, 2005
  10. Oct 3, 2005 #9


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    Energy conservation can sometimes lead you astray though. Take for example the alternate case I mentioned above, were the chain is all bunched up so that the links don't have to travel any appreciable distance across the table but instead just "flop off" one by one as the chain falls.

    This is actually a very interesting (and more difficult) problem for several reasons. Firstly because the moving mass is time varying so F=ma doesn't hold, secondly because it results in a harder to solve non-linear ode, and finally because a "conservation of energy" solution appears exceedingly simple yet fails spectacularly!

    The derivation of the basic equations for this case (with m denoting the mass per unit length as per my above post) is as follows.

    d/dt( mx v) = m g x

    [tex] \stackrel{\ }{x} \stackrel{..}{x} + (\stackrel{.}{x})^2 = g x[/tex]

    This seems a pretty hard DE to solve (if anyone knows the full solution let me know. As a time function I can only find a particular solution that doesn't solve the full dynamics).

    Fortunately you can re-cast the equation to a DE for v^2 in terms of x that is much easier. Letting z=v^2 and using the relation that a = 1/2 d/dx(v^2) you get the following :

    1/2 x z' + z = gx

    This equation has the solution of (after including initial conditions of v(0)=0 and x(0)=x_0) :
    [tex] v^2 = \frac{2g}{3} (x - x_0^3 / x^2) [/tex]

    If anyone is interested you can do a "conservation of energy" solution for this problem very readily and get the very different (and very wrong) result that :
    [tex]v^2 = g(x - x_0^2/x)[/tex]

    Here's a challenge, can anyone explain where the missing energy went? :surprised
    Last edited: Oct 3, 2005
  11. Oct 3, 2005 #10


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    The potential energy of the length of chain hanging over the edge is [itex]-mgx^2/2[/itex] if you use the surface of the table as your zero level of potential.
  12. Oct 4, 2005 #11


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    Yep, so if you take the initial overhang as [itex]x_0[/itex] then the change in potential energy is [tex]mgx^2/2-mgx_0^2/2[/tex]

    If you equate that to the change in KE you get,

    [tex]m x v^2 /2 = mgx^2/2-mgx_0^2/2[/tex]


    [tex]v^2 = g(x - x_0^2/x)[/tex].

    If you work out the speed of the chain at the 1 meter point (with 0.2 meter initial overhang as per the original problem) then you get v(1)=3.0672 m/s using this PE/KE formula whereas the correct solution is actually v(1)=2.5458 m/s (using the soln to the DE in my above post).

    So the puzzle is, where did the lost energy go ?
    Last edited: Oct 4, 2005
  13. Oct 4, 2005 #12


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    How can you be so certain the energy conservation method is in error? For example, in setting up your ODE, you haven't taken into account the fact that in your model new links falling over the edge instantly acquire the speed of the falling links. You will at least have to take this into account by including the reaction force on the falling links due to the inertia of those just being added.

    This will reduce the acceleration but it will be occur over a longer period of time. We won't know the outcome until your ODE is set up properly. Energy conservation doesn't have to deal with those complications.
  14. Oct 5, 2005 #13


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    Tide, the reaction force was implicit in the momentum equations I used to develop the DE. Take a look at that DE again (slightly rearranged)

    [tex]m x\stackrel{..}{x} = m x g - m v^2[/tex]

    Can you see that the last term on the RHS (the negative term) is the reaction force of the links being added!

    Let me show it explicitly. Taking [tex]\delta x[/tex] as the length of each link then the change in momentum of each added link is [tex]\delta p = m\, \delta x\, v[/tex]. (remember that I'm using symbol m as the mass per unit length throughout). Now the time taken to add each link (or the time between impluses if you prefer to look at it that way) is [tex]\delta t = \delta x / v[/tex].

    So the average rate of change of momentum for adding the links is [tex]F = \stackrel{.}{p} = m v^2[/tex], exactly as accounted for in the original DE.
  15. Oct 5, 2005 #14


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    No. Actually, the reaction force is [itex]-\frac {1}{2}v^2[/itex].

    Let's make it simple and start from scratch. I will use the symbol [itex]\lambda[/itex] to represent the mass per unit length of the chain so as not to confuse it with the total mass of the chain.

    The rate of change of momentum of the portion of chain already over the edge is

    [tex] \lambda x \frac {dv}{dt}[/tex]

    This portion of the chain is subject to two forces. One is the force of gravity which is just [itex]\lambda x g[/itex]. As is moves it "plucks" off links from the top of the table which are initially at rest and acquire speed v. So, as the chain moves through a distance of [itex]\Delta x[/itex] a portion of the table top chain, [itex]\lambda \Delta x[/itex], is accelerated by the falling chain.

    The falling chain does work on the table top chain such that [itex]F \Delta x = \frac {1}{2} \lambda \Delta x v^2[/itex] which brings the element [itex]\lambda \Delta x[/itex] up to speed. Therefore, the reaction force on the falling chain is [itex]-\frac {1}{2} \lambda v^2[/itex].

    This gives the equation of motion of the chain as

    [tex]x \frac {dv}{dt} = g x - \frac {1}{2} v^2[/tex]

    Solving this equation gives the same solution as the energy conservation method.

    The problem in your approach to the reaction force is that the [itex]v[/itex] in the denominator of [itex]\delta t = \delta x / v[/itex] is really the average speed of the newly accelerated links which is half of the maximum speed. Thus, you should really write [itex]\delta t = 2 \delta x / v[/itex] with [itex]v[/itex] representing the speed after acceleration over that time interval. This makes the reaction force [itex]-\lambda v^2 / 2[/itex] in agreement with my derivation and the conservation method.
  16. Oct 6, 2005 #15


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    Not true. Each link only has avaiable time of [tex]\Delta x /v[/tex] to come up to speed. During the time period of [itex]\Delta t = 2 \Delta x / v[/itex] two links must be pulled from the table!

    The dilema is very simple, the moving chain and the "new" link collide, after the collision they have essentually coalesced. Under such conditions it is simple not possible for both KE and momentum to be conserved. Momentun is conserved but some KE is lost in the collision.
    Last edited: Oct 6, 2005
  17. Oct 6, 2005 #16


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    If each link instantly goes from 0 speed to v then you're not conserving energy. The only dilemma is that you are going out of your way to prevent energy from being conserved and then expressing bewilderment that it is not conserved! :)
  18. Oct 7, 2005 #17
    Hi, I'm having problem trying to solve this problem as well and would really appreciate it if someone could point me in the right direction.
    Tide mentioned that tHe GPE of the mass hanging of the table is given by [itex]-mgx^2/2[/itex]. I was wondering how to arrive on that expression; I've tried everything, but my definition of everything is pretty limited, hah. Need some help pls.
  19. Oct 7, 2005 #18


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    Set up your coordinates such that x increases going DOWNWARD and set the zero potential to be at the table top. Then the gravitational potential of each element [itex]dx'[/itex] of the hanging chain is [itex]-\lambda x' g dx'[/itex]. (In other words, the gravitational potential decreases as you move downward!) Now integrate this from 0 to x to find the gravitational potential of the entire hanging chain.

    (NOTE: I'm using [itex]\lambda[/itex] to represent the mass per unit length of the chain instead of m.)
  20. Oct 7, 2005 #19

    Thanks! :smile:
  21. Oct 7, 2005 #20


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    No is it not necessary that it happens "instantly", it is just not possible for the falling chain and the new link to both come to the same speed while conserving both momentum and KE. The new link needs to be projected forward at a speed just a little less than 2v (twice the velocity of the falling chain - see note) to conserve KE. I'm not saying that this couldn't happen, this energy wouldn't nesseccarily be transfered forward to the remainder of the falling chain though.

    Note. Before arguing this point any further please go and review the basic physics of a moving mass having a co-linear collision with a stationary mass. In order for both KE and momentum to be conserved the mass that was initially stationary must have a velocity after collision of [tex]2u/(1+k)[/tex], where [tex]u[/tex] is the inital velocity of the moving mass and [tex]k[/tex] is the ratio of the initially stationary mass to the initially moving mass.
    Last edited: Oct 7, 2005
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