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Launch of a mass attached to rope with mass

  1. Jun 6, 2010 #1
    Vertical Launch of an Attached Mass

    1. The problem statement, all variables and given/known data
    A mass M attached to an end of a very long chain of mass per unit length [tex]\lambda[/tex]
    , is thrown vertically up with velocity [tex]v_{0}[/tex].
    Show that the maximum height that M can reach is:

    [tex]h=\frac{M}{\lambda}\cdot \left [ \sqrt[3]{1+\frac{3\cdot \lambda\cdot v_{o}^{2}}{2\cdot M\cdot g}}-1 \right ][/tex]

    and that the velocity of M when it returns to the ground is [tex]v=\sqrt{2\cdot g\cdot h}[/tex]

    2. Relevant equations

    [tex]F=\frac{dp}{dt}=\frac{dp}{dx}\cdot v[/tex]

    3. The attempt at a solution

    I start by setting up that the total mass at a position y is:
    [tex]M_{total}=M+\lambda\cdot y[/tex] and thus the momentum at any position is given by:

    [tex]p=(M+\lambda\cdot y)\cdot v[/tex] but I can't figure out an expression for v and using

    [tex]F=\frac{dp}{dt}[/tex] I get an differential equation I can't solve.

    Any help would be appreciated.
    Last edited: Jun 6, 2010
  2. jcsd
  3. Jun 6, 2010 #2


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    In this problem it is best to consider just the mass that is projected up. There are two forces acting on it downwards, the constant force Mg and the variable force λyg. Write Newton's Second Law, then use the standard transformation

  4. Jun 7, 2010 #3
    Hmm, this is kind of interesting. How do you get a cube root in your answer?
    Your integrals would be like ydy which gives you a square (at most). That would produce a square root in the final answer.

    Secondly, you should be able to solve this using an "energy" approach (or "work" if you like). I get the same answer doing it by using the transformation. Here is what I got:

    [tex]h=\frac{M}{\lambda}\cdot \left ( \sqrt{1+\frac{\lambda v_{o}^{2}}{Mg}}-1\right )[/tex]
  5. Jun 7, 2010 #4


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    I agree that the cube root is perplexing. Using the transformation is equivalent to using a "work-energy" approach. That's because an element of net work is


    At this point use the transformation to replace the acceleration

    dWNet=mv(dv/dy)dy=mv(dv)=d[(1/2)mv2] = dK

    In other words, the transformation has been used to derive the work-energy theorem.
  6. Jun 7, 2010 #5


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    I have got that cubic root, by assuming variable mass, using dp/dt=F, and changing to the variable m instead of t. And I assumed F=mg where m =M+yλ. But there must be some force between the segment of the rope already in air and the remaining part, so that F=mg does not seem correct.
    I would apply energy conservation instead, but it results in the formula cartonn30gel obtained. Any idea to resolve this controversy?

  7. Jun 7, 2010 #6
    Hi, thanks for the reply but in this case energy isn't conserved because there are inelastic collisions within the rope which dissipate some energy (my teacher told me). ehild can you please show me how you got to the cubic root (I think F=m*g seems correct)? I'm clueless about this one.
  8. Jun 7, 2010 #7


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    I do not see those inelastic collisions in the rope, but it can be true.

    Your start was good, show the differential equation you got. It is a bit tricky to solve it, but I guide you if I see what you have.

  9. Jun 7, 2010 #8
    Hi ehild, did you get a nonlinear equation?
  10. Jun 7, 2010 #9
    I got the following:

    [tex]\frac{dp}{dt}=m\cdot \frac{dv}{dt}+\frac{dm}{dt}\cdot v[/tex]

    [tex]\frac{dp}{dt}= M\cdot \frac{dv}{dt}+\lambda\cdot y\cdot \frac{dv}{dt}+\lambda\cdot v^2[/tex]

    [tex]\frac{dp}{dt}=F=-m\cdot g \ \ \ \mbox{Where m is equal to M +\lambda \cdot y}[/tex]

    From that I get to:

    [tex]M\cdot \frac{dv}{dt}+\lambda\cdot y\cdot \frac{dv}{dt}+\lambda\cdot v^2 = -\left(M\cdot g + \lambda\cdot y\cdot g \right)[/tex]

    Or using the dot notation:

    [tex]M\cdot \ddot{y}+\lambda\cdot y\cdot \ddot{y}+\lambda\cdot \dot{y}^2+M\cdot g+\lambda\cdot y\cdot g =0[/tex]

    How can I solve this equation? Thanks for the help
  11. Jun 7, 2010 #10


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    Yes, but I could transform it to a linear one.

  12. Jun 7, 2010 #11
    Is my equation right? How would you transform it to a linear one?
  13. Jun 7, 2010 #12


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    I see what you did. Eliminate time, use the transformation

    dp/dt = dp/dy dy/d = dp/dy v.

    Or it can be even simpler using the linear relationship m=M+λy, and choosing m as independent variable:

    dp/dt = dp/dm dm/dy dy/dt = dp/dm λ v.

    dp/dm = d(mv)/dm = v+ m dv/dm

    So you get a first order differential equation for v as function of y, the length of the rope in air:

    (v+m dv/dm) λ v = -mg

    I will denote by ' the derivation with respect to m :

    v^2+ mvv'=-mg/λ

    Introduce the new function z = 0.5v^2. Rewrite the differential equation for z: It is a linear first-order one. Solve with some of the standard methods. Rewrite it for v^2, find the integration constant from the condition that initially m=M and v=v0. You get the v(m) function. Find m when v is 0, that is, M is at its maximum height. Find h.

  14. Jun 7, 2010 #13


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    This is OK, but it is a second order DE, which does not contain the time: so it can be transformed into a first order DE by using y as independent variable and v as the dependent one.

  15. Jun 7, 2010 #14
    I got it thanks
  16. Jun 8, 2010 #15


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    The energy conservation question has been gnawing at me, so this is how I looked at it



    Put in Newton's Second Law

    [tex]pdp=-m^2gdy=-g(M+\lambda y)^2dy[/tex]

    Integrate over the appropriate limits and divide through by M to get

    [tex]\frac{p_{0}^{2}}{2M}=Mgh + \lambda gh^2+\frac{1}{3} \lambda^2gh^3[/tex]

    This form shows where the initial kinetic energy goes when the mass reaches maximum height. Only the first term is easily recognizable. The second term is twice what one would expect for the potential energy of the rope's CM. I have no intuition for the third term. This goes to show that energy conservation is not recommended for this type of problem. I am not convinced, however, that mechanical energy is not conserved.

    By the way, if the third degree equation for h is solved, it yields the correct answer. I didn't solve it, but I verified the solution by substitution.
  17. Jun 9, 2010 #16
    Since one could use this approach to prove that energy is conserved (as kuruman pointed out in post #4), it's a good idea not to use conservation laws to begin with.
    I also think this is the easiest way to solve this problem. Choosing momentum(p) and position(y) as variables to work with produces a separable differential equation, which is much easier to solve compared to other types of differential equations. However, it's not obvious what variables would produce the simplest differential equation when you first begin the problem.

    I'm wondering if we can use something similar in the "chain rolling off a table" problem. I'll try that and see what happens.
  18. Jun 9, 2010 #17


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    Kuruman, yours is a very elegant solution!

    The integral can be written as


    from where the maxium height is easy to find.

    As for energy, I think the problem is that we should not ignore elastic energy. In the real word, that mass can not get an initial velocity without building up some tension in the rope. This tension will accelerate the next piece to the common velocity, and we do not know anything how this happens. So the initial energy is not known. There is varying tension along the elevated chain, which also contributes to the mechanical energy.
    When falling back the tension is released as it is a free fall of the whole chain.

    Last edited: Jun 9, 2010
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