1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Launching a balloon

  1. Jul 7, 2009 #1
    The following came from a discussion about launching a balloon.

    Suppose you have a heavy coiled rope of total length, L, and constant linear density, MU. You take hold of one end of the rope and pull it vertically up with a force, F(t), so that the tip of the rope moves at a constant velocity, v.

    What is F(t), assuming the rope is so long that some of it remains coiled on the ground?

    In drawing our free-body diagram, two questions arise:
    1) Can you define our system to be just the vertical length of rope (of length, y(t)) --- see figures 1 & 2?
    2) If so, how do we calculate the force that the coiled portion exerts on the vertical section, F(coil on rope)? There must be a force, otherwise the coiled portion wouldn’t unwind.

    Thanks in advance.

    Attached Files:

  2. jcsd
  3. Jul 7, 2009 #2


    User Avatar

    Staff: Mentor

    Is this a homework question? Do you have any work to show us?

    As a hint, though, it may be useful to delete the velocity from the free body diagrams and consider the difference between the two static scenarios...
  4. Jul 7, 2009 #3
    No. It is not a homework question!
  5. Jul 7, 2009 #4


    User Avatar
    Homework Helper

    You'd know the rate of mass flow transitioning from not moving to moving upwards at the fixed velocity. There's a period of time where the speed of the unwinding rope is faster than the upward velocity, because it also has a horizontal component.

    The unknown is the amount of time it takes for each section of the coiled rope to transition into upwards movement, specifically, the relationship between acceleration and time of the rope as it unwinds.

    As time goes on, the mass of the rope moving vertically increases, while the mass of the rope sections in transition remains near constant, so the limit of this is simply the weight of the suspended rope as it moves at constant speed.

    You could simplify this by having the rope on a drum unwinding at constant speed, eliminating the transitional acceleration of the rope.
  6. Jul 8, 2009 #5
    What I was wondering about was that the system is not fixed. IE Can we define our system to be the vertical length of rope even though the length of rope increases?
  7. Jul 8, 2009 #6


    User Avatar
    Homework Helper

    Using a drum that the rope unwinds from approximates this, as the rope no longer experiences any linear acceleration as it spools off the drum, and if the drum is friction free, it's not adding any load to the system. The effective diamter decreases as layers of rope are peeled off, but this can be ignored to simplify the problem.
  8. Jul 11, 2009 #7
    Does this mean we could apply Newton's second law to the vertical section of the rope ("our system") as:

    F(t) - m(t)g - Force(coil on rope) = ma

    where we take Force(coil on rope) = 0

    and a = 0 (as rope rises at constant vertical velocity)
  9. Jul 11, 2009 #8


    User Avatar
    Homework Helper

    Yes, in which case, F(t) = m(t)g.
  10. Jul 11, 2009 #9
    OK. What if we took the WHOLE rope as our system. Afterall, we are free to define our system as we wish.

    For the sake of simplicity, go back to the original posting where the rope was coiled on the table.

    Then the rope is subject to a Normal force of
    where MU = the length density of the heavy rope and
    L-y(t) is the length of the coiled segment.

    Surely Newton’s Second Law must apply to our system.
    m*a = F(t) - y(t)*MU*g + N(t)

    Can we do this?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook