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A balloon lifting a heavy stone

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data
    A party balloon(of mass 0.0025kg when empty) is filled with helium to a volume of 0.0045m3. it is tied to a small stone of mass 0.015kg by a light string of length 1.5m to prevent it from flying away. a child holds the balloon at ground level and then releases it.
    air=1.20kg/m3 ρHe=0.178kg/m3)

    (a)how long does it take for the balloon to rise 1.5m,that is,for the string to become taut?

    (b)What is the velocity of the stone when it is lifted off the ground? you may assume the time needed to bring the stone to this velocity is very short(impulse approximation) once the string is taut, and the string remains taut throughout

    (c)How long after lifting off will the stone touch the ground again?(neglect air resistance and treat the stone as a particle)

    2. The attempt at a solution
    part 1 is relatively easy,i have mtotal=0.00330kg,net upward force=0.0206N,a=6.24m/s2, t=0.693s which agrees with the given answer

    not really understand part 2 as the net upward force acting on balloon is only 0.0206N,how can it then lift a stone with a weight of 0.147N?

    for part 3,if the stone can be lifted,why would it touch the ground again?
     
  2. jcsd
  3. Aug 19, 2012 #2
    I think it is the conservation of momentum for the part2, where

    m(of balloon)v=m(of balloon and stone)v
    v of balloon=at=4.32432, mass of balloon is 0.00330kg (as you calc-ed), mass of balloon+stone=0.0183kg, then you can solve for v.

    as for part3, hmm...
     
  4. Aug 19, 2012 #3
    Did you read the question?

    "it is tied to a small stone of mass 0.015kg by a light string of length 1.5m to prevent it from flying away."

    If the upward force of the balloon could lift the stone indefinitely, then the stone isn't doing its job.

    "(b)What is the velocity of the stone when it is lifted off the ground? you may assume the time needed to bring the stone to this velocity is very short (impulse approximation) once the string is taut, and the string remains taut throughout"

    If the balloon is floating and accelerating upward, it is gaining momentum. If the balloon just flies up, makes the string taught, and nothing further moved, that would imply that the momentum disappears. The momentum obviously doesn't disappear, and the question tells you that. It tells you to use the "impulse approximation." Impulse is instantaneous momentum change.

    Have you ever tried this as a kid? It really does lift the stone.
     
  5. Aug 20, 2012 #4
    No,maybe the stones i used to tie my balloons to were all too heavy for this effect to be noticeable.

    anyway,for part 3,my solution is as follows
    maximum momentum=mballoon x at=0.0143 kgm/s
    net downward force=0.1264N
    time needed to "cancel" the upward momentum=0.0143/0.1264=0.113s
    time needed to touch the ground again(achive the same downward momentum)=0.113x2=0.226s

    is my method correct?
     
  6. Aug 20, 2012 #5
    Your method for part 3 is correct. However, your method for part 1 is incorrect, which led to an incorrect answer. Next time, try to do the problem with variables as opposed to numbers. It looks cleaner, and it's easier to identify where you went wrong.

    In the setup of the problem, the balloon is tied to a string, which is tied to a stone. If the balloon is at ground level (the string is not taut), then the stone is on the ground, and there is no way for balloon to lift the stone. We assume that the string has no mass, so the only thing accelerating up is the balloon.

    The only moving piece is the balloon, so the rock does not contribute to the total momentum of the apparatus. Therefore, the acceleration and time you calculated for part 1 is wrong, because it uses the total mass, not the mass of the balloon.
     
    Last edited: Aug 20, 2012
  7. Aug 21, 2012 #6
    Actually in my solution for part 1,"total mass" means the combined mass of balloon and helium, i did exclude mass of the stone, that's why my answer agrees with the given answer.
    but thank you for your advice though
     
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