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Relative velocity

  1. Aug 15, 2009 #1
    1. The problem statement, all variables and given/known data

    In the figure: A balloon is attached to a ladder.
    The ladder stands vertical on the ground.
    The height of the ladder is 6m and there's a monkey at the base of the ladder

    Q) The mass of the balloon is 50kg and mass of the monkey is 25kg.
    The monkey starts to climb upwards with a constant speed of 0.5m/s

    1)What is the velocity of the balloon relative to earth ?
    2)What is the velocity of the monkey relative to earth ?
    3)What is the velocity of the monkey relative to the balloon?
    4)Find the timetaken to reach the top of the ladder
    5)What's the displacement of the monkey relative to the earth?
    6)What's the displacement of the balloon relative to the earth?
    7)What's the displacement of the balloon relative to the balloon?
    8)what's the increase in PE?
    9)What's the total work done by the monkey?

    2. Relevant equations

    3. The attempt at a solution

    1) [tex]\uparrow[/tex]V (m,b) = velocity of monkey relative to balloon = 0.5
    V(b,e) = v1 down
    V(m,e) = v

    V(m,e) = V(m,b) + V(b,e)
    V(m,e) = 0.5 - v1 = vel' of monkey relative to earth.

    But how do I find the magnitude of V(m,e) ?
    I think I can manage to do the other parts of the question,if only I knew how to solve for V(m,e) . :(

    Thanks in advance .
  2. jcsd
  3. Aug 15, 2009 #2


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    It's a bit more complicated than what you have written down. Suppose the monkey and balloon system are initially at rest. Then the monkey starts climbing up. Will the balloon move? How can you find its speed? (Think center of mass). The same considerations apply when the balloon is moving with velocity v1.
  4. Aug 15, 2009 #3
    Yes,when the monkey starts climbing,it exerts a downward force on the ladder causing the ladder to move down .

    I tried using centre of mass,but I end up with 2 unknowns.

    Here's what I did,
    Taking y as the reference distance from the top of the system,
    Initially,when the monkey's at the bottom of the ladder
    [25 * (6+y) + 50 * y/2 ]/75 = X --------(1) where X = COM

    When monkey starts climbing up,
    in 1 second,
    the balloon moves down by a distance v1,
    the monkey moves up by a distance 0.5-v1,sooo....

    [ 25 * (6-(0.5-v1)) + y) + 50 (y+v1) ] / 75 = X -----------(2)

    137.5 + 75v1 + 25y = 150

    Not sure how to proceed from here
  5. Aug 15, 2009 #4


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    Before we proceed any further, what does the picture show about the motion of the balloon? Is there a direction to the balloon's velocity v1? In other words is the balloon already moving when the monkey starts climbing?
  6. Aug 15, 2009 #5
    The balloon doesn't seem to be moving in the picture,but its just a rough sketch that our teacher put up on the board for us to copy.
    I could draw the pic, if needed.may take a few minutes
  7. Aug 15, 2009 #6


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    No need for a picture. If the balloon is initially at rest, when the monkey starts climbing, the centre of mass should remain at rest. Can you write an expression for VCM? (Note that the velocities of monkey and balloon must be relative to the Earth.) Then VCM = 0.
  8. Aug 15, 2009 #7
    Is VCM the velocity of center of mass?

    Here's the pic. anyway
    http://img441.imageshack.us/img441/3106/balloon.png" [Broken]
    Last edited by a moderator: May 4, 2017
  9. Aug 15, 2009 #8


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  10. Aug 15, 2009 #9
    Well,if that's the case,I'm sorry I don't know to write an expression for VCM.

    One more hint,please...?
  11. Aug 15, 2009 #10
    OK,I'm kinda desperate here.I have a physics exam the day after tomorrow and i have alot more to revise,but I can't seem to concentrate in my studies without solving this problem,so anybody know how i could find expressions for VCM?

  12. Aug 15, 2009 #11


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    [tex]V_{CM} = \frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}[/tex]
  13. Aug 15, 2009 #12
    Oh that's the conservation of momentum equation.i feel so stupid i completely missed it.

    Anyway,this is what i get
    [tex]\downarrow[/tex] m1v1 - m2v2 = 0
    50 * v1 = 25*v2
    v2 = 2v1

    2v1 = 0.5 - v1
    v1= 1/6

    Is this correct now?

  14. Aug 15, 2009 #13


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    The number is correct, but the sign is wrong. Velocity v1 is an algebraic quantity which might turn out positive or negative when you put in the numbers. You should start from

    m1v1 + m2v2=0,

    do the math and put in the numbers in the end. Then v1 will come out negative indicating that the balloon is moving down while the monkey is climbing up. Note that


    I think you can finish the rest now.
  15. Aug 15, 2009 #14
    Yeah.The rest are easy.

    Thanks for the help,sir.
    You're awesome!
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