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Acceleration of massless pulley and blocks

  1. Jan 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider a situation
    287dlrm.png

    I'm curious why is that the massless pulley's acceleration is just the avarage of acceleration of m1 and that of m2?
    ab=(a1+a2)/2

    2. Relevant equations
    Forces acting on pulley are Tc-2Ta=0 (because pulley is massless, force can't give it acceleration)
    Therefore tensions are Tc=2Ta
    But, Ta-m1g=m1a1
    and Ta-m2g=m2a2


    3. The attempt at a solution
    Intuitively it makes sense.But I can't find the equation that describes this relationship.
     
  2. jcsd
  3. Jan 25, 2015 #2

    mfb

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    Consider the position of B as function of the positions of the masses and the string lengths. There is no need to consider forces, this is a simple mechanical constraint.
     
  4. Jan 25, 2015 #3
    Does it have something to do with notion that string on the left side is twice the length that on the right?
     
  5. Jan 25, 2015 #4

    mfb

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    The absolute lengths of the strings is not relevant, they just have to stay constant.
     
  6. Jan 25, 2015 #5
    What I meant was that m1 is placed lower than m2.Anyway, I still don't get why ab=(a1+a2)/2 holds.Can you please write those equations?
     
  7. Jan 25, 2015 #6

    haruspex

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    Let the string lengths be L1, L2, L=L1+L2. Differentiate twice.
     
  8. Jan 25, 2015 #7
    @haruspex I don't know what you mean by that.What are L1 and L2 respectively? Why differentiate?
     
  9. Jan 25, 2015 #8

    haruspex

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    The lengths either side of the pulley, i.e. from pulley to m1 and pulley to m2.
    Because you are interested in accelerations.
     
  10. Jan 25, 2015 #9
    L''=L1''+L2'', this gives a=a1+a2 .I'm not sure how that helps. But based on initial picture, L1 =2L 2 ,but how this relates to acceleration?
    And why is L related to acceleration of a pulley?
     
  11. Jan 25, 2015 #10

    haruspex

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    L is the total length of the string, so what is L"?
    L1" and L2" represent relative accelerations of objects in the diagram. Can you figure out what they are?
     
  12. Jan 25, 2015 #11
    L1'' is acceleration of m1 and L2'' of m2.L'' is then aceleration of pulley,right?But how that gives desired relationship?
     
  13. Jan 25, 2015 #12

    haruspex

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    No. As I wrote, L1" and L2" are relative accelerations.
    Think of L1 as a relative height, i.e. the difference in height of two objects. Which two objects?
    As for L", does L vary?
     
  14. Jan 25, 2015 #13
    L is constant (string). So L'' is zero.
    But L1 and L2 do vary because of m1 is not equal m2.
    Then pulley goes distance L1 relative to m1, hence L1'' is acceleration of pulley relative to m1.Likewise, L2'' is acceleration of pulley relative to m2.Right?
     
  15. Jan 25, 2015 #14

    haruspex

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    Yes. Can you now get the result you are after?
     
  16. Jan 25, 2015 #15
    Unfortunately, I can't figure out how to put this to use.Considering I found out relations between acc. of pulley and acc. both m1 and m2 I should be able to relate those 3.
    So I just need equation that includes accelerations of those 3 objects. This however I know ab=(a1+a2)/2.
    But how do I get a1 and a2 from this stuff above?I understand conceptually but can't write it down.
     
  17. Jan 25, 2015 #16

    haruspex

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    I'm confused... I thought that was what you were trying to prove.
     
  18. Jan 25, 2015 #17
    Yes,indeed i was trying to prove that.But I don't know how a1 and a2 come up from what we discussed above.(or whole equation for that matter)
     
  19. Jan 25, 2015 #18

    haruspex

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    Taking all accelerations as positive upwards (say), write ab in terms of a1 and L1". Do the same for mass 2. Eliminate L1" and L2" using L''=L1''+L2".
     
  20. Jan 25, 2015 #19

    gneill

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    This might be a case where a couple of pictures can help to clarify the situation.

    Suppose we isolate the pulley B and the masses m1 and m2 as a system and have an observer ride along with it. The observer decides to count downward acceleration as positive and sees mass m1 accelerating downward with some acceleration +a1 with respect to the pulley, and mass m2 accelerating with acceleration -a1. The accelerations must have the same magnitude because the string has a constant length; however much or fast m1 moves, m2 must move by the same amount and at the same rate as m1 with respect to the pulley.

    Fig1.gif

    Now another observer in the fixed lab frame of reference looks at the situation (figure below) and notes that the pulley B system as a whole is accelerating downward with some acceleration a. He thus sees m1 accelerating downward with acceleration a + a1, and mass m2 upward with acceleration -a1 + a. What's the average of those two accelerations?

    Fig2.gif
     
  21. Jan 26, 2015 #20
    I have a trick through which you can easily find these types of constraint equations.
    For any closed system ##\sum { \vec { T } .\vec { a } }=0## holds. (T.a is the dot product of tension and acceleration )
    But there are some condition in which it does not holds.

    rrr4.png
    So
    ##-2Ta_{3}+Ta_{1}+Ta_{2}=0##

    So ##a_{3}=\frac{a_{1}+a_{2}}{2}##

    As acceleration of ##m_{3}## is ##a_{3}## so acceleration of pulley should also be ##a_{3}##.

    There are some limitations regarding the application of this concept. We can use this only when there is no angular motion between string and pulleys.
     
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