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Constrained motion problem

  1. Aug 1, 2013 #1
    1. The problem statement, all variables and given/known data
    If acceleration of A is 2m/s2 to the left and acceleration of B is 1m/s2 to left, then acceleration of C is (see attachment 1)
    A)1m/s2 upwards
    B)1m/s2 downwards
    C)2m/s2 downwards
    D)2m/s2 upwards

    Ans: A


    2. Relevant equations



    3. The attempt at a solution
    (see attachment 2, I hope the labels are clear)
    Let the radius of smaller pulleys be r and that of bigger pulleys be R. I have marked the lengths of different portions of string. L is the length of total string.
    [tex]L=x_1+2 \times \pi r+x_2+x_3+2\times \pi R/4+2x_4[/tex]
    Differentiating twice w.r.t time
    [tex]0=\ddot{x_1}+\ddot{x_2}+\ddot{x_3}+2\ddot{x_4}[/tex]
    Now I am confused as to what should be the value of ##\ddot{x_1}##. :confused:

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Aug 1, 2013 #2

    verty

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    I suggest expressing ##x_1## as a function of time, then differentiating twice.

    Without calculus, it is not as easy. I can't think of a useful hint to give in this case except to say, use your intuition.
     
    Last edited: Aug 1, 2013
  4. Aug 1, 2013 #3

    tiny-tim

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    Hi Pranav-Arora! :smile:

    Isn't x1 a function of x2 and x3 ? :wink:
     
  5. Aug 1, 2013 #4
    Yes but then how do I write ##x_1## in terms of ##x_2## and ##x_3##? Something like this: ##x_2+x_3+\text{some constant}##?
     
  6. Aug 1, 2013 #5

    tiny-tim

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    yes!!! :smile:

    (what's wrong with that?? :wink:)
     
  7. Aug 1, 2013 #6
    Okay so I get,
    [tex]0=2\ddot{x_2}+2\ddot{x_3}+2\ddot{x_4}[/tex]
    [tex]\ddot{x_4}=-(\ddot{x_2}+\ddot{x_3})[/tex]
    ##\because \ddot{x_2}=2 m/s^2## and ##x_2=\ddot{x_2}=1 m/s^2##
    [tex]\ddot{x_4}=-3 m/s^2[/tex]

    This is definitely wrong. :confused:
     
  8. Aug 1, 2013 #7

    tiny-tim

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    :wink:
     
  9. Aug 1, 2013 #8
    Can you please explain a bit more? I am still not able to understand what to do with that info. :(
     
  10. Aug 1, 2013 #9

    tiny-tim

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    so x2' = 2, and x3' = … ? :wink:
     
  11. Aug 1, 2013 #10

    haruspex

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    I assume you meant ##\ddot{x_3}=1 m/s^2##, but that's still wrong. Which way is B moving?
     
  12. Aug 1, 2013 #11

    PeterO

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    If A moves left while B is stationary, C would move up.

    If B moved left while A was stationary, C would move down.

    So the effects of leftward motion of A & B have opposite effects on C.
     
  13. Aug 2, 2013 #12
    Is that a typo or are you actually asking me to calculate the first derivative? :uhh:
     
  14. Aug 2, 2013 #13

    tiny-tim

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    sorry, i meant '' not '
     
  15. Aug 2, 2013 #14
    But that still doesn't help. Both the accelerations are in the same direction. Should I work on the signs of first derivatives?
     
  16. Aug 2, 2013 #15

    tiny-tim

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    But x2 is measured to the left, and x3 is measured to the right. :smile:
     
  17. Aug 2, 2013 #16
    So ##\ddot{x_2}=2 m/s^2## and ##\ddot{x_3}=-1 m/s^2##?

    But is it wrong to think about the signs of first derivative? I never specified any directions, like you said that x2 is measured to the right.
     
  18. Aug 2, 2013 #17

    tiny-tim

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    i don't see how it helps :confused:
    yes you did!

    it's clear from your diagram that x2 and x3 are always positive, and so must be measured from the fixed end, which is in the middle
     
  19. Aug 2, 2013 #18
    Okay, so in general, I have to specify the directions and measure distances from a fixed end. I guess I have to do some more practice on these problems. -.-'

    Anyways, I get ##\ddot{x_4}=-1 m/s^2##. Do I have to put my reasoning for the negative sign like this: Since ##x_4## is positive, the negative sign indicates that the block C moves upwards. Would that be enough?

    Thanks a lot tiny-tim! :smile:
     
  20. Aug 2, 2013 #19

    tiny-tim

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    It probably doesn't need any reasoning.

    If you want to provide a reason, just say that x4 is decreasing, and x4 is the depth below the main level :wink:
     
  21. Aug 2, 2013 #20
    Okay. Thanks again! :)
     
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