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Physics torque and pulley

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data

    The two blocks are connected by a massless rope that passes over a pulley. The pulley is 17 cm in diameter and has a mass of 2.6 kg . As the pulley turns, friction at the axle exerts a torque of magnitude 0.51 Nm.


    2. Relevant equations

    http://answers.yahoo.com/question/index?qid=20081114102306AA0xfMk
    i followed the advice given here but it doesnt seem to work

    3. The attempt at a solution

    i know how to do the problem without torque but I dont know how to add the torque in, i've been doing it for long time and haven't been getting the right answer
    thanks in advance
     
  2. jcsd
  3. Nov 15, 2008 #2

    Hootenanny

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    Whilst I haven't read through the entire response at yahoo, I did spot one obvisous error in your calculations
    Are you sure that you should multiply by the acceleration?
     
  4. Nov 15, 2008 #3
    the response on yahoo used that forumla, i dont know what i should use, should it be. thanks for helping
     
  5. Nov 15, 2008 #4

    Hootenanny

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    Indeed, the response from Yahoo does appear to make use of that formula. However, the formula contains a typo. The correct formula can be derived from the following kinematic equation

    [tex]h = v_0t +\frac{1}{2}at^2[/tex]
     
  6. Nov 15, 2008 #5
    but isnt the Vo = 0 zero since they are dropped from rest so solving for t so that should give you t = sqrt[2h/a], is that right now?
     
  7. Nov 15, 2008 #6
    but then how come in the yahoo answer, the time was 1.7s?? using this forumla i only get that t = around0.7 s? can u tell me what i am doing wrong please thanks so much
     
  8. Nov 16, 2008 #7

    Hootenanny

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    It would be much easier to point out your mistake if you could post your working.
     
  9. Nov 16, 2008 #8
    i did in the yahoo link. i replied to the answer with my own workings
     
  10. Nov 16, 2008 #9
    First let's get the symbols right:
    m1 = larger mass, 4 kg
    m2 = smaller mass, 2 kg
    T1 = rope tension on the m1 side
    T2 = rope tension on the m2 side
    r = radius of the pulley, 0.085 m
    mp = mass of pulley, 2.6 kg
    M = moment or torque due to friction at pulley spindle, 0.51 Nm
    J = polar moment of inertia of pulley, 0.5 mp r^2 = ... kg-m^2
    a = constant acceleration of the system, till m1 hits ground

    Now then, let's start;

    at the m1 side, gravity is causing tension T1 after catering to acceleration a, so:
    T1 = m1 g - m1 a ...........................................

    at the pulley, moment of the tension difference at the center caters to both the frictional torque as well as the angular acceleration of the pulley, which is a/r, hence:
    (T1 - T2) r = J a/r + M, or
    T1 - T2 = J a/r^2 + M /r = 0.5 mp a + M /r ......... 2)

    at the m2 side, the torque T2 is simply pulling up mass m2 against gravity at acceleration a, hence:
    T2 = m2 g + m2 a ........................................... 3)

    Now, for solving, the simplest way is to eliminate the tensions and get a simple equation in a only, as we are not interested in the other unknowns. The best way is to subtract 3) from 1) and equate that to 2) to eliminate T1 and T2

    Once you find a, it's a piece of cake to find the time required to go down a height of 1 m, starting from rest with the equation:
    h = 0.5 a t^2, or t = sqrt (2 h/a)

    1.58 m/s, 1.78 s
     
  11. Nov 16, 2008 #10
    the above is what i tried to follow, and by following that this is what i arrived at
    T1 - T2 = .5mpa +M/r = (m1-m2)g - (m1-m2)a
    plugging the numbers in i get:
    .5(2.6)a + .51/.085 = (2)(9.8) - 2a
    solving for a : a= 4.12 m/s^2
    solving for t:
    t = sqrt[2ah] = sqrt[2(1) /4.12] = 0.696s
    is this right? becuase the answer given was 1.78s thank you
     
  12. Nov 17, 2008 #11

    Hootenanny

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    You may want to re-check the highlighted term.
     
  13. Nov 17, 2008 #12
    T1 = m1 g - m1 a
    T2 = m2 g + m2 a
    then T1 - T2 = (m1-m2)g - (m1-m2)a doesnt it? i dont see what is wrong, can you please tell me?
     
  14. Nov 17, 2008 #13

    Hootenanny

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    Expanding the brackets:
    T1-T2 = m1g - m2g - m1a + m2a
     
  15. Nov 17, 2008 #14
    wouldn't that give the same answer
    T1 - T2 = (m1-m2)g - (m1-m2)a = (4-2)(g) - (4- 2)a = 2g - 2a
    is the same thing as 4g - 2g - 4a +2a = 2g - 2a
     
  16. Nov 17, 2008 #15

    Hootenanny

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    No. In the post above I expanded your incorrect expression. The correct expression should be

    T1-T2 = m1g - m2g - m1a - m2a
    =(m1 - m2)g - (m1 + m2)a

    Do you follow?
     
  17. Nov 17, 2008 #16
    oh yes, thank you so much! so the answer should be T1 - T1 = 2g - 4a = .5mpa +M/r
    2g - 6a = .5(2.4)a + .51/.085
    2g - 6a = 1.2a + 6
    a = 1.89
    t = 1.03s? does this look right?
     
  18. Nov 17, 2008 #17

    Hootenanny

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    You're close, just be careful with the mass of the pulley.
     
  19. Nov 17, 2008 #18
    oh i see the mass of the pulley is 2.6 so it should be .5(2.6) this is confusing because in my problem the actualy mass of the pulley is 2.4kg, so other than that everything else is right? thank you sooo much!
     
  20. Nov 17, 2008 #19

    Hootenanny

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    I haven't actually run through the numerical calculations, but the method is certainly correct, as are the initial equations.
     
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