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Energy Methods + Pulley + Friction, finding final velocity

  1. Mar 30, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    The pulley in the figure has radius R=0.160m and a moment of Inertia I_p = 0.560 kg*m^2. The rope does not slip on the pulley rim. As m_4 (4kg block) drops due to gravity, m_2 (2kg block) is dragged up a ramp inclined at theta = 65 degrees. The coefficient of kinetic friction for the block/ramp is 0.100. Use energy methods to calculate the speed of m_2 after m_4 has dropped 5.00 meters. (Hint the normal force on m_2 is m_2gcos(theta) and change in height is not the same for m_2 and m_4 although they are related to each other).

    2. Relevant equations


    3. The attempt at a solution
    So first, decrease in PE is equal to increase in KE.
    4*5*9.8 = 196 J.

    The frictional work done is 1.96*cos65*5 = 9.8cos(65)
    The force parrallel is 19.6sin(65)*5 = 98sin(65)
    Final KE of the system = 196-9.8cos(65)-98sin(65)

    Final KE's of the syem
    M_4 = 1/2*4*v^2
    M_2 = 1/2*2*v^2
    Pulley = 1/2*I*w^2 ==> using w=v/R ==> 10.938v^2
    Total KE = 13.938v^2

    So:
    13.938v^2 = 196-9.8cos(65)-98sin(65)
    Which results in a 2.72 m/s

    Should I have used the coefficient of friction instead of doing what I did? :O
    Is there anything wrong with this reasoning?
     
  2. jcsd
  3. Mar 30, 2015 #2

    Doc Al

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    Where's the figure?
     
  4. Mar 30, 2015 #3

    RJLiberator

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    Here it is, sorry.
     

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  5. Mar 30, 2015 #4

    Doc Al

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    To calculate the work done by friction, you need the friction force. And that requires the coefficient of friction.
     
  6. Mar 30, 2015 #5

    RJLiberator

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    I calculated it here The frictional work done is 1.96*cos65*5 = 9.8cos(65) which is from mass=2 gravity = 9.2 and u_k = 0.1 = 1.96.

    I suppose I had already used it.

    With that in play, does the reasoning/logic of this problem hold up?
    Namely: 13.938v^2 = 196-9.8cos(65)-98sin(65)
     
  7. Mar 30, 2015 #6

    Doc Al

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    You'll have an easier time keeping things straight if you viewed it as ##\Delta E = \Delta KE + \Delta PE = Work_{friction}##. (Realizing that the work done by friction is negative, of course.)
     
  8. Mar 30, 2015 #7

    RJLiberator

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    Also, when they state that the change in y is not the same for both of them, I assume this means that M_4 = 5 while m_2 you must use trigonometry as I did with 5cos(65) and 5sin(65). Correct?
     
  9. Mar 30, 2015 #8

    Doc Al

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    Yes. The change in height is relevant when finding the change in PE. The 5sin(65) is the change in height of m_2.
     
  10. Mar 30, 2015 #9

    RJLiberator

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    Hm. If I do that I insert the following:

    Change in KE = final - initial
    Final I found to be 13.938v_f^2
    Initially the system is at rest so KEi = 0.

    Change in PE
    Final = 4*-5*9.8 =-196J

    Force due to friction is 1.96*cos(65)*5 = 4.1416

    This nets me the answer of 3.78 = v_f :O

    I feel like I am doing something wrong with the change in PE ?
     
  11. Mar 30, 2015 #10

    Doc Al

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    That's the change in PE of the m_4 mass. What about the other mass?
     
  12. Mar 30, 2015 #11

    RJLiberator

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    It would then be 2*9.81*5sin65 = 17.76
    So -196+17.76 = -178.24

    Plugging this back into the equation: 14*x^2-178.24=4.1416

    I get x = 3.61
     
  13. Mar 30, 2015 #12

    Doc Al

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    Redo that bit of arithmetic.
     
  14. Mar 30, 2015 #13

    RJLiberator

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    Oi. You are absolutely right, I had foolishly forgotten about the 5 in 5sin(65).

    So 2*9.8*5sin(65) = 88.82.

    So -196+88.82 = -107.18

    13.938*x^2-107.18=4.1416

    = 2.82 m/s which is similar to my original answer of 2.72 :O

    Hm, I'm looking at both methods here and wondering why I am getting the difference of .1.
     
  15. Mar 30, 2015 #14

    RJLiberator

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    Admittedly, the way you suggested to do it was much more organized and to the point. It just made sense.
    I would assume that this method would produce the more exact result, being 2.82.
     
  16. Mar 30, 2015 #15

    Doc Al

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    Good. It also makes it easier to spot and correct errors.

    I haven't checked the arithmetic, but often such differences are due to round off errors.
     
  17. Mar 30, 2015 #16

    RJLiberator

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    Excellent. Thank you for your help this morning. You clarified this question well for me.
     
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