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Conservation of Energy spool

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

    Use conserbvation of Energy

    The spool has a mass of 50 kg and a raius of gratino k = .280 m. If the 20 kg block A is released from rest, determine the distance the block must fall inorder for the spool to have an angulr velocity w = 5 rad/s. Also what is the tension in the cord while the block is in motion.

    The spool has a radius of .3m, but where the cord is wrapped around it is a radius of .2m

    2. Relevant equations



    3. The attempt at a solution

    T1 + v1 = T2 + V2

    T & V 1 are both 0 because it starts from rest soooooo...

    I = mk2 = 50*.282 = 3.92

    0 = 1/2*20*v2 + 1/2*3.92*52 + Wy + 1/2*.25*s2

    0 = 59 + Wy + .14s2

    s = [tex]\sqrt{(-Wy - 59)/.14}[/tex]

    y = (.14s + 59)/W How do i find W
     
  2. jcsd
  3. May 15, 2009 #2
    ah W = mass*gravity

    W = 50(9.81) = 490.5

    so how do i get the y and the s
     
  4. May 17, 2009 #3
    can anyone give me a hint
     
  5. May 17, 2009 #4

    Redbelly98

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    I'm not following everything you did, so here are some questions for you:

    Which object is W the weight of?

    What does "Wy" represent in the energy equation?

    What does "1/2*.25*s2" represent in the energy equation?
     
  6. May 17, 2009 #5
    0 = .5mv2 + .5Iw2 + Wy + .5ks2

    I = mk2 = 50(.282) = 3.92

    0 = .5(20)v2 + .5(3.92)w2 + 20(9.81)y + .5(.28)s2

    oops, my W was wrong, should be 20(9.81)

    Wy represents the Potential Energy of Gravity, i would have to solve for y to find the distance the block falls

    .5*.25*s2 represents the elastic potential energy, i would have to solve for s to find the stretch of the rope

    0 = 10v2 + 1.96w2 + 196.2y + .14s2

    is this wrong, how do i finish from here
     
  7. May 17, 2009 #6

    Redbelly98

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    Where did you get the 0.25 from in that equation? It wasn't given in the problem statement.
     
  8. May 18, 2009 #7

    Redbelly98

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    Hint: forget about rope stretch in this problem. Assume the rope does not stretch.

    However, you will have to figure out how v and ω are related.
     
  9. May 18, 2009 #8
    the .25 should be the .280 = k, sorry i shoulda been more carfull
     
  10. May 18, 2009 #9
    I = 3.92, v = wr = 1, w = 5

    0 = 1/2(20)(12) + 1/2(3.92)(52) + 20(9.81)y

    y = 59/196.2 = .3007 m,

    Now i need to get the stretch of the rope, hint please
     
  11. May 18, 2009 #10

    Redbelly98

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    The problem asks for the tension in the cord, not the amount of stretch.
     
  12. May 18, 2009 #11

    Redbelly98

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    Units!

    k=0.28 m. It is the radius of gyration of the spool, not the spring constant of the cord.

    Spring constants have units of N/m, so this k=0.28m cannot be a spring constant.

    Looks good!
     
  13. May 18, 2009 #12
    On Wikipedia it says Tension (T) is .... T + mg = 0 where velocity is constant, and T + mg [tex]\neq[/tex] 0 where velocity is increasig

    T + 20(9.81) = 0 when T = -196.2 is that correct
     
  14. May 18, 2009 #13

    Redbelly98

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    Since the velocity is increasing, T + mg 0.

    Again, think in terms of energy: the work done on the brick, and the change in energy of the brick.
     
  15. May 18, 2009 #14
    Work done on brick

    Wy= 20(9.81)(.300713) = 59

    The energ goes from potentil to kinetic,

    is there an equation for this, i cant find one
     
  16. May 19, 2009 #15

    Redbelly98

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    (Work done on brick) = (Change in total energy of brick)
    = (Change in kinetic energy of brick) + (Change in potential energy of brick)​
     
  17. May 20, 2009 #16

    Redbelly98

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    FYI, you can treat "Work done on brick" as entirely due to the tension in the cord. So if you know the change in the brick's kinetic and potential energy, as well as the distance the it moved, you can solve for the cord tension.
     
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