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Rotation with friction

  1. Apr 1, 2004 #1
    A 4kg block resting on a horizontal ledge is attached to his right to a string that passes over a pulley and is attached to a hanging 2kg block. The coefficient of friction between the ledge and the 4kg block is 0.25. The pulley is a uniform disk of radius 8cm and mass 0.6kg. a) Find the speed of the 2kg block after it falls from the rest a distance of 2.5m. b) What is the angular velocity of the pulley at this time?


    Delta K + Delta U = 0
    Because Ki=Uf = 0
    ½(m+M)v^2+1/2(I,pulley)w^2-mgh=0
    ½(m+M)v^2+1/2(1/2MR^2)(v^2/R^2)-mgh=0

    so v = sqroot(2mgh/(M+m+1/2M,p))
    v = sqroot(2(2kg)(9.81)(2.5)/(4kg+2kg+1/2(0.6kg))) =3.95m/s


    b) w= v/r = (3.95m/s)/0.08m) = 49.5rad/s

    I did solve it without the 0.25 coefficient of fiction though may you help me please?
     
    Last edited: Apr 2, 2004
  2. jcsd
  3. Apr 1, 2004 #2

    NateTG

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    How much work does the friction do?
     
  4. Apr 1, 2004 #3
    49.5rad/s * 0.25?
     
  5. Apr 1, 2004 #4

    NateTG

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    No.

    Work = Force X Distance

    It will be an extra term in the energy equation.
     
  6. Apr 1, 2004 #5

    NateTG

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    The work done by friction is going to be force times distance.

    This would be easier to illustrate with a drawing, but it's going to be:
    [tex]F_f=mg\mu_k[/tex]
    so
    [tex]F_f=4 (9.81) 0.25=9.81[/tex]
    mow
    [tex]W_f=9.81*2.5[/tex]
    you can plug that into your energy equation:
    [tex]\frac{1}{2}(m_1+m_2)v^2+\frac{1}{2}I\omega^2+W_f-mgh=0[/tex]

    I think you've got the math under control from there.
     
  7. Apr 1, 2004 #6
    thank you very much, I got it.
     
  8. Apr 2, 2004 #7

    NateTG

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    In response to your post:

    [tex]E_{final}=E_{inital}[/tex]

    The work done by friction typically ends up as a temeperature change on the [tex]E_{final}[/tex] side. It's just one more thing that you need to account for in the energy equation
     
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