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Moment of Inertia Problem

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Two People are holding the ends of a uniform plank of lenth, l and mass, m. Show that if one person suddenly lets go, the load suppored by the other person suddenly drops from mg/2 to mg/4.

    2. Relevant equations

    Fnet=ma
    L=Iw'
    I=ml^2+I'


    3. The attempt at a solution

    I have done the first part of the problem. I am have trouble with the second part.

    Before Plank Falls:

    Fnet = ma = F1 + F2 -Fg = 0 (F1=F2, Fg=mg)

    2F1=mg
    F1=(mg)/2


    After Plank Falls:

    Fnet = ma = F1 - Fg = 0
    F1=mg
     
  2. jcsd
  3. Oct 27, 2008 #2

    tiny-tim

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    this plank is not constant

    Hi vsharma88! :smile:
    Nooo … the acceleration is not 0, is it? :wink:

    Hint: what is the angular acceleration? :smile:
     
  4. Oct 27, 2008 #3
    After the person has dropped the plank, the force becomes a torque.

    Fnet = N - mg = ma
    ma = (ml^2)/12 * w' - mg [(ml^2)/2 - for a thin uniform plank]

    I am now stuck on what to do with w'.
     
  5. Oct 28, 2008 #4

    tiny-tim

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    Hi vsharma88! :smile:

    (btw, you can't say "the force becomes a torque" … a force is a force, a torque is a torque … torque = force times distance :rolleyes:)

    Take moments about the end to find the angular acceleration.

    Then take moments about the centre to find the normal force (using that value of angular acceleration). :smile:
     
  6. Oct 29, 2008 #5
    hey tiny-tim i have taken the moment about the centre:

    [tex]\sum\tau[/tex]=0=ml^2/12*w'-mg*l/2
    w'=(gl/2)*(12/l^2)
    =(g6)/l

    I am now confused what to do next, you said to take moments about the centre. Is it just :

    [tex]\sum\tau[/tex]=0=ml^2/12*w' ?

    Thanks
     
  7. Oct 29, 2008 #6

    tiny-tim

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    Nooo … the weight has no moment about the centre … you should have the normal force, N, in there, instead. :wink:

    Then take moments about the end, and compare the two equations to get N/g. :smile:
     
  8. Oct 29, 2008 #7
    ok tiny-tim i think i got it, tell me if im right.

    Fnet = F1 - Fg

    Moment about End
    [tex]\sum[/tex]=N - (l/2)Fg=0
    0=Iw' - (l/2)(mg)
    0=(ml^2/12)w' - (l/2)(mg)

    w'= 6g/l

    Moment about center of mass
    [tex]\sum[/tex]= N*(l/2)
    =Iw' * (l/2)
    =(ml^2/12)(6g/l)*(l/2)
    =(1/4)mg
     
  9. Oct 29, 2008 #8

    tiny-tim

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    Hi vsharma88! :smile:

    Sorry, but this is wrong:
    and this is wrong:
    I'm going to bed now … :zzz:​
     
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