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Rotational momentum (previous problem)

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A thin rod of length 1.50m is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge. The mass of the rod may be ignored compared to the mass of an object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward.
    Find the angular speed of the rod just before it strikes the floor.


    2. Relevant equations

    KE=PE
    Iw*w/2=mgh where h=L/2=CG
    I=mr*r


    3. The attempt at a solution
    w=?
    r=1.5m=L
    g=6.50 rad/s*s
    h=0.75m (CG=L/2)
    s=2.36m
    angle=90degrees=pi/2 rad
    T=14.7


    mr*r*w*w/2=mgL/2
    Cancel out m
    r*r*w*w/2=gL/2
    now I have calcultated g = 6.50 rad/s*s
    so
    1.5*1.5*w*w/2=6.5*1.5/2
    1.125*w*w=4.875
    w*w=4.875/1.125
    w=2.00

    But the correct answer is 3.61 according to textbook.

    What am I doing wrong?
    Please help, I am so frustrated.
     
  2. jcsd
  3. Mar 18, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi anelmarx! Welcome to PF! :smile:
    Why are you using the centre of the rod (L/2)? The mass is all at the end of the rod.

    And how do you get g = 6.50 rad/s*s? g is length/time2 … radian is not a length.

    ok, that's tactics … now overall strategy …

    this is not a moment of inertia case, it's a perfectly simple ordinary conservation of energy case with a point mass …

    it's exactly the same as as a mass on a string (except upside down, of course) …

    just use mgh and 1/2 mv2

    the only relevance of the angular speed is that for some reason they've asked you to convert the ordinary speed into angular speed at the end :wink:
     
  4. Mar 18, 2009 #3
    Ok now I am really lost.

    I am studying online and do not have anyone to ask - lecturer not resonding:frown:.

    If I use v*v=v_o*v_o+2ax
    and a=-9.81m/s*s
    and x=1.5m
    then
    v*v=0+2*-9.81*1.5
    but then you cannot take the root of a negative?

    OMG I am just so frustrated
     
  5. Mar 18, 2009 #4

    tiny-tim

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    uhh? that's a constant acceleration equation … why are you using it? this isn't constant acceleration

    use mgh and 1/2 mv2 and conservation of energy
     
  6. Mar 18, 2009 #5
    Ok, think I might have figured it out:

    PE=KE
    mgh=mv*v/2
    m falls away so
    gh=v*v/2
    and g = 9.81
    so
    9.81*1.5*2=v*v
    v=5.43m/s
    then ang speed=v/r=5.43/1.5=3.61rad/s

    and to find ang acc:

    90degrees = pi/2 and s=r*pi/2=1.5*pi/2=2.36m
    so
    9.81m/s*((pi/2)rad/2.36m)=6.53rad/s
     
  7. Mar 18, 2009 #6

    tiny-tim

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    Fine, except better if you say PE lost = KE gained :wink:
    oh, you didn't say anything about angular acceleration in the original question …

    hmm … i'm not following your reasoning …

    where does π/2 come into it? :confused:

    either use general principles, or use torque (= moment of force) = moment of inertia times angular acceleration :wink:
     
  8. Mar 18, 2009 #7
    Wow! thank you so much for your help, I have been struggling with this for hours and hours.
     
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