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Thin rod standing upright tips over

  1. Aug 27, 2012 #1
    1. The problem statement, all variables and given/known data

    A uniform thin rod of mass m and length l is standing upright on a table, then given a tiny kick so it tips over. The lower end is fixed. Find angular velocity when it hits the table.


    2. Relevant equations



    3. The attempt at a solution

    So I applied conservation of energy.
    Ei=Ef (1)
    Ei=mgl/2
    Ef=1/2mvcm2 +1/2Iω2
    where vcm is the linear velocity of the CM and ω is the angular velocity of the rod when it hits the floor. vcm=ωl/2
    For a thin rod that rotates about one end
    I=ml2/3
    So substituting in equation (1) i get
    mgl/2=1/2(ωl/2)2 + 1/6m(lω)2
    which leads to
    ω2=12/7g/l

    But i 've checked the answer and it turns out is ω2=3g/l

    where am i wrong?
    thanks in advance for your help
     
    Last edited: Aug 27, 2012
  2. jcsd
  3. Aug 27, 2012 #2
    It is not a single point of mass at the centre.
    It is made up of infinite "dm" at different height from the table.
     
  4. Aug 27, 2012 #3

    ehild

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    The kinetic energy of a rigid body rotating around a fixed axis is simply 0.5 Iω2, where I is the moment of inertia with respect to the axis.

    You can also calculate the KE of a rigid body moving in plane as the KE of the CM + energy of rotation around the CM. In this case, you have to use the moment of inertia with respect to the CM.

    You mixed the two methods. As the bottom end of the rod is fixed you can consider it a fixed axis.

    ehild
     
  5. Aug 27, 2012 #4
    Ok, but basically what i do is saying that the energy at the beginning Ei is potential energy due to gravity. Then when the rod hits the table potential energy equals zero, so i only have kinetic energy which is a sum of translational and rotational energy. which is the translation of the center of mass plus the rotational energy of the body.
     
  6. Aug 27, 2012 #5
    Ok, now i get it. Thank you very much!
     
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