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Conservation of energy of falling chimney

  1. Aug 10, 2007 #1
    1. The problem statement, all variables and given/known data

    A tall, cylindrical chimney falls over when its base is ruptured. Assuming that the chimney remains intact before it hits the ground, using conservation of energy, calculate its angular speed [tex]\dot \theta{}[/tex] as a function of [tex]\theta[/tex] , the angle which the chimney makes with the vertical.

    2. Relevant equations

    Etotal = Ep+Ek

    I posted a question yesterday similar to this one. I really do not understand how to tackle this type of question. Any help would be appreciated.
    Last edited: Aug 10, 2007
  2. jcsd
  3. Aug 10, 2007 #2


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    Do they give any more details?

    The gravitational potential energy of the chimney is converted into rotational kinetic energy... have you learned about rotational kinetic energy and moments of inertia?
  4. Aug 10, 2007 #3
    I don't remember the formula for the rotational KE... but I gather the question is asking you to set the gravitational PE equal to rotational KE... then just isolate the angular speed term...
  5. Aug 10, 2007 #4


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    Yes, the difference in gravitational potential energy is the rotational kinetic energy.

    Calculate the total energy of the system at the beginning [tex]E_i[/tex] which is all gravitational potential energy initally... then at any angle you know by conservation of energy that

    [tex]E_i = E_r + E_g[/tex]. (the quantity on the right is rotational kinetic energy plus gravitational potential energy) You can use that to isolate the angular speed...
    Last edited: Aug 10, 2007
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