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Angular velocity of cylinder

  1. Dec 10, 2011 #1
    1. The problem statement, all variables and given/known data
    A uniform cylinder of radius 12 cm and mass 25 kg is mounted so as to rotate freely about a horizontal axis that is parallel to and 6.6 cm from the central longitudinal axis of the cylinder.
    If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

    2. Relevant equations



    3. The attempt at a solution
    I'm using conservation of energy..
    1/2 Iw^2= mgh
    1/4 (r^2)(w^2) = gh
    Then, solve for w
    And i got 13.40 rad/s
    But the answer is not correct.
    Can someone point out my mistake. Thank you.
     
  2. jcsd
  3. Dec 10, 2011 #2
    You must apply parallel-axis theorem to find the moment of inertia (since the axis doesn't pass through the center of mass). Take in account the displacement of the center of mass and the rotational and translational kinetic energies when applying conservation of energy. Remember this equation: vCM = R*angular speed (vCM = speed of the center of mass). :)
     
  4. Dec 10, 2011 #3
    I found moment of inertia using parallel-axis theorem.
    Then, I used the conservation of energy to solve for angular velocity.
    But I still got it wrong.

    I = Icom + MH^2
    solve for I.
    Then,
    1/2 mv^2 + 1/2 Iw^2 = mgh
    m(wr)^2 + Iw^2 = 2mgh
    w^2 = ( 2mgh) / ( I + mr^2)
    w = [( 2mgh) / ( I + mr^2)]^2

    Maybe I missed something.
     
  5. Dec 10, 2011 #4
    Try to ignore translation. It seems that there is just a rotation, since the axis doesn't move.
     
  6. Dec 10, 2011 #5
    ok . i ll try it.
     
  7. Dec 10, 2011 #6
    You are right. We need to exclude the translation.
     
  8. Dec 10, 2011 #7
    1/2mω^2=[itex]_{M}Δ[/itex]P
     
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