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Translational vs. Rotational Kinetic Energy

  1. Dec 17, 2008 #1
    Suppose I am trying to find the work done in bringing a resting cylinder to an angular speed of 8 rad/s.
    Why is it INCORRECT to find the corresponding tangential velocity at a point on the outer surface of the cylinder (using angular speed * radius = tangential speed) and use the translational (0.5mv^2) work-kinetic energy theorem?
    Why MUST we use the rotational version with I and angular speed?
    Thank you.
  2. jcsd
  3. Dec 17, 2008 #2

    Vanadium 50

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    Because there is rotational kinetic energy as well.
  4. Dec 17, 2008 #3

    Doc Al

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    Realize that the tangential velocity depends on the distance from the axis--the cylinder does not have a uniform tangential velocity. But if you're willing to add up the translational KE of each piece (dm) of the cylinder, that's just fine. (You'll get the same answer.)

    KE = Σ½dm v² = Σ½dm r²ω² = ½(Σdm r²)ω² = ½Iω²
    It's just much easier. :wink:
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