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Deriving work-kinetic energy theorem for rotational motion

  1. Aug 14, 2014 #1
    1. The problem statement, all variables and given/known data

    How to derive W=(.5*I)*((ωf^2)-(ωi^2)) from W=τ*(θ)

    2. Relevant equations

    The two equations seen above (big W is work; little w is rotational velocity, τ is torque, alpha=rotational acceleration).

    s=r*θ a=r*(alpha) (ωf^2)-(ωo^2)=2*(alpha)*(θ) F=r*τ F=ma

    3. The attempt at a solution

    W=T*(θ)=(F/r)*(s/r)=(r^-2)*F*s=(r^-2)*ma*s=(r^-2)*m*(r*alpha)((ωf^2)/(2*alpha))*r
    =0.5*m*ωf^2

    Obviously this isn't right. What went wrong?
     
  2. jcsd
  3. Aug 14, 2014 #2
    Are you allowed to use calculus?
     
  4. Aug 15, 2014 #3

    ehild

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    [/QUOTE]

    The formula in red is wrong. How is the torque defined?

    ehild
     
  5. Aug 15, 2014 #4
    response to ehlid

    How is it wrong? Force is torque times radius, is it not. Oh, I can't use scalar multiplication on this. Do I have to use the cross product: F=t x r?
     
  6. Aug 15, 2014 #5
    Or τ = I*α = I*a/r maybe...

    and you should think of your s as Δs
     
  7. Aug 15, 2014 #6

    ehild

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    It is not. Check your textbooks.

    ehild
     
  8. Aug 15, 2014 #7
    correction

    Shoot! I'm sorry!!! T=r*F. I got it backwards!!! If I plug in the right equation, the first equation will eventually turn into 0.5*m*(r^2)*(ωf^2)=I*(ωf^2).
     
    Last edited: Aug 15, 2014
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