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Work done by torque: wheel turning about a curb

  1. Nov 18, 2016 #1
    1. The problem statement, all variables and given/known data
    1.jpg

    2. Relevant equations
    work-kinetic energy theorem

    3. The attempt at a solution
    3.JPG
     
  2. jcsd
  3. Nov 18, 2016 #2

    TSny

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    You've assumed that it is possible for Wnet to equal zero when F is constant. Try to see why this can't be true.

    Can you figure out how much force is required to get the wheel to start to rise?
     
  4. Nov 18, 2016 #3

    haruspex

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    Further to TSny's hints...
    You write that you have to solve it by the Work-KE theorem. I suspect you have misunderstood the requirement. As TSny writes, you have no guarantee that ##\Delta KE=0##.
    You can solve statics problems using virtual work. Maybe that is what you are supposed to be using? But that method considers infinitesimal changes in position, not integrating over a substantial change.
     
  5. Nov 19, 2016 #4
    Ah, yes. Thank you. It would take a variable force to result in a final kinetic energy of 0. I know torque equilibrium yields the constant force required to offset the torque produced by gravity. Thank you again.
     
  6. Nov 19, 2016 #5
    Right. Thank you. I need to find the rotational kinetic energy of the wheel once it rises to the top of the curb.
     
  7. Nov 19, 2016 #6
    You assumed that ##ds=Rd\theta## but ##ds>0## and ##d\theta <0##. So it is wrong
    If you assumed that ##ds=-Rd\theta## I think you will get answer
     
  8. Nov 19, 2016 #7
    Isn't that angle = 30o ?
     
  9. Nov 19, 2016 #8

    haruspex

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    Taulant's error was to calculate a funny kind of average force (averaged over horizontal distance, which is not what is meant by "average force") necessary to provide the PE gain. Instead, the force to be found is the minimum constant force that will get it over the step. The next stage is to find the residual KE that results. That could be done by integration, but it is not necessary.
    Taulant set θ as the angle to the vertical, the 60o. That reduces as the wheel rises.
     
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