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Rolling Without Slipping

  1. Mar 29, 2012 #1
    1. The problem statement, all variables and given/known data

    A 2.7kg solid sphere (radius .2m) is released from rest at the top of a ramp and allowed to roll without slipping. The ramp is .5m high and 5.4 long.
    When the sphere reaches the bottom of the ramp what is the rotational kinetic energy and the translational kinetic energy.

    2. Relevant equations

    krotf=1/2(2/5mr^2)wf^2
    ktransf=1/2mvf^2

    3. The attempt at a solution

    krotf=1/2(2/5(2.7)(.2^2))wf^2
    ktransf= 1/2(2.7)vf^2
     
  2. jcsd
  3. Mar 29, 2012 #2

    tiny-tim

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  4. Mar 29, 2012 #3
    mg(.85)=Kf
    Kf=13J
     
  5. Mar 29, 2012 #4

    tiny-tim

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    what is .85 ? :confused:
     
  6. Mar 30, 2012 #5
    i meant .5 but I already know that kf=13 is correct
     
  7. Mar 30, 2012 #6

    tiny-tim

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    ahhh! :rolleyes:

    ok, you know the total KE is 13,

    so add the formulas for the two KEs, use v = ωr, and put the whole thing equal to 13 :wink:
     
  8. Mar 30, 2012 #7
    but how will that get me kinetic rotational and kinetic translational? i need each individually!
     
  9. Mar 31, 2012 #8

    tiny-tim

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    (just got up :zzz:)

    write out the equations, and see how far you get :smile:
     
  10. Mar 31, 2012 #9
    Hey,
    According to the law of conservation of mechanic energy the potential energy of the sphere transforms into kinetic and rotational energy of the sphere if we ignore the resistive forces. Therefore kinetic energy+rotational energy should equal to mgh=9.81(m/s^2)*2.7kg*0.5m=13.2435J. I'd use the equation E(pot.)=E(rotational)+E(kinetic)...mgh=0.5*J*w^2+0.5*m*v^2 to solve for w if you need to calculate both energies individually. Maybe I thought it too complicated, but please let me know if you know a better solution. :smile:
     
    Last edited: Mar 31, 2012
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