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Kinetic energy of a rolling hoop

  1. Nov 10, 2016 #1
    1. The problem statement, all variables and given/known data
    A 120 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on the hoop to stop it?

    2. Relevant equations
    I of hoop=MR^2

    3. The attempt at a solution
    KE=0.5*m*v^2+0.5*(mR^2)(v/R)^2
    =0.5*120kg*.24^2m/s+0.5*(120kg*R^2)(.24m/s^2/R^2)
    =6.912 J

    .
     
  2. jcsd
  3. Nov 10, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    Your calculation looks good. You might want to trim the significant digits in your result to match the given data. If this is a "show your work" type question rather than a web based "enter the solution" assignment, you might want to include a note as to why the total KE is the same as the work required.
     
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