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Finding Rotational Kinetic Energy of Sphere on Ramp

  1. Dec 19, 2012 #1
    1. The problem statement, all variables and given/known data
    If Inga, the Laboratory Assistant, rolls a spare head down a 4 m ramp because it was spherical and solid and too heavy at 4.5 kg at a speed of 4.5 m/s, what was its total kinetic energy?


    2. Relevant equations
    KE = (1/2) I ω^2
    I = (2/5) MR^2

    3. The attempt at a solution
    Basically, I'm not sure how to solve this without the sphere's radius. Any help, solutions, or hints would be highly appreciated.
     
  2. jcsd
  3. Dec 19, 2012 #2

    haruspex

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    Just go ahead with the algebra and you should find the radius cancels out. The rotational KE of a sphere is in fixed ratio to its linear KE, regardless of radius.
     
  4. Dec 19, 2012 #3
    So does this look correct:

    KE = (1/2)(4.5 kg)(4.5m/s)^2 + (1/2)(2/5)(4.5)r^2[(4.5m/s)/r]^2
    KE = 6.75 J + 18.225 J
    KE = 24.98 J
     
  5. Dec 19, 2012 #4
    Where'd that 6.75 J come from?
     
  6. Dec 19, 2012 #5
    Hmm..I too am now wondering the same.

    Is this better:

    KE = 45.453 J + 18.225 J
    KE = 63.69 J
     
  7. Dec 19, 2012 #6

    haruspex

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    Still a bit low. I get 45.56+18.23=63.79
     
  8. Dec 19, 2012 #7
    That's also what I'm getting. Joel, I think you're not plugging into your calculator correctly.
     
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