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Gravitational energy

  1. Dec 4, 2017 #1
    1. The problem statement, all variables and given/known data
    I have a 100 lb mass that is attached to a disk which is rotating freely about an axle. How do I calculate the gravitational energy contributed buy this 100 pound mass as it rotates from the 12 position to the six position

    2. Relevant equations
    No clue maybe. M x g x h?

    3. The attempt at a solution
    No clue
     
  2. jcsd
  3. Dec 4, 2017 #2

    BvU

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    Depends on the orientation of the axle .... but otherwise you are correct. Pay attention to the units !
     
  4. Dec 4, 2017 #3

    DrClaude

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    This is not acceptable, per PF rules. Using @BvU's help, your next post should be your attempt at a solution.
     
  5. Dec 4, 2017 #4
    Ok. My attempt would look like this:

    KE = 100lbs x 9.8ms2 x 10ft = 1354J

    But that is if the weight it's Free Falling, correct? This particular scenario involves the weight being attached to a disc that is 10 feet in diameter. The disc weighs approximately 100 pounds as well. So wouldn't those two variables be taken into consideration?
     
  6. Dec 4, 2017 #5

    BvU

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    I should think not.

    You still haven't told us about the orientation nor about the location of the axis of rotation :rolleyes:
     
  7. Dec 4, 2017 #6
    My apologies. The 100lbs is permanently mounted on the disc 5 feet from the axel ( or at the outer edge of the disk). As the disc rotates, the orientation of the weight changes respectively (as the weight is hard mounted to the disc which is free spinning).
     
  8. Dec 4, 2017 #7

    CWatters

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    Yes that's a relevant but incomplete equation. M*g*h = ??
     
  9. Dec 4, 2017 #8
    Ke = mgh. So, 100 pounds attached at the Outer Perimeter of a 10-foot diameter free Spinning Disk (which has a mass of 100 lb itself) would still be 100 times 9.8 times 10?
     
  10. Dec 4, 2017 #9

    BvU

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    Watch your units. Is ##g## 9.81 thingies in your non-SI units ?
     
  11. Dec 4, 2017 #10

    CWatters

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    Not KE. The change in Potential Energy (PE) = mgh

    What BvU said about units. g is only 9.81 m/s/s if you are working in SI units..

    Working in SI units the description becomes.... "45kg attached to the outer perimeter of a 3 meter diameter spinning disc"

    To rotate the disk so that the 45kg mass moves from the bottom/6 o'clock position to the top/12 o'clock position would require..
    ΔPE = 45 * 9.81 * 3 = 1324 Joules

    To rotate the disk so that the 45kg mass moves from the top/12 o'clock position to the bottom/6 o'clock position would require..
    ΔPE = 45 * 9.81 * -3 = -1324 Joules

    eg you get the same number of joules back again.

    Note: We're assuming the 100lbs/45kg is all lumped together in one place on the perimeter not distributed around the perimeter of the disk.
     
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