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Ke trans+ ke rotatational

  1. Mar 9, 2006 #1
    when i add ke translational + ke roatational, i dont get potential which i should, although i was wondering if maybe it was due to slipping although i thought it would still be closer than i got

    does this sound right?

    ke trans(0.002520831)+ke rotational (0.001008332) = 0.003529163
    and
    pe=0.014278472

    as you can see not very near it!

    is this right or have i gone wrong?

    the exp was rolling a ball bearing down a 5 degree slope and these calcs were from the 1m release mark. the radius is 0.00765m


    thanks

    to get the angular speed i used linear V/R (radius)

    i got linear v by divinding distance by time.
     
  2. jcsd
  3. Mar 9, 2006 #2
    If you are rolling a ball down a slope, the linear v will NOT be d/t because the ball is accelerating. d/t is average speed not instantaneous speed, which is what you need for your equations.

    -Dan
     
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