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AP Physics CH 5: Centripetal Motion

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  1. Oct 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Derive an expression for Vmin, the minimum speed the ball can have at point Z without leaving the circular path.


    2. Relevant equations
    V≤(x/t)
    V≤(2∏r)/(t)
    V=√(GM1)/(r)


    3. The attempt at a solution
    I set 2∏r = to x since its in a circle. i then had V≤(2∏r)/(t)
    then i had (2∏r)/(t)=√(GM1)/(r)
    in the end i got t=2∏√(r3)/(Gm1)
    it didnt make sense to me so i came here for help /:
     
  2. jcsd
  3. Oct 10, 2011 #2

    lightgrav

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    The key equation here are v.v=-a.r , and ΣF=ma .
    Is "z" the top of a circular loop? if so, then a fast ball would be pushed down by the track (helping mg to accelerate the ball's mass downward). too slow, and the radius of the ball's path, given by r = -v.v/g , would be too small to stay on the track. you want the radius of the ball's path to be the same as the Radius of the track.

    They want the speed, not the time.
     
  4. Oct 10, 2011 #3
    point "z" is about 90 degrees there are also points "M, P, and Q". "Q" is about 270 degrees. "P" is 180 and m is about 60.


    idk if that helps
     
    Last edited: Oct 10, 2011
  5. Oct 10, 2011 #4

    lightgrav

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    90 degrees? from where? from horizontal? then that is the TOP. are you trying to hide important aspects of the question from yourself?
     
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