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Centripetal Force of a particle

  1. Feb 17, 2008 #1
    This is regarding a particle traveling in a uniform circular motion around a "loop."

    I understand that at the very top of the loop, the particle experiences Fg (Force of Gravity) and also Fn (Normal Force) due to contact with the surface, but also experiences acceleration in the "-y" direction. My question is why is it considered minimum velocity when you set the Fn to zero to find the velocity at the very top? Using Newton's Second Law, Fnet, y = may

    -Fn - Fg = m (-ay)

    The way I am interpreting it is that there is really no Fn to support you at the very top because its pretty much Fg right?

    Can someone clarify this idea for me?
  2. jcsd
  3. Feb 17, 2008 #2

    Hi there,

    The way I will interprete is that at the top of the loop, it is very much the minimum velocity required for you to be suspended just by g acting downwards only. Any other forces, be it normal contact force or centripetal force, is there to prevent you from flying off the loop only.
  4. Feb 17, 2008 #3
    So, are you saying that if there was an additional force besides that of gravity itself, than the particle would fall of the loop at the top?
  5. Feb 18, 2008 #4

    Doc Al

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    Staff: Mentor

    Because the faster you go, the greater the acceleration and the greater normal force needed to keep you on the track. So you can set Fn = 0 to find the minimum speed that still just barely keeps you in contact with the track. If you go any slower, you will fall off the track and never make it through the loop.
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