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Vertical circular motion minimum speed proof

  1. Nov 21, 2009 #1
    i've read in a lot of places that the minimum speed fora body to "finish" a vertical circle is (rg)^0.5, but i've never understood why.. they assume the if the body passes the top, it will finish the whole circle.. but i can't actually see it... can someone explain/proof it to me?

    secondly, is the mentioned speed is for the bottom or for the top?
    this is the proof i've seen:
    (the top point)
    mg + N = ma
    a = v^2 / r

    assume N=0

    mg = mv^2 / r
    v = (rg)^0.5

    my problem is with this formula
    a = v^2 / r

    the acceleration is changing, thus this is true for momentary velocity.

    so v = (rg)^0.5
    must be for v of the top.. not the bottom.....

    sorry about my english :)
    Last edited: Nov 22, 2009
  2. jcsd
  3. Nov 21, 2009 #2
    v is the speed the object must have as it passes the top of the circle

    for example, if it was 0 the object would just drop vertically to the floor....
  4. Nov 21, 2009 #3
    fine, but why the body will necessarily complete the circle? (if the speed is such)

    p.s. so it is possible the body get to the top point at speed 0? or any thing below (rg)^0.5 ?
    Last edited: Nov 21, 2009
  5. Nov 22, 2009 #4


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    The top point is critical because v and a are minimal and Fg is acting fully to reduce N, therefore N is minmal. After passing the top v and a increase and Fg acts only partly to reduce N, so N increases too.

    No, it would fall off earlier.
  6. Nov 22, 2009 #5
    The equation for minimum speed as a function of angle is:

    v_min(ang) = (r*g*sin(ang))^0.5

    ang = 0 when the body reaches a quarter of the circle
    ang = pi/2 when the body reaches the top of the circle

    The equation for actual speed as a function of v_top (speed at the top of circle) and angle is: (solved using conservation of energy)

    v_act(v_top,ang) = (v_top^2 + 2*g*r*(1-sin(ang))^0.5

    For the case of v_top = (r*g)^0.5, the plot of these two equations looks like this:



    The top curve is actual speed. The bottom curve is minimum required speed. They meet at pi/2 (top of the circle) with the value of v = (r*g)^0.5. If you lowered the actual speed curve, the curves would intersect. The point at which they first intersect is the angle at which the body would fall. This is what "A.T." meant when he said "it would fall off earlier."
    Last edited: Nov 22, 2009
  7. Nov 23, 2009 #6
    another way is you know if it falls, then it'll fall along a parabola not along a circle. So just compare the shape of the given parabola that would form (at that instantaneous velocity) if the circle wasn't there.
  8. Nov 23, 2009 #7
    i've thought about it.. but that just a proof it won't fall at the top, but a moment after that, it has different velocity..
  9. Nov 23, 2009 #8
    I meant you could take the velocity anywhere on the cirlce and show whether it's fast enough at that point to stay on the circle or fall parabolically inside of it.
  10. Nov 23, 2009 #9


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    yes, a higher velocity, resulting in higher centripetal acceleration and higher normal force. so it cannot fall anymore, if it got that far.
  11. Nov 23, 2009 #10
    i've just thought about a simpler one..
    so the body reached the top.. let's take what happened so far, in reverse.. now the body has velocity of (rg)^0.5 to the other side, mg is still downward.. we KNOW the body will complete the half circle reversed motion, because we've just seen it complete it...
    now the second half of the circle, is exact the same as the first half reversed... so the body will necessarily complete the second half because we know it complete the first half reversed...
  12. Nov 23, 2009 #11


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    Yeah, but how do you know that it can get to the top with v = (rg)^0.5 ? :smile: The analysis by bluelava0207 shows that.
  13. Nov 24, 2009 #12
    i know it can get to the top at (rg)^0.5, according to what i've write in the first post
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