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THE CIRCLE HAS RETURNED(Help me please)

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data
    PICTURE: http://imageshack.us/photo/my-images/21/circls.png/
    A Force of gravity acts upon a ball on top a circle. The ball rolls a down the curve of the circle until a CERTAIN POINT. at this CERTAIN POINT the ball detaches from the circle and travels until it his the ground. What is the distance between the top of the CIRCLE and the point at which it detaches...

    GIVEN INFORMATION: Mo=Mas Of ball, D = Diameter Of Circle
    -There is NO FRICTION
    -Energy is conserved (only conservative forces acting-- Gravity)
    -

    2. Relevant equations
    Centrepetal Force = Inward force
    Conservation Of Energy
    This problem is similiar to BANKED CURVE!!!!!!!! (this is the only way that i know of an object can be 'forced' towards the center, which is due to the horizontal component of normal force)

    3. The attempt at a solution

    First i started by solving for forces on a banked curve...

    Fnety= FNSinθ - mg = 0
    Therefore FN = mg/Sinθ

    Fnetx = FNCosθ
    Fnetx = MgTanθ

    Now i used the Horizontal component (FNET X) and made it Equal to FC because FC is the horizontal inward force.

    Fc = Fnetx
    (Simplify)
    Vo=√[DgTanθ/2]

    Now i'm using the conservation of energy and comparing it's initial moment at the top where it has a Velocity of 0. And comparing it to the "CERTAIN POINT" at which it detaches.

    ET1 = mgD
    Et2 = mg(d-x) + 1/2mVo^2

    Et1 = Et2 (SIMPLIFY)
    X = DTanθ/4

    Now i'm left with a problem... The angle... I tried to solve for it by creating a Right angled triangle with a radius line mideway through the cirlce, and a radius line going toward the object... but yea no luck.. Any ideas anyone? My teacher for some reason seems to tell me i don't need an angle... Which leads me to think that it's not on a banked curve and i'm supposed to have an angle but it must... C.c SOME ONE HELP ME!!! i've had this problem for WEEKS but i'll never give it UP!!!!
    I Believe my solution is correct, due to the fact that the units match up. HOWEVER i don't know how to solve for Theta... THE ANGLE the projectile detaches at!!!! :(
    -Can some one give me some hints on how i can get this angle...
     
  2. jcsd
  3. Dec 20, 2011 #2

    Doc Al

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    Staff: Mentor

    My advice: Scrap the idea of using the banked curve solution to solve this problem.

    Instead, apply Newton's 2nd law to the ball. What's the condition that tells you when the ball just starts to lose contact?
     
  4. Dec 20, 2011 #3
    I would have to say... When the balls horizontal velocity is more than the balls vertical velocity...

    The normal force on the object would also have to be zero at the point it detaches as well..

    And is it impossible to find the angle cause i dont have enough info :(?
     
    Last edited: Dec 20, 2011
  5. Dec 20, 2011 #4

    Doc Al

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    I don't think so.

    That's the one. Express that mathematically.

    You have all the info needed to find the angle.
     
  6. Dec 20, 2011 #5
    omg. i think i love you if the way i'm thinking is correct...


    Okay so when i did the Y components on the banked curve for the force.

    I know FN is = 0 as you confirmed for me


    So, Fnety = FNSinθ - FG = 0

    there fore FN = Mg/Sinθ

    So since FN is 0

    0=mg/sinθ

    θ=mg/sin

    :)?
     
  7. Dec 20, 2011 #6
    pooh.. when i plug this in units don't match up.. I get a newton instead of meters.
    ..

    So if i do FN as components.. the only force acting on the ball is the conservative force of gravity... But how can this help me solve the angle?
     
  8. Dec 20, 2011 #7

    Doc Al

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    Staff: Mentor

    Please forget about the banked curve problem!

    OK.

    Nah.

    Consider forces in the radial direction. Apply Newton's 2nd law. (What's the acceleration in the radial direction?)
     
  9. Dec 20, 2011 #8
    okay so heres my next attempt then... And I'll let the banked curve idea go, as long as you promise to tell me why? or after i solved this explain how to do it using the banked curve?

    So i have my forces set as FN in the x direction, and FG in the y direction...

    I solved for Velocity at that point by FC = FN

    mv^2/r = 0
    V=sqrt [D/2m]

    But if i do this, do i consider the speed at the top of the circle also which is
    V=sqrt[dg/2]
     
    Last edited: Dec 20, 2011
  10. Dec 20, 2011 #9
    wait. i see i just use this velocity into the energy haha one sec.. let me write this up
     
  11. Dec 20, 2011 #10

    Doc Al

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    Staff: Mentor

    The only connection between the two is that they both involve centripetal acceleration.

    I don't quite understand your x and y axes. Two forces act on the ball: The normal force and the weight. Which way do they act?

    That's not true.

    Consider force components in the radial direction. (There are only two forces. What are their radial components?)
     
  12. Dec 22, 2011 #11
    Okay my bad i had to study for a math test.. so can you correct me plz if im wrong.

    So if consider the force components in the radial direction..

    I have Force of gravity FG, which is in the y component

    and a normal force in the x component..


    Now is this normal force on an angle in the radial direction? and if it is... this is the same problem as the banked curve, how am i to solve for this dang angle.


    http://imageshack.us/photo/my-images/14/circlesk.png/

    i gotta pass out.. but ill get ur feedback tmw hopefully.. uhh and also.. Does the fact that there are two normal forces? the normal force from the circle and the normal force from the Ball itself in opposite directions?
     
  13. Dec 22, 2011 #12
    Also my force diagram... http://postimage.org/image/7g18q7iip/ [Broken]
    correct me if im wrong here too.
     
    Last edited by a moderator: May 5, 2017
  14. Dec 22, 2011 #13

    Doc Al

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    The force of gravity is vertical, but the normal force is not generally horizontal.

    The normal force is in the radial direction.

    It's not the same problem.

    Once again: Apply Newton's 2nd law in the radial direction.
    All you care about are the forces acting on the ball. Only one normal force acts on the ball; it pushes the ball radially outward.
     
  15. Dec 22, 2011 #14

    Doc Al

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    Staff: Mentor

    Only two forces act on the ball. Get rid of everything else.

    Also: Indicate the angle of the ball's position, measured from the vertical. You'll need that angle when finding the radial components of the forces. (And when you set up the equations as I suggest you'll end up solving for that angle.)
     
    Last edited by a moderator: May 5, 2017
  16. Dec 23, 2011 #15
    OKAY FINALLY CHRISTMAS BREAK. Now i can invest all my time into solving this damned problem...

    http://imageshack.us/photo/my-images/835/newbitmapimage3o.png/

    Okay so here are my ideas... just let me know if i can't use them :O!(also your help is much appreciated through all of this. I'm sorry its taking me so long.. i just can't seem to grasp this for some reason :()

    okay so

    fnety= fncosa - mg = 0
    therefore fn = mg/cosa

    Fnetx = fnsina
    fnetx= mgtana (sina/cosa = tana)

    fnetx = mAx
    mAx = mgtana
    Ax = Gtana

    Now since i know my Ay = g

    i thought i could make a triangle using the ax and ay vectors.. but that didn't seem to work, i'm wondering why i can't do that?

    anyways, with the diagram, i noticed there is some type of relation between the x and the forces, but i still don't quite see how to relate it? a force with a length? :(?

    Perhaps my diagram is misleading?
     
  17. Dec 24, 2011 #16

    Doc Al

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    Staff: Mentor

    The diagram is OK except that you drew the point of contact at 90° from the vertical. Instead, imagine the point of contact at angle 'a' from the vertical.

    The acceleration isn't zero in the y-direction. It's sliding down the sphere, accelerating as it goes.

    Again, you're getting hung up by comparing it to the 'banked road' problem.

    One more time: Consider forces in the radial direction.
     
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