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Centripetal Force, Gravity and Normal

  1. Jun 20, 2014 #1
    1. The problem statement, all variables and given/known data
    image.png

    2. Relevant equations
    [itex]F_{c} = \frac{mv^{2}}{r}[/itex]


    3. The attempt at a solution

    I think the normal force would be the magnitude of vector sum of centripetal force and gravitational force. So I did:

    [itex]N = \sqrt{(\frac{mv^{2}}{r})^{2} + (mg)^{2}}[/itex]

    However, it isn't one of the answer choice. The actual answer is (c), and I have got no clue why it is (c).

    I could imagine why answer is:
    (a) Since mg is perpendicular to point Q, normal force is only [itex]\frac{mv^{2}}{r}[/itex]
    (b) if calculated from the highest tip of the circle...

    but where does [itex]2mg[/itex] comes from!?

    Thank in advance.
     
  2. jcsd
  3. Jun 20, 2014 #2

    CWatters

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    The -2mg part originally comes from the KE converted to PE.

    Try writing an equation for the energy of the ball at Q.
     
  4. Jun 20, 2014 #3
    Oh thanks, so we approach this problem using conservation of energy.

    let v = initial velocity

    We have this at Point Q:
    KE = KE' + PE..

    1/2mv^2 = 1/2mv'^2 + mgr.

    mv^2/r - 2mg = mv'^2/r

    We also have Force at point Q:
    Fn = Fc = mv'^2/r = mv^2/r - 2mg
     
  5. Jun 20, 2014 #4

    CWatters

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    Yes that how I got the book answer.
     
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