1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Centripetal Force, Gravity and Normal

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

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes that how I got the book answer.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Centripetal Force, Gravity and Normal