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!

Energy conservation and circular motion

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A ball with a mass m swings from a string of length L from a pivot. A distance D directly below the pivot there is a nail. The ball is released making an angle theta=pi/2 with the vertical. For this particular problem, there is no friction.

    Assuming the ball swings completely around the nail in a circle, what is the velocity of the ball at the top of the circle?

    2. Relevant equations

    F=m(v^2/r) for circular motion

    3. The attempt at a solution

    So as soon as the string hits the nail, the radius of circular motion will be r=(L-D). Therefore, the centripetal force will be m(v^2/(L-D)). When the ball is at the top of the circle, if the ball is to just make it over the centripetal force should be equal to the gravitational force, right?




    v= sqrt((L-D)*g)

    I think this is correct because if the velocity at the top of the circle were equal to zero, then it would not be able to complete the rest of the circle which is a necessary condition.

    However, it seems unsettling to me that I did not calculate the velocity of the ball right before the string hits the nail.

    Does my answer look correct?
  2. jcsd
  3. Sep 11, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, correct.
    only after it hits the nail.
    not necessarily, there still could be a tension in the string contributing to an increase in the centripetal force (centripetal force = T+ mg).
    this is not correct
    That should be unsettling, so why not settle yourself and calculate the velocity right before the string hits the nail? Then use conservation of energy to find V at the top.
  4. Sep 11, 2010 #3
    The velocity right before hitting the nail would be (1/2)m*v^2=mgh.

    Therefore, v=sqrt(2gh) which in this case is:




    (1/2)m*(2gL) = (1/2)m*v'^2 + mg*2(L-D)
    g*L=(1/2)v'^2 + 2g(L-D)
    v'^2 = 2[g*L - 2g(L-D)]


    v' = sqrt[2[g*L - 2g(L-D)]]

    v'=sqrt[2gL - 4gL + 4gD]

    v'=sqrt[4gD - 2gL] (velocity at the top of the circle)
    Last edited: Sep 11, 2010
  5. Sep 11, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  6. Sep 11, 2010 #5
    Cool. Thanks a lot PhanthomJay
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook