1. Not finding help here? Sign up for a free 30min 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!

Springs & circular motion

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A small block is launched on the fricntionless loop the loop shown. te sprink launcher has a sprink constant k

    a)find the velocity at the top of the loop as a function of the displacement x of the spring launcher from its equilibrium length

    b)find the minimum cvalude of x such that the block goes over the top in contact with the track

    c)find the normal force on the bock A as a function of x

    2. Relevant equations



    3. The attempt at a solution

    I believe i would have to use .5ks2, and then calculate the acceleration, and aply it to uniform circular motion, is this correct

    W = .5kx2

    K = .5mv2

    v = (2W/m).5 = (kx2/m).5
     
    Last edited: Nov 7, 2009
  2. jcsd
  3. Nov 8, 2009 #2
    ok pretend my first attemp never happened

    Part A
    TA = .5ks2
    UA = 0

    TB = .5mv2
    UB = -2Rmg

    .5ks2 = .5mv2 - 2Rmg

    v = ((ks2 + 4Rmg)/m).5

    Part B

    same as above, just solved for s

    s = ((mv2-4Rmg)/k).5

    Part C

    not really sure about the Normal Force, Does it still point up even if its on the top of the loop the loop
     
  4. Nov 8, 2009 #3

    rl.bhat

    User Avatar
    Homework Helper

    When the block is moving in the loop, two forces are acting. One centripetal force and the other the weight of the block. Mg*cosθ contributes to the normal reaction.
     
  5. Nov 9, 2009 #4
    so the normal is

    N = mgcos(180)

    do u agree with my other responses, they do not match what the answer is said to be

    Part B answer is xmin = (5mgR/k)1/2

    Part C answer is N(x) = (kx2/R) - 5mg

    where does that 5 mg come from
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook