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Homework Help: Maximum kinetic energy

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data
    An object is constrained by a cord to move in a circular path of radius 0.5m on a horizontal frictionless surface. The cord will break if its tension exceeds 16 N. What is the maximum kinetic energy?

    2. Relevant equations

    3. The attempt at a solution

    Really lost on this problem. Am I applying the wrong formulas?
  2. jcsd
  3. Dec 7, 2008 #2
    if its moving in a circular horizontal path the tension isn't mg its equal to the centripetal force (mv2/r)

    so im guessing you set the centripetal force to equal 16 to solve for v. After that you can find the KE
  4. Dec 7, 2008 #3
    thanks teckid.

    So a = v2/R

    so V = (a*R)1/2
    V = (9.8*.5)1/2
    V = 2.21

    m = T*R/(v2)
    m = 1.63

    So for KE, we have:
    KE = 1/2*m*v2
    KE = 1/2*1.63*2.212
    KE = 4 J

    I think that's right..
  5. Dec 7, 2008 #4
    No i think you're confusing horizontal and vertical... vertical suggest there is a downward force due to gravity (aka 9.81...). However, here the path is horizontal, like a merry-go-round, so there gravity does not effect the tension on a frictionless surface. since the tension must be 16 or less and the tension is equal to the centripetal force then:

    16 = mv2/r


    16 *radius * mass = v2
    (16*radius*mass).5 = v

    with this velocity you can find the max KE (.5MV2)

    You should not be using centripetal acceleration to find the velocity unless we know what that acceleration is.
  6. Dec 8, 2008 #5
    (16*radius*mass)^.5 = v

    But this leaves us with 2 unknown variables (mass and velocity)...
  7. Dec 8, 2008 #6


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    Homework Helper

    Not really.

    Maximum F = mv²/r = 16N

    But Kinetic Energy is ½mv²

    But you know that mv² = 16*r

    Which means your KE is merely ½*16*r
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