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!

Vertical Circular Motion help

  1. Oct 29, 2003 #1
    A child is playing with a ball on a string. The ball has mass 0.065 kg, the string has radius 1.2 m and assume that the string is massless. The child swings the ball on a string in a vertical circle. Halfway up the circle the speed of the ball is measured to be 8.7 m/s. What is the maximum tension in the string?

    I know that a=V^2/R

    F=MV^2/R

    And since I'm working in the work-energy chapter this week, I think maybe I could apply the work energy theorem to this problem, but not sure how or even how to start this problem.

    So help greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Oct 29, 2003 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You need to be able to calculate the velocity at each point. In order to use F= ma= mV2/R. Try using "conservation" of energy. You know the velocity "halfway up the circle" so you can calculate the kinetic energy. Calculate the potential energy since this is a vertical circle. Add to find the total energy. Subtracting off the potential energy at every other point lets you find the kinetic energy and so the velocity (squared) at each point.
     
  4. Oct 29, 2003 #3
    Um, what?
    Sorry, but I don't follow what you mean by this "Subtracting off the potential energy at every other point lets you find the kinetic energy and so the velocity (squared) at each point."

    -------------------------------------------------------------
    Anyhow, can someone check this work logic please?

    I figured out the kinetic energy at that point,
    Ke=1/5*m*v^2
    = .5*.065kg*(8.7m/s)^2
    = 2.46J

    PE= m*g*h=> .065kg*9.8*1.2m =.765 J.

    Total energy =3.225J.

    So now I need velocity.
    3.225J=.5*m*v^2
    v=9.96 m/s

    Now I know that F=mv^2/R
    so F=.065kg*(9.96)^2/1.2m =>5.37N

    So my max force or Tension rather is going to be 5.37N. Does that look right guys?
     
    Last edited: Oct 29, 2003
  5. Oct 29, 2003 #4
    hey m8, do you mean the circle is spinnning horizontally over his head or vertically next to him? i found out horizontally i think, but if its vertically spinning next to him let me know. i got like 4.47 Newtons, but thats if the ball is spinning above his head horizontally. lots of redundency in my reply eh?
     
    Last edited: Oct 29, 2003
  6. Oct 30, 2003 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Kylemadigan: did you notice that the title of this thread was "Vertical Circular Motion help"? Also the original post said "The child swings the ball on a string in a vertical circle. "

    Juntao: You did exactly what I said: except that you reasoned correctly that the greatest speed will be where the potential energy is 0- so you "subtracted off" 0! My point about "velocity (squared)" was that you once you got "3.225J=.5*m*v^2" you didn't really need to take the square root to get v: Since the formula for centripetal force also uses mv^2, you only need mv^2= 3.225/(.5)= 6.45.

    Then F= mv^2/r= 6.45/1.2= 5.375 N just as you got.
     
  7. Oct 30, 2003 #6
    For some reason, the answer is not working out with the computer. Hmm...maybe the circle isn't uniform?
     
  8. Oct 30, 2003 #7

    jamesrc

    User Avatar
    Science Advisor
    Gold Member

    It's probably not working because at the position the maximum tension occurs, the tension of the string must support the centripetal force AND the weight of the object. Try adding the weight of the object to your answer.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Vertical Circular Motion help
Loading...