Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Spinning ball

  1. Mar 18, 2005 #1
    have a look a the diagram
    heavy particle of mass m is tied to one end of a string which is of length l. The other end is attached to the ceiling. The particle forms uniform circular motion. Angular velocity of omega. String is angle theta. Ignore friction and mass of string.

    Find magnitude of string force S in terms of m, l, omega
    horizontal components [tex] F_{x} = F sin \theta = m \omega^2 l sin \theta [/tex]
    vertical components = [tex] F_{y} = F cos \theta = mg [/tex]
    total components [tex] F = m \sqrt{g^2 + \omega^4 l^2 sin^2 \theta) [/tex]
    the text book however omits the sin theta and the g part which gives [tex] F = ml \omega^2 [/tex]

    who is right??

    Attached Files:

  2. jcsd
  3. Mar 18, 2005 #2


    User Avatar
    Science Advisor

    The book is correct:
    {String Length} = L
    {Horizontal Radius of Circular Orbit} = r = L*sin(θ)
    {Centripetal Force} = m*v2/r =
    = m*ω2*r =
    = m*ω2*{L*sin(θ)} =
    = m*L*ω2*sin(θ)

    {String Tension} = T
    {Horizontal Component of String Tension} = T*sin(θ)

    At equilibrium:
    {Horizontal Component of String Tension} = {Centripetal Force}
    ::: ⇒ T*sin(θ) = m*L*ω2*sin(θ)
    ::: ⇒ T= m*L*ω2

  4. Mar 18, 2005 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Both answers are "correct" but the book's is the simpler and more useful one. Realize that your equation for the horizontal components ([itex] F sin \theta = m \omega^2 l sin \theta [/itex]) immediately implies the book's answer.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook