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!

Angular Velocity of a pivoting rod

  1. Mar 2, 2009 #1

    TG3

    User Avatar

    1. The problem statement, all variables and given/known data
    A uniform rod 5.8m long weighing 10kg is pivoted at its center and a small weight of mass 5.15 kg is rigidly attached to one end. You may neglect the size of the weight and assume it is located right at the end of the rod. It is held at a 37 degree angle at rest, then released.
    What is the angular velocity when the rod is vertical?

    2. Relevant equations
    After some work, I earlier solved that angular acceleration at it's release point is 1.6295 Don't know if that will be needed or not...
    Angular Velocity = Tangential Velocity /R
    VF^2 = VI^2 + 2A (X1-Xf)

    3. The attempt at a solution
    I dared to hope I could do this on my first try, but was disappointed. (Again.)
    5.8/2 = 2.9, so radius = 2.9
    sin37 2.9 = 1.745
    2.9-1.745 =1.1547 m of fall

    VF^2 = 0 + 2 (9.81) (1.1547)
    VF^2 = 22.65
    VF = 4.76

    4.76 / 2.9 = 1.64
    It looks so right.... but is so wrong. *sigh*
     
  2. jcsd
  3. Mar 3, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi TG3! :smile:
    Sorry, not following your proof at all :confused:

    using conservation of energy is right :smile:

    but what's happened to the mass, and the moment of inertia ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Angular Velocity of a pivoting rod
Loading...