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Angular velocity of a ball+rod falling through 90˚?

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data

    http://puu.sh/57VmL.png [Broken]

    A thin rod, 28.5cm long with a mass of 0.97kg, has a ball with diameter 7.81cm and mass 2.15kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top. The apparatus is free to pivot about the bottom end of the rod.
    (A) After it falls through 90°, what is its rotational kinetic energy?
    (B) What is the angular speed of the rod and ball after it has fallen through an angle of 90°?

    2. Relevant equations

    Ek = 0.5*I*ω2

    3. The attempt at a solution

    I solved part A successfully by finding the center of mass of the ball and rod and then solving for mgh.

    (A)

    Centre of mass:

    d = Σ(m*d) / Σm
    = [0.97*0.285/2 + 2.15*(0.285+0.0781/2)] / (0.97 + 2.15)
    = 0.26760657m

    Ek = Ep
    Ek = m*g*h
    = (0.97+2.15)*9.8*0.26760657
    = 8.1823 J


    But then when I try to solve B I get the wrong answer. Here's what I tried..

    (B)

    Ek = 0.5*I*ω^2
    8.1823 = 0.5*m*r22
    8.1823 = 0.5*(0.97+2.15)*0.2676065722
    ω = 8.55813 /s

    But this is wrong, and I'm not sure why.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 3, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Your initial potential energy looks good to me.

    For calculating the moment of inertia of the system, you can't lump all the mass of the system at the center of mass and just use mr2. (This is because moment of inertia depends on the square of the distance of the mass elements from the axis of rotation.) You'll need to treat the moment of inertia of the rod and ball separately.
     
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