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Rigid body dynamics question

  1. May 6, 2010 #1
    1. The problem statement, all variables and given/known data
    The rigid body assembly is made from a 15 Kg disk, an 8 Kg slender rod, and a small 4 Kg sphere. The radius of the disk is R = 0.4 m and the length of the rod is L = 1.2 m. The pivot O is at the center of the disk. The assembly is released from rest at the horizontal position shown. Find the angular velocity of the assembly when it rotates 90° down to the vertical position.
    http://img269.imageshack.us/img269/4544/problemtp.png [Broken]

    Uploaded with ImageShack.us

    2. Relevant equations
    COE: K1+UG1= K2+UG2
    Inertia equations for particle, rod, and disk.

    3. The attempt at a solution

    I Found Inertia for disk, rod, and sphere seperatedly, with I disk = 1.2, I rod = 1.28 , I sphere= 3.2
    Then I use reference point as the horizontal line it started, therefore UG1 are all zero. So my energy equation look like:
    0= K2+ UG2
    0= 1/2 (I)w^2 + UG2rod + UG2sphere
    0= 1/2 (5.68) w^2 + (-0.2) (8) (g) + (-0.8) (4) (g)
    I got 4.07 for w, angular velocity.
    The answer is 4.32. I dont see what I am doing wrong. I did Irod= Icm + md^2 for rod inertia since its parallel axis theorem.

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 6, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Show how you found the inertia of the sphere.
  4. May 6, 2010 #3
    Idisk= 1/2 ( MR^2) = (1/2) ( 15) (0.4) (0.4)= 1.2

    Irod= 1/12 ( ML^2)+ Md^2= 1/12 (8)(1.2)(1.2)+ 8(0.2)(0.2)= 1.28

    Isphere= MR^2= 4 ( 0.8)^2 = NOT 3.2

    I see haha thanks so much man I knew i messed up one place, i forgot to square the R for Inertia for sphere.
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