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Parallel Axis Theorem | Kinetic and Potential Energy

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data

    A rigid body is made of three identical rods, each with length L = 0.525 m, fastened together in the form of a letter H, as in Figure 10-56 below. The body is free to rotate about a horizontal axis that runs along the length of one of the legs of the H. The body is allowed to fall from rest from a position in which the plane of the H is horizontal. What is the angular speed of the body when the plane of the H is vertical?

    http://www.webassign.net/hrw/10-56.gif

    2. Relevant equations



    3. The attempt at a solution

    I_1 = 0

    I_2 = m/L∫x^2dx from 0 to L
    I_2 = (1/3)*mL^2

    I_3 = mL^2

    I_total = (4/3)mL^2

    E_Mec,Top = E_Mec, Bot

    U_T + K_T = U_B + K_B

    3*m*g*L + 0 = 0 + (1/2)*Iv^2

    3*m*g*L = 1.5*m*L^2*v^2

    2*g = L*v^2

    [(2*g)/L]^.5 = v

    v = 6.11 rad/s

    This is wrong I'm pretty sure that I'm calculating the Rotational Inertia incorrectly. I don't know how to use the parallel axis theorem, especially not on three continuous bodies. I know that the formula is:

    I = I_com+Mh^2

    When I use that to calculate the Inertia, I get:

    I_com = 3/2mL^2

    I = 3/2mL^2 + 3mL^2
    I = 4.5mL^2
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 24, 2013 #2

    tms

    User Avatar

    Why [itex]3mgL[/itex]? And on the right hand side, the usual symbol for angular speed is [itex]\omega[/itex], not [itex]v[/itex], which is usually taken to mean translational speed.
     
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