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Rotating rod Please Help

  1. Feb 5, 2006 #1
    A thin homogeneous rod hsa length L and mass M. Initially the rod is vertical at rest on a frictionless surface. The rod is given a slight push and begins to fall in the gravitational field.

    1) write an expression for the kinetic energy T of the rod as functions of M,L,g, theta and d theta/dt

    2) if y is the distance the center of mass drops. find d y/dt as a function of L, g and theta



    #1)i know the Kinetic energy of the rod is given by the equation

    T= 1/2 I w^2

    where w = omega = d theta/dt
    and I = M(L/2)^2

    so T = 1/2 M(L/2)^2 d theta/dt^2

    T = 1/8 M L^2 d theta/dt^2

    The answer is T = 1/8 M L^2 d theta/dt^2 (Sin^2theta + 1/3)

    i dont know where the (Sin^2theta + 1/3) comes from


    #2)


    i dont even know where to start on this one

    but the answer is Y prime = sintheta sqrt ( (3gL(1-costheta) / 4-3cos^2theta)


    any help would be greatly appreciated
     
  2. jcsd
  3. Feb 5, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That would be true if the rod were purely rotating, but it's also translating: the total KE is the sum of the translational KE (of the cm) and the rotational KE (about the cm).
     
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