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Homework Help: Natural frequency of a simple harmonic system

  1. Aug 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Derive the natural frequency [itex]f_n[/itex] of the system composed of two homogeneous circular cylinders, each of mass [itex]M[/itex], and the connecting link [itex]\text{AB}[/itex] of mass [itex]m[/itex]. Assume small oscillations.

    2. Relevant equations
    The sum of the kinetic + potential energies in an isolated system remains constant.

    An undamped, simple harmonic system has an equation of motion of the form [tex]\ddot{\theta} + \omega^{2}_n \theta = 0[/tex]

    3. The attempt at a solution
    The sum of the kinetic + potential energy when the system rotates by an angle [itex]\theta[/itex] measured as in the diagram counterclockwise is [tex]\sum E = V + T = mg r_0 (1 - \cos \theta ) + \tfrac{1}{2} m v_{AB}^{2} + 2 \times \tfrac{1}{2} M v_{O}^2 + 2 \times \tfrac{1}{2} I_O \dot{\theta}^2 = \text{const.}[/tex]

    where [itex]v_O[/itex] denotes the velocity of the centre of mass of the cylinders, [itex]\dot{\theta}[/itex] denotes the angular velocity of the system (and by extension the angular velocity of the cylinders) and [itex]v_{AB}[/itex] the velocity of the centre of mass of [itex]\text{AB}[/itex].

    [tex]E = mgr_0 (1 - \cos \theta ) + \tfrac{1}{2} m (r_0 \dot{\theta})^2 + M (r \dot{\theta})^2 + \tfrac{1}{2} M r^2 \dot{\theta}^2 = \text{const.}[/tex]
    [tex]\implies \frac{\text{d}E}{\text{d}t} = mg r_0 \sin \theta \dot{\theta} + m r^{2}_0 \dot{\theta} \ddot{\theta} + 3 Mr^2 \dot{\theta} \ddot{\theta} = 0[/tex]

    using small angle approximation and rearranging...

    [tex]\ddot{\theta} + \frac{mgr_0}{3Mr^2 + mr^{2}_0} \theta = 0 \implies \omega_n = \sqrt{\frac{mgr_0}{3Mr^2 + mr^{2}_0}} \implies f_n = \frac{1}{2\pi} \omega_n[/tex]

    The answer given in the book is [tex]f_n = \frac{1}{2\pi} \sqrt{\frac{mgr_0}{3Mr^2 + m(r-r_0)^{2}}}[/tex]

    Comparing the answers, my problem appears to be that I've calculated the kinetic energy of the link [itex]\text{AB}[/itex] incorrectly but I can't figure out why. My logic was that the angle [itex]\theta[/itex] is the angle between the vertical through the centre of the cylinder and the line connecting the centre of the cylinder and the end of the link [itex]\text{AB}[/itex], so the velocity of the end of [itex]\text{AB}[/itex] would be equal to [itex]\dot{\theta} r_0[/itex] (angular velocity x arm) and as [itex]\text{AB}[/itex] remains horizontal throughout, the velocity of the centre of mass of [itex]\text{AB}[/itex] would be the same as the velocity of its end.

    Last edited: Aug 6, 2015
  2. jcsd
  3. Aug 6, 2015 #2


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    Gold Member

    You could put in a few more steps to help a tired old duffer understand how you get each thing.

    It looks like you have ##v_{AB}## as ##r_0 \dot{\theta}##, and ##V_O## as ##r \dot{\theta}##. But you need to be sure that these are both in the same frame of reference. Are these both with respect to the ground?
  4. Aug 6, 2015 #3
    Ahaa, there's my problem. No, I've calculated [itex]v_{AB}[/itex] to be wrt the centre of the cylinder but [itex]v_O[/itex] is wrt the ground.
  5. Aug 6, 2015 #4
    Got the right answer when I take ##v_{AB}## wrt the ground. Thanks for your help!
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