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Simple harmonic motion of two solid cylinders attached to a spring

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data
    ) Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, massless rod. They are attached to a spring with force constant k using a frictionless ring around the axle. If the spring is pulled out and released, the cylinders will roll without slipping as they oscillate. Show that the period of their oscillation is T = 2∏√(3M/2k).

    2. Relevant equations
    solid cylinder: I=1/2MR^2

    3. The attempt at a solution
    Messing around with the equations has gotten me nowhere so far; I've only achieved redundancy by accidentally deriving other equations. The only way I see of making the moment of inertia relevant is by using torque, but trying to use that just has me going in circles (unintentional pun, haha). I'm stuck. Any advice would be appreciated.
  2. jcsd
  3. Dec 4, 2013 #2
    Try conservation of energy.
  4. Dec 4, 2013 #3
    I don't understand how that might help. It just seems to add more variables that I don't have.
  5. Dec 4, 2013 #4
    There is only one variable: the displacement of the cylinders from the equilibrium. Their angular velocity is kinematically related to the linear velocity of that displacement.

    The above consideration is applicable without energy, using just forces/torques, but that gets a bit more messier. But see for yourself.
  6. Dec 4, 2013 #5
    I'm sorry. I'm still totally stuck :(
  7. Dec 4, 2013 #6
    Let ##x## be the displacement of the axle from the equilibrium. ##v = \dot x##, the time derivative of ##x##, is the velocity of the axle. The cylinders are rolling without slipping. What is their angular velocity ##\omega##? What is the total kinetic energy of the cylinders? What is the potential energy of the spring?
  8. Dec 4, 2013 #7
    U = 1/2kx^2
    KE = 1/2mv^2 + 1/2Iω^2
    ω = v/R
  9. Dec 4, 2013 #8
    Very well. Now write down the equation for conservation of energy, and get rid of ##\omega## in it.
  10. Dec 4, 2013 #9
    1/2kx^2 = 1/2mv^2 + 1/2v^2/R^2
  11. Dec 4, 2013 #10
    Where is the momentum of inertia?

    When corrected, simplify that further.
  12. Dec 5, 2013 #11
    Oh, whoops, I'm sorry.
    1/2kx^2 = 1/2mv^2 + 1/2Iv^2/R^2
    Is I additive here? Would it simply be MR^2?
    That would make the equation simplify to 1/2kx^2 = mv^2.
  13. Dec 5, 2013 #12
    My interpretation of the problem is that M is the mass of two cylinders, not of one. Which means the equation is $$ \frac 1 2 kx^2 = \frac 1 2 \cdot \frac 3 2 M v^2 $$ Can you convert that to an SHM equation?
  14. Dec 5, 2013 #13
    YES. I think so. If R=2∏x (which would cancel out the x's and R's and give me the factor of 2∏ that I want on the right side) then I've got it. Is that true? If so, can you explain why?
  15. Dec 5, 2013 #14
    I do not understand what you are talking about. What R? It is already absent from the equation.

    Do you understand what simple harmonic motion means? What is its general equation?
  16. Dec 5, 2013 #15
    Well, I don't have any equations with v for simple harmonic motion. All I know is F=-kx, which leads to a=(-k/m)x, which gives us ω=√(k/m). I just converted v back to ω, which puts R back into the picture. Is there a simpler way?
  17. Dec 5, 2013 #16
    What happens if you differentiate both sides of the energy equation (#12) with respect to time?
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