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Spring cylinder system question

  1. Nov 16, 2006 #1
    So there's a spring attached to a wall and it's connected to the axle of a cylinder. The cylinder doesn't slip and the axle-spring attachment is frictionless. I'm suppose to solve for the frequency of the motion using newton's second law both translational and rotational form. I set up torque=(moment of inertia)*(angular acceleration) the problem is, the spring force will not exert anyi torque since it has a radius of zero so I'm not sure how to go about doing this problem. Any help would be appreciated! Thank you!
     
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  3. Nov 16, 2006 #2

    OlderDan

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    There is friction between the surface of the cylinder and whatever it is rolling on. It does not slip.
     
  4. Nov 16, 2006 #3
    Mhm, I know that, but we still have to incorporate the spring in some way but it exerts no torque so how do I incorporate the spring force?
     
  5. Nov 16, 2006 #4

    OlderDan

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    The spring force, combined with the surface friction gives you a net force that will accelerate the CM of the cylinder. The surface friction produces a torque about the CM. The no-slipping condition constrains the angular displacement about the CM to be proportional to the linear displacement of the CM.
     
  6. Nov 16, 2006 #5
    So if we make a force equation do we have friction force + spring force= mass*aceleration? Then we make a torque equation with only friction? Should I approach it that way then?
     
  7. Nov 16, 2006 #6

    OlderDan

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    That's it exactly.
     
  8. Nov 16, 2006 #7
    So after the two equations I should relate the accelerations since acceleration = angular accel*radius of cylinder? Now a new problem arises, I'm not sure how to relate these equation to the frequency since frequency is 1/period and period=2pi*r/v...am I suppose to take the integral of the combined equations?
     
  9. Nov 16, 2006 #8

    OlderDan

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    If you can get an equation that is of the form

    F = ma = -k'x
    a = -(k'/m)x

    where F is the net force, x is the displacement of the CM from equlibrium, and k' is some factor related to the spring constant and probably the dimensions of the cylinder, then you know the motion is going to be harmonic motion with frequency related in the usual way to k'/m.
     
  10. Nov 16, 2006 #9
    And if k/m is angular frequency squared, do I just plug it into the period equation to find frequency?
     
  11. Nov 16, 2006 #10

    OlderDan

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    Yes, but the k' is not going to be the spring constant k. It will be some other constant.

    If you are taking calculus you will know why this works. If you assume harmonic motion it follows that x is of the form

    x = A*cos(ωt + φ) where φ is established by the choice of time zero

    v = dx/dt = -A*ω*sin(ωt + φ)

    a = dv/dt = -A*ω²*cos(ωt + φ) = -ω²x

    so ω² = k'/m
     
  12. Nov 16, 2006 #11
    I see, alright so the second part asks us to solve this same idea using conservation laws. So if I used conservation of energy, we'll have to include work which is friction force*amplitude of the spring right? And so I'll have Energy initial+ work= Energy Final where energy initial would be potential energy of spring, rotational and kinetic energy at the initial distance and energy final would also be potential energy of spring, rotational and kinetic energy but the difference is in the displacement or x value of the spring? Then should I solve for velocity and put it into the period equation and ultimately find frequency?
     
  13. Nov 16, 2006 #12

    OlderDan

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    Work is a bit tricky when you have friction. If you rolled the cylinder without slipping down an incline, and wanted to find its velocity, what would you do?
     
  14. Nov 16, 2006 #13
    Oh so we can neglect friction because i believe the equation for rolling down an incline is mgh= translational energy+rotational energy. So we can eliminate work altogether then?
     
  15. Nov 16, 2006 #14

    OlderDan

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    You can eliminate frictional work because there is no slipping. You are including the work done by the spring in the calculation because you are expressing that work in terms of a potential energy. Work is being done on the cylinder by the spring, and then the cylinder does work on the spring. As long as no energy is dissipated, this exchange goes on forever.
     
  16. Nov 17, 2006 #15
    So for initial energy, I will have the potential energy of the spring and that will equal to the energy final which is just the translational and rotational energy of the cylinder correct?
     
  17. Nov 17, 2006 #16

    OlderDan

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    If by final you mean the equilibrium position, then yes.
     
  18. Nov 17, 2006 #17
    Well the problem is, the questions says the cylinder is going back and forth so it doesn't clearly state the initial position and final positions. Since I don't see the point of setting an energy equation at the ends of the motion where potential will end up equaling each other.... but anyway. Thank you so much for your help!
     
  19. Nov 17, 2006 #18

    OlderDan

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    Right. There is no final position. There are maximum displacement positions where the energy is all potential energy, and the equilibrium position where the energy is all kinetic energy. At all positions the sum of the kinetic and potential energies is constant.
     
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