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Period of a uniform plank with a spring

  1. Nov 13, 2006 #1
    Question:

    A uniform plank of mass m is pivoted at one end. A spring of force constant k is attached to the center of the plank, as shown in the figure. The height of the pivot has been adjusted so that the plank will be in equilibrium when it is horizontally oriented.
    [tex]I_{CM rod} = \frac{1}{12}mL^2[/tex]
    Find the period of small oscillation about
    the equilibrium point.

    Comments:

    How do I relate [tex]T = 2 \pi \sqrt{\frac{I}{mgh}}[/tex] to a spring constant? The answers to choose from look like: [tex]T = 2 \pi \sqrt{\frac{m}{k}}[/tex]

    thank you.
     
  2. jcsd
  3. Nov 13, 2006 #2
    Lizzyb: I don't think anyone has looked at this much because the question is a bit hard to understand. Can you include a picture of the system -- maybe with force diagrams? Where you given the following equations, or did you derive them?

    [tex]I_{CM rod} = \frac{1}{12}mL^2[/tex]

    [tex]T = 2 \pi \sqrt{\frac{I}{mgh}}[/tex]

    [tex]T = 2 \pi \sqrt{\frac{m}{k}}[/tex]

    At least -- What does each of them mean to you?

    What kind of things have you figured out about the system so far?:confused:
     
  4. Nov 13, 2006 #3
    Here is the diagram from the question:

    diagram

    The moment of inertia was in the question and I'm familiar with those other equations but how does one relate the moment of inertia with a value of k? it seems like one or the other.

    thanks for your response!
     
  5. Nov 13, 2006 #4
    Okay -- They gave you the moment of inertia for a rod fixed at an end.

    But can you explain how your first equation relating the period of oscillation to the moment of inertia came about?
     
  6. Nov 13, 2006 #5
    oh i don't know - that's one of the answers we can choose from; they all have k in the denomintor under a square root - there are different ratios around the [tex]\sqrt{\frac{m}{k}}[/tex].
     
  7. Nov 13, 2006 #6
    okay -- Next time be really clear about what you know and don't know...

    so I think what you have to do is set up a free body diagram for the system, that finds the forces on the system and the torque... probably when the system experiences a small displacement from equilibrium. This should probably give you some net force that you can relate to an effective spring constant, keff. This effective spring constant will really be a function of things like the spring constant k, the inertia of the rod I, the position of the spring relative to the pivot as function of L, etc. Your result for the period will then be:
    [tex]T=2\pi\sqrt{\frac{m}{k_{eff}}}[/tex]., which will then reduce somehow to having a different ratio in front of the original sqrt(m\k).
     
    Last edited: Nov 13, 2006
  8. Nov 13, 2006 #7

    andrevdh

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    What I think you should aim for is an equation of the form

    [tex]\alpha = -p^2\theta[/tex]

    where [tex]\alpha[/tex] is the angular acceleration of the plank and [tex]\theta[/tex] is its angular displacement (a sign convention is needed to distinguish between clockwise and anticlockwise angular displacements from equilibrium). This is equivalent to the usual

    [tex]a = -\omega ^2 x[/tex]

    from which the period is determined via [tex]\omega[/tex]
     
    Last edited: Nov 13, 2006
  9. Nov 13, 2006 #8

    OlderDan

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    [tex]T = 2 \pi \sqrt{\frac{I}{mgh}}[/tex]
    is the period of a physical pendulum with h being the distance from the pivot point to the center of mass and I being the moment of inertia about the pivot point. In the derivation of that equation, there is a torque tending to restore the pendulum to equilibrium, which in the small angle approximation is

    torque = -mghθ

    In your problem, the torque is provided by the spring. If you follow the derivation for the physical pendulum using the torque from the spring instead of the torgue resulting from a component of gravity, you should find your way to the result. You are given the moment of inertia about the CM. You will need it about the pivot point.

    http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html
     
  10. Nov 14, 2006 #9

    andrevdh

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    So in the derivation one just ignores the torque of the weight since it does not contribute to the restoration of the plank?
     
  11. Nov 14, 2006 #10

    OlderDan

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    The weight and the spring combined produce the equilibrium condition of horizontal plank. The combined force and torque is linear with displacement from equilibrium. This should be directly analogous to hanging a mass on a spring in a uniform gravitational field. Gravity moves the equilibrium position of the system, but the force is still proportional to the displacement from equilibrium, so the motion is still harmonic and independent of the force of gravity.
     
  12. Nov 15, 2006 #11

    andrevdh

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    I tried an unsuccessful attempt (my maths is very rusty) at it along the lines of my suggestion in #7:

    [tex]\Gamma = I \alpha[/tex]

    [tex]\alpha = \frac{1}{I} \Gamma[/tex]

    taking up as positive angular diplacements and considering a small upwards displacement (ignoring the fact that the forces will make slight angles with the vertical) and taking [tex]l[/tex] the length of the plank:

    [tex]\Gamma = -\frac{l}{2}(F_s + w)[/tex]

    for the restoring force of the spring:

    [tex]F_s = ky = \frac{kl}{2}\theta[/tex]

    the problem is therefore with the weight - it does not vary proportional with [tex]\theta[/tex]?
     
  13. Nov 15, 2006 #12

    OlderDan

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    The weight is constant. There is an equilibrium position of the plank (horizontal) where the net torque is zero. Any angular displacement of the plank from the equilibrium results in a net torque that is proportional to that angular displacement. In your torque equation, you would need an initial angle θ' from the position of an uncompressed spring to have zero torque at equilibrium. (The small angle approximation need not apply to finding this angle, but it does not hurt to assume it does.) All that matters is that any deviation from that angle (call it θ, so the total angle related to spring compression is θ' + θ) results in a net torque that is proportional to the deviation angle θ.

    [tex]\Gamma_o = -\frac{l}{2}(\frac{kl}{2}\theta ' + w) = 0[/tex]

    [tex]\Gamma = -\frac{l}{2}(\frac{kl}{2}\left[\theta ' + \theta \right] + w) = 0 -\frac{l}{2} \frac{kl}{2}\theta [/tex]
     
  14. Nov 15, 2006 #13

    andrevdh

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    Quite simple - I would'nt have thought about it.

    Thank you.

    Does this mean that the period of the plank for small amplitudes will be

    [tex]T = \frac{4\pi}{l\sqrt{k}}[/tex]
     
  15. Nov 15, 2006 #14

    OlderDan

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    I don't think so

    [tex]\Gamma = I \alpha = - \frac{kl^2}{4}\theta [/tex]

    [tex] \alpha = - \frac{kl^2}{4 I}\theta [/tex]

    The constant on the right is ω². Evaluate I and the length will divide out to give you a fraction of k/m
     
  16. Nov 16, 2006 #15

    andrevdh

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    Ok. So the moment of inertia of the plank about an axis throught its com parallel to its edge is

    [tex]I_c = \frac{m}{3}\left(\frac{l}{2}\right)^2[/tex]

    using the parallel axis theorem it comes to

    [tex]I_e = \frac{ml^2}{3}[/tex]

    about an axis along it short edge (like a dive board). The period will therefore be

    [tex]T = \frac{4\pi}{l}\sqrt{\frac{I}{k}}[/tex]

    which comes to

    [tex]T = 4\pi\sqrt{\frac{m}{3k}}[/tex]

    which is just slightly longer than the period for a mass suspended from a spring.
     
    Last edited: Nov 16, 2006
  17. Nov 16, 2006 #16

    OlderDan

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    I think that is correct. For a more direct comparison with a mass on a spring I would have left it as

    [tex]T = 2\pi\sqrt{\frac{4m}{3k}}[/tex]

    bit it's all the same.
     
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