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Simple Harmonic Motion of a Bent Rod

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data

    A uniform rod of mass m and length 2L is bent into 2 straight segments about its
    midpoint. The angle between the 2 segments is α. The bent rod is balanced on a sharp
    fulcrum. Find the angular frequency for small oscillations if the rod is displaced from
    equilibrium and released.

    know total mass of rod: m
    length: 2L
    angle between rods is: alpha

    2. Relevant equations

    x(t)=A cos((omega)(t)+(phi))

    3. The attempt at a solution

    I've tried to draw a diagram explaining the motion, and forces involved and I can tell that the "tension" or force that the rod applies at the center of mass of each length of the rod is responsible for the restoring force, however I am not sure how to treat this in order to write newton's second law so that I can solve the 2nd order differential equation for the angular frequency. Help is greatly appreciated!
  2. jcsd
  3. Nov 22, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Gravity supplies the restoring force.
    What torque does gravity exert about the pivot? What's the rotational inertia about the pivot?
  4. Nov 22, 2008 #3
    Aha, another one of these rotational problems. I thought I had escaped them.

    Gravity does indeed supply the restoring force and I fell into the same old physics student trap.

    Approaching this with Newton's second law for rotations, we get the sum of all the external torques= (Moment of Inertia)*(Angular Acceleration).

    I am aware that torque is: R (vector) X Force (vector), but I am still unsure as to how to label the angles. Lets say we displace it a small angle theta, I am unsure how to treat this. Qualitatively it is easy to see that the torque will be greater for the "upper" end of the bent rod when it is displaced, because the angle between the weight and the "lower" part of the rod is approaching parallel, but I am not sure how to notate this in quantitative terms.

    As for the moment of inertia, it is the sum of all of the individual mass particles and their radius from the axis of rotation squared. In this case I believe it would be ML^2/4. Now all I need is a way to represent those torques...

    (I)(d^2 (theta/dt) +??(theta)=0, where the ??'s actually give me the square of the angular velocity.

    Thanks for the quick, and very helpful response!
  5. Nov 22, 2008 #4

    Doc Al

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    Staff: Mentor

    For the purpose of calculating the torque, realize that gravity can be considered to act at the center of mass of the bent rod. Find the center of mass and its distance from the pivot.

    How did you arrive at this value? Hint: What's the moment of inertia of a single rod about one end?
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