Find the period of a small oscillation

  • #1
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Homework Statement


A rod of lenght ##2L## is bent at point of its middle so that the rods now created are in a upside V shape and the angle between them is ##120°##. The system oscilates. Find the expression of the period of oscilation.

Homework Equations


3. The Attempt at a Solution [/B]
I thought of one thing. Since the angle is ##120°## the system oscilates the same way as a normal pendulum rod of lenght ##L/2##. If that is true all that follows is the center of mass which for a rod would be ##L/2## i think. Therefore we have imagined a pendulum of lenght ##L/6##. Its period is ##T=2*\pi * \sqrt{\frac{L}{6g}}##. Is this correct?
 

Answers and Replies

  • #2
gneill
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A pendulum that is not a simple bob on a massless string does not behave in the same way. You need to look at the moment of inertia around the pivot point.

Look up: "Physical Pendulum". The Hyperphysics web site would be a good place to start.
 
  • #3
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Well it seems i need to find the center of mass. But that means that the center of mass for a poll of lenght ##L## is ##L/2##. I can think of the center of mass being located at point ##L/4## below the point ##O##. Is that correct?
 
  • #4
gneill
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Well it seems i need to find the center of mass. But that means that the center of mass for a poll of lenght ##L## is ##L/2##. I can think of the center of mass being located at point ##L/4## below the point ##O##. Is that correct?
Could be. Show how you worked it out.
 
  • #5
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The poll of lenght ##L## has a center of mass located at ##L/2##. ##(L/2)*cos60=L/4## but that gives the period of ##\pi*\sqrt{\frac{L}{g}}##. The solution in the book read ##T=4*\pi* \sqrt{\frac{L}{3g}}## How can this be? That means that the center of mass is at ##4L/3## No idea how...
 
  • #6
gneill
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What did you use for the moment of inertia of the pendulum about its pivot point?
 
  • #7
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What did you use for the moment of inertia of the pendulum about its pivot point?
I didnt. I should have used ##I=mL^2/3## but i dont know how to use it to get the period.
 
  • #8
haruspex
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I didnt. I should have used ##I=mL^2/3## but i dont know how to use it to get the period.
Do you understand how to write down and solve the differential equation for a pendulum? It involves angular acceleration and moment of inertia.
The formula for period is obtained from that.
 
  • #9
gneill
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I didnt. I should have used ##I=mL^2/3## but i dont know how to use it to get the period.
Did you investigate the term "physical pendulum" as I had suggested previously? The period of such a pendulum depends upon the moment of inertia about the pivot point and the distance of the pivot point from the center of mass.

Hint: when you formulate your moment of inertia, remember that your pendulum was originally a single rod of mass m.
 
  • #10
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I see now that the period of a physical pendulum is ##2*\pi*\sqrt{\frac{I}{mgL_{cm}}}##
I know that the moment of inertia of a rod is ##mL^2/3## but that inputed doesnt give correct answer. I need the center of mass. Im not sure what to do
 
  • #11
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I came up with something. The center of mass of these two rods is a rod of lenght ##L/2## staying vertical and oscilating. With above equation and inputing values of ##I=mL^2/3## ##L_{cm}= L/4## i get ##T= 4*\pi*\sqrt{\frac{L}{3g}}## correct?
 
  • #12
haruspex
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I came up with something. The center of mass of these two rods is a rod of lenght ##L/2## staying vertical and oscilating. With above equation and inputing values of ##I=mL^2/3## ##L_{cm}= L/4## i get ##T= 4*\pi*\sqrt{\frac{L}{3g}}## correct?
Looks right to me.
 

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