Physical pendulum thin rods problem

AI Thread Summary
The discussion revolves around calculating the frequency of oscillation for an L-shaped object formed by two identical thin rods. The center of mass is determined to be at a 45° angle between the rods, and the distance from the pivot is calculated as √2/2L. The moment of inertia for each rod is known, but the challenge lies in combining these to find the total moment of inertia for the oscillating object. A key realization occurs that the problem requires the ordinary frequency instead of angular frequency, which resolves the confusion regarding the calculations. The correct frequency of oscillation is ultimately identified as 1/4∏(√(6g/√(2L)).
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Homework Statement


Two identical thin rods, each with mass m and length L, are joined at right angles to form an L shaped object. This object is balanced on top of a sharp edge. If the L shaped object is deflected slightly, it oscillates. Find the frequency of oscillation.
Here is a picture:
http://www.luiseduardo.com.br/undulating/SHM/shmproblems_arquivos/image111.jpg
the correct answer is 1/4∏(√(6g/√(2L)).

Homework Equations


moment of inertia of a slender rod about one end: I = 1/3mL^2
ω=√(mgd/I)

The Attempt at a Solution


I think the center of mass is at 45° between the rods, √2/2L down from the pivot. Let me know if that is wrong. I think that would make d=√2/2L. So my main problem is in finding I. I know I for each rod, but I don't know how to use that to find I for the entire oscillating object.
 
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Moment of inertia is additive. If you know the moments of inertia of two bodies (with respect to one point), then the moment of inertia of the combined body (with respect to the same point) is the sum of the moments.
 
What do I do to take the angle into account?
 
Why would you? Are there any angles in the definition of the moment of inertia?
 
Well, no. But the answer I am getting is just a factor of pi/2 off from the correct answer, but I don't know where that is coming from.
 
What frequency are you supposed to find? Angular ##\omega## or ordinary ##f##?
 
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Oh my goodness, you're right. I've been looking for the angular frequency when this question calls for regular frequency. Thank you so much! I can't believe I wasted so much time on such a silly oversight :P
 

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