Calculating the moment of inertia in bifilar pendulums

I'm an A2 student, and I've been looking into how experimental and theoretical determined mass moments of inertia differ.

I came across a method (search Youtube for Measuring Mass Moment of Inertia - Brain Waves.avi) that uses the rotational time period of a bifilar pendulum using a rod to calculate through an equation of mass moment of inertia of a rod.

The equation given is $$I=\frac{MgT^2 b^2}{4\pi^2 L}.$$

Where

• $I$=inertia
• $M$=mass of rod
• $T$=rotational time period for one rotating oscillation
• $b$=length between where string attaches to rod and centre of gravity
• $L$=length of string that suspends rod

I've tried looking around on the internet, but I can't find out where the equation is derived from, and I've seen others that have 16 instead of 4 for the $\pi^2$. I've also seen something to do with Lagrangian mechanics?

• – Kyle Kanos Nov 18 '14 at 17:09
• Per my recent experience with Bifilar pendulums the period of oscillation depends on the initial conditions (amplitude, or speed through rest condition). Multiple measurements had to be taken in order to fit a parabola to find the period at zero amplitude. – ja72 Sep 22 '15 at 2:07

Here are some slides I put together. They may not be completely rigorous but I believe they may help you understand a simplified derivation:   • I have no problem with these slides. But please at least cite the source you have taken the slides from; even if it it is the case that you created those, please mention it explicitly. – user36790 Oct 28 '16 at 3:44

normally the whole length is take as $b$, but in the video it is $2b$ hence if you substitute it for $b 16 \pi^2$ formula you get $4b^2/16 \pi^2$ ... hence he got $4\pi^2$

protected by Qmechanic♦Oct 27 '16 at 21:36

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