I've been tasked with researching ways to find the gravitational acceleration g and its uncertainty together with the moment of inertia and and the moment of inertia's uncertainty. I have figured out everything with the equations provided except how to calculate the moment of inertia for the lamina pendulum. I will be using period and a length from the center of mass for my readings. The equation for period will be linearised. I just want to know how to calculate the moment of inertia about the center of mass.


2 Answers 2


There is no need to calculate the moment of inertia I, which could be quite difficult for an irregular lamina. You can find both I and g from the same experiment.

The period T of small amplitude oscillations of a compound pendulum is give by (see Source below) :
$(\frac{T}{2\pi})^2 = \frac{k^2+h^2}{gh}$
where k is the radius of gyration (related to moment of inertia through the centre of gravity) and h is the distance of the pivot from the CG.

This equation can be re-arranged as :
$h^2 = gh(\frac{T}{2\pi})^2 - k^2$.
So if you plot values of $h^2$ against $h(\frac{T}{2\pi})^2$ you will get a straight line $y=ax+b$ with slope $a=g$ and y intercept $b=-k^2$, from which you can find the moment of inertia about the CG :
$I = mk^2$
where m is the mass of the lamina.

So the procedure is :
1. Suspend the lamina freely from 3 corners in turn (each as far from the estimated CG as possible), marking a plumb line for each corner used, and identify where they cross (CG). Only 2 lines are needed, but the 3rd is a check on the accuracy of the other two. The lines should cross at about 60-90 degrees.
2. Time small amplitude oscillations when the lamina is suspended from each of the 3 corners; also when it is suspended from other points at various distances h from the CG. Make sure the lamina swings freely each time.
3. Carefully measure the distances h of each pivot from the CG and record alongside the corresponding values for T.
4. Plot corresponding values of $h^2$ against $h(\frac{T}{2\pi})^2$. Find the slope and intercept.

If you plot the data in a curve-fitting package and select a straight line fit, the program will give you not only the slope and intercept but also the error estimates for these, which you can use to get your errors in g and I.

If you don't have access to a curve-fitting package, search online for details about linear regression (statistics). You can do your own calculations in a spreadsheet - more accurate than making measurements from a graph.



Well, this will certainly depend of the shape and mass distribution of your pendulum. For instance, for a pendulum made of a rod and a thin cylindrical disk, the moment of inertia of the rod (about its own symmetry axis) would be $\frac{1}{12}M_rL^2$, and the moment of inertia of the disk $\frac{1}{2}M_sR^2$. If you have a solid sphere instead of the disk, then it'll be $\frac{2}{5}M_dR^2$, etc. You can easily find tables of different moments of inertia just by googling.

In any case, don't forget to apply Steiner's theorem when necessary!

  • $\begingroup$ Thank you, the problem is that this is a lab exam, I have been told that we will be given an irregular lamina. I've seen the apparatus and it was a triangular board, but they said this may change. I was also told to find the center of mass, which I know how to do, will finding that be beneficial to calculate the moment of inertia? $\endgroup$ May 18, 2016 at 9:17
  • $\begingroup$ I will have to use that theorem but to use it I still need to know I(cm) $\endgroup$ May 18, 2016 at 9:18
  • $\begingroup$ @TeyashArjun Oh, sorry, I misread your question. I suppose that if you know the mass of the lamina, you can still find a rough estimate of its density, then calculate the inertia moment by integrating over its area? Without any more precise information about the mass distribution, I'm afraid we can't do much more. $\endgroup$
    – dahemar
    May 18, 2016 at 9:32
  • $\begingroup$ The told me I will be given the mass, how will I then find the moment of inertia. I recall the integration formula but I've only ever done written problems with it, never anything practical, is it possible for you to explain the procedure I would use if I were given the mass. $\endgroup$ May 18, 2016 at 9:37
  • $\begingroup$ So what I will have is the center of mass position and the mass of the lamina $\endgroup$ May 18, 2016 at 9:37

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