# Moment of inertia?

Hi, i'm investigating a compound pendulum and was wondering if this isn't a stupid idea :tongue2:

I have a meter ruler with 2 flat cylinder 1 kilogramme weights sandwiched at the bottom of it securely fastened with a load of cellotape . The ruler has holes every 5 cm or so, and i found the centre of mass of the ruler is about 21cm.

I was planning to use the $$fn = 2\pi^-^1\sqrt(mgrIo^-^1)$$ and came to realise that i hadn't a clue what Io was, looked it up, and had a go at trying to find something to work it out. I thought if i found the moment of inertia for both the ruler and the weights separately and added them together it'd be ok

So i found quite a few formulas for rigid bodies, one of them looked like a general one but i didn't know how to do the integration, so i found this one for a ruler shaped object and thought hurray!

$$Icm = \frac{1}{12}mr(a^2 + b^2)$$

Then realised it was only for an axis running through the centre of mass, but i was going to have the axis at different spots, but then found this other formula which seemed to sort that out and thought hurray! Since all the holes drilled are along the middle then if i used a hole not running through the centre of mass, it'd be parallel to the axis that could run through the centre of mass right?

$$I = md^2 + Icm$$

And then i was dead tired, and thought it'd just be ok if i used the general

$$I = mr^2$$

For the weights, and so if i added both of those I values together would that give me a okish value for Io or is it all terribly wrong :uhh: ?

Cheers
Richy

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OlderDan
Homework Helper
You have all the ingredients you need. The idea of adding the moments of inertia for the stick and the "flat cylinders" (disks) will work for you. Finding the moments of inertia about axes that are not the center of mass of each object by using

$$I = md^2 + Icm$$

will work for you. You need to apply this for both the disk and the stick.

Doc Al
Mentor
sanitykey said:
I have a meter ruler with 2 flat cylinder 1 kilogramme weights sandwiched at the bottom of it securely fastened with a load of cellotape . The ruler has holes every 5 cm or so, and i found the centre of mass of the ruler is about 21cm.
How is the center of mass of the "meter ruler" at 21cm? (Or is it not a meterstick?) Also: What's the mass of the ruler compared to the 2 kg masses?

I was planning to use the $$fn = 2\pi^-^1\sqrt(mgrIo^-^1)$$ and came to realise that i hadn't a clue what Io was, looked it up, and had a go at trying to find something to work it out. I thought if i found the moment of inertia for both the ruler and the weights separately and added them together it'd be ok
Sure, but it's probably overkill, since the moment of inertia of the ruler is probably small compared to that of the masses.

So i found quite a few formulas for rigid bodies, one of them looked like a general one but i didn't know how to do the integration, so i found this one for a ruler shaped object and thought hurray!

$$Icm = \frac{1}{12}mr(a^2 + b^2)$$
It's good, but even more overkill. If you insist on including the rotational inertia of the ruler, just treat it as a thin rod.

Then realised it was only for an axis running through the centre of mass, but i was going to have the axis at different spots, but then found this other formula which seemed to sort that out and thought hurray! Since all the holes drilled are along the middle then if i used a hole not running through the centre of mass, it'd be parallel to the axis that could run through the centre of mass right?

$$I = md^2 + Icm$$
That's the parallel axis theorem. Good!

And then i was dead tired, and thought it'd just be ok if i used the general

$$I = mr^2$$

For the weights, and so if i added both of those I values together would that give me a okish value for Io or is it all terribly wrong
Sounds good to me.

But this is not much of a compound pendulum, since the mass is concentrated, not distributed. You can probably get away with treating it as a simple pendulum. In fact, do the analysis both ways and compare!

Oh sorry i meant with the weights already attatched the centre of mass appears to be about 21cm from the end that they're put, hopefully it'd be around 50cm if i took the weights off, unless i got a dodgy meterstick. Not sure what the meterstick mass is yet, i'll find out soon though! It's probably about 0.4 kilogrammes i reckon. Thanks a load for all the feedback Olderdan and Doc Al this is really helpful! I'll try comparing the analysis both ways too as you suggested, sounds like a good way to show improvement and maybe get a few more marks

Also just noticed another mistake they're not 1 killogramme they're 0.1 killogramme i guess that might change the situation more than i expected (only a 1800 gramme difference...doh!), i know it'd make the mass more distributed, do you reckon everything would still work alright?

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