Inertia

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A thin, uniform rod is bent into a square of side length a.
If the total mass is M , find the moment of inertia about an axis through the center and perpendicular to the plane of the square. Use the parallel-axis theorem.

Thanks for the brief explanation before,I got this as the answer but am unsure, so can someone please review it. Thanks, it would be appreciated.

I=1/12 M (2a^2)

• Inertia -

The moment of Inertia about the center of a thin rod of length L is ..

I=1/12 M L^2

Here the length is a/4, mass is M/4
I=1/12 (M/4)(a^2/16)

or you can calculate that.

Now, the Parallel axis theorm, the displacement is 1/2 a side, or a/8
The new I is
I=Icm + Mass*d^2
= above I for one rod + M/4*a^2/64
Then, you have four sides, so multiply it by 4

I=4(1/12 (M/4)(a^2/16))+4(Ma^2/4*64)

=1/12 * M a^2/16+ M a^2/64
= Ma^2 * 1/64 ( 4/3) which is nowhere near what you have. So check it.

I am having a great debate on this question and your others, wheather you are answer grazing, or just lost. I will assume for now lost, so I recommend this helpful book: Schaum's Outline Series, College Physics (or College Physics for Scientists and Engineers), both very inexpensive. So run down to the college bookstore, or Barnes Noble, and examine them. Many, many worked in detail physics problems in easy to read and understand format.

Check my work, I did it quickly typing. THere may well be an error.

• Inertia -

okay no i was lost but thank you i will go check it out.