Posted by **Mary** on Monday, November 3, 2008 at 1:40pm.

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.

Express your answer in terms of the variables M and a.

PLEASE HELP WITH A THOROUGH EXPLANATION PLEASE!!!

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 -
**bobpursley**, Monday, November 3, 2008 at 2:43pm
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

so you add them.

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 -
**Mary**, Monday, November 3, 2008 at 4:08pm
okay no i was lost but thank you i will go check it out.

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