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November 24, 2014

November 24, 2014

Posted by **Mary** on Saturday, October 18, 2008 at 10:49am.

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.

- physics -
**bobpursley**, Saturday, October 18, 2008 at 10:59amYou have four rod segments, in which you can use the parallel axis theorem.

I will be happy to check your work.

- physics -
**drwls**, Saturday, October 18, 2008 at 11:12amThe side length will be a/4, and the center of a side will be dispalced a/8 from the center of the square. The moment of inertia, I, will be four times the value for one of the sides. For that value, you need to use the parallel axis theorem.

The moment of inertia of a single side rotated about its center of mass is

Icm = (1/12)*(mass of side)*(length of side)^2

= (1/12)(M/4)(a/4)^2 = (Ma^2/768

When rotated about the center of the square, you must add

(M/4)(a/8)^2 = Ma^2/256

Add those two together to get

Ma^2/192, and multiply by 4.

I get (1/48)M a^2

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