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)

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.

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

To find the moment of inertia of the square rod about an axis through the center and perpendicular to the plane of the square, we can apply the parallel-axis theorem.

The parallel-axis theorem states that the moment of inertia of an object about an axis parallel to and at a distance 'd' from an axis through the center of mass is given by the sum of the moment of inertia about the center of mass and the product of the mass and the square of the distance 'd'.

In this case, the rod is bent into a square of side length 'a' and total mass 'M'. We need to find the moment of inertia about an axis through the center and perpendicular to the plane of the square.

First, let's find the moment of inertia of the square rod about an axis through its center of mass. The moment of inertia of a thin rod about an axis passing through its center and perpendicular to its length is given by the formula (1/12) * mass * length^2.

The length of each side of the square is 'a', so the moment of inertia of the square rod about an axis through its center of mass is (1/12) * M * a^2.

Now, let's apply the parallel-axis theorem. The distance between the original axis through the center of mass and the axis through the center of the square is zero since they coincide. Therefore, the moment of inertia about the final axis is simply the moment of inertia about the center of mass.

So, the moment of inertia about the final axis through the center and perpendicular to the plane of the square is (1/12) * M * a^2.

Therefore, the correct answer is I = (1/12) * M * a^2.

To find the moment of inertia about an axis through the center and perpendicular to the plane of the square, we can make use of the parallel-axis theorem. This theorem states that the moment of inertia of a body about any axis parallel to an axis through its center of mass is equal to the sum of the moment of inertia about the axis through its center of mass and the product of its mass and the square of the distance between the two axes.

In this case, let's consider the moment of inertia of the bent rod about an axis passing through its center and perpendicular to the plane of the square. We can divide the bent rod into four equal segments, each with a length of a/4. Each segment can be considered as a point mass located at its center, with a mass m = M/4.

The moment of inertia of each segment about the axis passing through its center is given by the formula for the moment of inertia of a point mass about an axis perpendicular to its motion, which is m * (a/4)^2 / 12 = m * a^2 / 192.

Now, we need to calculate the moment of inertia of each segment about the axis passing through the center of the square. Since the distance between the two axes is a/2, we can apply the parallel-axis theorem to find the moment of inertia of each segment about this axis: I' = m * (a/2)^2 = m * a^2 / 4.

Finally, we sum up the moment of inertia of each segment about the axis passing through the center of the square to get the total moment of inertia of the bent rod: I = 4 * (m * a^2 / 4) = m * a^2.

Substituting m = M/4 into the equation, we get: I = (M/4) * a^2 = 1/4 * M * a^2.

Therefore, the moment of inertia of the thin, uniform rod bent into a square of side length a about an axis through the center and perpendicular to the plane of the square is I = 1/4 * M * a^2, which differs from the answer you provided.