Moment of Inertia - parallel axis theorem

[ph123]

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.


According to Parallel Axis theorem:

I = I(cm) + Md^2

The distance across the diagonal of the square (corner to corner) is 1.4142...a (= to sqrt(2a^2)). The distance bisecting the square through the center of mass is a/2.

I = m(a/2)^2 + m(0.7071...a)^2
I = (1/4)ma^2 + (1/2)ma^2
I = (3/4)ma^2

This answer isn't right. I think I need to integrate since every point on the square is a different distance from the rotational axis through the center of mass. But there is not other axis specified in the problem, so I'm not sure as to which parallel axis I need to derive an expression for. Any ideas?

 

[tmplaya3]

You are off by a multiplicative factor but I am not sure what it is...

 

[qspeechc]

You want to find the moment of inertia about an axis through the centre, the calculation you did says you know what this is! The geometrical centre of a square is its centre of mass.

Do you have to use the parallel axis theorem? If not, I would consider each side of the square individually. Find the inertia about the centre of mass due to one side by integrating. Find the distance of a general point on the length as a function, then integrate. The other 4 sides should be the same by symmetry.

 

[highcoughdrop]

Alright. So, first of all, you should know that the moment of inertia of a rectangular plate, axis through the center of the plate is:

I = (1/12)*M*(a^2+b^2)


1. Simply, for a square it would just be (a^2+a^2) or (2a^2)
2. Then, we find moment of inertia from the parallel-axis theorem.
3. Finally, take the moment of inertia from the parallel-axis theorem and subtract.
(so, parallel-axis - moment of inertia)
4. That's it, that should be the answer.

 

[alphysicist]

Hi highcoughdrop,

Alright. So, first of all, you should know that the moment of inertia of a rectangular plate, axis through the center of the plate is:

I = (1/12)*M*(a^2+b^2)


1. Simply, for a square it would just be (a^2+a^2) or (2a^2)
2. Then, we find moment of inertia from the parallel-axis theorem.
3. Finally, take the moment of inertia from the parallel-axis theorem and subtract.
(so, parallel-axis - moment of inertia)
4. That's it, that should be the answer.

I don't think this is right. This is not a rectangular plate, it is a set of 4 thin rods. So in your formula we would have to set b=0 for each rod. Then use the parallel axis theorem for each rod and finally add the four results together.

If you believe your procedure gives the right answer anyways, please post some more details. I don't think I understand what you are doing in steps 2 and 3 for this problem.

 

[inky]

Actually all your answers are wrong.
Mass of each rod =M/4
moment of inertia for each rod through its centre =(1/12)(M/4)(a^2)=(M/48)a^2
Use parallel axis theorem for each rod
I=(M/48)a^2 + (M/4)(a/2)^2=(1/12)Ma^2
For 4rod(the whole system)
I=(4)(1/12)Ma^2=(1/3)Ma^2

I taught mant students this type of problems