# Rotational Motion Problem - 9

1. Dec 16, 2013

### coldblood

Hi friends,

The problem is as:

https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-ash3/q71/s720x720/1467427_1461727597387679_1141225220_n.jpg

Attempt -

https://fbcdn-sphotos-b-a.akamaihd.net/hphotos-ak-ash3/q71/s720x720/1472752_1461727710721001_1917071548_n.jpg

2. Dec 16, 2013

### haruspex

You treated the rod furthest from the axis as a point mass at its centre.
Also, I think you have confused the mass per rod with the total mass.

3. Dec 17, 2013

### coldblood

So what will with the farthest rod Should I take its moment of inertia as ml2/12(about center) + m(l√3/2)2

If I do so,

The total M.I. is found to be (3/2) ml2

Which gives radius of giration as l√(3/2)

4. Dec 17, 2013

### haruspex

Yes.
That's the other error I mentioned. If each rod has mass m then the entire structure has mass 3m. If its radius of gyration is k then its moment of inertia will be 3mk2.

5. Dec 17, 2013

### coldblood

Well, I did a mistake in reading the question also. I thought that I have to calculate the moment of inertia of the system about the axis parallel to the plane of the structure.

If I have to find the M.I. of the system about an axis parallel, So what would the M.I of the farthest rod?
In that case should I treat it as a point mass situated at the center of the rod?

6. Dec 17, 2013

Yes.