Rotational Motion of a Square Arrangement of Spheres

[totalmajor]

[SOLVED] Rotational Motion

Homework Statement

Four small spheres, each of which you can regard as a point of mass m = 0.170 kg, are arranged in a square d = 0.250 m on a side and connected by light rods (Fig. 9.27).

(a) Find the moment of inertia of the system about an axis through the center of the square, perpendicular to its plane (an axis through point O in the figure).

(b) Find the moment of inertia of the system about an axis bisecting two opposite sides of the square (an axis along the line AB in the figure).
wrong check mark kg·m2

(c) Find the moment of inertia of the system about an axis that passes through the centers of the upper left and lower right spheres and through point O.
wrong check mark kg·m2

http://irollerblade.org/pics/physics.JPG

Homework Equations

I=MR^2
I=I+MR^2 (Parallel axis theorem)

The Attempt at a Solution

Okay, so Rotational Motion was never my one of my favorite units, I always HATED doing it. I tried several things to get this problem right, but nothing worked!

I know about the center of mass equation too, but that just makes everything = 0!

Originally i tried working out the problem by finding the center of mass on both ends, then finding the moment of inertia through the center using the parallel axis theorem, which obviously didn't work.

Anybody have any suggestions?
Thanks

 

[mer584]

For part A I'm going to throw out an idea. I would try finding R by bisecting the square which would be d rt2. Then I would divide that in 2 to get the radius of each mass from the center O. Then you can treat each ball separately as a mass concentrated at the end of a weightless string from O and add them to find the I for the whole system. I dunno, give it a try maybe.

 

[totalmajor]

Hey I figured out A (thanks)
But B and C , I am a loss for

 

[Battlecruiser]

I think for Part B, you can use one half of D which is .125 as the r value and the m value is just .250. I'm not sure what moment of inertia equation you would use though, so you need to check that, but I don't think it would be MR^{}2 because that's for cylindrical shells.

 

[totalmajor]

Got it!
Thanks a lot!