# Moment of Inertia of a Wheel

1. Apr 14, 2014

### leeone

Twelve uniform, thin rods of mass and length are welded
together to form a “wheel” as shown in the figure. What is the
moment of inertia of this wheel for rotation around an axis through
its center and perpendicular to the plane of the wheel? The welds
contribute no mass to the wheel.

I understand the contributions from the spoke....but how would I get the contribution from the six rods around the circumference?

I think it can be done by noting I=1/3ML^2 for a rod with the pivot center at the end and I=1/12ML^2 when the pivot center at the middle.

If i try and integrate the from the center along the circumference I have

I=∫M/L (x^2)dx

but how would I get my limits of integration? I would have to integrate through an angle zero to a fixed distance from the center? SO ((3)^1/2)/2Lcos(Θ)dΘ integrate from 0 to (360)/6=60 degrees, correct?

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2. Apr 14, 2014

### TSny

Are you familiar with the parallel axis theorem? If so, can you use it to get I for one of the perimeter rods?

3. Apr 14, 2014

### leeone

That's what my first attempt was......but where would a convenient point be to compute the moment of inertia? If I move along one of the spokes to the outside I feel like I am in the same situation as I was before.

4. Apr 14, 2014

### TSny

The parallel axis theorem states $I = I_{cm} + Md^2$. What does $d$ represent here?

Can you find $d$ for one of the perimeter rods?

5. Apr 14, 2014

### leeone

would d be (L)root(3)/2.....(the vertical distance from the center?)

6. Apr 14, 2014

### TSny

Yes!

7. Apr 14, 2014

### leeone

I figured it out! I =(1/12l^2 +3/4L^2 )6 + (1/3Ml^2(6)=7mL^2

Thanks!