# Proof of parallel axis theorem.

1. Nov 10, 2013

### AakashPandita

Everybody says that the distance ,between the two axis, used in the formula, is perpendicular. But in the proof it was a hypotenuse. It was not perpendicular.

2. Nov 10, 2013

### SteamKing

Staff Emeritus
Be specific about which proof. Provide details.

3. Nov 10, 2013

### AakashPandita

4. Nov 10, 2013

### SteamKing

Staff Emeritus
The original axes and the transformed axes are parallel to one another. The theorem covers a displacement not only in the x direction, but the y direction as well. I don't see what your question is about.

5. Nov 10, 2013

### AakashPandita

$$I=\int r^2 dm = \int [(x-a)^2 + (y-b)^2]dm$$

$$I= \int (x^2 + y^2) dm - 2a \int x dm - 2b \int y dm + \int (a^2 + b^2) dm$$

$$I = Icom + Mh^2$$

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