# 平行軸定理

$I_{C}\,\!$ 代表剛體對於質心軸的轉動慣量、$M\,\!$ 代表剛體的質量、$d\,\!$ 代表另外一支直軸 z'-軸與質心軸的垂直距離。那麼，對於 z'-軸的轉動慣量是

$I_{z'}=I_{C}+Md^2\,\!$

$I_z = I_x + Ad^2\,\!$

## 進階理論

$\mathbf{I} = \begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz}\end{bmatrix}\,\!$

$I_{xx}\ \stackrel{\mathrm{def}}{=}\ \int\ y^2+z^2\ dm\,\!$
$I_{yy}\ \stackrel{\mathrm{def}}{=}\ \int\ x^2+z^2\ dm\,\!$
$I_{zz}\ \stackrel{\mathrm{def}}{=}\ \int\ x^2+y^2\ dm\,\!$

$I_{xy}=I_{yx}\ \stackrel{\mathrm{def}}{=}\ - \int\ xy\ dm\,\!$
$I_{xz}=I_{zx}\ \stackrel{\mathrm{def}}{=}\ - \int\ xz\ dm\,\!$
$I_{yz}=I_{zy}\ \stackrel{\mathrm{def}}{=}\ - \int\ yz\ dm\,\!$

$I_{xx}=I_{G,xx}+m(\bar{y}^2+\bar{z}^2)\,\!$
$I_{yy}=I_{G,yy}+m(\bar{x}^2+\bar{z}^2)\,\!$
$I_{zz}=I_{G,zz}+m(\bar{x}^2+\bar{y}^2)\,\!$
$I_{xy}=I_{yx}=I_{G,xy} - m\bar{x}\bar{y}\,\!$
$I_{xz}=I_{zx}=I_{G,xz} - m\bar{x}\bar{z}\,\!$
$I_{yz}=I_{zy}=I_{G,yz} - m\bar{y}\bar{z}\,\!$

a) 參考右圖 ，讓 $(x\,',\ y\,',\ z\,')\,\!$$(x,\ y,\ z)\,\!$ 分別為微小質量 $dm\,\!$ 對質心 G 與原點 O 的相對位置：

$y=y\,'+\bar{y}\,\!$$z=z\,'+\bar{z}\,\!$

$I_{G,xx}=\int\ y\,'\,^2+z\,'\,^2\ dm\,\!$
$I_{xx}=\int\ y^2+z^2\ dm\,\!$

\begin{align} I_{xx}&=\int\ (y\,'+\bar{y})^2+(z\,'+\bar{z})^2\ dm\\ &=I_{G,xx}+m(\bar{y}^2+\bar{z}^2)\ . \\ \end{align}\,\!

b) 依照慣性張量的慣性積定義方程式 ，

$I_{G,xy}= - \int\ x\,'y\,'\ dm\,\!$
$I_{xy}= - \int\ xy\ dm\,\!$

\begin{align} I_{xy}&= - \int\ (x\,'+\bar{x})(y\,'+\bar{y})\ dm \\ &=I_{G,xy} - m\bar{x}\bar{y}\ . \\ \end{align}\,\!

## 實例

$I_G =\begin{bmatrix} \frac{1}{12} m (w^2 + h^2) & 0 & 0 \\ 0 & \frac{1}{12} m (h^2 + d^2) & 0 \\ 0 & 0 & \frac{1}{12} m (w^2 + d^2)\end{bmatrix}\,\!$

$I_{xx} =\frac{1}{12} m (w^2 + h^2) +m \left(\left(\frac{w}{2}\right)^2 + \left(\frac{h}{2}\right)^2\right)\,\!$
$I_{yy} =\frac{1}{12} m (h^2 + d^2) +m \left(\left(\frac{h}{2}\right)^2 + \left(\frac{d}{2}\right)^2\right)\,\!$
$I_{zz} =\frac{1}{12} m (w^2 + d^2) +m \left(\left(\frac{w}{2}\right)^2 + \left(\frac{d}{2}\right)^2\right)\,\!$
$I_{xy}= - m\left(\frac{w}{2}\right)\left(\frac{d}{2}\right)= - \frac{mwd}{4} \,\!$
$I_{xz}= - m\left(\frac{h}{2}\right)\left(\frac{d}{2}\right)= - \frac{mhd}{4} \,\!$
$I_{yz}= - m\left(\frac{w}{2}\right)\left(\frac{h}{2}\right)= - \frac{mwh}{4} \,\!$

$I_G =\begin{bmatrix} \frac{1}{3} m (w^2 + h^2) & - \frac{1}{4}mwd & - \frac{1}{4}mhd \\ - \frac{1}{4}mwd & \frac{1}{3} m (h^2 + d^2) & - \frac{1}{4}mwh \\ - \frac{1}{4}mhd & - \frac{1}{4}mwh & \frac{1}{3} m (w^2 + d^2)\end{bmatrix}\,\!$

## 參考文獻

• Beer, Ferdinand; E. Russell Johnston, Jr., William E. Clausen (2004). Vector Mechanics for Engineers. 7th edition. USA: McGraw-Hill, ISBN 0-07-230492-8