#### Question

Consider a rigid plane body or "lamina," such as a flat piece of sheet metal, rotating about a point O in the body. If we choose axes so that the lamina lies in the xy plane, which elements of the inertia tensor I are automatically zero? Prove that

$I_{zz} = I_{xx} + I_{yy}.$

#### Solution

Verified#### Step 1

1 of 2We are considering a rigid plane body or “lamina,” such as a ﬂat piece of sheet metal that lies in xy plane.

Density of the metal sheet can be written as:

$\rho(x,y,z)=\sigma(x,y)\delta(z)$

If we integrate term above, we can get full mass of the sheet:

$\int{\sigma(x,y)\delta(z) dxdydz}=\int{\sigma(x,y)dxdy}\int{\delta(z)dz}=\int{\sigma(x,y)dxdy}=\mu$

Diagonal inertia tensors are given by:

$I_{xx}=\int{\sigma(x,y) (y^2+z^2)\delta(z)dxdydz}=\int{\sigma(x,y)\cdot y^2dxdy}$

$I_{yy}=\int{\sigma(x,y) (x^2+z^2)\delta(z)dxdydz}=\int{\sigma(x,y)\cdot x^2dxdy}$

$I_{zz}=\int{\sigma(x,y) (x^2+y^2)\delta(z)dxdydz}=\int{\sigma(x,y)\cdot (x^2+y^2)dxdy}$

We can see that :

$\boxed{I_{zz}=I_{xx}+I_{yy}}$

The only non diagonal element that is not equal to zero is:

$I_{xy}=-\int{\sigma(x,y) x\cdot y\delta(z)dxdydz}=-\int{\sigma(x,y)\cdot xy\cdot dxdy}$

All other non diagonal elements are equal to zero:

$\boxed{I_{xz}=I_{yz}=0}$