Moment of inertia, x and y coordinates

[kottur]

Homework Statement

Consider a rigid body that is a thin, plane sheet of arbitrary shape. Take the body to lie in the xy-plane and let the origin O of coordinates be located at any point within or outside the body. Let $I_{x}$ and $I_{y}$ be the moments of inertia about the x- and y-axes, and let $I_{O}$ be the moment of inertia about an axis through O perpendicular to the plane.

1. By considering mass elements $m_{i}$ with coordinates $(x_{i},y_{i})$, show that $I_{x}+I_{y}=I_{O}$. This is called the perpendicular-axis theorem. Note that point O does not have to be the center of mass.

2. For a thin washer with mass M and with inner and outer radii $R_{1}$ and $R_{2}$, use the perpendicular-axis theorem to find the moment of inertia about an axis that is in the plane of the washer and that passes through its center. You may use the information in Table 9.2 in the textbook (which is University physics with modern physics, 13th ed.) Express your answer in terms of the given quantities.

3. Use the perpendicular-axis theorem to show that for a thin, square sheet with mass M and side L, the moment of inertia about any axis in the plane of the sheet that passes through the center of the sheet is $\frac{1}{2}ML^{2}$. You may use the information in Table 9.2 in the textbook.

Homework Equations

$I_{P}=I_{cm}+Md^{2}$

The Attempt at a Solution

The answer to 1. is just the proof of the theorem isn't it? I can use this website: http://hyperphysics.phy-astr.gsu.edu/hbase/perpx.html

The answer in 2. is $I=\frac{1}{4}M(R_{1}^{2}+R_{2}^2)$ because I can see the answers to odd-numbered questions in the textbook but I don't know how to get there.

 

[lippyka]

The answer in 3 is \frac{1}{2}ML^2. I know that for a thin, square sheet the moment of inertia about an axis passing through the center of the sheet will be \frac{1}{2}ML^2 but I don't know how to prove it using the perpendicular-axis theorem.