Update: (2014/03/17) : Based on the comment by the OP, saying the first answer was not what was asking for. Feeling duty bounded, this second attempt tries to address the OP's concern again, which is the outer edges (of each tooth) should be arcs with center at (0,0) like the inner two circles, the first attempt was simply a rounded corner for each rectangle tooth.
Note: The blue and red lines for demonstration can be removed by replacing the \draw with \path in line 24 and 31.

Code:
\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{positioning,intersections,calc}
\begin{document}
\begin{tikzpicture}
\def \rotor {
[name path=curve] (0, 0) circle (2)
(0, 0) circle (1)
}
\startscope[even odd rule]
\foreach \x in {1, ..., 6}
{
\path[rotate=(\x-1)*60] (0,0) -- ++ (-10:3)
(0,0) -- ++ (10:3)
(0,0) -- ++(-10:3) arc (-10:10:3); % generate the arc section of a tooth
}
\draw[fill=lightgray] \rotor;
\foreach \x in {1,...,6}
{ \draw[red,rotate=(\x-1)*60,name path=line \x] (0,0) -- ++ (-20:3); % Adjust this angle -10,-20, ... for all kind of tooth
\path[name intersections={of=curve and line \x, by={isect \x}}];
\draw[rotate=(\x-1)*60,fill=lightgray]
(isect \x) -- (-10:3) arc (-10:10:3) -- (20:2); % Adjust the last coordinate accordingly
}
\foreach \x in {1,...,6}
{ \draw[blue,rotate=(\x-1)*60,name path=line \x] (0,0) -- ++ (20:3); % Adjust this angle 10, 20, ... for all kind of tooth
\path[name intersections={of=curve and line \x, by={isect \x}}];
\draw[rotate=(\x-1)*60,fill=lightgray]
(isect \x) -- (10:3) arc (10:-10:3) --(-20:2); % Adjust the last coordinate accordingly
}
\stopscope
\end{tikzpicture}
\end{dcument}
Two appaoches are studied. The left one is done without using intersection package whilst the second one on the right is with help from intersection to find the insersecion of the tooth bases

Code
\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{positioning,intersections,calc}
\begin{document}
% -- Do not use intersection package
\begin{tikzpicture}
\def \rotor {
(0, 0) circle (1)
(0, 0) circle (2)
}
\startscope[even odd rule]
\foreach \x in {1, ..., 6}
{
\draw[rotate=(\x-1)*60,rounded corners]
(1.5, -0.3) rectangle (3, 0.3);
}
\draw[fill=lightgray] \rotor;
\foreach \x in {1,...,6}
{ %\path[rotate=(\x-1)*60,name path=line \x] (1.5,-0.3) -- (3,-0.3);
%\path[name intersections={of=curve and line \x, by={isect \x}}];
\draw[rotate=(\x-1)*60,fill=lightgray,rounded corners]
(1.97,-0.3) -- (3,-0.3) -- (3, 0.3) -- (1.97,0.3); % knowing that it is cloe dotse to 2, so manually to determine 1.97,
%(isect \x) -- (3,-0.3) -- (3, 0.3) -- ($(isect \x)+(0,0.6)$);
}
\stopscope
\end{tikzpicture}
% -- Use of insetection package
\begin{tikzpicture}
\def \rotor {
(0, 0) circle (1)
[name path=curve] (0, 0) circle (2)
}
\startscope[even odd rule]
\foreach \x in {1, ..., 6}
{
\draw[rotate=(\x-1)*60,rounded corners]
(1.5, -0.3) rectangle (3, 0.3);
}
\draw[fill=lightgray] \rotor;
\foreach \x in {1,...,6}
{ \path[rotate=(\x-1)*60,name path=line \x] (1.5,-0.3) -- (3,-0.3);
\path[name intersections={of=curve and line \x, by={isect \x}}];
\draw[rotate=(\x-1)*60,fill=lightgray,rounded corners]
(isect \x) -- (3,-0.3) -- (3, 0.3) -- ($(isect \x)+(0,0.6)$);
}
\stopscope
\end{tikzpicture}
\end{document}