Base on your description, is tis the system you are talking about?
First thing: Having just a single blade is just a waste, I recommend having a few more blades on it (more "cutting" per revolutions of the disc, or more "cutting" in the same time frame)
Now, you need to find the first thing: cutting force, for example, based on experience, experiments or simply a value out of nowhere (it is acceptable for freshman projects, as long as it "feels right"). For example, you know that you can cut a slash of beef with a force of 100N. Due to your design, the blade is 0.1m from the center of the rotation (or from the centre of the shaft). Therefore, the torque is
Converting 30 RPM into rad per second (basic unit), we have
$\omega = \pi rad/s$, or approximatly 3.14159 rad/s
The power is equal to the product of torque and angular velocity, or
$P=\omega*T = 3.14159 * 10 = 31.4159W$
This is the power output of the motor (with the gearbox included). The mechanical efficiency of the gearbox should be high, say 95%, which means the actual output of the motor (before the gearbox) is 33.069W, or just 33W because I hate long decimals. Continue to extrapolate the input of the motor, again we assume a 95% efficiency, leads to a 34.8W motor.
Now you have it, a motor of 33W is "just enough" to cut the meat as you want. Now, you have to look for the technical data online, seeing which is the best motor for you (power is the closest match, range, size, weight...). You check for its speed limit, say it delivers 35W at 1600RPM. Now, you want to have an actual output of 30RPM (after gearbox). Therefore, the ratio must be
So, for a summary, with a crapload of assumption, you need a ~35W motor and a gearbox with 53.33 ratio.
Note, a 53.33 decreasing ratio means that gonna be huge! Still, I make a lot of assumption here, which means "making an ass out of u and me".
The main idea is just pure Math (secondary school math), so it should not be too hard... Hope it helps