OK...we have a little bit of confusion still going here. Graywolf, the moment of inertia is more or less as you have described it except for your specific case. In fact, a wheel is a combination of the characteristics of a hollow cylinder, and a series of rods rotated about their midpoints, or if you prefer, something partway between a hollow cylinder and a solid disk. In any case, I=(1/2)mr^2 is not an accurate depiction of what happens with a car's wheel, as that is the solid cylinder case.
If one really wanted to calculate the moment of inertia of a car's wheel, it would be a sum of the different parts, i.e. for a simple 6 spoke wheel with solid spokes and a solid rim, the moment of inertia would be something like:
I = I(spokes) + I(rim) = 6*(1/3)ML^2 + mr^2 where L is the length of a single spoke (from the hub out to the rim of the wheel), M is the mass of a single spoke, r is the radius of the cylindrical part of the wheel, and m is the mass of the cylindrical bit. Then of course you have to add the tire onto that, and that comes to another term of the form, mr^2 but with different masses and different radii. Of course this is still quite inaccurate, as any wheel you might find on a car is a much more complicated shape than this simple depiction. If you really wanted to calculate the moment of inertia from a theoretical standpoint, you'd have to consider the hub, the bearings, the driveshaft, the reciprocating masses of the engine, the brake disk and so on and so forth.
As for torque being simply inertia times the angular accel, that's also not entirely accurate. The sum of all torques present is the product of the angular acceleration and the moment of inertia. Thus, for the case of a wheel on the road, you have to account for the torque applied by the coefficient of static friction, as well as that of the brakes, driveshafts, reciprocating masses of the engine etc...etc...etc...
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