Chinook Helicopter Torque The Chinook Helicopter has 2 rotors to counteract the torque generated by spinning the blade. 
Theoretically, could you use a smaller "back" rotor that is farther away from the main rotor to achieve the same result, ie no twisting?
 A: The separation between the rotors does not actually matter. What matters is that the torque exerted by each of the motors on the respective rotor be the same and in opposite directions. Those torques then add vectorially and cancel out to give zero net torque on the helicopter.
It's not clear to me exactly what you mean by "small". It is indeed possible to have a rotor with shorter blades, but it will need to be driven faster to provide the same torque. This will typically be associated with a smaller power consumption and less lift coming from that rotor.
In general, though, it is the symmetry between the rotors which makes the configuration work. While in practice they won't be exactly the same, they will be very similar to each other, and straying far from this configuration will be very impractical unless you have some stringent requirements on your 'secondary' motor, as well as radical new engines for it to match.
A: Assuming a mass moment of inertia of $I_1$ for the main rotor and $I_2$ for the secondary rotor, and a coefficient of drag of $\beta_1$ and $\beta_2$ respectively then the torque on the rotor shafts are
$$ T_1 = I_1 \dot \Omega + \beta_1 \Omega^2 
\\ T_2 = I_2 (\gamma \dot \Omega) + \beta_2 (\gamma \Omega)^2 $$
where $\Omega$ is the main rotor speed, and $\gamma$ the gearing ratio for the small rotor.
To make those equal you need gearing of $$\gamma = \frac{I_1}{I_2}$$ and coefficient of drag $$\beta_2 = \beta_1 \frac{I_2^2}{I_1^2}$$
Since $I_1 > I_2$ this requires the drag to be $ \beta_2 \gg \beta_1 $ which is a) hard to do with a small rotor, and b) very inefficient.
FYI - The total torque on the motor is going to be 
$$ T_E = \left( I_1 +  I_2 \gamma^2\right) \dot \Omega + \left(\beta_1 + \beta_2 \gamma^3\right) \Omega 
\\ = \left(I_1 + \frac{I_1^2}{I_2} \right) \dot \Omega + \left( 1 + \frac{I_1}{I_2} \right) \beta_1 \Omega^2 $$
A: The interchangeability of rotors for tourque might be nearly impossible because of the ma in f=ma.The rotor speeds would have to continuously vary ideally.Now if it were just a question of lift it would be in aronautical engineering. But then again as they say ,nothing is really impossible,you just have to find a way of doing it.
