If the helicopter's main rotor has a rad us of
\[r\]
and rotates \[n\]
times per second, the total area swept out in 1 second is \[\pi r^2 n\]
. If the air has downwards speed \[v\]
then volume of air swept down per second is \[\pi r^2 nv\]
.The mass of air swept down per second is
\[\pi r^2 n \rho v \]
.The momentum of air swept down per second is
\[\pi r^2 n \rho v^2 \]
.The relationship between the speed of the rotor and the speed of the displaced air will depend on the angle the rotor makes with the horizontal. If the speeds are equal, then
\[v= 2 \pi r n\]
and the momentum of the air swept downwards becomes \[\pi r^2 n \rho (2 \pi r n)^2 =4 \pi^3 n^3 \rho r^4\]
.By Newton;s Second Law of Motion the change in momentum per second is equal to the force exerted. Hence if
\[M\]
is the mass of the helicopter, the force of gravity is \[Mg\]
Equating the force of gravity on the helicopter with the change in momentum per second of the displaced air gives
\[Mg= \pi r^2 n \rho (2 \pi r n)^2 =4 \pi^3 n^3 \rho r^4 \rightarrow n= ( \frac{Mg}{4 \pi^3 r^4 \rho})^{1/3}\]