Question

A helicopter has a mass $m$ and maintains its height by imparting a downward momentum to a column of air defined by the slipstream boundary as shown in the figure. The propeller blades can project downward air at speed $v$, where the pressure in the stream below the blades is atmospheric and the radius of the circular cross section of the slip stream is $r$. Neglect any rotational energy of the air, the temperature rise due to air friction and any change in air density $\left(\rho \right)$. Then

• A. $v=\frac{1}{r}\sqrt{\frac{mg}{\pi \rho }}$ for equilibrium
• B. $v=\frac{2}{r}\sqrt{\frac{mg}{\pi \rho }}$ for equilibrium
• C. Power of engine $\frac{mg}{2r}\sqrt{\frac{mg}{\pi \rho }}$ for equilibrium
• D. Power of engine $\frac{mg}{r}\sqrt{\frac{mg}{\pi \rho }}$ for equilibrium

Answer 1

Answer: (A), (C)

The correct options are
A $v=\frac{1}{r}\sqrt{\frac{mg}{\pi \rho }}$ for equilibrium
C Power of engine $\frac{mg}{2r}\sqrt{\frac{mg}{\pi \rho }}$ for equilibrium

Increase in momentum of the air in time $t=\rho Avt\times v=\rho v^{2}At$
Force = rate of change of momentum and the force due to air column must balance weight of the helicopter.
$mg=\frac{\rho v^{2}At}{t}$
$v=\sqrt{\frac{mg}{A\rho }}=\frac{1}{r}\sqrt{\frac{mg}{\pi \rho }}$ (A)

Kinetic energy imparted to the air column $K=\frac{p^2}{2m}=\frac{(\rho v^{2}At{)}^2}{2\rho Avt}=\frac{1}{2}\rho \pi r^2{v}^{3}t$

Power of the engine
$=\frac{K}{t}=\frac{1}{2}\rho \pi r^2{v}^3=\frac{mg}{2r}\sqrt{\frac{mg}{\pi \rho }}$ (C)