Question

Fan of a remote operated model helicopter of mass $m$ has diameter $d$. Denoting density of air by $\rho$ and acceleration of free fall by $g$, deduce expression for the minimum power required for take-off.

Answer 1

$\frac{2(mg)^{3/2}}{d\sqrt{\rho \pi}}$

Answer 2

1. For take-off, the upward lift force must equal the helicopter's weight $mg$.

2. The lift force is generated by pushing air downwards. Using a simplified model where the force exerted by the fan on the air is $F = \rho A v^2$, where $A$ is the fan area and $v$ is the speed of the air pushed downwards.

3. The fan area is $A = \pi (d/2)^2 = \pi d^2/4$.

4. Setting lift equal to weight: $\rho (\pi d^2/4) v_{min}^2 = mg$.

5. Solve for the minimum air speed $v_{min} = \sqrt{\frac{4mg}{\rho \pi d^2}} = \frac{2}{d}\sqrt{\frac{mg}{\rho \pi}}$.

6. The power required to push the air downwards is $P = F v = (\rho A v^2) v = \rho A v^3$.

7. The minimum power is $P_{min} = \rho A v_{min}^3$.

8. Substitute $A$ and $v_{min}$: $P_{min} = \rho (\pi d^2/4) \left(\frac{2}{d}\sqrt{\frac{mg}{\rho \pi}}\right)^3 = \rho \frac{\pi d^2}{4} \frac{8}{d^3} \left(\frac{mg}{\rho \pi}\right)^{3/2} = \frac{2\pi}{d} \frac{(mg)^{3/2}}{(\rho \pi)^{3/2}} = \frac{2\pi}{d} \frac{(mg)^{3/2}}{\rho^{3/2} \pi^{3/2}} = \frac{2}{d} \frac{(mg)^{3/2}}{\rho^{1/2} \pi^{1/2}} = \frac{2(mg)^{3/2}}{d\sqrt{\rho \pi}}$.

Answer: The expression for the minimum power required for take-off is $\frac{2(mg)^{3/2}}{d\sqrt{\rho \pi}}$.