Question

During a rescue operation, a 53oo-kg helicopter hovers about a fixed point. The helicopter blades send air downward with a speed of 62 m/s. What mass of air must pass through the blades every second to produce enough thrust for the helicopter to hover?

Answer 1

Remember that we can express change in momentum, the impulse as the integral of Force with respect to (wrt) time,
\[ \Delta p = I = \int F \ dt \]
That in turn means that we can also express Force as the derivative of p wrt time
\[ F = \frac{dp}{dt} \]
From this we can recover the usual expression, Newton's law, F=ma when mass is constant because
\[ F = \frac{dp}{dt} = \frac{d \ }{dt} (mv) = m \frac{d \ }{dt} v = ma \]

But in the case of the helicopter, what we are measuring is the impulse of the thrust from the air moved by the propeller, call it \( F_p \) and m is what is changing and v is constant, hence
\[ F_p = \frac{d \ }{dt} (mv) = v \frac{dm}{dt} \]

Now, if the aircraft is hovering, the force from the thrust of the propeller must exactly equal the gravitational force, \( F_g = Mg \), where M is the mass of the helicopter.

Hence \( F_g = F_p \), i.e.,
\[ Mg = v \frac{dm}{dt} \]
You now just need to solve for dm/dt.

Answer 2

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Or to put it in simpler terms without derivatives,
Force x time = Impulse = Change in momentum = Change in (mass x velocity):
\[ F \Delta t = I = \Delta p = v \Delta m \]
Hence
\[ F = v \frac{\Delta m}{\Delta t} \]