- #1
FBayer
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So, I got a challenge question from my physics teacher but despite attacking it from different angles I’m not quite sure I’ve gotten the correct solution, in fact most of it was just shady assumptions which I’m not sure were supposed to be made, so I’d like to ask whether this is anywhere near correct and, if it isn’t, be provided with some pointers as to how to do this.
The question reads:
A helicopter hovers in still air well above the ground. If the helicopter has a mass of 4000 kg, and rotor blades of diameter 10 m, what is the velocity of air projected downwards by the blades? (g = 9.8 ms-1, density of air = 1.3 kgm-3)
So, first thing I modeled the hovering helicopter as a cylinder. With blade diameter 10 (= radius of 5), the covered area is 25π, or roughly 78.5 m².
Now, we know that F = ṁv where ṁ is flow rate and v is velocity, as proven by the units: kgs-1 × ms-1 = kgms-2 = N.
The flow rate can be described as m/t, and as m = ρV where ρ is density and V is volume, flow rate ṁ = ρV/t.
We can substitute this into our earlier equation while rearranging, such that:
[itex]\vec{v} = \frac{\vec{F}}{\dot{m}} = \frac{\vec{F}}{\frac{\rho V}{t}} = \frac{\vec{F}t}{\rho V}[/itex]
And here’s where it gets shady. I don’t know how to get a volume if the height isn’t given, and the same deal goes for time, so I just assume they're all 1.
So, 4000 kg × 9.8 ms-2 = 39200 kgms-2, multiplied by 1 second is 39200 kgms-1.
Area, ~78.5m², multiplied by 1 m = 78.5m³.
Now:
[itex]\frac{39200\: kg\, m\, s^{-1}}{78.5\: m^{3} \times 1.3\: kg\, m^{-3}} = \frac{39200\: kg\, m\, s^{-1}}{102.05\: kg} = 384\: ms^{-1}[/itex]
However, I made the assumptions that I can just use the area for the volume and the force for what is essentially momentum/impulse (Ft), which leads me to believe I probably did it wrong...so, how's it supposed to be done?
The question reads:
A helicopter hovers in still air well above the ground. If the helicopter has a mass of 4000 kg, and rotor blades of diameter 10 m, what is the velocity of air projected downwards by the blades? (g = 9.8 ms-1, density of air = 1.3 kgm-3)
So, first thing I modeled the hovering helicopter as a cylinder. With blade diameter 10 (= radius of 5), the covered area is 25π, or roughly 78.5 m².
Now, we know that F = ṁv where ṁ is flow rate and v is velocity, as proven by the units: kgs-1 × ms-1 = kgms-2 = N.
The flow rate can be described as m/t, and as m = ρV where ρ is density and V is volume, flow rate ṁ = ρV/t.
We can substitute this into our earlier equation while rearranging, such that:
[itex]\vec{v} = \frac{\vec{F}}{\dot{m}} = \frac{\vec{F}}{\frac{\rho V}{t}} = \frac{\vec{F}t}{\rho V}[/itex]
And here’s where it gets shady. I don’t know how to get a volume if the height isn’t given, and the same deal goes for time, so I just assume they're all 1.
So, 4000 kg × 9.8 ms-2 = 39200 kgms-2, multiplied by 1 second is 39200 kgms-1.
Area, ~78.5m², multiplied by 1 m = 78.5m³.
Now:
[itex]\frac{39200\: kg\, m\, s^{-1}}{78.5\: m^{3} \times 1.3\: kg\, m^{-3}} = \frac{39200\: kg\, m\, s^{-1}}{102.05\: kg} = 384\: ms^{-1}[/itex]
However, I made the assumptions that I can just use the area for the volume and the force for what is essentially momentum/impulse (Ft), which leads me to believe I probably did it wrong...so, how's it supposed to be done?