its question 11 part dii) Sorry that you have to follow this through via a link! It might ask you to login BUT just log in as a guest and it will work!
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A2 pastpaper helicopter question watch
- Thread Starter
- 20-01-2014 20:34
- 24-01-2014 04:11
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- 24-01-2014 22:43
a) If the helicopter is at a fixed height, then there is no resultant force because it's velocity is constant. Zero meters per second, in this case. The lift force must therefore be equal in magnitude and opposite in direction to the force downwards due to gravity.
F=mg = 5300 x 9.8 = 51940 N
b) Rate of change in momentum is the same as force. You may have seen the equation F=(mv-mu)/t where v means the final velocity and u the initial. The rate of change of momentum of the air providing the lift force upwards must be equal and opposite to force on the helicopter due to gravity.
F=(mv-mu)/t = -mg = --5.2x10^4 N = 5.2x10^4 N.
mg in this case is negative, because force is a vector quantity. The question states the force in the upward direction is positive, so force down (mg) must be negative.
c)i) m = rho x volume.
volume = area swept out by blades (pi x r^2) x vertical length of air that has passed the rotor blades, h.
h = velocity of air x time elapsed
time elapsed is one second, h x 1 = v therefore h = v
m = rho x pi x r^2 x v
That's all for now.