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Homework Statement
This is not my homework question, I was asked to help with it, but I've been out of the engineering field for many years now. Here's the question:
Starting from rest, a 2500 kg helicopter accelerates straight up at a constant 1.7 m/s2. What is the helicopter's height at the moment its blades are providing an upward force of 29 kN? The helicopter can be modeled as a 2.6-m -diameter sphere, and air resistance is not negligible.
Homework Equations
Gravity:
F_g = m g
Drag:
C_d = F_d/(1/2 ρ u^2 A) |
F_d | drag force
C_d | drag coefficient
ρ | mass density
u | characteristic speed
A | frontal area
I looked up the drag coefficient (https://is.gd/qpe40K) of a sphere, and found two figures:
C_d = 0.1 (laminar)
C_d = 0.45 (turbulent)
So, I chose the turbulent figure (0.45).
The Attempt at a Solution
We know that the two forces will add up to 29 kN at a certain point, so:
F_g + F_d = 29 kN
F_d = 29 kN - F_g
= 29 kN - m g
= 29 kN - 2500 kg * 9.8 m/s^2
= 4.5 kN
The standard density of air at sea level is:
ρ = 1.204 kg/m^3
The frontal area of a sphere is:
A = pi * r^2
= pi * (2.6 m / 2)^2
= 5.309 m^2
So we have everything except the vertical speed of the helicopter, which we can find by rearranging the drag coefficient formula, and we get:
C_d = F_d/(1/2 ρ u^2 A)
u^2 = F_d/(1/2 ρ C_d A)
u = sqrt(F_d/(1/2 ρ C_d A))
= sqrt(4.5 kN/(0.5 * 1.204 kg/m^3 * 0.45 * 5.309 m^2))
= 56 m/s
Now to find the final height of the helicopter, we have to use and rearrange the acceleration-velocity-distance formula:
v_f^2 = v_i^2 + 2 a d |
v_f | final speed
a | acceleration
v_i | initial speed
d | distance
v_i = 0
v_f = u = 56 m/s
a = 1.7 m/s^2
d = h = ?
h = (v_f^2 - v_i^2)/ 2a
= (56 m/s)^2 / 2 (1.7 m/s^2)
Therefore,
h = 883 m
Were my procedures okay? Any critiques?