You are hanging onto the end of a long rope, the other end of which is attached to a Coast Guard helicopter. Model yourself as a particle of mass
M = 52.8 kg
with a diameter equal to 0.540 m. The density of the air is
ρ = 1.29 kg/m3.
Assume the drag coefficient between you and the air is
C = 0.495.
Now, at time
t = 0,
the helicopter begins to accelerate horizontally with an acceleration
a = 2.80 m/s2.
(Assume an initial velocity of 35.5 m/s.)
(a) If we ignore the effect of the air drag force, what is the angle of the rope with the vertical?
°
(b) Suppose in this case where the helicopter accelerates we also consider the drag force due to the air. Derive an expression for the angle made by the rope with the vertical as a function of time. (Do not include units in your answer.)
tan θ =
In order to answer the first part of the question and find the angle of the rope with the vertical, we need to consider the forces acting on the particle. Ignoring air drag force, the only forces acting on the particle are gravity and tension in the rope.
(a) Ignoring air drag force:
The tension in the rope will have two components, one vertical and the other horizontal. The vertical component of tension will balance the weight of the particle, while the horizontal component will provide the necessary centripetal force to keep the particle moving in a circle.
Given:
Mass of particle, M = 52.8 kg
Acceleration, a = 2.80 m/s^2
Initial velocity, V = 35.5 m/s
Diameter of particle, d = 0.540 m
The force balance in the vertical direction can be written as:
T * cosθ = Mg
Where:
T = tension in the rope
θ = angle of the rope with the vertical
Mg = weight of the particle
And the force balance in the horizontal direction can be written as:
T * sinθ = Mac
Where:
Mac = centripetal force = M * V^2 / d
Dividing the two equations, we get:
tanθ = Mac / Mg
Substituting the given values and solving the equation will give us the angle θ.
(b) Including air drag force:
In this case, we need to consider the additional force due to air drag. The particle experiences a drag force given by the equation:
Drag force = (1/2) * ρ * C * A * V^2
Where:
ρ = density of air = 1.29 kg/m^3
C = drag coefficient = 0.495
A = cross-sectional area of the particle = π * (d/2)^2
V = velocity of the particle
The drag force acts opposite to the direction of motion, so we need to include it in our force balance equation in the horizontal direction. This means that our equation for tension in the horizontal direction becomes:
T * sinθ - Drag force = Mac
Dividing this equation by the equation for tension in the vertical direction, we get:
(tanθ - (1/2) * ρ * C * A * V^2 / Mg = Mac / Mg)
Substituting the given values and solving this equation will give us the expression for the angle made by the rope with the vertical as a function of time, tanθ.