The aerodynamic drag acting of the carbon-fibre component in the free fall (for example).
This is just a perspective, it applies more to the falling helicopter
In the context of F1 – Imagine that the new component in the vertical wind tunnel has to tested.
The component is released from the rest so that the engineers can study how quadratic air resistance affects the motion under the influence of the gravity.
Because the F1 aero components will experience drag forces that the scale with the square of the velocity ( just like an F1 car the high speeds)
The mathematical description:
m dv/dt = mg – kv^2
The substitution used
mg = ka^2
upon solving we get
v (x) = a* sqrt( 1 – e^(2kx/m))
v(x) is the speed of the falling aero component, analogous to how an F1 car’s speed is limited by the drag on a straight.
a = sqrt(mg/k) plays a role of terminal velocity, similar to how an F1 car reaches a maximum top speed on a long straight when engine power = aerodynamic drag power.
The exponential term shows how the component rapidly accelerates at first, but the drag force ramps up quadratically, limiting the eventual speed.
The exponential term shows how the component rapidly accelerates at first, but the drag force ramps up quadratically, limiting the eventual speed.
This explanation mirrors how the Formula One car accelerates with huge acceleration at low speeds and gradually diminishing the gains as drag rises and when hitting the top speed it plateaus due to air resistance.
This particular example might not be used in the Formula One in the terms of the dynamics of the car but it could be helpful to mimic the scenarios during the testing for validating through the verified fundamental physics of the free falling body.
_ Just for the implementation of the one of the questions of the Engineering Mathematics for learning through in the perspective of Formula One car _
Find the code below ( simulated on the Spyder environment)
# Example 12.12 - falling body with quadratic air resistance
# Equation : m* dv/dx = m*g -k*v^2 where mg = k*a^2
import sympy as sp #symbolic mathematics
import numpy as np
import matplotlib.pyplot as plt
# defining symbols
x,v,m,g,k,a = sp.symbols('x,v,m,g,k,a', postive = True)
# Given mg = k*a^2
relation = sp.Eq(m*g, k*a**2)
# Differential equation m*v*dv/dx = m*g - k*v^2
# => v/(a^2 - v^2)dv = (k/m) dx
eq = sp.Eq(v/ (a**2 - v**2), (k/m) *sp.diff(x))
# Integrate both the sides
lhs = sp.integrate (v/ (a**2 - v**2),v)
rhs = sp.integrate (k/m,x)
integrated_eq = sp.Eq(lhs,rhs +sp.Symbol('C'))
print ('Integrated Equation:')
sp.pprint(integrated_eq)
# Applying the initial conditions: when c=0, v=0
C_val = sp.solve(integrated_eq.subs({v:0,x:0}),sp.Symbol('C'))[0]
print ("\nConstant of integration C=",C_val)
# substitute C back
final_eq = integrated_eq.subs(sp.Symbol('C'),C_val)
print("\nFinal equation:")
sp.pprint(final_eq.simplify())
# Simplify and solve for v as function of x
simplified = sp.simplify(final_eq.rewrite(sp.log))
display(simplified)
# Optional numeric example
m_val = 1.0
g_val = 9.81
k_val = 0.2
a_val = np.sqrt(m_val*g_val/k_val)
#compute v vs x numerically
x_vals = np.linspace(0,10,200)
v_vals = [a_val*np.sqrt(1-np.exp(-2*k_val*x/m_val)) for x in x_vals]
# plot
plt.plot(x_vals, v_vals, label = "Velocity vs Distance")
plt.axhline(a_val, color ='r', linestyle='--',label =f"Terminal velocity = {a_val:.2f} m/s")
plt.xlabel("Distance fallen(x)")
plt.ylabel("Velocity (v)")
plt.title("Body falling with quadratic air resistance")
plt.legend()
plt.grid(True)
plt.show()