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DYNAMICS PROBLEMS

1. The mass of each wheel is (m)/(16). Kinetic friction force acting is thus given by:
  f_k = μ_kN
  
  ⇒ f_k = μ_k(m)/(16)g
2. Since the coefficient of rolling friction is zero,
  f_r = 0
Kinetic friction is constant throughout and
ω_f = ω_i + αt
In the slipping phase V_l = ω_fr, ω_i = 0 and Iα = μ_kMgr. Substituting these values in the above equation gives

(4) (V_l)/(r) = 0 + (μ_kmg.r)/(16I)t
Therefore, time taken from the point of touchdown for the wheels to start rolling is given by
(5) t = (16V_lI)/(μ_kmgr²)
From equation 5, the time to achieve rolling is inversely proportional to coefficient of kinetic friction. So it will take more time on an icy runway. The frictional force that accelerates the wheel to the rolling speed is smaller. Hence more time will be taken. For the same reason, the stopping distance will also be greater.
Force balance for the entire aircraft,
(6) 16f = Ma
Force balance for each wheel,
(7) τ − fr = Iα
While still rolling,
α = (a)/(r)
Then, substituting the values of f and α, yields
(8) τ − (Ma)/(16).r = (Ia)/(r)
Thus,
τ = (Ma)/(16).r + (Ia)/(r)

(9) ⇒ a = (τ)/(⎡⎣(Mr)/(16) + (I)/(r)⎤⎦)
Now substituting this value of acceleration into force balance equation for the entire aircraft gives,
(10) f = (ma)/(16) = (τ)/(r + (16I)/(Mr))
Now, for f < μ_sN, where N = (mg)/(16)

(11) ⇒ τ < μ_smg⎡⎣(r)/(16) + (I)/(Mr)⎤⎦
This is the value of the optimal torque required for braking the wheel.