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The mass of each wheel is (m)/(16). Kinetic friction force acting is thus given by: fk = μkN⇒ fk = μk(m)/(16)g
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Since the coefficient of rolling friction is zero, fr = 0
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The mass of each wheel is (m)/(16). Kinetic friction force acting is thus given by:
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Kinetic friction is constant throughout and ωf = ωi + αtIn the slipping phase Vl = ωfr, ωi = 0 and Iα = μkMgr. Substituting these values in the above equation gives
Therefore, time taken from the point of touchdown for the wheels to start rolling is given by(5) t = (16VlI)/(μkmgr2) - 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.
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Force balance for the entire aircraft, (6) 16f = MaForce balance for each wheel,(7) τ − fr = IαWhile still rolling,α = (a)/(r)Then, substituting the values of f and α, yields 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, Now, for f < μsN, where N = (mg)/(16)
This is the value of the optimal torque required for braking the wheel.
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