Question

An airliner lands with a speed of 47.0 m/s. Each wheel of the

plane has a radius of1.25 m and a moment of inertia of
110kgm2.
Attouchdown the wheels begin to spin under the action of
friction.Each wheel supports a weight of 1.40 104
N,and the wheels attain their angular speed in 0.47 s while rolling
without slipping. What is thecoefficient of kinetic friction
between the wheels and the runway?Assume that the speed of the
plane is constant.

Answer 1

coefficient of kinetic friction= 0.502

Step-by-step explanation:

Given Data

Speed 47.0 m/s

radius=1.25 m

moment of inertia=110 kg×m²

weight=1.40×10⁴N

time=0.47 s

To find

The Coefficient of Kinetic Friction

Solution

First to find initial angular speed in radian per second

angular speed w₀ = 47 / 1.25 = 37.6 rad /s.......as w₀=speed/radius

v= u+at

we have to find angular acceleration

as v=0 so

0 = 37.6 + 0.47a

a = -80 rad /s² shows it moves in anti clock wise direction

torque = Moment of Inertia * angular acceleration

t= 110 * 80

t= 8800 Nm

Now for friction force

friction force=t/radius

f f = 8800 / 1.25

f f= 7040 N

And finally to find coefficient of kinetic friction between the wheels and the runway

coefficient of kinetic friction=friction force/weight

coefficient of kinetic friction= 7040 / 1.4×10⁴

coefficient of kinetic friction= 0.502