This problem requires some thoughts... An airline lands with a speed of 50 m/sec

Question

This problem requires some thoughts...

An airline lands with a speed of 50 m/sec. Each wheel of the plane has a radius of 1.25 m and a moment of inertia of 110 kg.m^2. At touchdown, the wheels begin to spin under the action of friction. Each wheel supports a weight of 1.4E4 N, and the wheels attain an angular speed of rolling without slipping in a time of 0.48 sec. What is the coefficient of friction between the wheels and the runway? Assume the plane keeps the same speed (a = 0 m/sec^2).

Hints
Draw a diagram of the wheel. Put all the forces on it. There will be three vectors:

The weight, mg, originating from the center of the wheel toward the ground, the Normal vector opposite to the weight, and the friction force vector, originating from the lower rim of the wheel in the opposite direction of the motion.

Use the torque equation:

Torque = I *angular acceleration

Tau = I (alpha)

That is,

f*r = I (alpha)

where f is the friction, r is radius, I moment of inertia, and alpha is the angular acceleration.

Use rotational kinematics to find alpha. Use f = uN, N=mg and combine all!

Explanation / Answer

Let the coefficient of friction = The radius of the wheel, r = 1.25 m The speed of the wheel, v = 50 m/s The moment of inertia of the wheel, I = 110 kg m^2 The weight supported by the wheel, F = 1.4 * 10^4 N = 14000 N The time taken to attain angular speed of rolling without slipping, t = 0.48 sec The torque is given by the formula                    = F * r = 14000 * 1.25 = 17500 N m Angular momentum, L = 0.48 * 17500 = 8400 Js But the angular momentum, L = I = I(v/r) = 4400 So, 8400 = 4400 From the above, = 0.524 So the coefficient of the friction, = 0.524 So, 8400 = 4400 From the above, = 0.524 So the coefficient of the friction, = 0.524

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