The 747 being plane is landing at a speed of 70 ms^(-1). Before touching the ground, the wheels are not rotating. How long a skid mark do the wing wheels leave (assume their mass is 100 kg which is distributed uniformly, radius is 0.7 m, and the coefficient of friction with the ground is 0.5)?

The 747 being plane is landing at a speed of 70 ms^(-1). Before touchi

1.	The 747 being plane is landing at a speed of 70 ms^(-1). Before touching the ground, the wheels are not rotating. How long a skid mark do the wing wheels leave (assume their mass is 100 kg which is distributed uniformly, radius is 0.7 m, and the coefficient of friction with the ground is 0.5)?
Answer» Solution :the tyres of the plane will leave a SKID mark if the speed of the tyres in contact with ground is lesser than the velocity of the plane. The condition for this is `v gt omega` (When the tyre attained an ANGULAR velocity of V/R) The tyres will stop the skidding and starts the rolling. The forces acting on the wheel after the plane touches down are, N - Normal force W-weigh The wheel is not accelerating means `N=omega` The torque about the centre of the wheel is `tau=RF= mu omegaR` The angular acceleration is `ALPHA=(tau)/(1)=(mu omegaR)/((1)/(2) mR^(2))=(2MU omega)/(mR)` ACCORDING to equation of motion, time taken to stop the skidding by the wheel is, `omega= omega_(0)+alphat+alphat=(V)/(R)` `:.t=(V)/(alphaR)=(V)/(R)((mR)/(2mu omega))=(mV)/(2mu omega)` `t=(mV)/(2mu omega)=(100xx70)/(2xx0.5xx232xx10^(3))=0.03s` The six mark will have a length of `l=vt=70xx0.03=2.1m` ote: The 747 is resting on the runway, supported by 16 wheels under the wing, and 2 under the nose total length is 68.63 m. The normal force experienced by plane through its 16 wheels is c = 232 KN