1) An airplane traveling at 515 km/hr needs to reverse its course. The pilot decides to accomplish this by banking the wings at an angle of 40 degrees.
A) Find the time needed to reverse course.[Hint: Assume an aerodynamic "lift" force that acts perpendicularly to the flat wings].
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The lift force L , when the wings are at the given angle, has a vertical component L cos(angle) . This equals the weight of the plane mg, so
L cos(angle) = m g
L = mg/cos(angle).
To do the (circular) bend, the horizontal component of the lift force provides the centripetal force:
m v^2 / R = L sin( angle)
This means that
R = m v^2 /(L sin(angle) )
= v^2 / ( g tan(angle) )
where in the last step I substituted the expression for L from the weight balance above.
So knowing the radius of the circle along which the plane turns around, the time is half a circle's length divided by the speed:
t = pi R / v
t = pi v /( g tan(angle) )
t = 3.1415 * (515/3.6 m/s)/(9.81 m/s^2 * tan(40))
= 54.6 s