Will an Airplane Maintain Uniform Circular Motion While Banking?

[pb23me]

Homework Statement

General question:Decide wether an airplane will continue to fly in uniform circular motion while banking on a turn.

Homework Equations

m(v²/r)
$$\Sigma$$F_x= Lsin$$\theta$$=m(v²/r)
$$\Sigma$$F_y=Lcos$$\theta$$-mg=0

The Attempt at a Solution

my question is if the centripetal force m(v²/r)
is greater than the force of lift in th x direction Lsin$$\theta$$ is that when it will no longer continue to fly in circular motion? also is there an amount of lift in the x direction that would be too much causing the plane to go in? If so then how do i know how much is the right amount to keep the plane in uniform circular motion?

 

[venkatg]

Yes it can fly in a circle until it exhausts all fuel.

By banking you know that the horizontal component provides the centripetal force to negotiate a circle. Yes, it reduces the vertical lift and so the airplane has to pitch up a bit so that the angle of attack is increased to compensate for the lost lift. Also since it flies sideways, drag is created which reduces the airspeed and so power has to be increased a bit. But a bank angle cannot exceed 30 degrees in a 737 series.

 

[pb23me]

Ok but on a word problem asking if the force of lift is enough to keep the plane in curcular motion...how do i decide if it is enough?

 

[pb23me]

would i set it equal to the equation for centripetal force m(v²/r)? and if it -force of lift- is smaller then its not enough?

 

[venkatg]

Ok but on a word problem asking if the force of lift is enough to keep the plane in curcular motion...how do i decide if it is enough?

It depends on what is radius of the circle needed and speed of the airplane.
Sharper turns are better done at low speeds because the centripetal force is lower. At low speeds, list is developed by extending flaps and slats. since for a give radius of turn, the centripetal force is much lower, less lift is stolen to provide centripetal force.

Also better experience for passengers.

V^2/r = L * sin(theta)

Initially a small bank angle won't decrease much the overall lift. The banking itself is caused by increasing lift on one wing and decreasing on the other. So upto a certain bank angle (say 15 deg), centripetal force can be provided. At high bank angles, overall lift starts decreasing faster and raising the nose may cause a stall.

 

[venkatg]

would i set it equal to the equation for centripetal force m(v²/r)? and if it -force of lift- is smaller then its not enough?

Assume bank angle is theta.
Assume lift force is L normal to the wings after banking

Balancing vertical forces of flight:

Lcos(theta) = mg

Balancing horizontal forces,

Lsine(theta) = mv^2/r

Now tan(theta) = v^2/rg

For a given theta and small r, v has to be small