physics problem

Question

When a plane flies in a circular curve, its wings are tipped (see diagram) so that the "lift" force, which is exerted perpendicularly to the bottom of the wings, has a component towards the center of the circle. If a plane is flying with a speed of 441 km/h and its wings are tilted at an angle of 40° to the horizontal, what is the radius of the circle in which the plane is flying?

Answer 1

Assuming the aircraft is maintaining a constant altitude and is reasonably close to the planet Earth's surface, we can then surmise that the gravitational pull on the plane is equivalent to the mass of the plane times 9.8 m/s^2, also known as the value g.

The force from the lift must fully counteract this pull to maintain altitude. What more, the plane is at an angle so the lift force will also provide a horizontal acceleration which curves the path of the plane. As such, the force of lift (Fl), can be broken down into a component in the vertical direction (which counteracts gravity) and a component in the horizontal direction (that causes the plane to circle).

With the knowledge of the plane being at 40°, we can break down the lift through the use of this angle. This leads to the vertical component being Fl*cos(40°) and the horizontal being Fl*sin(40°). Even with an unknown mass of the plane (m), we can say that:

m*g=Fl*cos(40°)

Fh=Fl*sin(40°)

With Fh being the horizontal force causing circular motion. What more, given the identity of such forces relating them to velocity and radius of circulation, we can say that:

Fh=Fl*sin(40°)=mv^2/r

We now have enough information to solve for r.

mg/cos(40°) = Fl

mv^2/(Fl*sin(40°)=r

mv^2/(mg*sin(40°)/cos(40°) ) = r

Cancel out the mass, substitute the trigonometric identity to get...

v^2/( g * tan(40°) ) = r

Let us convert 441 km/h (the velocity) to m/s: v = 441 km/h *(1/60) h/min * (1/60) min/s *1000 m/km = 122.5 m/s

And now we can put all our numbers in the above equation to get our answer:

(122.5 m/s)^2 / ( 9.8 m/s^2 * tan(40°) ) = 1824.87 m ∼ 1.82 km