# Circular Motion airplane

## Homework Statement

A pilot with mass m fles an airplane at a speed of v in a turn of radius r. Prove that angle of the wings of the airplane to the horizontal is tanQ=(VxV)/gr

## The Attempt at a Solution

This was asked on my first exam last week.
Since it's tangent. All we need to get is the force, result will equal to (FxcosQ)/(FxSinQ)=(VxV)/gr?

I'm not getting anywhere with this? Can someone help me?

dynamicsolo
Homework Helper
The issue we are concerned with here is how we get the plane to move along a circle of radius r. Start with thinking about the plane flying with the wings level (in a horizontal plane). There are two forces acting on the plane in the vertical direction: its weight, Mg, and the "lift", L, supplied by the wings. With the plane in level flight, these vertical forces must balance.

Now we want the plane to make a turn, so we "bank" the wings by an angle Q. So the "lifting force" will be off-vertical by an angle Q; since it must still balance the weight of the plane vertically, its magnitude will change to L' . (How is it related to Mg?)

This altered lifting force now has a horizontal component as well. This component of L', which will equal L' sin Q , is what supplies the centripetal force to pull the plane into a circular path (for the interval of the wing-banking). So we have

L' sin Q = M·(v^2)/r .

Try things from there.

BTW, a similar argument can be used to explain why a cyclist (leg- or motor-powered) must "lean into the turn" when they want to go round a corner...

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oh, I think i got it.

So we got our force FnXsinQ=m.a=m.(vXv)/r, also the other component is "mg", so

when you divide those two, you will get mvv/gr.

thanks a lot.

dynamicsolo
Homework Helper
So we got our force FnXsinQ=m.a=m.(vXv)/r, also the other component is "mg", so

when you divide those two, you will get mvv/gr.

I'm presuming that what you are calling 'Fn' corresponds to the lift force; there is no normal force for an aviation problem. The lift force when the plane is banked by an angle Q will be L' = mg/cos Q .

In your last sentence, shouldn't the mass m have divided out?