How can I calculate the angle of an airplane's wings during a circular motion?

AI Thread Summary
To calculate the angle of an airplane's wings during circular motion, the relationship tan(Q) = (v^2)/(gr) is derived from the balance of forces acting on the plane. In level flight, the lift force must counteract the weight of the airplane, but when banking occurs, the lift force's vertical component must still equal the weight while its horizontal component provides the necessary centripetal force. The equation L' sin(Q) = M(v^2)/r is established, where L' is the adjusted lift force when the wings are banked. The discussion clarifies that the lift force can be expressed as L' = mg/cos(Q), leading to the conclusion that mass m cancels out when deriving the final relationship. Understanding these force dynamics is crucial for accurately calculating the wing angle during turns.
Thiendrah
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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


Homework Equations





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?
 
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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.
 
Thiendrah said:
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?
 
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