Need Help- Centripetal plane problem

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    Centripetal Plane
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An airplane flying in a horizontal circle at 420 km/h with wings tilted at 35° requires the calculation of the circle's radius. The speed converts to 116.67 m/s, and the centripetal force is derived from the aerodynamic lift, expressed as Fc = Fnsin35°. By applying the equation a = v^2/r and substituting the necessary values, the radius is calculated to be approximately 2421.59 meters. A free body diagram was suggested to aid in visualizing the forces acting on the airplane. The discussion emphasizes the relationship between speed, tilt angle, and radius in circular motion dynamics.
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Homework Statement


An airplane is flying in a horizontal circle at a speed of 420 km/h. If the wings of the plane are tilted 35° to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an aerodynamic lift that is perpendicular to the wing surface.



Homework Equations



a=v^2/r
Fc = Fnsin35°

The Attempt at a Solution


So first i looked at what was given:
V= 420 km/h = 116.67 m/s
r= ?

Since i know that the plane wings are tilted 35° to the horizontal, i broke the Fn into two parts the Fn pointing towards the centre of the circle being Fnsin35°

So:
Fc= Fnsin35°
ma=mgsin35°
a=gsin35°
(v^2/r) = gsin35°
(116.67m/s^2)^2/(9.8m/s^2)(sin35°)=r
r=2421.59 m
 
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Lolagoeslala said:

Homework Statement


An airplane is flying in a horizontal circle at a speed of 420 km/h. If the wings of the plane are tilted 35° to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an aerodynamic lift that is perpendicular to the wing surface.

Homework Equations



a=v^2/r
Fc = Fnsin35°

The Attempt at a Solution


So first i looked at what was given:
V= 420 km/h = 116.67 m/s
r= ?

Since i know that the plane wings are tilted 35° to the horizontal, i broke the Fn into two parts the Fn pointing towards the centre of the circle being Fnsin35°

So:
Fc= Fnsin35°
ma=mgsin35°
a=gsin35°
(v^2/r) = gsin35°
(116.67m/s^2)^2/(9.8m/s^2)(sin35°)=r
r=2421.59 m
Start by drawing a free body diagram for the airplane.
 
SammyS said:
Start by drawing a free body diagram for the airplane.

i did this is it...

http://s1176.beta.photobucket.com/user/LolaGoesLala/media/fff.jpg.html
 
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