The Problem:

An aircraft executes a horizontal loop at a speed of $720$ kmph (or $200$ m/s) with its wings banked at an angle of $15^{\circ}$. What is the radius of the loop.

My Confusion: Unlike the case of cars on banked roads executing a loop (which have the aid of friction and normal forces to provide the centripetal force), there seems to be no force (except air resistance) that aids the plane to execute a loop. So how does an aircraft maintain a loop horizontally?

  • $\begingroup$ What about he horizontal component of the lift? $\endgroup$ – Farcher Feb 9 at 7:14
  • $\begingroup$ Well, obviously, a force which is perpendicular to the wings is present. This is the same force which makes the plane fly. $\endgroup$ – Photon Feb 9 at 7:15
  • $\begingroup$ @farcher I see, that can be a viable reason! But my book tells me that the only relation needed to find the radius of the loop is $\tan \theta = v^2/rg$. How is there no component of the lift in it as well? $\endgroup$ – Apekshik Panigrahi Feb 9 at 7:23
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    $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$ – Ben Crowell Feb 9 at 21:50
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    $\begingroup$ @ja72 That is exactly what I found surprising. Two completely different scenarios (well not completely, but you get the point), but the exact same equation works! Physics is truly marvelous! $\endgroup$ – Apekshik Panigrahi Feb 10 at 7:32

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$$\tan \left( \beta \right) =\dfrac {f_{z}}{f_g}=\dfrac {m\dfrac {v^{2}}{r}}{m\cdot g}\tag 1$$

Where $f_z$ Is the centrifugal forces ,$f_g$ Is the weight forces and $r$ is the radius of the loop.

$\begin{aligned}r=\dfrac {v^{2}}{\tan \left( \beta \right) \cdot g}= \dfrac {200^2}{10\cdot \tan \left( \dfrac {15\cdot \pi }{180}\right) }\simeq 14.9 \end{aligned} \quad [km]$

  • $\begingroup$ I see... because we know the angle of banking, we need to only consider the weight of plane. How each and every force acting on the plane causes the banking angle to be so doesn't need to be analyzed. That's really cool! Thanks! $\endgroup$ – Apekshik Panigrahi Feb 9 at 14:23
  • $\begingroup$ For high school level physics that answer is correct, but in real life, there are aerodynamic side loads to consider also. In order to achieve a constant yaw rate, ailerons are deflected causing a slight slide slip which changes the force balance in the radial & tangential direction. $\endgroup$ – ja72 Feb 10 at 17:27

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