Question

An aircraft in level flight completes a circular turn in 100 seconds.
1) What is the radius of the circular turn?
2) What is the angle of banking, if the velocity of aircraft is 40 m/s?

Answer 1

Hint: For finding the radius of circular turn, we can find the angular velocity of the aircraft from the time period given to us. Then by using the relation between angular velocity and the linear velocity we can obtain the radius of the path. For the angle of banking, the tangent of the angle is equal to the square of linear velocity divided by radius and acceleration due to gravity.

Formula used:
The relation between angular velocity and the time period of an object moving in a circular path is given as
$\omega = \dfrac{{2\pi }}{T}$
The relation between the angular velocity of the object, the linear velocity and the radius of the circular path is given as
$v = r\omega $
The angle of banking is given in terms of the linear velocity and the radius of circular path by the following relation.
$\tan \theta = \dfrac{{{v^2}}}{{rg}}$

Complete answer:
We are given an aircraft which is in a level flight. The time period of the aircraft for completing one circular turn is given as
$T = 100s$
From the time period, we can calculate the angular velocity of the aircraft which is given as
$\omega = \dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{{100}} = \dfrac{\pi }{{50}}rad/s$
We are also given the linear velocity of the aircraft whose value is
$v = 40m/s$
Now by using the relation between the angular velocity and the linear velocity, we can obtain the radius of the circular turn in the following way.
$
  v = r\omega \\
   \Rightarrow r = \dfrac{v}{\omega } = \dfrac{{40}}{{\pi /50}} = \dfrac{{2000}}{\pi }m = 636.6m \\
 $
Now in order to find the angle of banking for the aircraft, we have the following relation.
$
  \tan \theta = \dfrac{{{v^2}}}{{rg}} \\
   \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{{v^2}}}{{rg}}} \right) \\
 $
Inserting the values of linear velocity and the radius of the circular path and also taking $g = 10m/{s^2}$, we get
$\theta = {\tan ^{ - 1}}\left( {\dfrac{{{{\left( {40} \right)}^2}}}{{\left( {\dfrac{{2000}}{\pi }} \right) \times 10}}} \right) = {\tan ^{ - 1}}\left( {0.251} \right) = 14.09^\circ $

Therefore, we have the radius of the path as 636.6m and the angle of banking as 14.09$^\circ $.

Note:
The angle of banking is the angle subtended by the wings of the plane and the horizontal direction. When the aircraft is flying horizontally then the angle of bank is zero while if it is moving in a vertical direction, then this angle is 90$^\circ $.