PHYS 2001
Example 01
Example 02
Example 03
Example 04
Example 05
Example 06
Example 07



Circular Motion - Example 04



Fig 4

4. An airplane is flying in a horizontal circle at a speed of 480 kmh-1. (Fig. 4). If its wings are tilted at angle 40° 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.


Fig 4

Information given:
v = 480 kmh-1= (480 X 1000 / 3600) ms-1= 133 ms-1
θ = 40°

(a) If the object moves in a circle it must have an umbalanced force (Fc) towards the centre of the circle called the centirpetal force.

Fc = mv2/r

There are 2 physical forces on the car
1. The weight of the plane W = mg acting in the vertical plane and at an angle θ = 40°.
2. The Lift by the wing F acting at a right angle to the wing.

F can be resolved into F v and F h

Fv = F Cos θ and Fh = F Sin θ

This road is designed so that when the vehicle velocity is 60 kmh-1 (16.7 ms-1) the centripetal force is generated ONLY by the normal reaction i.e. there is no friction between the road and the vehicle.

Since the car is not moving vertically then by Newton's first law:

F v = W
Fv = mg
Now Fv = F Cos θ

So F Cos θ = mg ---Eq1

Centripetal force (Fc) is the unbalanced Fh = F Sin θ

Fh = F Sin θ

F Sin θ = mv2/r ---Eq2

Dividing E2 by Eq1 we get:

F Sin θ / F Cos θ = (mv2/r) / mg

Tan θ = (v2/rg)

Tan 40° = (133 ms-1)2/[(r) (-9.81 ms2)

r = (133 ms-1)2 / [(tan 40°) (-9.81 ms2)

r = 2.14 X 103 m



Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.