Boomerangs                                           Dr Hugh Hunt, Cambridge, June 2000           

A boomerang does funny things because it is in fact a gyroscope.  Aerodynamic forces generate a twisting moment which cause the 'gyroscope' to precess  and to move on a circular path. 

Let us examine the forces acting on a boomerang of radius  a .  The centre of the boomerang is moving at a constant forward speed  V  and the boomerang is spinning with angular velocity  w  as shown in the diagram.  The 'top' end A is moving faster than  V  with speed  V+aw  and the 'bottom' end B is moving slower with speed  V–aw .  A wing generates more lift when it is moving faster so point A is generating more lift than point B.

 

The two forces  FA  and FB  can be represented by a single force  F  and a single couple  C .  With this simple representation of the forces acting on the boomerang we can give two reasons why it moves on a circular path:
1.  A constant centripetal force  F  produces circular motion with velocity  V  on a radius  R  :
            F = mV2/R                                            eq. 1
2.  A constant couple C  acting on a gyroscope spinning at angular velocity  w   causes steady precession at rate  W   :
            C = JwW                                              eq. 2
with m  & J   the boomerang's mass & moment of inertia.

 

If the rate of precession  W  exactly corresponds to the angular velocity of circular motion, then the boomerang stays tangential to the flight path as shown.  This gives an equation relating  V  to  W
            V = RW                                                            eq. 3

The aerodynamic lift force  L  acting on an airfoil of area  A  moving at speed  v  in air with density  r  is given by 

            L = rv2 CL A                                                 eq. 4

where  CL  is defined as the lift coefficient.  It can be shown by integrating the lift force over the area of a cross-shaped boomerang that the net lift force  F  and aerodynamic couple  C   are given by

            F = r(V2+(aw)2) CL As                       eq. 5
and       C = rVa2w CL As                               eq. 6

where  As = pa2  is the swept area of the boomerang, and  V , w  and  a   are the velocity, spin speed and radius of the boomerang as before.

boomerang on circular flight path

From equations 2, 3 and 6, we find that the radius R  of the circular flight path is independent of spin speed  w   and forward velocity  V, and that it is a constant for a given boomerang:

                        R =                                                                            eq. 7

For the case of a cross-shaped boomerang, J = ma2  and equations 1, 5 & 7 can be arranged to give

                        aw = V                                                                                           eq. 8

which defines the 'flick-of-the-wrist' needed to make the boomerang fly properly.

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For more information see      http://www.eng.cam.ac.uk/~hemh1/boomerangs.htm