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My Ideas


           All the information provided in this page are all my own views and observations. Don't take them for granted. My intention here is to give you an idea of looking at things in a different way. I am not responsible if you have encountered into any sort of argument with your friends or with any one. These ideas which I am presenting here are not approved by any scientific committee. I hope you enjoy going through this page. I'll be very happy if this page makes you think. Do send your suggestions. The content of this page is copy write to me.   



The motion of boomerang is the most interesting in universe. I interpret the motion of boomerang to be something like this. First of all the body of boomerang is aerodynamical. Now a days we have different shapes of boomerangs. Here I'll give you only the theoretical approach to this motion.

Everybody agrees that a boomerang as two wings (we also have boomerangs with 3 or 4 wings or any number of wings). I have considered the basic shape of boomerang. Infact boomerangs with any number of wings will again come down to the basic shape of 2 wings, that is considering the wings two at a time discretely. For my theoretical approach I made some assumptions. Some of them are:

Here, the force is applied on one of the wings. For a person standing on earth, the motion looks to be taking place in a non-inertial frame or accelerated frame of reference. When a person stands in the plane of motion at the center, the motion appears to be taking place in a inertial frame.


I am sorry for not giving a proper figure having description. I hope you go through the description below to understand the figure. The figure is a manually drawn one, so it may not exactly resemble that of a boomerang. I think it might almost resemble the shape of a boomerang.

Let x, y, z be the three axes which the lines in the above fig. represents. There are infinite number of particles lying on the wings of the boomerang which all take part in the motion of the boomerang. Now let us find the equation for the motion of boomerang.

First step would be to find the position of center of mass. We can arrive at the conclusion that when a force is applied on the boomerang's wing, it is equal to the same force applied directly on the center of mass. when this happens we can say that it is equal to a solid body in motion.

As I already mentioned there are infinite particles lying on the whole body of boomerang having a mass m(i). The position of each particle with respect to the co-ordinate axes can be represented by x, y, z axes. Every particle will have a unique position represented by ( x(i), y(i), z(i) ). 
For continuous mass distributed bodies. Let 'dm' be the small mass of an element or particle on the boomerang. Let X, Y, Z be the co-ordinates of this small which by taking the appropriate limits for integrals give the position of all the particles. Here  the notation I use for various mathematical symbols are....

So, we can write the co-ordinates has      X = (1/M) Int (x dm) ; Y = (1/M) Int (y dm) ; Z = (1/M) Int (z dm).

Now, suppose the i'th particle is having a mass m(i) and the position vector r(i) with respect to the inertial frame. Then each of the particle on the boomerang's body will experience a force due to the other (n-1) particles excluding the force due to the particle itself. It will also experience an external force i.e., the force with which it is thrown or the force supplied by us on boomerang's wing or blade. The acceleration experienced by each of the particle is

 VEC( A(i) ) = 1/m(i) [ Sig j not = i ( VEC F(ij) + VEC F(i)ext ) ] ...... ( According to Newton's second law ) 

The meaning of the above equation is that the acceleration of the particle is equal to (1/m(i)) times summation of the force on the particle due to other particles and the external force. j is not equal to i, because excluding the force experienced by the particle itself.

Therefore from above equation, it gives

         m(i) * (VEC A(i)) =   Sig j not = i ( VEC F(ij) + VEC F(i)ext ) 

         m(i) * (VECA(i)) = VEC F(ext) ........( Because the intended forces on the particle will sum upto zero )

Therefore we have

         Sig[m(i)*(VEC v(i))] = M*Rcm     giving     Sig[m(i)*(VEC v(i))] = M*(VEC Acm)    

Possibility----> (1)

         VEC(Acm) = VEC(F ext)/M  ........... ( 1 )

F ext is produced by us. The center of mass takes a motion which is not exactly circular. But from my assumption stated earlier, I take it to be circular. When I say it takes a circular path, I mean not the rotation of the boomerang about it's own axis but the translation motion in forward direction is circular. That is way the boomerang is coming back to the person who throws it. The acceleration components which are involved in circular motion are radial acceleration A(r) and tangential acceleration A(t).

                 A(r) = - Sqr(V)/r        where 'r' is the radius of circular path, V is velocity of boomerang. 

                 A(t) = d(Vcm)/dt       where Vcm = d(Xcm)/dt.

Acceleration towards the center of mass is  A = Sqrt( Sqr(A(t)) + Sqr(A(r)) )

Possibility------> (2)                               A   = Sqrt( Sqr(Sqr(V)/r) + Sqr(dVcm/dt) )

My Interpretation

                               When we apply force on one wing of the boomerang or rather when we throw a boomerang, the entire force (I mean true or real force (RF) ) acts on the center of mass of the boomerang. Here I introduce or name the force which is actually acting in the opposite direction to the original or initial force has Pseudo force (PF). This is the force which is actually making the boomerang come back to the person who is throwing.


The two arrows pointing into the intersection of 3 lines represent real force and a pseudo force. The resultant of these two forces is the one coming out from the center. This resultant force ( F(RES) ) will always try to pull the boomerang towards the center. 

This Pseudo force is initially zero when the boomerang is thrown. It acts like an imaginary or unreal force when the initial motion of the boomerang starts. It comes into picture as the motion of the boomerang progresses. That is it turns into a real force. At any point of time in it's motion the resultant force on the boomerang is the sum of real and pseudo forces. When it actually begins to take a turn, the pseudo force matches with that of the real force but the resultant force which is coming out of the center of the two wings is not zero since they are at an angle other than 180 degrees. The real force begins to seize as it approaches the person who has thrown it. This time the pseudo force is the only one which will act.

Therefore,        VEC (F (RES) ) = Mod( VEC (F (PF))) + Mod( VEC (F (RF))).
  RES - Resultant force.
  PF   - Pseudo force.
  RF   - Real force.

 Newton's laws work here as it is inertial frame. 

F (RF) = M*(Sqr VEC Vcm) / r         ;       F (PF) = -M*Acm*S     Where 'S' is a parameter introduced 

which will govern the pseudo force to become real force. It's value ranges from 0 to 1. This 'S' parameter is called Srikanth's parameter. ( Srikanth is my name, I guess you are laughing :) ). 

Therefore,      F (RES) = M*(Sqr VEC Vcm) / r  +  Mod( M*Acm*S) Where 'S' tends to 0 to 1.

When S=0, F (PF) = 0 and the resultant force is only the real or user supply external force. 
   i.e., F (RES) = F (PF).

When S=1, F (PF) = 1 and the resultant force is the pseudo force which turned real. i.e., F (RES) = F (PF).

Here the angle of turn ,we get from

              Tan #t = Sqr(Vcm) / (r*g) and the angle is obtained by arch tan of Sqr(Vcm) / (r*g).

At the extreme point both forces balance each other, but still resultant is present and not zero, as they (two forces are at an angle.

              F (RES) = Mod(VEC (F (RF)) + Mod(Sqr(VEC(PF))

Now using vector addition formula, we get VEC(F (RES))

Now we can apply all newton's laws here for motion of boomerang.

VEC(F RES) = Sqr[ Sqr(m*Vcm) + Sqr(M*Acm*S - 2*(M*Sqr(Vcm))/r - 2*(M*SQr(VCm) * (M*Acm * S) * Cos#t.

So, from the above we can calculate many things using newton's laws. This is what I thought of. I hope You send your valuable suggestions. watch out for my universe theory which is interesting which supports theory of rebirth.