# Should angular momentum change or it should not.?

Let us consider a body with an arbitrary shape. Let its center of mass be assigned coordinates (o,o,z) with respect to an origin.Now let us consider two equal and opposite forces acting on the body in such a way that they do not share their line of action and their lines of action do not pass through the center of mass.

Now I first propose that the net torque acting on a given system about any given point is simply equal to the net torque acting on its center of mass about that point.

I attempted to prove it in the following manner;

Let us consider a system of mass with the center of mass having a position vector R w.r.t the origin (the point about which I am considering the angular momentum and torque).Now the angular momentum ,L, of the whole system can be broken into two parts, i.e. it can be expressed as the vector sum of the angular momentum of the center of mass w.r.t to the origin, denotated by A, and the angular momentum of the system about the center of mass,denotated by Z.

so L = A+Z. ..........(1)

Now let an external force F act on the system.

Now we can write that F= mA, Where A is the acceleration the center of mass and m is the total mass of the system.

Now the torque acting on the center of mass about the point = R X F. .............(2)

Now let us consider the torque acting on the system about the center of mass.

As the center of mass is accelerating any mass in the mentioned system will experience a torque due to the pseudo force.

Hence the net torque acting on the whole system about the center of mass = K=sigma (Ri X miA).

where Ri denotes the position vector of a point mass about the center of mass and mi denotes its mass.

Hence K = {sigma ( mi. Ri)} X A.

Now sigma (mi.Ri) = position vector the center of mass w.r.t to the center of mass= zero vector.

Hence K= Zero vector.

Now by differentiating bothe sides of (1) w.r.t time T,and interpreting in terms of physics we have

torque acting on the system about the origin= torque acting on center of mass about origin + torque acting on the system about the center of mass.

= R X F + K = R X F , since K= zero vector.

Hence I proved what I stated.

Now let us come back to the original question. In that question as the center of mass is not accelerating owing to the action of balanced forces, hence no torque is acting on C.O .M and as such the angular momentum of the body should not change.

But practically speaking these forces generate a couple and tend to change the angular momentum of the body.

So Where was I wrong ?

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If you apply a force to the center of mass then the torque is 0 (r=0, rXF=0), it will make is translate and not rotate. A 0 in one side of the eq will kill the problem, imo. – Helder Velez Sep 28 '11 at 15:35
@Helder Velez- First of all thanks for being the first one to comment on the question after such a long drought of replies. Could you kindly tell me what it means by imo(you used it at the end of your sentence)? could you specify torque on "which body" and "about which point" ? – Primeczar Sep 28 '11 at 15:41
".. propose. the net torque acting .. about any given point is simply equal to the net torque acting on its COM about that point." A torque, see fig. is perpendicular of the plane of rotation and is applyed at the point of rotation then you must clarify if 'symply equal' is about an applyed vector or what else. Two vectors applyed in distinct points can not be 'equal'. May be including a simple drawing could make the problem clear. – Helder Velez Sep 28 '11 at 16:13
I'm sorry @Primeczar if I didn't read your question with more attention but in this moment my mind is devoted to quite different problems. May be in another occasion. Sorry. – Helder Velez Sep 28 '11 at 16:48
I think you are trying too hard ! Work back to basic definitions, and try to keep your discussion with as few degrees of freedom as possible. This is classic Luke Skywalker stuff : ditch the variables and equations, instead hold true to the basic definitions. Clarity will follow. – Well3 Oct 13 '14 at 20:06