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Obtain the relation between angular momentum of a particle and torque acting on it. |
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Answer» Solution :The TOTAL angular moment of a system of particles is the angular moment of individual particles. For a systm of n particles, `vecL=vec(l_(1))+vec(l_(2))+vec(l_(3))+....vec(l_(n))` `=underset(i=1)overset(n)sumvec(l_(i))` where `i=1,2,3,….n` and `vec(l_(i))=vec(r_(i))xxvec(p_(i))` where `vec(r_(i))` is the position vector of the `i^(th)` particle with RESPECT to a given origin and `vec(p_(i))` is the linear momentum of the particle. The total angular momentum of the system of particle as `therefore vecL=underset(i=1)overset(n)sumvec(l_(i))=underset(i=1)overset(n)sumvec(r_(i))xxvec(p_(i))` Differentiating w.r.t. time `therefore (dvecL)/(dt)=underset(i=1)overset(n)sumvec(tau_(i))....(1) [because vec(r_(i))xxvec(p_(i))=vec(tau_(i))]` `therefore vectau=underset(i=1)overset(n)sumvec(r_(i))xxvec(F_(i))[because vectau=vecrxxvecF]` Here, `vec(F_(i))` is the force on the `i^(th)` particle is the vector sum of external forces `F_("iext")` acting on the particle and the INTERNAL forces `F_("i(int)")` exerted on it by the other particles of the system. `therefore vec(F_(i))=vecF_(i("ext"))+vecF_(i("int"))` `therefore vectau=vectau_(("ext"))+vectau_(("int"))` where `vectau_((ext))=sumvec(r_(i))xxvecF_(i(ext))` and `vec(tau)_("(int)")=sumvec(r_(i))xxvecF_(i("int"))` but the forces between any two particles of the system are equal and opposite, but also that these forces are directed along the line joiningthe two particles. The contribution of the internal forces to the total torque on the system is zero. So `vectau_("(int)")=0` and therefore `vectau=vectau_("(ext)")` From eqn. (1), `(dvecL)/(dt)=vectau_(("ext"))` Hence, the time rate of the total angular momentum of a system of particles about a point (taken as the origin of our frame of reference) is equal to the sum of the external torques EQUATION `(dvecL)/(dt)=vectau_(("ext"))` is the rotational ANALOGUE of `(dvecp)/(dt)=vec(F_(ext))`. |
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