  Question

Find the components along the x, y, z axes of the angular momentum $$l$$ of a particle, whose position vector is $$r$$ with components x, y, z and momentum is $$p$$ with components $$p_x, p_y$$ and $$p_z$$. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Solution

Linear momentum of particle, $$\vec{p}=p_x\hat{i}+p_y\hat{j}+p_z\hat{k}$$Position vector of the particle, $$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$Angular momentum, $$\vec{l}=\vec{r}\times \vec{p}$$$$\implies l_x\hat{i}+l_y\hat{j}+l_z\hat{k}=\hat{i}(yp_z-zp_y)-\hat{j}(xp_z-zp_x)+\hat{k}(xp_y-yp_x)$$Therefore on comparison of coefficients,$$l_x=yp_z-zp_y$$$$l_y=zp_x-xp_z$$$$l_z=xp_y-yp_x$$The particle moves in the x-y plane. Hence the z component of the position vector and linear momentum vector becomes zero.$$z=p_z=0$$Thus $$l_x=0$$$$l_y=0$$$$l_z=xp_y-yp_x$$Thus when particle is confined to move in the x-y plane, the angular momentum of particle is along the z-direction.PhysicsNCERTStandard XI

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