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Q7.6  Find the components along the x,y,z  axes of the angular momentum 1  of a particle, whose position vector is r with components x,y,z  and momentum is  p with componentsp_{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.

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Linear momentum of particle is given by :

                                                          \overrightarrow{p}\ =\ p_x \widehat{i}\ +\ p_y \widehat{j}\ +\ p_z \widehat{k}

And the angular momentum is :

                                                           \overrightarrow{l}\ = \overrightarrow{r}\times \overrightarrow{p}

                                                                  = \left ( x \widehat{i}\ +\ y \widehat{j}\ +\ z \widehat{k} \right ) \times \left ( p_x \widehat{i}\ +\ p_y \widehat{j}\ +\ p_z \widehat{k} \right )

                                                                  =\begin{vmatrix} i &j &k \\ x &y &z \\ P_x &P_y &P_z \end{vmatrix}

                                                                  =\ \widehat{i}\left ( yp_z - zp_y \right )\ -\ \widehat{j}\left ( xp_z - zp_x \right )\ +\ \widehat{k}\left ( xp_y - yp_x \right )

When particle is confined to x-y plane then z = 0 and  pz = 0.

When we put the value of z and pz in the equation of linear momentum then we observe that only the z component is non-zero.

 

Posted by

Devendra Khairwa

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