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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 px, py and 'p_z`. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.
Solution
lx = ypz – zpy
ly = zpx – xpz
lz = xpy –ypx
Linear momentum of the particle,`vecp = p_x hati + p_y hatj + p_z hatk`
Position vector of the particle, `vecr = xhati + yhatj + zhatk`
Angular momentum, `hatl = hatr xx hatp`
=`(xhati + yhatj + zhatk) xx (p_x hati + p_y hatj + p_z hatk)`
`=|(hati,hatj,hatk),(x,y,z), (p_x, p_y,p_z)|`
`l_xhati + l_yhatj + l_z hatk = hati (yp_z - zp_y) - hatj(xp_z - zp_x) + hatk (xp_y - zp_x)`
Comparing the coefficients of `hati, hatj, hatk` we get:
`((l_x = yp_z - zp_y),(l_y = xp_z -zp_x),(l_z = xp_y - yp_x))}...(i)`
The particle moves in the x-y plane. Hence, the z-component of the position vector and linear momentum vector becomes zero, i.e.,
z = pz = 0
Thus, equation (i) reduces to:
`((l_x=0),(l_y=0),(l_z=xp_y -yp_x))} `
Therefore, when the particle is confined to move in the x-y plane, the direction of angular momentum is along the z-direction.
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