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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 Pz. Show that If the Particle Moves Only in the X-y Plane the Angular Momentum Has Only a Z-component. - Physics

Find the components along the x, y, z axes of the angular momentum of a particle, whose position vector is with components xyand momentum is with components pxpy and 'p_z`. Show that if the particle moves only in the x-plane the angular momentum has only a z-component.

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Solution 1

lx = ypz – zpy

l= 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-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-plane, the direction of angular momentum is along the z-direction.

Solution 2

We know that angular momentum `vecl` of a particle having position vector `vecr` and momentum `vecp` is given by

`vecl = vecr xx vecp`

But vecr = [xveci + yhatj + zveck], where x, y,z are the component of `vecr and vecp = [p_xveci + p_yhatj + p_zhatk]`

`:. vecl  = vecr xx vecp  = [x hati + yhatj + zhatk] xx [p_xhati + p_y hatj + p_z hatk]`

or `(l_xhati + l_yhatj + l_zhatk) `= `|(hati ,hatj ,hatk),(x,y,z), (p_x,p_y,p_z)|`

`=(yp_z - zp_y)hati + (zp_x - xp_z)hatj + (xp_y - yp_x) hatk`

From this relation we conclude that

`l_x = yp_z - zp_y, l_y = zp_x - xp_z, l_z = xp_y- yp_x`

if the given particle moves only in the x - y plane then z = 0 and 'p_z = 0' and hence

'vecl = (xp_y - yp_x)hatk` which is onl;y the z- component of hatl

it means that for a particle moving only in the x-y plane, the angular momentum has only the z-component

  Is there an error in this question or solution?
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NCERT Class 11 Physics Textbook
Chapter 7 System of Particles and Rotational Motion
Q 6 | Page 178
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