Ncert Exercises & Solutions

Question 7. 6. 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.

$l_{x} = {yp}_{z} - {zp}_{y} l_{y}$

$= {zp}_{x} - {xp}_{z} l_{z}$

$= {xp}_{y} - {yp}_{x}$

Linear momentum of the 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,

$\vec{l} = \vec{r} \times \vec{p}$

$= (x \hat{i} + y \hat{j} + z \hat{k}) \times (p_{x} \hat{i} + p_{y} \hat{j} + p_{z} \hat{k})$

$= |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_{x} & p_{y} & p_{z} \end{matrix}|$

$l_{x} \hat{i} + l_{y} \hat{j} + l_{z} \hat{k} = \hat{i} ({yp}_{z} - {zp}_{y}) - \hat{j} ({xp}_{z} - {zp}_{x}) + \hat{k} ({xp}_{y} - {zp}_{x})$

Comparing the coefficients of

$\hat{i}, \hat{j}, and \hat{k},$ we get:

$\begin{array}{r} l_{x} = {yp}_{x} - {zp}_{y} \end{array}\}$

$l_{y} = {xp}_{z} - {zp}_{x}$

$l_{z} = {xp}_{y} - {yp}_{x} . . . . . . . . . . . . (i)$
The particle moves in the x-y plane. Hence, the z-component of the position vector
and linear momentum vector becomes zero, i.e.,

$z = p_{z} = 0$

Thus, equation (i) reduces to:

$\begin{array}{r} l_{x} = 0 \end{array}\}$

$l_{y} = 0$

$l_{z} = {xp}_{y} - {yp}_{x}$

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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