Find the components along the x, y, z axes ...

Systems of Particles and Rotational Motion

Question : 6 of 33

Marks: +1, -0

Find the components along the x, y, z axes of the angular momentum

\vec{L}

of a particle, whose position vector is

\vec{r}

with components x, y, z and momentum is

\vec{p}

with components

p_x, p_y \text{ and } p_z

. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component.

Solution:

(a) Angular momentum,

\vec{L} = \vec{r} \times \vec{p}

It is a vector quantity and its direction is given by right hand rule for vector product. As

\vec{r}

and

\vec{p}

here lie in xoy plane, so

\vec{L}

acts along Z axis.

In cartesian co-ordinates,

and

\begin{array}{l} \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \\ \vec{p} = p_x \hat{i} + p_y \hat{j} + p_z \hat{k} \end{array}

...(ii)

∴ From (i) and (ii), we get

\vec{L} = (x \hat{i} + y \hat{j} + z \hat{k}) \times (p_x \hat{i} + p_y \hat{j} + p_z \hat{k})

= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix}

L_x \hat{i} + L_y \hat{j} + L_z \hat{k} = \hat{i}(y p_z - z p_y) + \hat{j}(z p_x - x p_z) + \hat{k}(x p_y - y p_x)

On comparing, we get

\begin{array}{l} L_x = y p_z - z p_y \\ L_y = z p_x - x p_z \\ L_z = x p_y - y p_x \end{array} \quad \text{...(iii)}

Equation (iii) gives the required components of

L

along

x, y

and

z

axes.

(b) As the particle moves in

x-y

plane, then

z=0

and

p_z = 0

Hence,

\vec{L} = \hat{k} (x p_y - y p_x) = L_z \hat{k}

Hence angular momentum has only z-component.

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