Question

The z component of the angular momentum of a particle whose position vector is $\overrightarrow{r}$ with components x,y and z and linear momentum is $\overrightarrow{P}$ with components px, py and pz is

• A. xpy – ypx
• B. ypz-zpy
• C. zpx — xpz
• D. xpy + ypx

Answer 1

Answer: (A)

xpy – ypx

Answer 2

Here, $\overset{\rightarrow}{r} = x\widehat{i} + y\widehat{j} + z\widehat{k}$

$\overset{\rightarrow}{P} = P_{x}\widehat{i} + P_{y}\widehat{j} + P_{z}\widehat{K}$

Let $\overset{\rightarrow}{L} = L_{x}\widehat{i} + L_{y}\widehat{j} + L_{z}\widehat{K}$ …… (i)

As $\overset{\rightarrow}{L} = \overset{\rightarrow}{r} \times \overrightarrow{p}$

$= \widehat{i}(yp_{Z} - zp_{y}) + \widehat{j}(zp_{x} - xp_{Z}) + \widehat{k}(xp_{y} - yp_{x})$.(ii)

Comparing the coefficients of $\widehat{i},\widehat{j}$and $\widehat{k}$in Eqs. (i) and (ii), we get

$L_{x} = yp_{Z} - zp_{Y}$

$L_{y} = zp_{x} - xp_{z}$

$L_{z} = xp_{y} - yp_{x}$