__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.

__Solution__:*l*

_{x}=

*yp*

_{z}–

*zp*

_{y}

*l*

_{y }= z

*p*

_{x}– x

*p*

_{z}

*l*

_{z}=

*xp*

_{y}–

*yp*

_{x}

Linear momentum of the particle,

**p**= p

_{x}

**i**+ p

_{y}

**j**+ p

_{z}

**k**

Position vector of the particle, r = x

**i**+ y

**j**+ z

**k**

Angular momentum,

**l = r X p**

= (x

**i**+ y

**j**+ z

**k**) X (p

_{x}

**i**+ p

_{y}

**j**+ p

_{z}

**k**)

=

**i j k**

p

_{x}p

_{y}p

_{z}

_{}l

_{x}

**i**+ l

_{y}

**j**+ l

_{z}

**k**=

**i**(yp

_{z}- zp

_{y}) -

**j**(xp

_{z}- zp

_{}

_{x}) +

**k**(xp

_{y}- zp

_{x})

Comparing the coefficients of

**i, j**and

**k**, we get:

l

_{x}= yp

_{z}- zp

_{y}

l

_{y}= xp

_{z}- zp

_{x}

l

_{z}= xp

_{y}- yp

_{x}

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:

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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