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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. - CBSE (Science) Class 11 - Physics

ConceptTorque and Angular Momentum

#### Question

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.

#### 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?

#### APPEARS IN

NCERT Solution for Physics Textbook for Class 11 (2018 to Current)
Chapter 7: System of Particles and Rotational Motion
Q: 6 | Page no. 178

#### Video TutorialsVIEW ALL [2]

Solution 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. Concept: Torque and Angular Momentum.
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