Question

Find the components along x,y and z axes of the angular momentum vector L of a particle, whose position vector is vector r with components x,y,z and momentum is vector 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-acomponent.

Answer 1

Let OX, OY and OZ be the three matually perpendicular axes. Consider a particle at point P with position vector vector(OP = r) in X-Y plane.

Let θ = angle made by its linear momentum vector(P and r) produced. If vector L be its angular momentum. Then

vector (L = r x P)    ...(i)

It is a vector quantity and its direction is given by right hand rule for vector product. As vector r and p here lie OXY plane, so vector L acts along Z axis.

In cartesian co-ordinates,

vector r = xi + yj + zk

and vector P = p_xi + p_yj + p_zk   ...(ii)

∴ From (i) and (ii), we get

vector L = (xi + yj + zk) x (p_xi + p_yj + p_zk)

or, L_xi + L_yj + L_zk = i(yp_z - zp_y) + j(zp_x - xp_z) + k(xp_y - yp_x)

On comparing, we get

equ. (iii) gives the required component of vector L along X,Y and Z axes.

(b) We know that the torque experienced by a particle moving in XY plane is given by

τ_Z = xF_y - yF_x    ....(1)

Where τ_z = component of torque acting along Z axis on the particle moving in XY plane.

Let m = mass of the particle having its vector v in XY plane

s.t. v_x and v_y be its velocity components along X and Y axes respectively

∴ According to Newton's Second law of motion

∴ from (3) and (4), we get

τ_z = m d/dt (xv_y - yv_x)

= d/dt (mv_y - ym_x)     ....(5)

Now as p_y = mv_y and p_x = mv_x     ....(6)

∴ From (5) and (6), we get

τ_z = d/dt(xp_y - yp_x) = d/dt L_z  (by using equation (iii) or (a) part)

∴ τ_z = dL_z/dt     ....(7)

Hence from equation (7), we conclude that a particle moving in XY plane has only one component of L i.e., L_z acting along z-axis