Angular momentum in Cartesian coordinates is calculated by writing the position vector and momentum vector in
component form, then applying the cross product.
\[
\vec L=\vec r\times\vec p.
\]
For a particle of mass \(m\) and velocity \(\vec v\),
\[
\vec p=m\vec v.
\]
Therefore,
\[
\vec L=m(\vec r\times\vec v).
\]
1. Position vector relative to the origin
Angular momentum is always measured about a chosen origin or pivot. If the particle is at
\[
\vec r_{\mathrm{particle}}=(x,y,z)
\]
and the chosen origin is
\[
\vec r_O=(x_O,y_O,z_O),
\]
then the position vector used in the cross product is
\[
\vec r=\vec r_{\mathrm{particle}}-\vec r_O.
\]
2. Linear momentum in coordinates
If
\[
\vec v=(v_x,v_y,v_z),
\]
then the momentum vector is
\[
\vec p=m\vec v.
\]
Component by component,
\[
p_x=mv_x,
\qquad
p_y=mv_y,
\qquad
p_z=mv_z.
\]
3. Cross product components
Let
\[
\vec r=(x,y,z),
\qquad
\vec p=(p_x,p_y,p_z).
\]
Then
\[
\vec L=\vec r\times\vec p.
\]
The Cartesian components are
\[
L_x=yp_z-zp_y,
\]
\[
L_y=zp_x-xp_z,
\]
\[
L_z=xp_y-yp_x.
\]
These formulas come directly from the determinant form
\[
\vec r\times\vec p
=
\begin{vmatrix}
\hat{\imath} & \hat{\jmath} & \hat{k}\\
x & y & z\\
p_x & p_y & p_z
\end{vmatrix}.
\]
4. Angular momentum for a system of particles
For several particles, compute each particle's angular momentum separately and add the vectors:
\[
\vec L_{\mathrm{total}}=\sum_i \vec L_i.
\]
Since
\[
\vec L_i=\vec r_i\times\vec p_i,
\]
the total is
\[
\vec L_{\mathrm{total}}
=
\sum_i \vec r_i\times\vec p_i.
\]
Vector addition is component-wise:
\[
L_{x,\mathrm{total}}=\sum_i L_{x,i},
\]
\[
L_{y,\mathrm{total}}=\sum_i L_{y,i},
\]
\[
L_{z,\mathrm{total}}=\sum_i L_{z,i}.
\]
5. Magnitude and direction
Once the components are known, the magnitude is
\[
|\vec L|=\sqrt{L_x^2+L_y^2+L_z^2}.
\]
The direction is perpendicular to the plane formed by \(\vec r\) and \(\vec p\), following the right-hand rule.
Point your fingers along \(\vec r\), curl toward \(\vec p\), and your thumb gives the direction of \(\vec L\).
6. Special 2D case
If all motion is in the \(x\)-\(y\) plane, then
\[
z=0,\qquad p_z=0.
\]
The \(x\)- and \(y\)-components vanish:
\[
L_x=0,\qquad L_y=0.
\]
Only the \(z\)-component remains:
\[
L_z=xp_y-yp_x.
\]
Since \(p_x=mv_x\) and \(p_y=mv_y\),
\[
L_z=m(xv_y-yv_x).
\]
If \(L_z>0\), angular momentum points out of the page in the \(+\hat{k}\) direction. If \(L_z<0\), it points into
the page in the \(-\hat{k}\) direction.
7. Link with torque
Net external torque changes angular momentum:
\[
\vec\tau_{\mathrm{net}}=\frac{d\vec L}{dt}.
\]
Over a short time interval \(\Delta t\), a constant torque gives
\[
\Delta\vec L=\vec\tau_{\mathrm{net}}\Delta t.
\]
Therefore,
\[
\vec L_{\mathrm{new}}=\vec L_{\mathrm{old}}+\vec\tau_{\mathrm{net}}\Delta t.
\]
8. Units
Momentum has units
\[
\mathrm{kg\,m/s}.
\]
Since angular momentum is position times momentum, its SI unit is
\[
\mathrm{m}\cdot\mathrm{kg\,m/s}
=
\mathrm{kg\,m^2/s}.
\]
9. Worked example
A particle has
\[
m=1.5\ \mathrm{kg},
\]
position
\[
\vec r=(2,3,0)\ \mathrm{m},
\]
and velocity
\[
\vec v=(4,-5,0)\ \mathrm{m/s}.
\]
First calculate momentum:
\[
\vec p=m\vec v.
\]
\[
\vec p=1.5(4,-5,0).
\]
\[
\vec p=(6,-7.5,0)\ \mathrm{kg\,m/s}.
\]
Now calculate the angular momentum components:
\[
L_x=yp_z-zp_y=(3)(0)-(0)(-7.5)=0.
\]
\[
L_y=zp_x-xp_z=(0)(6)-(2)(0)=0.
\]
\[
L_z=xp_y-yp_x=(2)(-7.5)-(3)(6).
\]
\[
L_z=-15-18=-33\ \mathrm{kg\,m^2/s}.
\]
Therefore,
\[
\vec L=(0,0,-33)\ \mathrm{kg\,m^2/s}.
\]
The negative \(z\)-component means the angular momentum points into the page.
Key idea: in coordinates, angular momentum is a vector built from the cross product of position and momentum.
