The angular momentum of a particle measures the rotational effect of its linear momentum about a chosen origin.
For a particle at position \(\vec r\) with momentum \(\vec p\), the angular momentum is the cross product
\[
\vec L=\vec r\times\vec p.
\]
Since \(\vec p=m\vec v\),
\[
\vec L=m(\vec r\times\vec v).
\]
1. Linear momentum
The linear momentum of a particle is
\[
\vec p=m\vec v.
\]
If
\[
\vec v=(v_x,v_y,v_z),
\]
then
\[
\vec p=(mv_x,mv_y,mv_z).
\]
2. Cross product formula
For
\[
\vec r=(x,y,z),
\qquad
\vec p=(p_x,p_y,p_z),
\]
the angular momentum vector is
\[
\vec L=
\vec r\times\vec p.
\]
Component by component,
\[
L_x=yp_z-zp_y,
\]
\[
L_y=zp_x-xp_z,
\]
\[
L_z=xp_y-yp_x.
\]
Equivalently, using velocity,
\[
L_x=m(yv_z-zv_y),
\]
\[
L_y=m(zv_x-xv_z),
\]
\[
L_z=m(xv_y-yv_x).
\]
3. 2D angular momentum
In many introductory problems, the particle moves in the \(x\)-\(y\) plane. Then \(z=0\), \(v_z=0\), and only the
\(z\)-component of angular momentum remains:
\[
L_z=m(xv_y-yv_x).
\]
If \(L_z>0\), the 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.
4. Magnitude formula
The magnitude of angular momentum is
\[
|\vec L|=|\vec r\times\vec p|.
\]
This can also be written as
\[
|\vec L|=rp\sin\phi,
\]
where \(\phi\) is the angle between \(\vec r\) and \(\vec p\).
In component form,
\[
|\vec L|=\sqrt{L_x^2+L_y^2+L_z^2}.
\]
5. Direction and the right-hand rule
The direction of \(\vec L=\vec r\times\vec p\) is perpendicular to the plane containing \(\vec r\) and \(\vec p\).
Use the right-hand rule:
- Point your fingers along \(\vec r\).
- Curl your fingers toward \(\vec p\).
- Your thumb points in the direction of \(\vec L\).
The order matters:
\[
\vec r\times\vec p=-(\vec p\times\vec r).
\]
6. Units
Position has units of meters and momentum has units of \(\mathrm{kg\,m/s}\). Therefore angular momentum has units
\[
\mathrm{m}\cdot\mathrm{kg\,m/s}
=
\mathrm{kg\,m^2/s}.
\]
7. Solving inverse problems in 2D
In 2D,
\[
L_z=m(xv_y-yv_x).
\]
This can be rearranged to solve for one unknown if the other quantities are known. For example, solving for \(v_y\):
\[
v_y=\frac{L_z/m+yv_x}{x}.
\]
Solving for \(v_x\):
\[
v_x=\frac{xv_y-L_z/m}{y}.
\]
Solving for \(x\):
\[
x=\frac{L_z/m+yv_x}{v_y}.
\]
Solving for \(y\):
\[
y=\frac{xv_y-L_z/m}{v_x}.
\]
These formulas require the denominator in the chosen expression to be nonzero.
8. Worked example
A particle has mass
\[
m=2\ \mathrm{kg},
\]
position
\[
\vec r=(3,0,0)\ \mathrm{m},
\]
and velocity
\[
\vec v=(0,8,0)\ \mathrm{m/s}.
\]
First find momentum:
\[
\vec p=m\vec v.
\]
\[
\vec p=2(0,8,0)=(0,16,0)\ \mathrm{kg\,m/s}.
\]
Now compute the cross product:
\[
\vec L=\vec r\times\vec p.
\]
\[
\vec L=(3,0,0)\times(0,16,0).
\]
The \(z\)-component is
\[
L_z=xp_y-yp_x.
\]
\[
L_z=(3)(16)-(0)(0)=48\ \mathrm{kg\,m^2/s}.
\]
Therefore,
\[
\vec L=(0,0,48)\ \mathrm{kg\,m^2/s}.
\]
Since \(L_z>0\), the angular momentum points out of the page in the \(+\hat{k}\) direction.
Key idea: angular momentum is largest when \(\vec r\) and \(\vec p\) are perpendicular, and zero when they are
parallel.
