AcademyKinematics
Academy
Angular Momentum
Level 1 - Math II (Physics) topic page in Kinematics.
Principle
Angular momentum measures rotational motion about a chosen origin. For a particle, it is the cross product of position and linear momentum.
Torque measures how a force changes angular momentum. About a fixed origin, the time derivative of angular momentum equals the net torque.
Notation
\(\mathbf r\)
position vector from the chosen origin to the particle
\(\mathrm{m}\)
\(\mathbf p\)
linear momentum
\(\mathrm{kg\,m\,s^{-1}}\)
\(\mathbf F\)
force
\(\mathrm{N}\)
\(\mathbf L\)
angular momentum about the origin
\(\mathrm{kg\,m^{2}\,s^{-1}}\)
\(\boldsymbol\tau\)
torque about the origin
\(\mathrm{N\,m}\)
\(\times\)
vector cross product
Method
Step 1: Choose the origin
Angular momentum and torque are measured about an origin. Changing the origin can change \(\mathbf r\), \(\mathbf L\), and \(\boldsymbol\tau\).
Step 2: Compute angular momentum
For a particle with position \(\mathbf r\) and momentum \(\mathbf p\):
Angular momentum
\[\mathbf L=\mathbf r\times\mathbf p\]
Step 3: Compute torque
For a force \(\mathbf F\) applied at position \(\mathbf r\):
Torque
\[\boldsymbol\tau=\mathbf r\times\mathbf F\]
Step 4: Relate torque to angular momentum change
For a fixed origin, use the angular momentum balance:
Angular momentum balance
\[\frac{d\mathbf L}{dt}=\boldsymbol\tau\]
Start with angular momentum
\[\mathbf L=\mathbf r\times\mathbf p\]
Differentiate the cross product
\[\frac{d\mathbf L}{dt}=\frac{d\mathbf r}{dt}\times\mathbf p+\mathbf r\times\frac{d\mathbf p}{dt}\]
Use momentum parallel to velocity
\[\frac{d\mathbf r}{dt}\times\mathbf p=\mathbf v\times m\mathbf v=\mathbf 0\]
Use Newton's second law
\[\frac{d\mathbf L}{dt}=\mathbf r\times\mathbf F=\boldsymbol\tau\]
Rules
Angular momentum magnitude
\[|\mathbf L|=|\mathbf r||\mathbf p|\sin\theta\]
Torque magnitude
\[|\boldsymbol\tau|=|\mathbf r||\mathbf F|\sin\theta\]
Conservation condition
\[\boldsymbol\tau_{\text{net}}=\mathbf 0\Longrightarrow\frac{d\mathbf L}{dt}=\mathbf 0\]
- The direction of \(\mathbf r\times\mathbf p\) follows the right-hand rule.
- Only the component of momentum perpendicular to \(\mathbf r\) contributes to angular momentum magnitude.
- Zero net torque about an origin means angular momentum about that origin is constant.
- Torque depends on the point about which it is calculated.
Examples
Question
Let
\[\mathbf r=2\mathbf i\]
m and \[\mathbf p=3\mathbf j\]
\[kg m s^{-1}\]
Find \[\mathbf L\]
Answer
Compute the cross product:
\[\mathbf L=(2\mathbf i)\times(3\mathbf j)=6\mathbf k\]
\[kg m^2 s^{-1}\]
Checks
- Always state the origin for angular momentum and torque.
- Do not replace a cross product by ordinary multiplication; direction matters.
- Check whether the force line of action passes through the origin. If it does, torque about that origin is zero.
- Use the right-hand rule for the sign and direction of cross products.