Angular momentum IB DP Physics Study Notes - 2025 Syllabus
Angular momentum IB DP Physics Study Notes
Angular momentum IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
- that an extended body rotating with an angular speed has an angular momentum $L$ as given by $L = I \omega$
- that angular momentum remains constant unless the body is acted upon by a resultant torque
- that the action of a resultant torque constitutes an angular impulse $\Delta L$ as given by
$\Delta L = \tau \Delta t = \Delta (I \omega)$ - the kinetic energy of rotational motion as given by
$E_k = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$
Standard level and higher level: There is no standard level content in A.4
Additional higher level: 7 hours
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Angular Momentum of a Rotating Extended Body
An extended body rotating with angular velocity \( \omega \) possesses angular momentum \( L \), defined as:
\( L = I \omega \)
Where:
- \( L \) = angular momentum (in kg·m²/s)
- \( I \) = moment of inertia of the object (kg·m²)
- \( \omega \) = angular velocity (rad/s)
Angular momentum is a vector quantity and its direction follows the right-hand rule. It is the rotational analogue of linear momentum \( p = mv \).
Example:
A solid cylinder (moment of inertia \( I = 0.4 \, \text{kg·m}^2 \)) spins with angular velocity \( \omega = 10 \, \text{rad/s} \). Find its angular momentum.
▶️ Answer/Explanation
Use: \( L = I \omega = 0.4 \times 10 = \boxed{4.0 \, \text{kg·m}^2/\text{s}} \)
Conservation of Angular Momentum
If no external (resultant) torque acts on a system, its total angular momentum remains constant.
This is known as the principle of conservation of angular momentum:
\( L_{\text{initial}} = L_{\text{final}} \Rightarrow I_1\omega_1 = I_2\omega_2 \)
This principle is crucial in analyzing closed systems such as:
- Spinning figure skaters pulling arms inward to spin faster
- Collapsing stars (supernovae or neutron stars)
Only external torques can change the total angular momentum of a system.
Example:
A skater with moment of inertia \( 3.0 \, \text{kg·m}^2 \) is spinning at \( 2.0 \, \text{rad/s} \). She pulls in her arms and reduces her moment of inertia to \( 1.0 \, \text{kg·m}^2 \). What is her new angular velocity?
▶️ Answer/Explanation
Use conservation of angular momentum:
\( I_1\omega_1 = I_2\omega_2 \Rightarrow 3.0 \times 2.0 = 1.0 \times \omega_2 \)
\( \omega_2 = \frac{6.0}{1.0} = \boxed{6.0 \, \text{rad/s}} \)
Angular Impulse and Change in Angular Momentum
A net torque acting over a time interval causes a change in angular momentum.
This product is called the angular impulse:
\( \Delta L = \tau \Delta t = I \Delta \omega \)
Where:
- \( \Delta L \) = change in angular momentum (kg·m²/s)
- \( \tau \) = average torque (N·m)
- \( \Delta t \) = time interval (s)
This is the rotational analogue of linear impulse: \( F \Delta t = \Delta p \).
If \( \tau = 0 \), then \( \Delta L = 0 \): angular momentum is conserved.
Example:
A torque of \( 5.0 \, \text{N·m} \) acts on a wheel for \( 4.0 \, \text{s} \). The wheel has a moment of inertia of \( 2.0 \, \text{kg·m}^2 \). Find the change in angular momentum and angular velocity.
▶️ Answer/Explanation
Step 1: Angular impulse:
\( \Delta L = \tau \Delta t = 5.0 \times 4.0 = \boxed{20 \, \text{kg·m}^2/\text{s}} \)
Step 2: Change in angular velocity:
\( \Delta \omega = \frac{\Delta L}{I} = \frac{20}{2.0} = \boxed{10 \, \text{rad/s}} \)
Rotational Kinetic Energy
A rotating rigid body possesses kinetic energy due to its motion.
This energy is called rotational kinetic energy and is given by:
\( E_k = \frac{1}{2} I \omega^2 \)
Where:
- \( E_k \) = rotational kinetic energy (Joules)
- \( I \) = moment of inertia (kg·m²)
- \( \omega \) = angular speed (rad/s)
It can also be expressed using angular momentum \( L \):
\( E_k = \frac{L^2}{2I} \)
This form is useful when angular momentum is known instead of angular speed.
Example:
A flywheel with moment of inertia \( 0.50 \, \text{kg·m}^2 \) rotates at \( 8.0 \, \text{rad/s} \). Calculate its rotational kinetic energy. Also compute it using angular momentum.
▶️ Answer/Explanation
Method 1: Using \( \frac{1}{2} I \omega^2 \)
\( E_k = \frac{1}{2} \cdot 0.50 \cdot (8.0)^2 = 0.25 \cdot 64 = \boxed{16 \, \text{J}} \)
Method 2: Using \( \frac{L^2}{2I} \)
\( L = I \omega = 0.50 \cdot 8.0 = 4.0 \, \text{kg·m}^2/\text{s} \)
\( E_k = \frac{(4.0)^2}{2 \cdot 0.50} = \frac{16}{1.0} = \boxed{16 \, \text{J}} \)