AP Physics 1- 6.3 Angular Momentum and Angular Impulse- Study Notes- New Syllabus
AP Physics 1-6.3 Angular Momentum and Angular Impulse – Study Notes
AP Physics 1-6.3 Angular Momentum and Angular Impulse – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Angular Momentum
- Angular Impulse
- Angular Impulse–Angular Momentum Relationship
Angular Momentum
The magnitude of the angular momentum of a rigid system about a specific axis is:
\( L = I \omega \)
- \( I \) = moment of inertia of the system about the axis
- \( \omega \) = angular velocity
Magnitude for a Particle/Object:
The magnitude of the angular momentum of an object about a given point is:
\( L = m v r \sin \theta \)
- \( m \) = mass of the object
- \( v \) = linear speed of the object
- \( r \) = perpendicular distance between the point (axis) and line of motion
- \( \theta \) = angle between \( \vec{r} \) and \( \vec{v} \)
Relationship with torque:
\( \tau_{net} = \dfrac{dL}{dt} \)
Effect of Axis Selection:
- The angular momentum of an object depends on the chosen axis or reference point.
- Changing the axis changes the distance \( r \), which directly changes the measured angular momentum.
- In many problems, the axis is chosen at the center of mass or at a point where torque analysis is simpler.
Key Dependence for Straight-Line Motion:
- For an object moving in a straight line, its angular momentum relative to a point depends on:
- Its mass \( m \)
- Its speed \( v \)
- The distance \( r \) from the point to the line of motion
- The angle \( \theta \) between \( \vec{r} \) and \( \vec{v} \)
Example:
A 2 kg ball is moving in a straight line with a speed of 4 m/s. At a given instant, the ball passes a point that is 0.5 m away (perpendicular distance) from the line of motion. Find the angular momentum of the ball about that point.
▶️Answer/Explanation
Using the formula: \( L = m v r \sin \theta \)
Here, \( m = 2 \, \text{kg}, \; v = 4 \, \text{m/s}, \; r = 0.5 \, \text{m}, \; \theta = 90^\circ \; \Rightarrow \sin 90^\circ = 1 \)
\( L = 2 \cdot 4 \cdot 0.5 \cdot 1 = 4 \, \text{kg·m}^2/\text{s} \)
Final Answer: The angular momentum of the ball about the point = 4 kg·m²/s.
Example:
A skater is spinning with an angular velocity of 2 rad/s and moment of inertia 5 kg·m². She pulls her arms in, reducing her moment of inertia to 2 kg·m². Find her new angular velocity.
▶️Answer/Explanation
Conservation of angular momentum: \( I_1 \omega_1 = I_2 \omega_2 \)
\( 5 \cdot 2 = 2 \cdot \omega_2 \)
\( \omega_2 = \dfrac{10}{2} = 5 \, \text{rad/s} \)
Final Answer: Her angular velocity increases to 5 rad/s.
Example:
A uniform disc of mass 4 kg and radius 0.5 m is rotating about its central axis with angular velocity 10 rad/s. Find its angular momentum.
▶️Answer/Explanation
Moment of inertia of disc: \( I = \tfrac{1}{2} M R^2 = \tfrac{1}{2} \cdot 4 \cdot (0.5^2) = 0.5 \, \text{kg·m}^2 \)
Angular momentum: \( L = I \omega = 0.5 \cdot 10 = 5 \, \text{kg·m}^2/\text{s} \)
Final Answer: Angular momentum = 5 kg·m²/s.
Angular Impulse
Angular impulse is defined as the product of the torque exerted on an object or rigid system and the time interval during which the torque is exerted. It is the rotational analogue of linear impulse and describes how torque changes the angular momentum of a system.
Key Points:
- Angular impulse is a vector quantity and has the same direction as the torque exerted on the object or system (right-hand rule).
- It represents the effect of torque applied over time, just as linear impulse represents force applied over time.
- Angular impulse changes a system’s angular momentum.
Relevant Equations:
For constant torque:
$ J_\tau = \tau \Delta t $
General case:
$ J_\tau = \int_{\;t_1}^{t_2} \tau \, dt $
Example :
A torque of 8 N·m is applied on a wheel for 5 seconds. Find the angular impulse and the change in angular momentum.
▶️Answer/Explanation
Using formula: \( J_\tau = \tau \Delta t \)
\( J_\tau = 8 \cdot 5 = 40 \, \text{N·m·s} \)
Since \( J_\tau = \Delta L \), the change in angular momentum is 40 N·m·s.
Example :
A torque increases linearly from 0 to 12 N·m over 4 seconds. Find the angular impulse on the system.
▶️Answer/Explanation
Angular impulse = Area under torque–time graph.
Graph is a triangle with base = 4 s, height = 12 N·m.
Area = \( \tfrac{1}{2} \cdot 4 \cdot 12 = 24 \, \text{N·m·s} \)
Final Answer: Angular impulse = 24 N·m·s.
Angular Impulse–Angular Momentum Relationship
The change in angular momentum of an object or rigid system is directly related to the angular impulse delivered to it. This is the rotational analogue of the linear impulse–momentum theorem
Main Points:
- The magnitude of the change in angular momentum can be described by comparing the magnitudes of the final and initial angular momenta of the system: $ \Delta L = L_f – L_i $
- A rotational form of the impulse–momentum theorem states: $ J_\tau = \Delta L $ where \(J_\tau\) is angular impulse, and \(\Delta L\) is the change in angular momentum.
- The angular impulse exerted on an object or rigid system is equal to the change in angular momentum of that object or rigid system.
- This theorem is a direct result of the rotational form of Newton’s Second Law when rotational inertia is constant: $ \tau_{net} = I \alpha $
- The net torque exerted on an object is equal to the slope of the angular momentum–time graph.
- The angular impulse delivered to an object is equal to the area under the torque–time graph.
Relevant Equations:
Angular Impulse:
$ J_\tau = \tau \Delta t \quad \text{(constant torque)} $
$ J_\tau = \int_{t_1}^{t_2} \tau \, dt \quad \text{(general case)} $
Impulse–Momentum Theorem (Rotational Form):
$ J_\tau = \Delta L $
Example :
A wheel of rotational inertia \( I = 4 \, \text{kg·m}^2 \) is at rest. A constant torque of \( \tau = 6 \, \text{N·m} \) is applied for 3 seconds. Find the angular momentum gained by the wheel.
▶️Answer/Explanation
Angular impulse: \( J_\tau = \tau \Delta t = 6 \cdot 3 = 18 \, \text{N·m·s} \).
By theorem: \( J_\tau = \Delta L \).
Thus, change in angular momentum: \( \Delta L = 18 \, \text{kg·m}^2/\text{s} \).
Example :
A torque increases linearly from 0 to 10 N·m over 4 seconds on a rigid body. Find the change in angular momentum.
▶️Answer/Explanation
Angular impulse = Area under torque–time graph (triangle).
Area = \( \tfrac{1}{2} \cdot 4 \cdot 10 = 20 \, \text{N·m·s} \).
So, change in angular momentum: \( \Delta L = 20 \, \text{kg·m}^2/\text{s} \).
Example:
A disk of moment of inertia \( I = 5.0\ \text{kg·m}^2 \) has a net torque \( \tau(t) \) acting on it as shown:
Find the magnitude and sense of the angular velocity at \(t = 5.0\ \text{s}\).
▶️Answer/Explanation
Rotational impulse (net) equals change in angular momentum:
\( J_{\text{rot}}^{\text{net}} = \Delta L = I\omega_f – I\omega_i \).
- \(J_{\text{rot}}^{\text{net}}\) = area under the torque–time graph (areas above time-axis are positive — CCW; below are negative — CW).
Triangle from \(t=0\) to \(t=3\ \text{s}\) with peak \(10\ \text{N·m}\):
\( A_1 = \dfrac{1}{2}\times 3 \times 10 = 15\ \text{N·m·s} \).
Rectangle from \(t=3\) to \(t=5\ \text{s}\) at \(-5\ \text{N·m}\):
\( A_2 = (-5)\times 2 = -10\ \text{N·m·s} \).
Net impulse:
\( J_{\text{rot}}^{\text{net}} = 15 + (-10) = 5\ \text{N·m·s} \).
Initial angular velocity is clockwise → choose CCW positive: \( \omega_i = -2.0\ \text{rad/s} \).
- Apply \( J_{\text{rot}}^{\text{net}} = I\omega_f – I\omega_i \).
Substitute values:
\( 5 = 5\omega_f – 5(-2) \)
\( 5 = 5\omega_f + 10 \)
\( 5\omega_f = -5 \)
\( \omega_f = -1.0\ \text{rad/s} \).
- \( \boxed{\omega_f = -1.0\ \text{rad/s}} \) → final angular speed \(1.0\ \text{rad/s}\) clockwise.