Angular Momentum of Objects

Magnitude of Angular Momentum

Angular momentum represents the rotational equivalent of linear momentum, describing an object's tendency to continue rotating at a constant rate unless acted upon by an external torque.

For a rigid object rotating about a fixed axis, the magnitude of angular momentum is calculated using:

$L = I\omega$

Where:

• $L$ is the angular momentum (measured in kg·m²/s)
• $I$ is the moment of inertia (measured in kg·m²)
• $\omega$ is the angular velocity (measured in rad/s)

This equation shows that objects with larger moments of inertia or faster rotation rates possess greater angular momentum. For example, a spinning figure skater with arms extended (larger moment of inertia) has more angular momentum than when their arms are pulled in, assuming the same angular velocity.

Angular Momentum About a Point

When dealing with objects that aren't necessarily rotating about a fixed axis, we calculate angular momentum using the cross product of position and linear momentum vectors:

$\vec{L} = \vec{r} \times \vec{p}$

Where:

• $\vec{L}$ is the angular momentum vector
• $\vec{r}$ is the position vector from the reference point to the object
• $\vec{p}$ is the linear momentum vector ($m\vec{v}$) of the object

The choice of reference point significantly affects the calculated angular momentum. For an object moving in a straight line, its angular momentum about a point depends on:

• The perpendicular distance from the reference point to the object's path
• The object's mass
• The object's speed
• The angle between the position vector and velocity vector

The magnitude of angular momentum in this case can be calculated as: $L = rmv\sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{v}$.

Angular Impulse from Torque

Definition of Angular Impulse

Angular impulse measures the cumulative effect of torque applied over a period of time, analogous to how linear impulse relates to force.

Mathematically, angular impulse is expressed as:

$\text{Angular impulse} = \int \tau \, dt$

Where:

• $\tau$ is the torque
• $dt$ is the differential time element

For constant torque, this simplifies to: $\text{Angular impulse} = \tau \Delta t$

This concept helps us understand how torques change the rotational motion of objects over time. For instance, a longer push on a merry-go-round creates more angular impulse than a brief push with the same force.

Direction of Angular Impulse

The direction of angular impulse aligns with the direction of the torque causing it. In three dimensions, this follows the right-hand rule: if you curl the fingers of your right hand in the direction of rotation, your thumb points in the direction of the angular impulse vector.

Graphical Representation of Impulse

Angular impulse can be visualized and calculated as the area under a torque-time graph. This graphical approach is particularly useful when torque varies over time.

For a torque that changes with time, the total angular impulse equals the area under the torque vs. time curve:

$\text{Angular impulse} = \int_{t_1}^{t_2} \tau \, dt = \text{Area under } \tau \text{ vs. } t \text{ curve}$

Change in Angular Momentum

Magnitude of Angular Momentum Change

The change in angular momentum is found by comparing the final and initial angular momenta:

$\Delta L = L - L_0$

Where:

• $\Delta L$ is the change in angular momentum
• $L$ is the final angular momentum
• $L_0$ is the initial angular momentum

This calculation helps determine how much an object's rotation has been affected by applied torques. For example, when a spinning top slows down due to friction, its angular momentum decreases over time.

Impulse-Momentum Theorem for Rotation

The rotational form of the impulse-momentum theorem connects angular impulse to the change in angular momentum:

$\Delta L = \int_{t_1}^{t_2} \tau \, dt$

This fundamental relationship states that the angular impulse delivered to an object equals the change in its angular momentum. It can be derived from Newton's second law for rotation:

$\tau_{\text{net}} = \frac{dL}{dt} = I\frac{d\omega}{dt} = I\alpha$

For cases with constant moment of inertia, integrating both sides with respect to time yields the impulse-momentum theorem for rotation.

Torque and Angular Momentum Graphs

The relationship between torque and angular momentum can be understood graphically:

• The net torque exerted on an object equals the slope of the angular momentum vs. time graph
• The angular impulse delivered to an object equals the area under the torque vs. time graph

These graphical interpretations provide visual insights into how torques affect rotational motion over time. A steeper slope on an angular momentum-time graph indicates a larger torque, while a larger area under a torque-time graph indicates a greater change in angular momentum.

Practice Problem 1: Angular Momentum Calculation

Solution

To find the angular momentum, we need to use the equation $L = I\omega$.

First, let's calculate the moment of inertia: $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 2.0 \text{ kg} \times (0.30 \text{ m})^2 = \frac{1}{2} \times 2.0 \text{ kg} \times 0.09 \text{ m}^2 = 0.09 \text{ kg·m}^2$

Now we can calculate the angular momentum: $L = I\omega = 0.09 \text{ kg·m}^2 \times 5.0 \text{ rad/s} = 0.45 \text{ kg·m}^2/\text{s}$

Therefore, the angular momentum of the disk is 0.45 kg·m²/s.

Practice Problem 2: Angular Impulse and Change in Angular Momentum

Solution

We can solve this problem using the impulse-momentum theorem for rotation.

First, let's calculate the angular impulse: $\text{Angular impulse} = \tau \Delta t = 15 \text{ N·m} \times 3.0 \text{ s} = 45 \text{ N·m·s}$

According to the impulse-momentum theorem, this equals the change in angular momentum: $\Delta L = L - L_0 = 45 \text{ N·m·s}$

Since the wheel starts from rest, $L_0 = 0$, so: $L = 45 \text{ N·m·s}$

We know that $L = I\omega$, so: $\omega = \frac{L}{I} = \frac{45 \text{ N·m·s}}{2.0 \text{ kg·m}^2} = 22.5 \text{ rad/s}$

Therefore, the wheel's final angular velocity is 22.5 rad/s.

Practice Problem 3: Angular Momentum About a Point

Solution

When an object moves in a straight line, its angular momentum about a point is: $L = rmv\sin\theta$

Where:

• $r$ is the distance from the reference point to the object
• $m$ is the mass of the object
• $v$ is the speed of the object
• $\theta$ is the angle between $\vec{r}$ and $\vec{v}$

Since the path is perpendicular to the line connecting the observer and the ball, $\theta = 90°$ and $\sin\theta = 1$.

Therefore: $L = rmv\sin\theta = 3.0 \text{ m} \times 0.5 \text{ kg} \times 4.0 \text{ m/s} \times 1 = 6.0 \text{ kg·m}^2/\text{s}$

The angular momentum of the ball about the observer is 6.0 kg·m²/s.