Angular Momentum and Torque in Rotational Motion
- When dealing with extended rigid bodies, forces can cause both linear and rotational motion.
- To understand this, we need to explore angular momentum, torque, and their real-world applications.
Angular Momentum: A Measure of Rotational Motion
Angular momentum
Angular momentum is the rotational equivalent of linear momentum. More precisely, it is the product of its moment of inertia and its angular velocity.
For a rigid body rotating about a fixed axis, it is defined as:
$$
L = I \omega
$$
where:
- $L$ is the angular momentum.
- $I$ is the moment of inertia.
- $\omega$ is the angular velocity.
Angular momentum is a vector quantity, but in this course, we focus on its magnitude.
Conservation of Angular Momentum
Conservation of angular momentum
Angular momentum is conserved unless an external torque acts on the system.
- Imagine a figure skater spinning with her arms extended.
- As she pulls her arms in, her moment of inertia decreases.
- To conserve angular momentum, her angular velocity increases, causing her to spin faster.
A figure skater is spinning with her arms extended. Her moment of inertia with arms extended is $I_1 = 5.0 \, \mathrm{kg \cdot m^2}$, and her initial angular velocity is $\omega_1 = 2.0 \, \text{rad s}^{-1}$. She pulls her arms in, reducing her moment of inertia to $I_2 = 2.0 \, \mathrm{kg \cdot m^2}$.
What is her final angular velocity $\omega_2$?
Solution
- The angular momentum of the system is conserved, so: $$L_1 = L_2$$
- Where angular momentum $L$ is: $$L = I \cdot \omega$$
- Thus: $$I_1 \cdot \omega_1 = I_2 \cdot \omega_2$$
- Solving for $\omega_2$: $$\omega_2 = \frac{I_1 \cdot \omega_1}{I_2}$$
- Substitute the values: $$\omega_2 = \frac{5.0 \cdot 2.0}{2.0}$$ $$\omega_2 = 5.0 \, \text{rad s}^{-1}$$
- Result: Final angular velocity is $\omega_2 = 5.0 \, \text{rad s}^{-1}$
Torque and Angular Acceleration
Torque
Torque is the rotational equivalent of force. It measures the ability of a force to cause an object to rotate.
Torque and Angular Acceleration
Torque is directly related to angular acceleration through the equation:
$$
\tau = I \alpha
$$
where:
- $\tau$ is the torque.
- $I$ is the moment of inertia.
- $\alpha$ is the angular acceleration.
This equation is the rotational analogue of Newton’s second law, $F = ma$.
The unit of torque is the Nm (Newton-meter).
A solid disk of radius $R = 0.5 \, \mathrm{m}$ and mass $M = 4.0 \, \mathrm{kg}$ is initially at rest. A force $F = 10.0 \, \mathrm{N}$ is applied tangentially at the edge of the disk.
Calculate the angular acceleration $\alpha$ of the disk.
Solution
Step 1: Moment of Inertia of the Disk
- The moment of inertia $I$ for a solid disk rotating about its central axis is given by: $$I = \frac{1}{2} M R^2$$
- Substitute the values: $$I = \frac{1}{2} (4.0) (0.5)^2$$ $$ = \frac{1}{2} (4.0) (0.25) = 0.5 \, \mathrm{kg \cdot m^2}$$
Step 2: Torque Acting on the Disk
- The torque $\tau$ is calculated as: $$\tau = F \cdot R$$
- Substitute the values: $$\tau = (10.0) (0.5) = 5.0 \, \mathrm{N \cdot m}$$
Step 3: Angular Acceleration
- Using the rotational form of Newton's second law: $$\tau = I \cdot \alpha$$
- Rearranging for $\alpha$: $$\alpha = \frac{\tau}{I}$$
- Substitute the values: $$\alpha = \frac{5.0}{0.5} = 10.0 \, \text{rad s}^{-2}$$
Result:
- Angular acceleration is $\alpha = 10.0 \, \text{rad s}^{-2}$
Angular Impulse: Changing Angular Momentum
Angular impulse
Angular impulse is a concept in rotational motion that describes the change in angular momentum of a rigid body due to the application of torque over a period of time.
It is mathematically expressed as:
$$\Delta L = \tau \Delta t$$
where:
- $\Delta L$ is the change in angular momentum.
- $\tau$ is the torque applied.
- $\Delta t$ is the time duration over which the torque acts.
- Angular impulse is the rotational analogue of linear impulse, which relates force to the change in linear momentum.
- Just as a force applied for a certain time changes the linear momentum of an object, a torque applied for a certain time changes the angular momentum of a rigid body.
- If no external torque acts on a system, $\tau = 0$, and angular momentum remains conserved ($\Delta L = 0$).
- However, when an external torque is applied over time, the angular impulse explains how the angular momentum changes.
A wheel of moment of inertia $I = 0.2 \, \mathrm{kg \cdot m^2}$ is initially at rest. A constant torque of $0.5 \, \mathrm{N \cdot m}$ is applied for $4.0 \, \mathrm{s}$.
What is the angular momentum of the wheel after this time?
Solution
- Using the angular impulse formula: $$\Delta L = \tau \Delta t$$
- Substitute the values: $$\Delta L = (0.5)(4.0) = 2.0 \, \mathrm{kg \cdot m^2 \ s^{-1}}$$
- The angular momentum of the wheel after $4.0 \, \mathrm{s}$ is $2.0 \, \mathrm{kg \cdot m^2 \ s^{-1}}$.
