Problem 34 - Essential Calculus Early Transcendentals

Question

If a particle with mass \(m\) moves with position vector \(\mathbf{r}(t),\) then its angular momentum is defined as \(\mathbf{L}(t)=m \mathbf{r}(t) \times \mathbf{v}(t)\) and its torque as \(\boldsymbol{\tau}(t)=m \mathbf{r}(t) \times \mathbf{a}(t)\). Show that \(\mathbf{L}^{\prime}(t)=\boldsymbol{\tau}(t) .\) Deduce that if \(\boldsymbol{\tau}(t)=\mathbf{0}\) for all \(t\) then \(\mathbf{L}(t)\) is constant. (This is the law of conservation of angular momentum.)

Step-by-Step Solution

Answer

If torque \( \boldsymbol{\tau}(t) = \mathbf{0} \), angular momentum \( \mathbf{L}(t) \) is constant.

Step 1: Understanding Angular Momentum

The angular momentum of a particle is given by \( \mathbf{L}(t) = m \mathbf{r}(t) \times \mathbf{v}(t) \), where \( *\times* \) represents a cross product. Here, \( m \) is the mass, \( \mathbf{r}(t) \) is the position vector, and \( \mathbf{v}(t) \) is the velocity vector.

Step 2: Calculating the Derivative

Differentiate \( \mathbf{L}(t) = m \mathbf{r}(t) \times \mathbf{v}(t) \) with respect to time \( t \). Apply the product rule for differentiation to cross products:\[ \frac{d}{dt} [ \mathbf{r}(t) \times \mathbf{v}(t) ] = \mathbf{r}'(t) \times \mathbf{v}(t) + \mathbf{r}(t) \times \mathbf{v}'(t). \]Substitute \( \mathbf{v}(t) = \mathbf{r}'(t) \) and \( \mathbf{a}(t) = \mathbf{v}'(t) \) (since velocity is the derivative of position, and acceleration is the derivative of velocity):\[ \mathbf{L}'(t) = m[ \mathbf{v}(t) \times \mathbf{v}(t) + \mathbf{r}(t) \times \mathbf{a}(t) ] \].

Step 3: Simplifying the Expression

Note that \( \mathbf{v}(t) \times \mathbf{v}(t) = \mathbf{0} \) because the cross product of any vector with itself is zero. Thus, the expression simplifies to:\[ \mathbf{L}'(t) = m \mathbf{r}(t) \times \mathbf{a}(t) \].Recognize that this is exactly \( \boldsymbol{\tau}(t) \), the torque.

Step 4: Understanding Torque and Angular Momentum Relationship

We've shown that \( \mathbf{L}'(t) = \boldsymbol{\tau}(t) \). This indicates that the rate of change of angular momentum is equal to torque.

Step 5: Deducing Conservation of Angular Momentum

If \( \boldsymbol{\tau}(t) = \mathbf{0} \) for all \( t \), then from \( \mathbf{L}'(t) = \boldsymbol{\tau}(t) \), we have \( \mathbf{L}'(t) = \mathbf{0} \). This implies that \( \mathbf{L}(t) \) is constant over time, which is the principle of conservation of angular momentum.

Key Concepts

Torque

Torque is a concept closely related to rotations and dynamics of a particle or object. It can be thought of as a rotational force. Torque is responsible for changes in the angular momentum of a system.
In simple terms, torque is what causes objects to rotate. It is defined as the cross product of the position vector \( \mathbf{r}(t) \) and the acceleration vector \( \mathbf{a}(t) \).
This is expressed in the equation:\[ \boldsymbol{\tau}(t) = m \mathbf{r}(t) \times \mathbf{a}(t) \]

• Position Vector \( \mathbf{r}(t) \): Represents the location of the particle related to a reference point.
• Acceleration Vector \( \mathbf{a}(t) \): Represents how quickly the velocity of the particle is changing with time.
• Mass \( m \): The mass of the particle, providing the scale for the amount of torque.

Understanding torque is essential in the study of rotational dynamics because it governs how the angular momentum changes over time.

Conservation of Angular Momentum

The principle of conservation of angular momentum is a fundamental concept in physics. It states that if no external torque acts on a system, the total angular momentum remains constant. This is a pivotal rule in rotational dynamics.
This principle becomes clearer if you consider the equation for the time derivative of angular momentum:\[ \mathbf{L}'(t) = \boldsymbol{\tau}(t) \]

• If \( \boldsymbol{\tau}(t) = \mathbf{0} \): No net external torque is acting on the particle.
• Then, \( \mathbf{L}'(t) = \mathbf{0} \): The angular momentum is not changing over time and is therefore constant.

This is particularly important in systems like closed and isolated systems (e.g., planets orbiting a star), where no external forces result in no change in angular momentum.
Conservation laws also greatly simplify the analysis of many mechanical systems because they allow us to predict the motion without necessarily knowing all the forces involved.

Cross Product

The cross product is a vector operation used to find a vector perpendicular to two given vectors in three-dimensional space. The cross product of vectors \( \mathbf{A} \) and \( \mathbf{B} \) is denoted by \( \mathbf{A} \times \mathbf{B} \).A few points about it:

• Magnitude: The magnitude of \( \mathbf{A} \times \mathbf{B} \) is given by the product of the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \) and the sine of the angle \( \theta \) between them: \( |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta \).
• Direction: The direction of \( \mathbf{A} \times \mathbf{B} \) is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \).
• Zero Vectors: If the vectors are parallel or if one of them is a zero vector, the cross product will be zero.

The cross product is significant in defining torque and angular momentum because it mathematically encapsulates how these quantities depend on both the point of application and the direction of forces. In our case, the torque and angular momentum equations utilize the cross product to link position, velocity, and acceleration for describing rotational motion.

Product Rule for Differentiation

The product rule is a fundamental technique in calculus for differentiating the product of two functions. When working with derivatives of cross products, the product rule helps us differentiate products of vector quantities.
Given two differentiable vector functions \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \), the product rule for differentiation can be applied as:\[ \frac{d}{dt} [ \mathbf{u}(t) \times \mathbf{v}(t) ] = \mathbf{u}'(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}'(t) \]

• First Term: The derivative of the first function cross multiplied with the second function.
• Second Term: The first function cross multiplied with the derivative of the second function.

This rule is critical when differentiating angular momentum, \( \mathbf{L}(t) = m \mathbf{r}(t) \times \mathbf{v}(t) \), because you must account for changes in both the position and velocity vectors over time when finding \( \mathbf{L}'(t) \).
By correctly applying the product rule, we see how torque emerges naturally from differentiating angular momentum, linking rotational dynamics with changes in motion.