Problem 40 - 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 \(\tau(t)=m \mathbf{r}(t) \times \mathbf{a}(t) .\) Show that \(\mathbf{L}^{\prime}(t)=\tau(t)\) .Deduce that if \(\tau(t)=0\) for all \(t\) then \(\mathbf{L}(t)\) is constant. This is the Iaw of conservation of angular momentum.\()\)

Step-by-Step Solution

Answer

Angular momentum's derivative equals torque. If torque is zero, angular momentum is constant.

Step 1: Express Angular Momentum's Time Derivative

First, we need to find the time derivative of the angular momentum vector \( \mathbf{L}(t) = m \mathbf{r}(t) \times \mathbf{v}(t) \). To do this, apply the product rule for differentiation to the cross product:\[ \frac{d}{dt} (\mathbf{r}(t) \times \mathbf{v}(t)) = \frac{d \mathbf{r}(t)}{dt} \times \mathbf{v}(t) + \mathbf{r}(t) \times \frac{d \mathbf{v}(t)}{dt} \]

Step 2: Substitute Velocity and Acceleration

Substitute \( \frac{d \mathbf{r}(t)}{dt} = \mathbf{v}(t) \) and \( \frac{d \mathbf{v}(t)}{dt} = \mathbf{a}(t) \) into the expression:\[ \mathbf{v}(t) \times \mathbf{v}(t) + \mathbf{r}(t) \times \mathbf{a}(t) \]Note that the cross product of any vector with itself is \( \mathbf{v}(t) \times \mathbf{v}(t) = \mathbf{0} \).

Step 3: Evaluate the Expression

This simplification leaves us with:\[ \mathbf{0} + \mathbf{r}(t) \times \mathbf{a}(t) = \mathbf{r}(t) \times \mathbf{a}(t) \]

Step 4: Relate to Torque

Recall the definition of torque: \( \tau(t) = m \mathbf{r}(t) \times \mathbf{a}(t) \). Notice that:\[ \frac{d\mathbf{L}(t)}{dt} = m (\mathbf{r}(t) \times \mathbf{a}(t)) = \tau(t) \]Hence, we have shown that \( \mathbf{L}'(t) = \tau(t) \).

Step 5: Deduce the Conservation of Angular Momentum

If \( \tau(t) = 0 \) for all time \( t \), then \( \mathbf{L}'(t) = 0 \) which implies that \( \mathbf{L}(t) \) is constant. This confirms the law of conservation of angular momentum: If no external torque acts on the system, the angular momentum remains constant.

Key Concepts

Angular Momentum

Angular momentum is an important concept in physics, especially when dealing with rotating objects. It is the rotational equivalent of linear momentum. Imagine a particle moving in a circular path or another form of rotation. The angular momentum, denoted as \( \mathbf{L}(t) \), is calculated using the formula: \( \mathbf{L}(t) = m \mathbf{r}(t) \times \mathbf{v}(t) \). Here, \( m \) is the mass of the particle, \( \mathbf{r}(t) \) is the position vector, and \( \mathbf{v}(t) \) is the velocity vector.

Angular momentum depends on:

• The mass of the object
• How far the object is from the axis of rotation, represented by the position vector \( \mathbf{r}(t) \)
• The speed at which the object rotates, given by the velocity vector \( \mathbf{v}(t) \)

The cross product between \( \mathbf{r}(t) \) and \( \mathbf{v}(t) \) ensures that angular momentum is a vector quantity, having both a magnitude and a direction.

Torque

Torque is to rotational motion what force is to linear motion. It measures the tendency of a force to rotate an object about an axis. The torque \( \tau(t) \) is given by the equation \( \tau(t) = m \mathbf{r}(t) \times \mathbf{a}(t) \), where \( \mathbf{a}(t) \) is the acceleration vector.

Torque can be understood as:

• Dependent on the distance from the pivot point, as represented by \( \mathbf{r}(t) \)
• Influenced by the acceleration, which changes the speed of rotation
• The cross product which again indicates that torque is a vector

When no external torque is applied to a system (\( \tau(t) = 0 \)), the angular momentum remains constant. This principle is the foundation of the conservation of angular momentum, which states that, in an isolated system, the angular momentum does not change.

Cross Product

The cross product is a mathematical operation that takes two vectors and produces another vector perpendicular to the plane of the input vectors. It is significant in calculating both angular momentum and torque.

For vectors \( \mathbf{a} \) and \( \mathbf{b} \):

• The cross product, \( \mathbf{a} \times \mathbf{b} \), creates a vector orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).
• The direction of the resulting vector follows the right-hand rule.
• The magnitude is given by \( \|\mathbf{a}\| \|\mathbf{b}\| \sin \theta \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).

In the context of angular momentum, the cross product captures how the position and velocity vectors interact to determine the rotation.

Differentiation

Differentiation in calculus refers to finding the rate at which one quantity changes concerning another. In our context, it helps examine how the angular momentum \( \mathbf{L}(t) \) changes over time. The derivative of angular momentum aligns with the definition of torque.

To differentiate angular momentum:

• We use the product rule for derivatives since angular momentum involves a cross product of two time-dependent vectors, \( \mathbf{r}(t) \) and \( \mathbf{v}(t) \).
• By differentiating each part and applying the product rule: \( \frac{d}{dt}(\mathbf{r}(t) \times \mathbf{v}(t)) \), we ensure the time derivative respects both vectors' time dependencies.
• From the differentiation, we derive \( \mathbf{L}'(t) = \tau(t) \), proving that the rate of change of angular momentum equals the applied torque.

Thus, differentiation provides a powerful tool to connect change in motion to the forces causing it, highlighting the relationship between concepts.