Problem 28 - Advanced Engineering Mathematics

Question

Suppose \(m\) is the mass of a moving particle. Newton's second law of motion can be written in vector form as $$ \mathbf{F}=m \mathbf{a}=\frac{d}{d t}(m \mathbf{v})=\frac{d \mathbf{p}}{d t} $$ where \(\mathbf{p}=m \mathbf{v}\) is called linear momentum. The angular momentum of the particle with respect to the origin is defined to be \(\mathbf{L}=\mathbf{r} \times \mathbf{p}\), where \(\mathbf{r}\) is its position vector. If the torque of the particle about the origin is \(\boldsymbol{\tau}=\mathbf{r} \times \mathbf{F}=\mathbf{r} \times d \mathbf{p} / d t\) show that \(\tau\) is the time rate of change of angular momentum.

Step-by-Step Solution

Answer

Torque is the time rate of change of angular momentum: \(\boldsymbol{\tau} = \frac{d\mathbf{L}}{dt}\).

Step 1: Identify the Given Information

We are given that the torque \(\boldsymbol{\tau}\) is defined as \(\mathbf{r} \times \mathbf{F}\), where \(\mathbf{F}\) is the force vector, and \(\mathbf{F} = \frac{d \mathbf{p}}{d t}\). The angular momentum \(\mathbf{L}\) is given as \(\mathbf{L} = \mathbf{r} \times \mathbf{p}\), where \(\mathbf{p} = m \mathbf{v}\) is the linear momentum.

Step 2: Differentiate Angular Momentum

To find if torque is the time rate of change of angular momentum, we need to differentiate \(\mathbf{L} = \mathbf{r} \times \mathbf{p}\) with respect to time \(t\): \[ \frac{d\mathbf{L}}{dt} = \frac{d}{dt}(\mathbf{r} \times \mathbf{p}) \]

Step 3: Apply Product Rule for Differentiation

Using the product rule for differentiation of a cross product, we write: \[ \frac{d}{dt}(\mathbf{r} \times \mathbf{p}) = \frac{d\mathbf{r}}{dt} \times \mathbf{p} + \mathbf{r} \times \frac{d\mathbf{p}}{dt} \] Here, \(\frac{d \mathbf{r}}{d t} = \mathbf{v}\) is the velocity of the particle, and \(\frac{d \mathbf{p}}{d t} = \mathbf{F}\) is the force exerted on the particle.

Step 4: Simplify Using kinematics

Substituting \(\mathbf{v}\) and \(\mathbf{F}\) into the differentiated expression, we get: \[ \frac{d\mathbf{L}}{dt} = \mathbf{v} \times \mathbf{p} + \mathbf{r} \times \mathbf{F} \] Since the cross product \(\mathbf{v} \times \mathbf{p}\) is zero because \(\mathbf{v}\) and \(\mathbf{p}\) are parallel, we simplify this to: \[ \frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F} \]

Step 5: Conclude that Torque is the Time Rate of Change of Angular Momentum

From step 4, we have shown that \(\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \mathbf{F}\), which is the definition of torque \(\boldsymbol{\tau}\). Therefore, \[ \boldsymbol{\tau} = \frac{d\mathbf{L}}{dt} \]This confirms that torque is indeed the time rate of change of angular momentum.

Key Concepts

Newton's Second Law

Newton's second law is fundamental in understanding the dynamics of objects. It states that the force acting on an object is equal to the rate of change of its momentum over time. In equation form, it's expressed as:

• \( \mathbf{F} = m \mathbf{a} \)
• \( \mathbf{F} = \frac{d\mathbf{p}}{dt} \)

Here, \( \mathbf{F} \) is the force applied, \( m \) is the mass, \( \mathbf{a} \) is the acceleration, and \( \mathbf{p} = m \mathbf{v} \) is the linear momentum. This law implies that the force on a particle is responsible for changing its motion. When mass is constant, we can use this relation to understand how velocity changes under different forces. It forms the foundation to analyze more complex interactions like torque and angular momentum.

Torque

Torque is a measure of the rotating effect of a force about a specific point or axis and is crucial in the study of angular dynamics. It's expressed in vector form as:

• \( \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \)

Here, \( \boldsymbol{\tau} \) represents torque, \( \mathbf{r} \) is the position vector, and \( \mathbf{F} \) is the force applied. The cross product here indicates that torque depends on both the magnitude of the force and the distance from the point of rotation, as well as the angle between the force and the position vector.
In simpler terms, torque is what causes objects to spin or rotate. If you push a door close to the hinge, it requires more effort than pushing it at the handle. This is because there's less torque near the hinge. Torque's relationship with angular momentum explains rotational motions.

Linear Momentum

Linear momentum is a key concept to grasp when discussing motion. Simply put, it's the product of an object's mass and velocity:

• \( \mathbf{p} = m \mathbf{v} \)

This vector quantity points in the same direction as velocity. Linear momentum helps understand how force affects motion.
For instance, a heavy truck moving at a certain speed has more momentum than a small car moving at the same speed due to its larger mass. Changes in an object's momentum are directly the result of forces acting upon it, aligning with Newton's second law. This concept makes it much easier to calculate effects in collisions where forces are involved.

Cross Product

The cross product is essential to calculating both torque and angular momentum. It involves two vectors and results in a new vector that is perpendicular to the plane containing the two original vectors. For two vectors \( \mathbf{A} \) and \( \mathbf{B} \), it is defined as:

• \( \mathbf{A} \times \mathbf{B} = \|\mathbf{A}\| \|\mathbf{B}\| \sin\theta \ \mathbf{n} \)

Where \( \theta \) is the angle between \( \mathbf{A} \) and \( \mathbf{B} \), and \( \mathbf{n} \) is a unit vector perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \). This is why for applications like torque, if the force and position vectors are parallel, the torque is zero because \( \sin 0 = 0 \).
The cross product highlights the rotational impact of forces, and is pervasive in physics when determining how vectors influence each other in 3D space.