Vector Products and Angular Quantities

Vector Product Of Two Vectors

In physics, we often encounter quantities that are products of two vectors, but where the result is itself a vector, not a scalar (like the dot product). This type of product is called the vector product or cross product.

The vector product is particularly useful in describing rotational motion, magnetism, and other phenomena where the orientation and magnitude of resulting quantities depend on the orientation and magnitudes of the input vectors.

---

Definition Of Vector Product

The vector product of two vectors $\vec{A}$ and $\vec{B}$, denoted by $\vec{A} \times \vec{B}$, is defined as a vector $\vec{C}$ that satisfies the following properties:

---

Magnitude of the Vector Product

The magnitude of the vector product $\vec{A} \times \vec{B}$ is given by:

$ |\vec{A} \times \vec{B}| = AB \sin\theta $

where $A$ and $B$ are the magnitudes of vectors $\vec{A}$ and $\vec{B}$ respectively, and $\theta$ is the angle between the two vectors ($0^\circ \le \theta \le 180^\circ$). The magnitude is maximum when $\vec{A}$ and $\vec{B}$ are perpendicular ($\theta = 90^\circ, \sin 90^\circ = 1$), and zero when $\vec{A}$ and $\vec{B}$ are parallel or anti-parallel ($\theta = 0^\circ$ or $180^\circ, \sin 0^\circ = \sin 180^\circ = 0$).

---

Direction of the Vector Product

The direction of the vector product $\vec{C} = \vec{A} \times \vec{B}$ is perpendicular to the plane containing both vectors $\vec{A}$ and $\vec{B}$. There are two possible directions perpendicular to a plane; the specific direction is determined by the right-hand rule.

Right-Hand Rule: To find the direction of $\vec{A} \times \vec{B}$, imagine placing the two vectors tail-to-tail. Point the fingers of your right hand in the direction of the first vector ($\vec{A}$). Curl your fingers towards the second vector ($\vec{B}$) through the smaller angle $\theta$. Your outstretched thumb will then point in the direction of the vector product $\vec{C} = \vec{A} \times \vec{B}$.

(Image Placeholder: A right hand with fingers pointing along vector A, curling towards vector B, and the thumb pointing perpendicular to the plane containing A and B in the direction of A x B.)

---

Properties of the Vector Product

• Non-Commutative: The order of the vectors matters. $\vec{B} \times \vec{A} = -(\vec{A} \times \vec{B})$. The magnitude is the same, but the direction is opposite.
• Distributive over addition: $\vec{A} \times (\vec{B} + \vec{C}) = (\vec{A} \times \vec{B}) + (\vec{A} \times \vec{C})$.
• Scalar Multiplication: $k(\vec{A} \times \vec{B}) = (k\vec{A}) \times \vec{B} = \vec{A} \times (k\vec{B})$, where $k$ is a scalar.
• Vector Product of Parallel Vectors: If $\vec{A}$ is parallel to $\vec{B}$ ($\theta=0^\circ$ or $180^\circ$), $\vec{A} \times \vec{B} = 0$. This means the vector product of a vector with itself is zero: $\vec{A} \times \vec{A} = 0$.

---

Vector Product in Cartesian Coordinates

If vectors $\vec{A}$ and $\vec{B}$ are given in terms of their components in a Cartesian coordinate system ($\hat{i}, \hat{j}, \hat{k}$ are unit vectors along x, y, z axes):

$ \vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} $

$ \vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k} $

The vector product $\vec{A} \times \vec{B}$ can be calculated using the determinant of a matrix:

$ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} $

$ \vec{A} \times \vec{B} = (A_yB_z - A_zB_y)\hat{i} - (A_xB_z - A_zB_x)\hat{j} + (A_xB_y - A_yB_x)\hat{k} $

Or, by remembering the cyclic properties of the unit vectors: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$, and $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{k} \times \hat{j} = -\hat{i}$, $\hat{i} \times \hat{k} = -\hat{j}$, and $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$.

Answer:

Angular Velocity And Its Relation With Linear Velocity

When a rigid body rotates about a fixed axis or a particle moves in a circular path, its motion can be described using angular quantities. Angular velocity and angular acceleration are the rotational equivalents of linear velocity and linear acceleration.

---

Angular Velocity

Consider a particle moving in a circular path of radius $r$ around a fixed centre, or a rigid body rotating about a fixed axis. As the particle or body rotates, its position changes by an angle. Angular displacement ($\Delta\theta$) is the angle through which a point or line rotates about a fixed axis. It is measured in radians.

Angular velocity ($\vec{\omega}$) is the rate of change of angular displacement with respect to time. For motion in a plane, we can define angular speed ($\omega$) as $ \omega = \frac{d\theta}{dt} $. For a rigid body rotating about a fixed axis, all particles in the body have the same angular speed.

For rotational motion, angular velocity is treated as a vector quantity. The magnitude of the angular velocity vector is the angular speed $\omega$. The direction of the angular velocity vector is along the axis of rotation and is determined by the right-hand rule:

Right-Hand Rule for Angular Velocity: Curl the fingers of your right hand in the direction of rotation. Your outstretched thumb points in the direction of the angular velocity vector $\vec{\omega}$.

For a rigid body rotating about the z-axis counterclockwise, the angular velocity vector $\vec{\omega}$ is directed along the positive z-axis ($\vec{\omega} = \omega \hat{k}$).

---

Relation between Linear Velocity and Angular Velocity

Consider a particle P moving in a circle of radius $r$ with angular velocity $\vec{\omega}$ about the origin. The linear velocity of the particle is $\vec{v}$. The position vector of the particle from the origin is $\vec{r}$.

The magnitude of the linear velocity of the particle is $v = r\omega$.

The linear velocity vector $\vec{v}$ is always tangential to the circular path and perpendicular to the radius vector $\vec{r}$ and the angular velocity vector $\vec{\omega}$. The direction of $\vec{v}$ is such that $\vec{r}$, $\vec{v}$, and $\vec{\omega}$ form a right-handed system (though usually we relate $\vec{v}$ to $\vec{\omega}$ and $\vec{r}$).

The relationship between the linear velocity vector $\vec{v}$, the angular velocity vector $\vec{\omega}$, and the position vector $\vec{r}$ (from the axis of rotation to the particle) is given by the vector product:

$ \vec{v} = \vec{\omega} \times \vec{r} $

where $\vec{r}$ is the position vector of the particle from any point on the axis of rotation. If the origin is on the axis of rotation, $\vec{r}$ is the position vector of the particle relative to the origin.

Let's check the magnitude: $|\vec{v}| = |\vec{\omega} \times \vec{r}| = \omega r \sin\theta$. Since $\vec{\omega}$ is along the axis of rotation and $\vec{r}$ is from the axis to the particle, the angle $\theta$ between $\vec{\omega}$ and $\vec{r}$ is $90^\circ$ (assuming $\vec{r}$ is perpendicular to the axis, i.e., the particle is not on the axis). So, $v = \omega r \sin(90^\circ) = \omega r$. This matches the magnitude relationship.

The direction of $\vec{\omega} \times \vec{r}$ using the right-hand rule points tangentially in the direction of the particle's motion for counterclockwise rotation (when $\vec{\omega}$ is upwards). This confirms the vector relationship.

---

Angular Acceleration

When the angular velocity of a rotating body changes with time, it undergoes angular acceleration ($\vec{\alpha}$). Angular acceleration is the rate of change of angular velocity.

$ \vec{\alpha} = \frac{d\vec{\omega}}{dt} $

Angular acceleration is also a vector quantity. Its direction is along the axis of rotation. If the angular speed is increasing, the angular acceleration vector is in the same direction as the angular velocity vector. If the angular speed is decreasing (deceleration), the angular acceleration vector is in the opposite direction to the angular velocity vector.

The units of angular acceleration are radians per second squared (rad/s$^2$).

---

Relation between Linear Acceleration and Angular Acceleration

For a particle moving in a circular path, its linear acceleration has two components: tangential acceleration ($\vec{a}_t$) and centripetal acceleration ($\vec{a}_c$). The total linear acceleration is $\vec{a} = \vec{a}_t + \vec{a}_c$.

The tangential acceleration is the rate of change of the magnitude of the linear velocity: $a_t = \frac{dv}{dt}$. Since $v = r\omega$, $a_t = \frac{d(r\omega)}{dt} = r \frac{d\omega}{dt} = r\alpha$, where $\alpha$ is the magnitude of angular acceleration. In vector form, $\vec{a}_t$ is tangential to the path and perpendicular to $\vec{r}$. The relationship is:

$ \vec{a}_t = \vec{\alpha} \times \vec{r} $

The centripetal (or radial) acceleration is directed towards the centre of the circle and is responsible for changing the direction of the linear velocity vector. Its magnitude is $a_c = \frac{v^2}{r} = r\omega^2$. In vector form, it's directed radially inwards, opposite to $\vec{r}$. Using vector product, the centripetal acceleration can be expressed as:

$ \vec{a}_c = \vec{\omega} \times \vec{v} = \vec{\omega} \times (\vec{\omega} \times \vec{r}) $

This gives $\vec{a}_c = -\omega^2 \vec{r}$ which is directed radially inwards (opposite to $\vec{r}$).

The total linear acceleration of the particle is:

$ \vec{a} = \vec{a}_t + \vec{a}_c = (\vec{\alpha} \times \vec{r}) + (\vec{\omega} \times \vec{v}) $

$ \vec{a} = (\vec{\alpha} \times \vec{r}) + \vec{\omega} \times (\vec{\omega} \times \vec{r}) $

For rigid body rotation, these relationships hold for every particle at a distance $r$ from the axis of rotation (where $\vec{r}$ is the perpendicular vector from the axis to the particle). Particles on the axis of rotation ($r=0$) have zero linear velocity and zero linear acceleration.

Torque And Angular Momentum

In translational motion, force is the quantity that causes acceleration, and linear momentum ($\vec{p} = m\vec{v}$) is the quantity that is conserved in the absence of external forces. In rotational motion, the corresponding quantities are torque and angular momentum.

---

Moment Of Force (Torque)

When a force is applied to an object, it can cause the object to translate. However, if the object is constrained to rotate about a fixed axis (like a door on hinges) or is free to rotate, the force can also cause it to rotate or change its rotational motion. The effectiveness of a force in causing rotation depends not only on the magnitude of the force but also on where it is applied relative to the axis of rotation or pivot point.

The moment of force or torque ($\vec{\tau}$) is the rotational analogue of force. It is a measure of how effectively a force causes or changes rotational motion.

---

Definition of Torque

For a force $\vec{F}$ acting at a point with position vector $\vec{r}$ relative to a chosen reference point (the pivot or origin), the torque ($\vec{\tau}$) about that reference point is defined as the vector product of the position vector and the force vector:

$ \vec{\tau} = \vec{r} \times \vec{F} $

where:

• $\vec{r}$ is the position vector from the pivot point to the point where the force is applied.
• $\vec{F}$ is the force vector.
• $\vec{\tau}$ is the torque vector.

The magnitude of the torque is given by:

$ |\vec{\tau}| = |\vec{r} \times \vec{F}| = r F \sin\theta $

where $r = |\vec{r}|$, $F = |\vec{F}|$, and $\theta$ is the angle between $\vec{r}$ and $\vec{F}$.

The term $r\sin\theta$ is the perpendicular distance from the pivot point to the line of action of the force. This perpendicular distance is often called the lever arm or moment arm ($d_{perp} = r\sin\theta$). So, the magnitude of torque can also be written as:

$ \tau = F \cdot (r\sin\theta) = F \cdot d_{perp} $

Alternatively, $F\sin\theta$ is the component of the force perpendicular to the position vector $\vec{r}$ ($F_{perp} = F\sin\theta$). Only this perpendicular component of the force contributes to the torque. So, $\tau = r \cdot (F\sin\theta) = r \cdot F_{perp}$. A force acting parallel to the position vector ($\theta=0^\circ$ or $180^\circ$) produces zero torque ($\sin 0^\circ = \sin 180^\circ = 0$).

The direction of the torque vector $\vec{\tau}$ is perpendicular to the plane containing $\vec{r}$ and $\vec{F}$, determined by the right-hand rule applied to $\vec{r} \times \vec{F}$. For rotation about a fixed axis, the torque vector is often directed along the axis of rotation.

The SI unit of torque is Newton-metre (N·m). Although N·m is dimensionally equivalent to Joules (unit of work), torque is a vector quantity associated with rotation, while work is a scalar quantity associated with energy transfer. It's generally best to use N·m for torque to avoid confusion.

Example: Opening a door. You apply a force $\vec{F}$ at the handle, which is at a position $\vec{r}$ from the hinges (the axis of rotation). The torque $\vec{\tau} = \vec{r} \times \vec{F}$ causes the door to rotate. Applying the force perpendicular to the door at the farthest point from the hinges provides the maximum torque for a given force magnitude, because $r$ is maximized and $\sin\theta = \sin(90^\circ) = 1$.

(Image Placeholder: A rigid object pivoted at an origin O. A force F is applied at point P with position vector r from O. The angle theta between r and F is shown. The line of action of F is drawn, and the perpendicular distance (lever arm) from O to the line of action is indicated as r sin theta.)

---

Angular Momentum Of A Particle

Just as linear momentum is a measure of an object's tendency to continue moving in a straight line, angular momentum is a measure of an object's tendency to continue rotating.

For a single particle, angular momentum is defined relative to a specific point (the origin or pivot).

---

Definition of Angular Momentum of a Particle

The angular momentum ($\vec{L}$) of a particle of mass $m$ moving with linear velocity $\vec{v}$ relative to an origin is defined as the vector product of the particle's position vector $\vec{r}$ (relative to the origin) and its linear momentum $\vec{p}$: ($ \vec{p} = m\vec{v} $).

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

$ \vec{L} = \vec{r} \times (m\vec{v}) = m (\vec{r} \times \vec{v}) $

where:

• $\vec{r}$ is the position vector of the particle from the origin.
• $\vec{p}$ is the linear momentum of the particle.
• $\vec{L}$ is the angular momentum of the particle relative to the origin.

The magnitude of the angular momentum is:

$ |\vec{L}| = |\vec{r} \times m\vec{v}| = rmv \sin\phi $

where $\phi$ is the angle between $\vec{r}$ and $\vec{v}$.

The direction of the angular momentum vector $\vec{L}$ is perpendicular to the plane containing $\vec{r}$ and $\vec{v}$, determined by the right-hand rule for vector products ($\vec{r} \times \vec{v}$).

The SI unit of angular momentum is kilogram-metre squared per second (kg·m$^2$/s). Another unit is Joule-second (J·s), as $1 \text{ J} = 1 \text{ N} \cdot \text{m}$, so $1 \text{ J} \cdot \text{s} = (1 \text{ N} \cdot \text{m}) \cdot \text{s} = (1 \text{ kg} \cdot \text{m/s}^2 \cdot \text{m}) \cdot \text{s} = 1 \text{ kg} \cdot \text{m}^2/\text{s}$.

Angular momentum is zero if the particle is at the origin ($\vec{r}=0$), if its velocity is zero ($\vec{v}=0$), or if its velocity vector is parallel or anti-parallel to its position vector ($\vec{v}$ points directly towards or away from the origin, $\sin\phi=0$). For a particle moving in a circle of radius $r$ with speed $v$, the velocity $\vec{v}$ is perpendicular to the position vector $\vec{r}$ from the centre, so $\phi=90^\circ$, and the magnitude of angular momentum relative to the centre is $L = rmv$. The direction is along the axis of rotation.

---

Relationship between Torque and Angular Momentum

Just as the net force acting on a particle is equal to the rate of change of its linear momentum ($\vec{F}_{net} = d\vec{p}/dt$), the net torque acting on a particle (relative to an origin) is equal to the rate of change of its angular momentum (relative to the same origin).

Taking the time derivative of the angular momentum $\vec{L} = m(\vec{r} \times \vec{v})$:

$ \frac{d\vec{L}}{dt} = \frac{d}{dt} [m(\vec{r} \times \vec{v})] = m \frac{d}{dt} (\vec{r} \times \vec{v}) $

Using the product rule for vector differentiation, $\frac{d}{dt}(\vec{r} \times \vec{v}) = (\frac{d\vec{r}}{dt} \times \vec{v}) + (\vec{r} \times \frac{d\vec{v}}{dt})$.

We know $\frac{d\vec{r}}{dt} = \vec{v}$, and $\frac{d\vec{v}}{dt} = \vec{a}$ (linear acceleration).

$ \frac{d\vec{L}}{dt} = m [(\vec{v} \times \vec{v}) + (\vec{r} \times \vec{a})] $

The vector product of a vector with itself is zero: $\vec{v} \times \vec{v} = 0$.

$ \frac{d\vec{L}}{dt} = m (\vec{r} \times \vec{a}) $

From Newton's Second Law for a particle, $\vec{F}_{net} = m\vec{a}$, where $\vec{F}_{net}$ is the net force on the particle.

$ \frac{d\vec{L}}{dt} = \vec{r} \times (m\vec{a}) = \vec{r} \times \vec{F}_{net} $

The term $\vec{r} \times \vec{F}_{net}$ is the net torque ($\vec{\tau}_{net}$) acting on the particle relative to the same origin as $\vec{r}$.

$ \vec{\tau}_{net} = \frac{d\vec{L}}{dt} $

This is the rotational analogue of Newton's Second Law for a particle. It states that the net torque acting on a particle is equal to the rate of change of its angular momentum.

This relationship is fundamental and extends to systems of particles and rigid bodies, where the net external torque on the system equals the rate of change of the system's total angular momentum.