CONTENT
6.1.1 What kind of motion can a rigid body have?
6.2 Centre of mass
6.3 Motion of centre of mass
6.4 Linear momentum of a system of particles
6.5 Vector product of two vectors
6.6 Angular velocity and its relation with linear velocity
6.7 Torque and angular momentum
6.8 Equilibrium of a rigid body
6.9 Moment of inertia
6.10 Kinematics of rotational motion about a fixed axis
6.11 Dynamics of rotational motion about a fixed axis
6.12 Angular momentum in case of rotation about a fixed axis
6.1 Introduction
Here is an engaging summary based on the introductory sections of
🌀 Beyond the "Point": When Objects Get Real
Until now, most of physics treated everything—from cars to planets—as a single, tiny point mass. It’s a neat trick that makes math easier, but it doesn't tell the whole story. What happens when a body has a real shape? What happens when it spins, wobbles, or rolls?
Welcome to the world of Extended Bodies and Rigid Body Dynamics.
🧱 What is a "Rigid Body"?
In a perfect physics world, a rigid body is an object with a perfectly definite and unchanging shape. No matter how hard you push it, the distance between any two points inside it remains exactly the same.
While real objects (like a wooden block or a steel beam) might deform slightly under pressure, we treat them as rigid to understand their core movements without getting bogged down by every tiny vibration.
6.1.1 What kind of motion can a rigid body have?
🏃♂️ 6.1.1 The Two Faces of Motion
A rigid body isn't limited to just moving from point A to point B. It generally exhibits two primary types of motion:
1. Pure Translational Motion
Imagine a rectangular block sliding down a smooth inclined plane.
-
The Rule: Every single particle in the block moves with the same velocity at any given instant.
-
The Visual: If you draw an arrow on the front, middle, and back of the block, those arrows stay perfectly parallel and move at the same speed.
2. Rotational Motion
Now, imagine a ceiling fan or a potter’s wheel. The object stays in one place, but it's definitely moving!
-
The Rule: Every particle moves in a circle whose centre lies on a fixed line called the axis of rotation.
-
The Twist: Unlike translation, particles at different distances from the axis move at different speeds. A point on the edge of a fan blade travels much faster than a point near the center!
🎢 The Best of Both Worlds: Rolling
The most common motion we see—like a cylinder rolling down a hill—is actually a combination. It is translating (moving down the hill) while simultaneously rotating (spinning around its center).
Did you know? In a perfectly rolling cylinder, the point of contact with the ground is actually instantaneously at rest ($v = 0$)!
⚖️ The "Secret Sauce": Center of Mass
To make sense of these complex movements, we use the Center of Mass. It’s the unique point where the entire mass of the body seems to be concentrated. By tracking this one point, we can solve the "translation" part of the puzzle, while the "rotation" part happens around it.
6.2 Centre of mass
Based on the
6.2 Centre of Mass (CM)
The Centre of Mass is a theoretical point where the entire mass of a system of particles (or a rigid body) can be considered to be concentrated for the purpose of describing its translational motion.
1. System of Two Particles
For a simple system consisting of two particles with masses $m_1$ and $m_2$ located at positions $x_1$ and $x_2$ along a straight line (the x-axis), the position of the centre of mass ($X$) is given by:
-
Equal Masses: If $m_1 = m_2$, the CM lies exactly midway between them ($X = \frac{x_1 + x_2}{2}$).
-
Origin Choice: If we place the origin at the centre of mass itself ($X = 0$), then $m_1x_1 + m_2x_2 = 0$.
2. System of $n$ Particles
For a system of $n$ particles with masses $m_1, m_2, ... m_n$ at positions $x_1, x_2, ... x_n$ along a line, the CM is:
Where $M$ is the total mass of the system.
3. Centre of Mass in Three Dimensions
In a 3D coordinate system, the position of each particle $i$ is defined by a position vector $\vec{r}_i = x_i\hat{i} + y_i\hat{j} + z_i\hat{k}$. The position vector of the centre of mass $\vec{R}$ is:
This can be broken down into individual coordinates:
-
$X = \frac{1}{M} \sum m_i x_i$
-
$Y = \frac{1}{M} \sum m_i y_i$
-
$Z = \frac{1}{M} \sum m_i z_i$
4. Centre of Mass for Continuous Bodies
For real-world rigid bodies (like a rod, sphere, or disc), we cannot count individual atoms. Instead, we treat the body as a continuous distribution of mass. We divide the body into infinitesimal elements of mass $dm$ located at position $(x, y, z)$:
-
$X = \frac{1}{M} \int x \, dm$
-
$Y = \frac{1}{M} \int y \, dm$
-
$Z = \frac{1}{M} \int z \, dm$
5. Symmetry and CM Location
For bodies with uniform mass density and geometric symmetry, the CM usually coincides with the geometric centre:
| Shape | CM Location |
|---|---|
| Thin Rod | Middle point of the rod |
| Cylinder / Sphere | Geometric center (axis/center point) |
| Ring / Disc | Geometric center |
| Homogeneous Cube | Point of intersection of diagonals |
Note: The centre of mass does not necessarily have to lie inside the body. For example, the CM of a hollow ring or a hollow sphere lies in the empty space at the center.
Finding the Centre of Mass (CM) of a complex shape like an L-shaped
lamina is a classic application of the principles discussed in
📐 The Problem: L-Shaped Lamina
An L-shaped lamina is a thin, flat plate with uniform mass density. To find its centre of mass, we treat it as a composite body made up of simpler geometric shapes (usually rectangles or squares) whose individual centres of mass are already known.
🛠️ Step 1: Divide the Lamina
The easiest way to analyze an L-shape is to divide it into three identical squares of side $L$ (or two rectangles).
-
Let’s assume the lamina is made of three squares, each of side $1 \text{ m}$ and mass $m$.
-
Since the lamina is uniform, the mass of each section is proportional to its area.
📍 Step 2: Establish a Coordinate System
To calculate the position, we must place the lamina on an $x-y$ graph.
-
Place the "inner corner" of the L-shape at the origin $(0,0)$.
-
Square 1 (Bottom-Left): Center at $C_1(0.5, 0.5)$
-
Square 2 (Bottom-Right): Center at $C_2(1.5, 0.5)$
-
Square 3 (Top-Left): Center at $C_3(0.5, 1.5)$
⚖️ Step 3: Identify Masses
Since the lamina is uniform:
-
Mass of Square 1 = $m$
-
Mass of Square 2 = $m$
-
Mass of Square 3 = $m$
-
Total Mass ($M$) = $3m$
🧮 Step 4: Apply the CM Formula
Using the formula for a system of particles from
For the X-coordinate ($X$):
For the Y-coordinate ($Y$):
🎯 Summary of Results
-
Final Position: The Centre of Mass is located at $(\dfrac{5}{6}, \dfrac{5}{6})$.
-
Observation: Notice that the CM lies along the line of symmetry (the $45^\circ$ diagonal). Because the shape is symmetrical about the line $y = x$, the $X$ and $Y$ coordinates must be equal.
💡 Key Takeaways for Your Exams
-
Symmetry is your friend: If a body has an axis of symmetry, the CM must lie on that axis.
-
Uniformity: For uniform bodies, you can replace "mass" with "area" in your calculations ($X = \dfrac{\sum A_i x_i}{\sum A_i}$).
-
Negative Area Method: You can also solve this by treating the L-shape as a large $2 \times 2$ square ($Area = 4$) and "subtracting" a missing $1 \times 1$ square ($Area = -1$) from the corner.
Based on the
📝 Example 6.3: The Question
Find the centre of mass of a uniform L-shaped lamina (a thin flat plate) with dimensions as shown in the figure. The total mass of the lamina is 3 kg.
🚀 Step-by-Step Solution
Step 1: Divide the Lamina
To solve this easily, we divide the L-shape into three identical squares, each of side $1 \text{ m}$.
-
Since the total mass is $3 \text{ kg}$ and the lamina is uniform, each square has a mass:
$$m_1 = m_2 = m_3 = 1 \text{ kg}$$
Step 2: Locate Individual Centres of Mass
By symmetry, the centre of mass of each square is at its geometric center. Based on the coordinate system (where the bottom-left corner of the L-shape is the origin $O(0,0)$):
-
Square 1 (Bottom-Left): $C_1 = (\dfrac{1}{2}, \dfrac{1}{2})$
-
Square 2 (Bottom-Right): $C_2 = (\dfrac{3}{2}, \dfrac{1}{2})$
-
Square 3 (Top-Left): $C_3 = (\dfrac{1}{2}, \dfrac{3}{2})$
Step 3: Calculate the X-coordinate ($X$)
Using the formula $X = \dfrac{\sum m_i x_i}{M}$:
Step 4: Calculate the Y-coordinate ($Y$)
Using the formula $Y = \dfrac{\sum m_i y_i}{M}$:
🎯 Final Answer
The centre of mass of the L-shaped lamina is located at $(\dfrac{5}{6}, \dfrac{5}{6})$.
Insight: Notice that the CM lies on the line of symmetry $y = x$. This makes sense because the distribution of mass is identical on both sides of that diagonal line.
6.3 Motion of centre of mass
Based on
6.3 Motion of Centre of Mass
Once we have defined the Centre of Mass (CM), we can analyze how this point moves when forces are applied to a system of particles. This section bridges the gap between particle mechanics and the mechanics of complex systems.
1. Velocity of the Centre of Mass ($\vec{V}$)
We start with the relation for the position vector of the CM for $n$ particles:
By differentiating this equation with respect to time ($t$), and assuming the masses $m_i$ are constant, we get the velocity of the centre of mass:
-
$\vec{V} = \dfrac{d\vec{R}}{dt}$ is the velocity of the CM.
-
$\vec{v}_i = \dfrac{d\vec{r}_i}{dt}$ is the velocity of the $i^{th}$ particle.
2. Acceleration of the Centre of Mass ($\vec{A}$)
Differentiating the velocity equation again with respect to time gives us the acceleration of the centre of mass:
-
$\vec{A} = \dfrac{d\vec{V}}{dt}$ is the acceleration of the CM.
-
$\vec{a}_i = \dfrac{d\vec{v}_i}{dt}$ is the acceleration of the $i^{th}$ particle.
3. Newton’s Second Law for a System of Particles
From Newton’s Second Law, the force on each particle is $\vec{F}_i = m_i\vec{a}_i$. Substituting this into the acceleration equation:
Here, the total force is the sum of all forces acting on all particles. These forces are of two types:
-
Internal Forces: Forces exerted by the particles on each other. According to Newton’s Third Law, these occur in equal and opposite pairs and cancel out in the vector sum.
-
External Forces ($\vec{F}_{ext}$): Forces exerted on the particles by bodies outside the system.
Thus, the equation simplifies to:
The Core Principle: The centre of mass of a system of particles moves as if all the mass of the system were concentrated at that point and all external forces were applied directly to it.
4. Physical Significance
-
Independence from Internal Motion: To determine the motion of the CM, we don't need to know anything about the internal forces or how the individual particles are moving/rotating relative to each other.
-
Simplification: This allows us to treat complex objects (like a spinning football or an exploding rocket) as a single point mass located at the CM when analyzing translational motion.
5. Illustrative Example: Projectile Explosion
Consider a projectile following a parabolic path that explodes in mid-air.
-
The forces causing the explosion are internal forces.
-
The only external force is gravity ($Mg$).
-
Because internal forces don't affect the CM's motion, the CM of the fragments will continue to follow the same parabolic trajectory it would have taken if the explosion never happened.
Summary Table
| Concept | Formula | Key Insight |
|---|---|---|
| Momentum | $\vec{P} = M\vec{V}$ | Total momentum is mass times CM velocity. |
| Force | $\vec{F}_{ext} = M\vec{A}$ | Only external forces cause CM acceleration. |
| Internal Forces | $\sum \vec{F}_{int} = 0$ | They cancel out and do not affect CM motion. |
6.4 Linear momentum of a system of particles
Welcome to Section 6.4! In the previous sections, we learned that the Centre of Mass (CM) moves as if the entire mass of the system were concentrated there. Now, we’re going to look at the "hidden power" behind that motion: Linear Momentum.
This section explains why a spinning gymnast, an exploding firework, and even a pair of orbiting stars follow predictable paths.
🚀 6.4 Linear Momentum of a System of Particles
1. Defining the "Total" Momentum
For a single particle, momentum is simply $\vec{p} = m\vec{v}$. But for a system of $n$ particles, the total linear momentum ($\vec{P}$) is the vector sum of the momenta of all individual particles:
⚡ The "Magic" Link: $\vec{P} = M\vec{V}$
From our study of the
The total momentum of a system is equal to the total mass ($M$) times the velocity of its centre of mass ($\vec{V}$).
⚖️ Newton’s Second Law (System Edition)
If we differentiate the total momentum with respect to time ($t$):
Wait... we already know from
This tells us that only external forces can change the total momentum of a system. Internal forces (particles pushing each other) might change the speed of individual parts, but they never change the total momentum.
🛡️ The Law of Conservation of Momentum
This is the "Holy Grail" of mechanics. If the total external force acting on a system is zero ($\vec{F}_{ext} = 0$), then:
The Meaning: If no outside force interferes, the total momentum of the system is conserved. Consequently, the velocity of the Centre of Mass remains constant.
🌌 Real-World Deep Dives
A. The Radioactive "Fly-Apart" (Radium Decay)
Imagine a moving Radium nucleus. It suddenly decays into a Radon nucleus and an alpha particle.
-
The Physics: The decay is caused by internal forces.
-
The Result: Even though the two new particles fly off in different directions, their Centre of Mass continues to move along the exact same straight-line path the original Radium was taking.
B. The Dance of the Binary Stars
In space, many stars exist in pairs (binary systems) orbiting each other.
-
Since there are no external forces in deep space, their Centre of Mass moves like a free particle (straight line at constant speed).
-
If you look at them from the "CM frame" (as if you were standing on the CM), you would see the stars simply moving in perfect circles around you.
📝 Summary Checklist
-
[ ] Total Momentum: $\vec{P} = M\vec{V}$
-
[ ] Force Link: The rate of change of total momentum is equal to the external force ($\vec{F}_{ext}$).
-
[ ] Conservation: If $\vec{F}_{ext} = 0$, $\vec{P}$ is constant.
-
[ ] CM Integrity: Internal forces cannot change the velocity of the Centre of Mass.
Next Step:
Ready to add some "twist" to the story? Would you like to explore
Section 6.5, where we introduce the
Vector Product (Cross Product)—the essential math tool for
understanding
This study guide breaks down Section 6.4 of the
🚀 6.4 Linear Momentum of a System of Particles
1. Defining Momentum for a Single Particle
Before looking at a group, remember that for a single particle of mass $m$ and velocity $\mathbf{v}$, the linear momentum $\mathbf{p}$ is:
According to Newton’s Second Law, the force acting on that particle is the rate of change of its momentum:
2. Momentum of a System ($n$ Particles)
For a system containing $n$ particles, the total linear momentum $\mathbf{P}$ is the vector sum of the momenta of all individual particles:
The Big Discovery:
Comparing this to the definition of the Centre of Mass (CM), we find a beautiful simplification:
$$\mathbf{P} = M\mathbf{V}$$Where $M$ is the total mass and $\mathbf{V}$ is the velocity of the Centre of Mass.
Key Insight: The total momentum of a system is as if the entire mass of the system were concentrated at the CM and moving with its velocity.
⚖️ Conservation of Linear Momentum
If we differentiate the total momentum with respect to time, we relate it to the external force:
The Law of Conservation:
If the total external force acting on the system is zero ($\mathbf{F}_{ext} = 0$):
-
$\dfrac{d\mathbf{P}}{dt} = 0$
-
$\mathbf{P} = \text{Constant}$
[!IMPORTANT]
When the external force is zero, the total linear momentum is conserved. This also means the velocity of the Centre of Mass ($\mathbf{V}$) remains constant, even if the particles inside are colliding or exploding!
🔍 Real-World Examples
A. The Exploding Projectile
Imagine a projectile exploding mid-air. The explosion is caused by internal forces.
-
The Fragments: Fly off in complicated directions.
-
The Centre of Mass: Because gravity is the only external force (which hasn't changed), the CM continues to follow the exact same parabolic path it was on before the explosion.
B. Radioactive Decay
Consider a Radium nucleus decaying into Radon and an Alpha particle:
-
If the Radium was moving, the two new particles fly off such that their CM continues on the original path.
-
If the Radium was at rest, the two particles fly off in opposite directions (back-to-back) so the CM stays perfectly still.
C. Binary Stars
In deep space, two stars (a binary system) rotate around each other.
-
In the Laboratory Frame, they look like they are moving in messy, wobbling paths.
-
In the Centre of Mass Frame, they move in simple, perfect circles around a stationary point.
📝 Quick Summary Table
| Feature | Single Particle | System of Particles |
| Momentum | $\mathbf{p} = m\mathbf{v}$ | $\mathbf{P} = \sum \mathbf{p}_i = M\mathbf{V}$ |
| Newton's 2nd Law | $\mathbf{F} = \dfrac{d\mathbf{p}}{dt}$ | $\mathbf{F}_{ext} = \dfrac{d\mathbf{P}}{dt}$ |
| If $\mathbf{F}_{ext} = 0$ | Velocity is constant | CM Velocity ($\mathbf{V}$) is constant |
6.5 Vector product of two vectors
This study guide is based on Section 6.5
of the
6.5 Vector Product of Two Vectors
1. Definition
The vector product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is a third vector $\mathbf{c}$. It is written as:
The properties of this new vector $\mathbf{c}$ are:
-
Magnitude: $c = ab \sin \theta$, where $a$ and $b$ are magnitudes and $\theta$ is the angle between them ($0 \leq \theta \leq 180^\circ$).
-
Direction: $\mathbf{c}$ is perpendicular to the plane containing both $\mathbf{a}$ and $\mathbf{b}$.
2. Determining Direction: The Right-Hand Rule
To find the direction of $\mathbf{c}$, use the Right-Hand Thumb Rule:
-
Open your right palm and point your fingers in the direction of the first vector ($\mathbf{a}$).
-
Curl your fingers toward the second vector ($\mathbf{b}$) through the smaller angle.
-
Your stretched thumb points in the direction of the product vector $\mathbf{c}$.
🛠️ Key Properties of Vector Products
A. Non-Commutativity
Unlike scalar products, the order matters. If you swap the vectors, the direction reverses:
Note: They have the same magnitude, but opposite directions.
B. Distributive Law
The vector product is distributive over addition:
C. Self-Product
The cross product of a vector with itself is always a null vector ($\mathbf{0}$) because the angle $\theta = 0^\circ$ and $\sin 0 = 0$:
📐 Cross Products of Unit Vectors ($\hat{i}, \hat{j}, \hat{k}$)
For the orthogonal unit vectors used in Cartesian coordinates:
-
Same vectors: $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \mathbf{0}$
-
Cyclic Order (Positive): Moving clockwise ($i \to j \to k$):
-
$\hat{i} \times \hat{j} = \hat{k}$
-
$\hat{j} \times \hat{k} = \hat{i}$
-
$\hat{k} \times \hat{i} = \hat{j}$
-
-
Anti-Cyclic Order (Negative): Moving counter-clockwise:
-
$\hat{j} \times \hat{i} = -\hat{k}$
-
$\hat{k} \times \hat{j} = -\hat{i}$
-
$\hat{i} \times \hat{k} = -\hat{j}$
-
🧮 Calculation using Components (Determinant Form)
If $\mathbf{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$ and $\mathbf{b} = b_x\hat{i} + b_y\hat{j} + b_z\hat{k}$, the cross product is easiest to find using a determinant:
Expanded Form:
Physics Applications
The vector product is essential for describing rotation. Two primary quantities defined this way are:
-
Torque (Moment of Force): $\mathbf{\tau} = \mathbf{r} \times \mathbf{F}$
-
Angular Momentum: $\mathbf{l} = \mathbf{r} \times \mathbf{p}$
-
Linear Velocity (in rotation): $\mathbf{v} = \mathbf{\omega} \times \mathbf{r}$
Example Problem
Find the vector product of $\mathbf{a} = 3\hat{i} - 4\hat{j} + 5\hat{k}$ and $\mathbf{b} = -2\hat{i} + \hat{j} - 3\hat{k}$.
Solution:
Using the determinant:
In the context of
Just as a force causes an object to accelerate in a straight line, a torque causes an object to undergo angular acceleration (to start spinning, stop spinning, or change its RPM).
1. The Three Factors of Torque
The physical "strength" of a torque depends on three specific variables. If you change any one of these, the "turning effect" changes:
-
Magnitude of Force ($F$): The harder you push, the more torque you generate.
-
Lever Arm / Position Vector ($r$): This is the distance from the pivot point (axis of rotation) to the point where the force is applied. This is why doorknobs are placed far from the hinges—increasing $r$ makes it much easier to open the door.
-
The Angle ($\theta$): Torque is most effective when the force is applied perpendicularly ($90^\circ$) to the lever arm.
The Mathematical Formula
In vector terms, torque ($\boldsymbol{\tau}$) is the cross product of the position vector and the force vector:
The magnitude is given by:
2. Visualizing "Line of Action"
To understand the physical meaning, it helps to look at the moment arm (or perpendicular distance).
Torque = Force $\times$ Perpendicular distance from the axis.
If you pull a wrench directly toward the bolt (an angle of $0^\circ$), the "line of action" passes through the pivot. Even if you pull with 1,000 Newtons of force, the torque is zero because there is no perpendicular distance. The bolt will not turn.
3. Direction and the "Twist"
Torque is a pseudovector. Physically, it doesn't point in the direction the object moves; it points along the axis of rotation.
-
Right-Hand Rule: If you curl the fingers of your right hand in the direction the object is rotating, your thumb points in the direction of the torque vector.
-
Interpretation: A torque pointing "up" (out of the page) generally represents a counter-clockwise twist, while a torque pointing "down" represents a clockwise twist.
4. Real-World Physical Analogies
| Action | Why Torque Matters |
| Opening a Door | Pushing near the hinges (small $r$) requires massive force. Pushing the handle (large $r$) requires very little. |
| Using a Wrench | A longer wrench handle allows you to loosen a "frozen" bolt because it multiplies your input force into a higher torque. |
| Seesaws | An adult can balance a child by sitting closer to the pivot (center). The adult's large weight $\times$ small distance = the child's small weight $\times$ large distance. |
Comparison: Force vs. Torque
| Linear Motion (Force) | Rotational Motion (Torque) |
| Changes linear velocity ($v$) | Changes angular velocity ($\omega$) |
| $\mathbf{F} = m\mathbf{a}$ | $\boldsymbol{\tau} = I\boldsymbol{\alpha}$ (where $I$ is Moment of Inertia) |
| Measured in Newtons (N) | Measured in Newton-meters (N·m) |
6.6 Angular velocity and its relation with linear velocity
This study material is based on
6.6 Angular Velocity and its Relation with Linear Velocity
In a rigid body rotating about a fixed axis, every particle moves in a circle. This circle lies in a plane perpendicular to the axis and has its centre on the axis.
1. Defining Angular Velocity ($\omega$)
As a particle $P$ rotates about a fixed axis, it describes an angular displacement $\Delta \theta$ in a time interval $\Delta t$.
-
Average Angular Velocity: $\dfrac{\Delta \theta}{\Delta t}$
-
Instantaneous Angular Velocity ($\omega$): The limit as $\Delta t$ approaches zero.
$$\omega = \dfrac{d\theta}{dt}$$ -
Key Property: In a rigid body, every particle has the same angular velocity at any given instant, even though their linear velocities may differ based on their distance from the axis.
2. Relation between Linear and Angular Velocity (Scalar)
For a particle at a perpendicular distance $r$ from the fixed axis, the magnitude of its linear velocity $v$ is given by:
-
For particles on the axis, $r = 0$, therefore $v = 0$. This confirms the axis is stationary.
3. The Vector Relationship
Angular velocity is a vector quantity. For rotation about a fixed axis, the vector $\vec{\omega}$ lies along the axis of rotation.
-
Direction: Determined by the Right-Hand Screw Rule. If the fingers of your right hand curl in the direction of rotation, your thumb points in the direction of $\vec{\omega}$.
-
Vector Formula: The linear velocity $\vec{v}$ of a particle with position vector $\vec{r}$ is the vector product of $\vec{\omega}$ and $\vec{r}$:
$$\vec{v} = \vec{\omega} \times \vec{r}$$
Note: $\vec{v}$ is always perpendicular to both $\vec{\omega}$ and $\vec{r}$, directed along the tangent to the circle described by the particle.
4. Angular Acceleration ($\alpha$)
Similar to how linear acceleration is the rate of change of linear velocity, angular acceleration is the rate of change of angular velocity.
-
Formula: $\vec{\alpha} = \dfrac{d\vec{\omega}}{dt}$
-
If the axis of rotation is fixed, the direction of $\vec{\omega}$ remains constant, and the equation simplifies to the scalar form:
$$\alpha = \dfrac{d\omega}{dt}$$
Summary Table: Translational vs. Rotational Variables
| Linear Motion | Rotational Motion | Relation |
| Displacement ($s$) | Angular Displacement ($\theta$) | $s = r\theta$ |
| Velocity ($v$) | Angular Velocity ($\omega$) | $v = r\omega$ |
| Acceleration ($a$) | Angular Acceleration ($\alpha$) | $a = r\alpha$ (tangential) |
6.7 Torque and angular momentum
This study material is based on
6.7 Torque and Angular Momentum
In translational motion, a force is required to produce linear acceleration. In rotational motion, we require a "turning effect" to produce angular acceleration. This effect is known as Torque or Moment of Force.
1. Moment of Force (Torque)
Torque is the rotational analogue of force. For a rigid body rotating about a fixed axis (like a door on its hinges), the effectiveness of a force depends not just on its magnitude, but also on where and in what direction it is applied.
-
Definition: For a particle at position vector $\vec{r}$ relative to an origin $O$, acted upon by a force $\vec{F}$, the torque $\vec{\tau}$ is defined as the vector product:
$$\vec{\tau} = \vec{r} \times \vec{F}$$ -
Magnitude: $\tau = rF \sin \theta$ (where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$).
-
It can also be written as $\tau = r_{\perp}F$ (where $r_{\perp}$ is the perpendicular distance from the origin to the line of action of the force) or $\tau = rF_{\perp}$.
-
-
Direction: Perpendicular to the plane containing $\vec{r}$ and $\vec{F}$, determined by the Right-Hand Screw Rule.
-
SI Unit: Newton-metre (Nm). Note that while its dimensions ($ML^2T^{-2}$) are the same as work/energy, torque is a vector, whereas work is a scalar.
2. Angular Momentum of a Particle
Angular momentum is the rotational analogue of linear momentum, often referred to as the moment of momentum.
-
Definition: For a particle of mass $m$ and linear momentum $\vec{p}$ at position $\vec{r}$, the angular momentum $\vec{l}$ is:
$$\vec{l} = \vec{r} \times \vec{p}$$ -
Magnitude: $l = rp \sin \theta = mvr \sin \theta$.
-
Vanishing Conditions: Angular momentum is zero if:
-
The particle is at the origin ($r = 0$).
-
The linear momentum is zero ($p = 0$).
-
The particle's path passes through the origin ($\theta = 0^\circ$ or $180^\circ$).
-
3. Relation Between Torque and Angular Momentum
There is a fundamental relationship between torque and the rate of change of angular momentum, which is the rotational analogue of Newton’s Second Law ($\vec{F} = \dfrac{d\vec{p}}{dt}$).
By differentiating $\vec{l} = \vec{r} \times \vec{p}$ with respect to time $t$:
Since $\dfrac{d\vec{r}}{dt} = \vec{v}$ and $\vec{p} = m\vec{v}$, the first term $(\vec{v} \times m\vec{v})$ becomes zero. Substituting $\dfrac{d\vec{p}}{dt} = \vec{F}$, we get:
Key Principle: The time rate of change of the angular momentum of a particle is equal to the torque acting on it.
4. Angular Momentum for a System of Particles
For a system of $n$ particles, the total angular momentum $\vec{L}$ is the vector sum of the angular momenta of the individual particles:
For a rigid body rotating about a fixed axis, the total torque $\vec{\tau}_{ext}$ acting on the system is equal to the rate of change of the total angular momentum:
| Linear Concept | Rotational Concept | Mathematical Relation |
| Force ($\vec{F}$) | Torque ($\vec{\tau}$) | $\vec{\tau} = \vec{r} \times \vec{F}$ |
| Momentum ($\vec{p}$) | Angular Momentum ($\vec{l}$) | $\vec{l} = \vec{r} \times \vec{p}$ |
| Newton's 2nd Law | Rotational 2nd Law | $\vec{\tau} = \dfrac{d\vec{L}}{dt}$ |
Based on
6.7.3 Conservation of Angular Momentum
The Law of Conservation of Angular Momentum is the rotational equivalent of the Law of Conservation of Linear Momentum. It describes how the rotational state of a system remains unchanged if no external "turning force" is applied.
1. The Core Principle
From the relation between torque ($\vec{\tau}_{ext}$) and angular momentum ($\vec{L}$):
If the total external torque acting on a system is zero ($\vec{\tau}_{ext} = 0$), then:
Definition: If the total external torque on a system of particles is zero, then the total angular momentum of the system is conserved (remains constant in both magnitude and direction).
2. Component Form
Since angular momentum is a vector, its conservation implies that each of its three scalar components along the $x, y,$ and $z$ axes remains constant:
-
$L_x = K_1$
-
$L_y = K_2$
-
$L_z = K_3$
(where $K_1, K_2, K_3$ are constants).
3. Relation with Moment of Inertia ($I$)
For a rigid body rotating about a fixed axis, the angular momentum is given by $L = I\omega$. If no external torque acts on the body:
This means if the Moment of Inertia ($I$) of a system decreases (e.g., mass moves closer to the axis), the Angular Velocity ($\omega$) must increase to keep $L$ constant.
4. Practical Examples
-
A Spinning Ice Skater: When a skater pulls their arms and legs inward, their moment of inertia ($I$) decreases. To conserve angular momentum, their angular velocity ($\omega$) increases, making them spin faster.
-
Planetary Motion: Planets move faster when they are closer to the Sun in their elliptical orbits because the distance ($r$) decreases, and to keep $L = mvr$ constant, the velocity ($v$) must increase.
-
Diving: A diver curls their body into a "tuck" position after jumping. This reduces their moment of inertia, allowing them to complete more somersaults (higher $\omega$) before straightening out to enter the water.
5. Illustrative Problem (NCERT Example 6.6)
Problem: Show that the angular momentum of a single particle moving with constant velocity remains constant.
Explanation: 1. Consider a particle moving in a straight line with velocity $v$.
2. Its angular momentum about an origin $O$ is $l = r \times p = m(r \times v)$.
3. The magnitude is $l = mvr \sin \theta$.
4. In the figure
5. Since the particle moves in a straight line at constant speed, both $v$ and the perpendicular distance $OM$ stay constant. Therefore, $l$ is conserved.
Summary Comparison
| Law | Linear Motion | Rotational Motion |
| Quantity | Linear Momentum ($\vec{p}$) | Angular Momentum ($\vec{L}$) |
| Cause of Change | External Force ($\vec{F}_{ext}$) | External Torque ($\vec{\tau}_{ext}$) |
| Condition | If $\vec{F}_{ext} = 0$ | If $\vec{\tau}_{ext} = 0$ |
| Conservation | $\vec{p} = \text{constant}$ | $\vec{L} = \text{constant}$ |
6.8 Equilibrium of a rigid body
This study material is based on
6.8 Equilibrium of a Rigid Body
A rigid body is in mechanical equilibrium if its state of motion (both translational and rotational) does not change with time. This means the body has neither linear acceleration nor angular acceleration.
1. The Two Conditions for Equilibrium
For a body to be in complete mechanical equilibrium, it must satisfy two distinct conditions:
A. Translational Equilibrium
The vector sum of all external forces acting on the body must be zero.
-
Result: The total linear momentum of the body remains constant.
-
If the body is at rest, it stays at rest; if it is moving, it continues with a constant velocity.
B. Rotational Equilibrium
The vector sum of all external torques acting on the body (about any point) must be zero.
-
Result: The total angular momentum of the body remains constant.
-
This means the body has no angular acceleration.
Note: If a body is in translational equilibrium, the total torque is independent of the choice of origin. You can calculate torque about any point, and the result will be the same.
2. Scalar Components of Equilibrium
In a 3D space, these two vector equations provide six independent scalar equations:
| Equilibrium Type | X-axis | Y-axis | Z-axis |
| Translational | $\sum F_x = 0$ | $\sum F_y = 0$ | $\sum F_z = 0$ |
| Rotational | $\sum \tau_x = 0$ | $\sum \tau_y = 0$ | $\sum \tau_z = 0$ |
For coplanar forces (forces acting in a single plane, e.g., the $xy$ plane), we only need three equations:
-
$\sum F_x = 0$
-
$\sum F_y = 0$
-
$\sum \tau_z = 0$ (Torque about an axis perpendicular to the plane).
3. Partial Equilibrium
A body can be in one type of equilibrium but not the other:
-
Translational but NOT Rotational: Occurs when a Couple acts on the body. A couple consists of two equal and opposite forces with different lines of action.
-
Example: Turning a bottle cap or a compass needle in a magnetic field. The net force is zero, but the net torque is not.
-
-
Rotational but NOT Translational: Occurs when parallel forces act in the same direction at different points.
-
Example: Pushing both ends of a rod in the same direction. The net torque about the center might be zero, but the net force is not.
-
4. Principles of Moments (The Lever)
An ideal lever is a light rod pivoted at a point called the fulcrum. It is a classic application of mechanical equilibrium.
-
The Principle: For a lever in rotational equilibrium about the fulcrum:
$$\text{Load} \times \text{Load Arm} = \text{Effort} \times \text{Effort Arm}$$$$F_1 d_1 = F_2 d_2$$ -
Mechanical Advantage (M.A.): This is the ratio of the load to the effort.
$$M.A. = \frac{F_1}{F_2} = \frac{d_2}{d_1}$$If the effort arm ($d_2$) is longer than the load arm ($d_1$), $M.A. > 1$, meaning a small effort can lift a large load.
5. Summary Checklist for Problems
When solving equilibrium problems (like a ladder leaning against a wall or a beam balance):
-
Draw a Free Body Diagram (FBD) showing all forces and their points of application.
-
Apply $\sum \vec{F} = 0$: Resolve forces into horizontal and vertical components.
-
Apply $\sum \vec{\tau} = 0$: Choose a convenient point as the origin (usually where an unknown force acts) to simplify calculations.
This study material is based on
6.8.2 Centre of Gravity
The Centre of Gravity (CG) is the specific point in a body through which the entire weight of the body acts. It is the point of balance where the total gravitational torque on the body is zero.
1. Definition and Physical Significance
If you try to balance an irregular-shaped object (like a cardboard) on a narrow tip (like a pencil), you will find a unique point $G$ where the object remains horizontal and stable.
-
At this point ($G$): The upward reaction force from the tip is equal and opposite to the total weight ($Mg$) of the object.
-
Mechanical Equilibrium: The body is in both translational equilibrium (forces cancel out) and rotational equilibrium (torques cancel out).
2. Mathematical Derivation
A rigid body is composed of many small particles with masses $m_1, m_2, ... m_n$. Each particle experiences a gravitational force $m_ig$ acting downwards.
If $\vec{r}_i$ is the position vector of the $i^{th}$ particle relative to the CG, the torque due to gravity on that particle is:
For the body to be in rotational equilibrium, the total gravitational torque about the CG must be zero:
Since $\vec{g}$ is constant for all particles in a uniform gravitational field, we can write:
3. CG vs. Centre of Mass (CM)
While the terms are often used interchangeably in everyday physics, they represent different concepts:
-
Centre of Mass (CM): Depends only on the distribution of mass. It is the point where $\sum m_i\vec{r}_i = 0$.
-
Centre of Gravity (CG): Depends on the distribution of weight (gravity). It is the point where the total gravitational torque is zero.
Key Rule: The CG coincides with the CM only if the body is in a uniform gravitational field (where $\vec{g}$ does not vary from one part of the body to another). For very large objects (like a mountain or a tall skyscraper), the CM and CG may slightly differ because $\vec{g}$ changes with height.
4. How to Determine the CG Experimentally
For an irregular flat object (like a piece of cardboard), you can find the CG using the Plumb Line Method:
-
Suspend the object from a point (A) near its edge. The CG will lie somewhere on the vertical line ($AA_1$) passing through the point of suspension.
-
Suspend the object from a different point (B). Draw a second vertical line ($BB_1$).
-
The intersection point of these lines is the Centre of Gravity ($G$).
5. Summary Table
| Property | Centre of Mass (CM) | Centre of Gravity (CG) |
| Depends on | Mass distribution | Weight/Gravity distribution |
| Formula | $\sum m_i\vec{r}_i = 0$ | $\sum \vec{\tau}_g = 0$ |
| Stability | Relates to linear inertia | Relates to rotational balance |
| Coincidence | Always exists | Coincides with CM in uniform $\vec{g}$ |
6.9 Moment of inertia
This study material is based on
6.9 Moment of Inertia
In translational motion, mass ($m$) is the measure of inertia—the resistance of a body to a change in its state of linear motion. In rotational motion, this role is played by the Moment of Inertia ($I$), also known as Rotational Inertia.
1. Kinetic Energy of a Rotating Body
To understand $I$, consider a rigid body rotating about a fixed axis with angular velocity $\omega$. Every particle $i$ of mass $m_i$ at a distance $r_i$ from the axis moves with a linear velocity $v_i = r_i\omega$.
The total Kinetic Energy ($K$) of the body is the sum of the kinetic energies of all its particles:
Since $\omega$ is constant for all particles, we factor it out:
The term in the brackets is defined as the Moment of Inertia ($I$):
Thus, the rotational kinetic energy is:
2. Physical Significance
-
Rotational Analogue of Mass: Just as $K = \frac{1}{2}mv^2$ for translation, $K = \frac{1}{2}I\omega^2$ for rotation. $I$ resists changes in rotational motion.
-
Dependence: Unlike mass, which is a fixed property, the moment of inertia of a body depends on:
-
The mass of the body.
-
Its shape and size.
-
The distribution of mass about the axis of rotation.
-
The position and orientation of the axis of rotation.
-
3. Radius of Gyration ($k$)
The mass distribution of a body can be represented by a single distance $k$ from the axis. If the entire mass $M$ of the body were concentrated at this distance, its moment of inertia would remain $I$.
-
Definition: The distance from the axis of rotation to a point where the total mass of the body could be concentrated without changing its moment of inertia.
-
Unit: Metre (m).
4. Moments of Inertia for Common Bodies
The
| Body | Axis | Moment of Inertia (I) |
| Thin Ring (Radius $R$) | Perpendicular to plane at centre | $MR^2$ |
| Thin Ring (Radius $R$) | Along any diameter | $\frac{1}{2}MR^2$ |
| Thin Rod (Length $L$) | Perpendicular to rod at mid-point | $\frac{1}{12}ML^2$ |
| Circular Disc (Radius $R$) | Perpendicular to disc at centre | $\frac{1}{2}MR^2$ |
| Hollow Cylinder ($R$) | Axis of cylinder | $MR^2$ |
| Solid Cylinder ($R$) | Axis of cylinder | $\frac{1}{2}MR^2$ |
| Solid Sphere ($R$) | Along any diameter | $\frac{2}{5}MR^2$ |
5. Practical Application: The Flywheel
A flywheel is a heavy disc with a large moment of inertia used in engines (like steam or automobile engines).
-
Function: Because of its high $I$, it resists sudden changes in rotational speed.
-
Benefit: It ensures a smooth ride by preventing jerky motions and allowing gradual changes in speed.
Summary Table: Linear vs. Rotational Comparison
| Feature | Linear Motion | Rotational Motion |
| Inertia | Mass ($m$) | Moment of Inertia ($I$) |
| Velocity | $v$ | $\omega$ |
| Kinetic Energy | $\frac{1}{2}mv^2$ | $\frac{1}{2}I\omega^2$ |
| Momentum | $p = mv$ | $L = I\omega$ |
6.10 Kinematics of rotational motion about a fixed axis
This study guide is based on the NCERT Physics Part-I material for
6.10 Kinematics of Rotational Motion about a Fixed Axis
When a rigid body rotates about a fixed axis, every particle of the body moves in a circle. To understand this motion, we use variables that are exactly analogous to those we use in linear (translational) motion.
1. Key Variables and Their Analogies
In rotational motion about a fixed axis, we only need one variable to describe the position because there is only one degree of freedom.
| Linear Motion (Translation) | Rotational Motion (Fixed Axis) | Analogy Relationship |
| Displacement ($x$) | Angular Displacement ($\theta$) | $x = r\theta$ |
| Velocity ($v$) | Angular Velocity ($\omega$) | $v = r\omega$ |
| Acceleration ($a$) | Angular Acceleration ($\alpha$) | $a = r\alpha$ |
| Time ($t$) | Time ($t$) | Identical |
Note: For a fixed axis, we don't need to treat $\omega$ and $\alpha$ as vectors because the direction of the axis does not change.
2. Kinematic Equations for Constant Acceleration
Just as we have three main equations for constant linear acceleration, we have three corresponding equations for constant angular acceleration ($\alpha$):
-
Final Angular Velocity:
$$\omega = \omega_0 + \alpha t$$(Analogous to $v = v_0 + at$)
-
Angular Displacement:
$$\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$$(Analogous to $x = x_0 + v_0 t + \frac{1}{2}at^2$)
-
Velocity-Displacement Relation:
$$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$$(Analogous to $v^2 = v_0^2 + 2a\Delta x$)
Where:
-
$\theta_0$ = Initial angular position at $t=0$
-
$\omega_0$ = Initial angular velocity
-
$\alpha$ = Constant angular acceleration
3. Step-by-Step Problem Solving
To solve problems in rotational kinematics, follow these steps:
-
Identify the Given Values: Look for $\omega_0$, $\omega$, $t$, $\theta$, or $\alpha$.
-
Check Units: Ensure angular speed is in $rad/s$. If given in $rpm$ (revolutions per minute), convert it:
$$\omega = \frac{2\pi \times \text{rpm}}{60}$$ -
Choose the Equation: Pick the equation that contains your "knowns" and your "unknown."
Example Walkthrough (Based on Example 6.11)
Scenario: A motor wheel increases from $1200\text{ rpm}$ to $3120\text{ rpm}$ in $16\text{ seconds}$.
-
Step 1: Convert $rpm$ to $rad/s$.
-
$\omega_0 = 1200 \times \frac{2\pi}{60} = 40\pi\text{ rad/s}$
-
$\omega = 3120 \times \frac{2\pi}{60} = 104\pi\text{ rad/s}$
-
-
Step 2: Find $\alpha$ using $\omega = \omega_0 + \alpha t$.
-
$104\pi = 40\pi + \alpha(16)$
-
$64\pi = 16\alpha \Rightarrow \alpha = 4\pi\text{ rad/s}^2$
-
4. Summary of Concepts
-
Rotational Inertia: Just as mass resists linear motion, Moment of Inertia ($I$) resists rotational motion.
-
Flywheels: In machinery, a "flywheel" uses a large moment of inertia to resist sudden changes in speed, ensuring a smooth ride and preventing jerky motions.
6.11 Dynamics of rotational motion about a fixed axis
Based on the
6.11 Dynamics of Rotational Motion about a Fixed Axis
In linear motion, we use Newton’s Second Law ($F = ma$) to understand how force causes acceleration. In rotational motion, we use an analogous law involving Torque ($\tau$) and Moment of Inertia ($I$).
1. Work Done by a Torque
When a force $F$ acts on a rigid body rotating about a fixed axis (like the z-axis), only the component of the force in the plane perpendicular to the axis contributes to the rotation.
-
The Formula: The work done ($dW$) by a torque ($\tau$) during an angular displacement ($d\theta$) is:
$$dW = \tau d\theta$$ -
Analogy: This is exactly like the linear work formula $dW = F dx$.
-
Key Condition: We assume the axis is fixed. Any torque component perpendicular to the axis is cancelled out by "forces of constraint" from the bearings or supports holding the axis.
2. Rotational Power
Power is the rate at which work is done. By dividing the work formula by time ($dt$), we get:
Since $\frac{d\theta}{dt} = \omega$ (angular velocity), the power in rotational motion is:
Analogy Check: Just as $P = Fv$ in translation, $P = \tau \omega$ in rotation.
3. Newton’s Second Law for Rotation
The most fundamental equation in rotational dynamics relates torque to angular acceleration ($\alpha$):
-
Derivation Logic: 1. Work done ($dW$) increases the Kinetic Energy ($K = \frac{1}{2}I\omega^2$).
2. Rate of work ($P = \tau \omega$) must equal the rate of change of K.E. ($\frac{dK}{dt}$).
3. $\frac{d}{dt}(\frac{1}{2}I\omega^2) = I\omega \frac{d\omega}{dt} = I\omega \alpha$.
4. Equating the two: $\tau \omega = I \omega \alpha$, which simplifies to $\tau = I \alpha$.
4. Comparison Table: Linear vs. Rotational Dynamics
Use this table to quickly memorize the relationships:
| Linear Motion | Rotational Motion (Fixed Axis) | Formula Relation |
| Force ($F$) | Torque ($\tau$) | $\tau = r \times F$ |
| Mass ($M$) | Moment of Inertia ($I$) | $I = \sum m_i r_i^2$ |
| Law: $F = Ma$ | Law: $\tau = I \alpha$ | Direct Analogy |
| Work: $W = \int F dx$ | Work: $W = \int \tau d\theta$ | Direct Analogy |
| Power: $P = Fv$ | Power: $P = \tau \omega$ | Direct Analogy |
| Momentum: $p = Mv$ | Angular Momentum: $L = I\omega$ | Direct Analogy |
5. Solved Example (Based on Example 6.12)
Problem: A cord is wound around a flywheel of mass 20 kg and radius 20 cm ($0.2\text{ m}$). A steady pull of 25 N is applied. Find the angular acceleration.
-
Find Torque ($\tau$):
$$\tau = \text{Force} \times \text{Radius} = 25\text{ N} \times 0.2\text{ m} = 5.0\text{ Nm}$$ -
Find Moment of Inertia ($I$): For a solid wheel/disc:
$$I = \frac{1}{2}MR^2 = \frac{1}{2}(20)(0.2)^2 = 0.4\text{ kg}\cdot\text{m}^2$$ -
Find Acceleration ($\alpha$):
$$\alpha = \frac{\tau}{I} = \frac{5.0}{0.4} = 12.5\text{ rad/s}^2$$
6. Important Definitions to Remember
-
Flywheel: A heavy wheel with a large moment of inertia ($I$). Because $\alpha = \frac{\tau}{I}$, a large $I$ means the wheel resists changes in speed, preventing "jerky" motions in engines.
-
Torque: The rotational equivalent of force; it is the "turning effect" produced by a force.
Work done by Torque
Based on the
1. Understanding Work in Rotational Motion
In linear motion, work is done when a force causes a displacement ($dW = F \cdot ds$). Similarly, in rotational motion, work is done when a torque ($\tau$) causes an angular displacement ($\theta$).
For a rigid body rotating about a fixed axis (e.g., the $z$-axis), we only consider the components of torque along that axis, as they are the ones responsible for rotation.
The Analogy Table
| Linear Motion | Rotational Motion |
| Displacement ($x$) | Angular displacement ($\theta$) |
| Force ($F$) | Torque ($\tau$) |
| Work ($dW = F \cdot ds$) | Work ($dW = \tau \cdot d\theta$) |
2. Mathematical Derivation
Consider a particle $P_1$ in a rigid body rotating about a fixed axis. The particle moves in a circle of radius $r_1$.
-
Displacement: If the particle rotates through a small angle $d\theta$, the linear distance traveled along the arc is:
$$ds_1 = r_1 d\theta$$ -
Work Done by Force: If a force $F_1$ acts on the particle at an angle $\phi_1$ to the tangent, the work done is:
$$dW_1 = F_1 \cdot ds_1 = F_1 (r_1 d\theta) \cos \phi_1$$ -
Relation to Torque: The magnitude of the torque $\tau_1$ produced by this force is $r_1 F_1 \sin \alpha_1$ (where $\alpha_1$ is the angle between the force and the radius vector). This simplifies to the tangential force times the radius:
$$dW_1 = \tau_1 d\theta$$ -
Total Work: For a rigid body, every particle undergoes the same angular displacement $d\theta$. Thus, the total work done by all external torques is:
$$dW = \tau d\theta$$
3. Power in Rotational Motion
Power is defined as the rate at which work is performed. By dividing the work equation by time ($dt$), we derive the formula for instantaneous power ($P$):
Since $\frac{d\theta}{dt} = \omega$ (angular velocity), the equation becomes:
Note: This is directly analogous to the linear power formula $P = Fv$.
4. Key Considerations
-
Fixed Axis Simplification: We assume the axis of rotation is maintained by constraint forces. Therefore, we only calculate torques that lie along the axis of rotation.
-
Forces Parallel to Axis: Forces acting parallel to the axis of rotation produce a torque perpendicular to the axis and do not contribute to the rotational work about that specific axis.
-
Perfectly Rigid Bodies: In a perfectly rigid body, there is no internal motion; therefore, all work done by external torques results in a change in the body's rotational kinetic energy.
Example Calculation
If a constant torque of 20 Nm is applied to a wheel, causing it to rotate through 10 radians, the total work done is:
6.12 Angular momentum in case of rotation about a fixed axis
Based on the
1. Introduction to Angular Momentum ($L$)
The angular momentum of a system of particles is defined as the moment of momentum. For a single particle, it is given by the cross product of its position vector ($r$) and its linear momentum ($p$):
In rotational motion about a fixed axis (e.g., the $z$-axis), we examine how individual particle contributions sum up to the total angular momentum of a rigid body.
2. Derivation for a Rigid Body
For a rigid body rotating about a fixed $z$-axis with angular velocity $\omega$:
A. Individual Particle Contribution
Consider a typical particle of mass $m_i$ at a perpendicular distance $r_i$ from the axis.
-
Velocity: The linear velocity is $v_i = \omega r_i$.
-
Angular Momentum Component: While the total angular momentum vector $\mathbf{l}_i$ of a particle may not be parallel to the axis, its component along the $z$-axis ($l_{iz}$) is:
$$l_{iz} = m_i r_i^2 \omega$$
B. Total Angular Momentum ($L_z$)
To find the total angular momentum along the axis, we sum the contributions of all particles:
Since $\sum m_i r_i^2$ is the Moment of Inertia ($I$) of the body about the axis:
3. Symmetry and the Vector $\mathbf{L}$
The relationship between the total angular momentum vector $\mathbf{L}$ and the angular velocity vector $\boldsymbol{\omega}$ depends on the symmetry of the body:
-
Symmetric Bodies: For bodies symmetric about the axis of rotation (like a cylinder or sphere rotating about its geometric axis), the components of angular momentum perpendicular to the axis cancel out. In this case, $\mathbf{L}$ is parallel to $\boldsymbol{\omega}$:
$$\mathbf{L} = I \boldsymbol{\omega}$$ -
Non-Symmetric Bodies: If the axis of rotation is not a symmetry axis, $\mathbf{L}$ and $\boldsymbol{\omega}$ are generally not parallel.
4. Conservation of Angular Momentum
According to the
Practical Examples
| Example | Mechanism |
| Swivel Chair | Stretching arms increases $I$, causing $\omega$ to decrease to keep $I\omega$ constant. |
| Circus Acrobat | Tucking the body in reduces $I$, significantly increasing rotation speed ($\omega$). |
| Figure Skater | Pulling arms inward results in a faster spin due to the conservation of $L$. |
5. Summary of Analogies
| Linear Motion | Rotational Motion |
| Mass ($M$) | Moment of Inertia ($I$) |
| Linear Momentum ($p = Mv$) | Angular Momentum ($L = I\omega$) |
| Force ($F = \frac{dp}{dt}$) | Torque ($\tau = \frac{dL}{dt}$) |
| Conservation: $p = \text{const}$ if $F=0$ | Conservation: $L = \text{const}$ if $\tau=0$ |
