6.1 Introduction
Chapter 6 — Systems of Particles and Rotational Motion
In the summer of 1918, a young figure skater in Europe stunned her audience with a move that seemed to defy physics itself. She began spinning slowly on the ice with her arms stretched wide, then suddenly pulled them close to her chest. In an instant, her slow, graceful turn exploded into a blur — she was now spinning so fast that the crowd gasped. She had used no extra push, no fresh energy from the ice. She had simply moved her arms. How could this be?
Around the same time, engineers designing the first heavy diesel engines were struggling with a very different problem. Their engines shook violently and ran unevenly, threatening to tear themselves apart. The solution, they discovered, was to attach a massive metal disc — a flywheel — to the shaft. Once spinning, the flywheel stubbornly resisted any change in its motion, smoothing out the engine’s jerks into steady, reliable power. The same physical idea that let a skater spin faster by pulling in her arms was, in another form, taming the roar of industrial machines.
Until now, we have treated every moving object as a single point — a tiny particle with no size and no ability to spin. But real bodies have size, shape, and the ability to rotate. In this chapter, we go beyond the point particle to study systems of many particles and rigid bodies that turn about an axis. We will discover how mass is distributed, how forces produce twisting effects, and why a spinning body has a “memory” for its motion.
Figure to come
Fig. 6.0 – A split illustration showing (left) a figure skater pulling her arms in and spinning faster on ice, and (right) a large industrial flywheel on a rotating shaft, with a curved arrow showing rotation.
By the end of this chapter, you will be able to answer each of these questions using the physics of centre of mass, torque, moment of inertia, and angular momentum.
In all the earlier chapters, we studied the motion of a single particle. A particle, in physics, is an idealised object — a point mass with no size at all. This was a very convenient starting point. Even when we studied bodies of finite size — a ball rolling on a track, a car on a road, a block sliding down a slope — we quietly treated each of them as if they were single particles. We spoke of “the velocity of the car” as though the car were a single point.
This shortcut worked for a while, but it is not the full picture. Every real object we meet in daily life — a book, a wheel, a cricket bat, a planet — has a definite size and shape. It is made up of a very large number of particles held together. Such a body is called an extended body.
For many problems, treating an extended body as a single particle is simply not enough. A spinning top, a rolling ball, and a rotating fan cannot be understood by pretending they are point particles — they have parts that move differently from one another at the same instant. In this chapter, we go beyond the single-particle picture and study the motion of extended bodies.
The key idea we will use is that any extended body is, at heart, a system of particles. We will first study how such a system moves as a whole, and we will discover a very special point associated with every system — the centre of mass. Understanding the motion of the centre of mass will unlock much of the physics of extended bodies.
A large class of extended bodies can be treated as rigid bodies. Roughly speaking, a rigid body is one that keeps its shape and size no matter what forces act on it.
Strictly speaking, no real body is perfectly rigid, because every material deforms a little under force. A steel beam bends slightly under a heavy load, a football flattens for an instant when kicked, and even the Earth bulges at its equator due to its own rotation. But in most everyday situations, such deformations are so small that we can safely ignore them.
Objects like wheels, tops, steel beams, molecules, and planets can, for most practical purposes, be treated as rigid bodies. We can ignore the tiny twists, bends, and vibrations they might undergo and study only their overall motion.
6.1.1 What kind of motion can a rigid body have?
The best way to answer this question is to look at some familiar examples of rigid bodies in motion and see how they behave.
Pure translational motion
Consider a rectangular block sliding down a smooth inclined plane without any sideways movement, as shown in Fig. 6.1. We treat the block as a rigid body. As it slides down, every single particle of the block — the top edge, the bottom edge, the centre — moves in exactly the same way. At any given instant, they all have the same velocity, pointing along the slope.
Figure to come
Fig. 6.1 – A rectangular block sliding down an inclined plane; two marked points P₁ and P₂ inside the block have arrows of equal length and direction showing they have the same velocity.
This is the simplest kind of motion a rigid body can have. It is called pure translation.
You can think of pure translation as the whole body being carried along without any twisting or turning — as if it were a single point moving through space.
Motion that is more than translation — the rolling cylinder
Now consider a very different situation: a solid metallic or wooden cylinder rolling down the same inclined plane, as shown in Fig. 6.2. Like the block, the cylinder as a whole shifts from the top of the incline to the bottom. So it certainly seems to have translational motion.
Figure to come
Fig. 6.2 – A solid cylinder rolling down an inclined plane, with four points P₁, P₂, P₃, P₄ marked on it, each with an arrow showing a different velocity; the point of contact P₃ at the bottom has zero velocity.
But look more carefully. If we mark different points on the cylinder — one at the top, one at the side, one at the point where the cylinder touches the incline — and we look at their velocities at the same instant, we find that they are not the same. Each marked point has a different velocity, both in size and direction.
Most strikingly, the point of contact between the cylinder and the incline — the point momentarily touching the ground — has zero velocity at any instant, provided the cylinder is rolling without slipping. Yet the cylinder as a whole is clearly moving down the slope. How can this be?
The answer is that the cylinder is not in pure translation. Its motion is translation plus something else. That “something else” is what we need to uncover next.
Rotation — the axis of rotation
To understand the “something else,” let us look at a rigid body that has been forced to have no translational motion at all. The simplest way to do this is to fix the body along a straight line so that it cannot shift as a whole. The only motion left for such a body is then a turning motion — a rotation.
The line about which the body rotates is called its axis of rotation.
Examples of rotation about a fixed axis are all around us. A ceiling fan rotates about the vertical rod holding it. A potter’s wheel spins about a vertical axle. A giant wheel at a fair and a merry-go-round in a park also rotate about fixed axes. These are shown in Fig. 6.3.
Figure to come
Fig. 6.3 – (a) A ceiling fan rotating about its vertical rod; (b) a potter’s wheel spinning about a vertical axle. Both axes marked with dashed lines and curved arrows showing rotation.
What characterises rotation?
Now let us look closely at what happens to each particle of a rigid body rotating about a fixed axis. Take the z-axis as the fixed axis, as shown in Fig. 6.4.
Figure to come
Fig. 6.4 – A blob-shaped rigid body rotating about the z-axis, with two particles P₁ and P₂ tracing circles of radii r₁ and r₂ in planes perpendicular to the z-axis, with centres C₁ and C₂ on the axis; a third particle P₃ lying exactly on the axis is shown stationary.
Pick any particle P₁ of the body, sitting at a perpendicular distance \(r_1\) from the axis. As the body rotates, this particle moves along a circle. The circle has radius \(r_1\) and its centre C₁ lies on the axis of rotation. Importantly, this circle lies in a plane that is perpendicular to the axis.
Take another particle P₂ at a different perpendicular distance \(r_2\) from the axis. It too traces a circle — but a circle of radius \(r_2\), with its centre C₂ on the axis, lying in a different plane (also perpendicular to the axis).
Different particles trace circles of different sizes, in different planes, but all these planes are perpendicular to the same axis of rotation.
What about a particle that lies on the axis itself, like P₃ in Fig. 6.4? For it, the perpendicular distance from the axis is \(r = 0\). Such a particle doesn’t move at all — it stays where it is while the body rotates around it. This makes sense: the axis of rotation is, by definition, the set of points that stay fixed during the rotation.
When the axis itself moves — precession
So far, the axis of rotation has been fixed in space. But this is not the only way a rigid body can rotate. Sometimes the axis itself moves.
A classic example is a spinning top, shown in Fig. 6.5(a). Suppose the top is spinning in place on a smooth floor, so its tip touches only one point on the ground and does not slide from there. Even though the tip is fixed, the top’s axis of rotation slowly sweeps out a cone about the vertical. This slow, sweeping motion of the axis is called precession.
Figure to come
Fig. 6.5(a) – A spinning top with its tip fixed at point O on the ground; the top’s own spin axis is tilted from vertical and traces out a cone about the vertical direction; curved arrows indicate both the spin and the precession.
Notice a subtle but important point: in a precessing top, only the point of contact with the ground is truly fixed — not a whole line. The axis of rotation passes through this fixed point at every instant, but the axis itself is changing its direction.
Another everyday example is the oscillating table fan or pedestal fan, shown in Fig. 6.5(b). Its blades spin about their own axis, but this axis itself swings from side to side about a vertical direction through the pivot O. The pivot point O is fixed, but the axis of the blades is not.
Figure to come
Fig. 6.5(b) – An oscillating table fan on a stand. The blades are shown spinning about their horizontal axis; a curved double arrow shows the axis of rotation itself oscillating side-to-side about a vertical direction through the pivot O.
In both these examples — the spinning top and the oscillating fan — one point of the rigid body is fixed, but no full line is fixed. This is a more general case of rotation than rotation about a fixed axis.
For most of this chapter, we will study only the simpler and special case where a whole line (the axis) stays fixed. So unless we say otherwise:
Coming back to the rolling cylinder
Now we can return to the puzzle of the rolling cylinder in Fig. 6.2. As it rolls down the incline, it is both shifting from top to bottom (translation) and turning about its own central axis (rotation). The cylinder is doing both at the same time.
This is why different particles of the rolling cylinder had different velocities — the translation part adds the same velocity to every particle, but the rotation part adds different velocities depending on where each particle sits relative to the central axis. At the bottom of the cylinder, the rotational velocity happens to point opposite to the translational velocity, and if the cylinder is rolling without slipping, the two exactly cancel — giving zero velocity at the point of contact.
So the “something else” we were looking for is rotational motion. Rolling = translation + rotation.
Fig. 6.6(a) and Fig. 6.6(b) make this point very clearly. Both figures show the same body moving along the same overall path (translational trajectory). In Fig. 6.6(a), the motion is pure translation — every particle moves the same way, and an arrow drawn on the body always points in the same direction. In Fig. 6.6(b), the motion is translation combined with rotation — the body’s overall path is the same, but the body has also turned, so the arrow drawn on it points in different directions at different instants.
Figure to come
Fig. 6.6 – (a) A body moving along a curved path Tr₁ in pure translation; at three positions, the arrow marked OP on the body keeps the same orientation. (b) The same body moving along an identical path Tr₂, but now with rotation added; at the same three positions, the arrow OP points in three different directions.
The final picture
We are now ready to state the complete answer to our starting question — what kind of motion can a rigid body have?
In the rest of this chapter, we will develop the tools needed to describe all these motions quantitatively. We will start with the concept of centre of mass, which will let us cleanly separate the translational part from any rotational or internal motion — a technique that will run through the entire chapter.
6.2 Centre of Mass
We now introduce one of the most powerful ideas in the physics of extended bodies — the centre of mass. As promised at the end of the last section, this single point will let us handle the motion of any system of particles or rigid body in a very clean way.
Let us build up the idea step by step, starting with the simplest possible system: just two particles.
Two particles on a line
Take two particles and place them on a straight line. Choose this line to be the x-axis. Let the two particles be at distances \(x_1\) and \(x_2\) from some origin O, with masses \(m_1\) and \(m_2\) respectively, as shown in Fig. 6.7.
Figure to come
Fig. 6.7 – Two particles of masses m₁ and m₂ placed on the x-axis at distances x₁ and x₂ from origin O; the centre of mass C is marked at distance X between them.
The centre of mass of this two-particle system is defined as the point C that lies at a distance X from O, where
\[X = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \quad \text{...(6.1)}\]
Here X is measured from the same origin O, and \(x_1\), \(x_2\) are the distances of the two particles from O. The quantity X in Eq. (6.1) can be thought of as a mass-weighted average of \(x_1\) and \(x_2\). It is not a simple arithmetic mean — each position is weighted by the mass at that position, so the heavier particle “pulls” the centre of mass closer to itself.
To see this clearly, take the special case where the two particles have equal masses, \(m_1 = m_2 = m\). Then Eq. (6.1) becomes
\[X = \frac{m x_1 + m x_2}{2m} = \frac{x_1 + x_2}{2}\]
which is just the arithmetic mean of the two positions.
More particles on a line
Suppose we have \(n\) particles of masses \(m_1, m_2, \dots, m_n\) lying along a straight line, which we again take as the x-axis. By a natural extension of Eq. (6.1), the position of the centre of mass is defined as
\[X = \frac{m_1 x_1 + m_2 x_2 + \dots + m_n x_n}{m_1 + m_2 + \dots + m_n} = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i} \quad \text{...(6.2)}\]
Here \(x_1, x_2, \dots, x_n\) are the distances of the \(n\) particles from the origin, and X is measured from the same origin. The Greek letter \(\sum\) (sigma) is a compact way of writing a sum — it just means “add up all these terms as \(i\) runs from 1 to \(n\).”
The denominator is simply the total mass of the system:
\[\sum m_i = M\]
where M is the total mass. So the centre of mass formula can also be written as
\[X = \frac{\sum m_i x_i}{M}\]
Three particles in a plane
Now let us step off the line. Take three particles which do not all lie on a single straight line. We can then set up an x-y plane containing all three, and describe their positions by coordinates \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\). Let their masses be \(m_1\), \(m_2\), and \(m_3\).
The centre of mass C now has both an x-coordinate and a y-coordinate, given by
\[X = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3} \quad \text{...(6.3a)}\]
\[Y = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3} \quad \text{...(6.3b)}\]
Both formulas are again mass-weighted averages, one for each coordinate.
For particles of equal mass, \(m = m_1 = m_2 = m_3\), these simplify to
\[X = \frac{m(x_1 + x_2 + x_3)}{3m} = \frac{x_1 + x_2 + x_3}{3}\]
\[Y = \frac{m(y_1 + y_2 + y_3)}{3m} = \frac{y_1 + y_2 + y_3}{3}\]
These are the coordinates of a very familiar point from geometry — the centroid of the triangle formed by the three particles.
General case — n particles in space
The results (6.3a) and (6.3b) generalise naturally to a system of \(n\) particles, no longer confined to a line or even to a plane, but distributed anywhere in three-dimensional space. The centre of mass is then a point with coordinates \((X, Y, Z)\), where
\[X = \frac{\sum m_i x_i}{M} \quad \text{...(6.4a)}\]
\[Y = \frac{\sum m_i y_i}{M} \quad \text{...(6.4b)}\]
\[Z = \frac{\sum m_i z_i}{M} \quad \text{...(6.4c)}\]
Here \(M = \sum m_i\) is the total mass of the system. The index \(i\) runs from 1 to \(n\). The mass of the \(i^\text{th}\) particle is \(m_i\), and its position is given by the coordinates \((x_i, y_i, z_i)\).
One vector equation instead of three
Three separate scalar equations (6.4a, 6.4b, 6.4c) can be combined into a single, more compact equation using vectors. Let \(\vec{r}_i\) be the position vector of the \(i^\text{th}\) particle, and let \(\vec{R}\) be the position vector of the centre of mass:
\[\vec{r}_i = x_i \hat{i} + y_i \hat{j} + z_i \hat{k}\]
\[\vec{R} = X \hat{i} + Y \hat{j} + Z \hat{k}\]
Then all three scalar equations (6.4a–c) collapse into
\[\vec{R} = \frac{\sum m_i \vec{r}_i}{M} \quad \text{...(6.4d)}\]
The sum on the right-hand side is a vector sum. This is a nice example of the economy vectors give us — three equations become one.
An immediate and useful consequence: if we choose the origin of our coordinate system to be at the centre of mass itself, then \(\vec{R} = 0\), and from Eq. (6.4d),
\[\sum m_i \vec{r}_i = 0\]
That is, when position vectors are measured from the centre of mass, the mass-weighted sum of all position vectors is zero. We will use this fact later.
From particles to rigid bodies — continuous distributions
A rigid body — a metre stick, a flywheel, a coin — is really a system of an enormous number of tightly packed particles (atoms and molecules). In principle, Eqs. (6.4a)–(6.4d) apply directly. But because the number of particles is so vast, we cannot actually add them one by one.
Since the particles are packed extremely closely together, we can treat the body as a continuous distribution of mass. We slice the body up into a large number \(n\) of small mass elements \(\Delta m_1, \Delta m_2, \dots, \Delta m_n\). We take the \(i^\text{th}\) element \(\Delta m_i\) to be located around the point \((x_i, y_i, z_i)\). Then the centre of mass coordinates are approximately
\[X = \frac{\sum (\Delta m_i) x_i}{\sum \Delta m_i}, \quad Y = \frac{\sum (\Delta m_i) y_i}{\sum \Delta m_i}, \quad Z = \frac{\sum (\Delta m_i) z_i}{\sum \Delta m_i}\]
As we make \(n\) larger and larger (more and more slices) and each \(\Delta m_i\) smaller and smaller, these expressions become more and more accurate. In the limit, the sums become integrals:
\[\sum \Delta m_i \to \int dm = M\]
\[\sum (\Delta m_i) x_i \to \int x \, dm\]
\[\sum (\Delta m_i) y_i \to \int y \, dm\]
\[\sum (\Delta m_i) z_i \to \int z \, dm\]
Here M is the total mass of the body. The centre of mass coordinates for a continuous body are then
\[X = \frac{1}{M} \int x \, dm, \quad Y = \frac{1}{M} \int y \, dm, \quad Z = \frac{1}{M} \int z \, dm \quad \text{...(6.5a)}\]
The vector form equivalent to these three scalar equations is
\[\vec{R} = \frac{1}{M} \int \vec{r} \, dm \quad \text{...(6.5b)}\]
If we choose the centre of mass itself as the origin of our coordinate system, then \(\vec{R} = 0\), and so
\[\int \vec{r} \, dm = 0\]
or equivalently
\[\int x \, dm = \int y \, dm = \int z \, dm = 0 \quad \text{...(6.6)}\]
The power of symmetry
We often need to find the centre of mass of familiar regular shapes — rings, discs, spheres, uniform rods, and so on. If such a body is homogeneous — meaning its mass is distributed uniformly throughout its volume — we can find the centre of mass without doing any integration at all. Symmetry does the work for us.
Let us see this with a very simple example — a thin, uniform rod.
Consider a thin uniform rod (whose thickness is much smaller than its length). Place the origin at the geometric centre of the rod, and take the x-axis along its length, as shown in Fig. 6.8.
Figure to come
Fig. 6.8 – A thin uniform rod lying along the x-axis with its geometric centre at the origin O; two small equal mass elements dm are marked symmetrically, one at position +x and the other at −x.
Because the rod is uniform, for every small mass element \(dm\) located at some position \(x\), there is a matching mass element of equal mass \(dm\) located at \(-x\) — the position obtained by reflecting \(x\) through the origin. This is called reflection symmetry.
The contribution of one such element to the integral \(\int x \, dm\) is \(x \, dm\); the contribution of its mirror partner is \((-x)\, dm\). These two contributions cancel exactly. Adding up all such pairs, the entire integral \(\int x \, dm\) is zero.
But from Eq. (6.6), the point about which \(\int x \, dm = 0\) is precisely the centre of mass. So the centre of mass of a uniform thin rod coincides with its geometric centre. Symmetry alone gave us the answer.
The same symmetry argument applies to homogeneous rings, discs, spheres, and even thick rods with circular or rectangular cross-sections. For all such bodies, for every mass element \(dm\) at a point \((x, y, z)\), we can always find an equal mass element at the point \((-x, -y, -z)\). This means the origin (at the geometric centre) is a point of reflection symmetry for these bodies. As a result, all the integrals in Eq. (6.5a) come out to zero, and:
Solved Examples
6.3 Motion of Centre of Mass
We now have a clear definition of the centre of mass for any system of particles or rigid body. But so far, this centre of mass has only been a point — a location. Its real power comes from asking: how does this point move?
If we can find a simple rule for the motion of the centre of mass, we will have a powerful tool. It will let us handle any system — no matter how complicated its internal motion — by focusing on this one special point.
Building the equation of motion
Let us start with Eq. (6.4d) from the previous section:
\[\vec{R} = \frac{\sum m_i \vec{r}_i}{M}\]
Multiplying both sides by the total mass \(M\), we get
\[M\vec{R} = \sum m_i \vec{r}_i = m_1 \vec{r}_1 + m_2 \vec{r}_2 + \dots + m_n \vec{r}_n \quad \text{...(6.7)}\]
This equation just re-states the definition of the centre of mass, but in a form we can now differentiate with respect to time.
First differentiation — velocity of the centre of mass
Differentiating both sides of Eq. (6.7) with respect to time \(t\):
\[M \frac{d\vec{R}}{dt} = m_1 \frac{d\vec{r}_1}{dt} + m_2 \frac{d\vec{r}_2}{dt} + \dots + m_n \frac{d\vec{r}_n}{dt}\]
The rate of change of a position vector is a velocity. So this becomes
\[M \vec{V} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n \quad \text{...(6.8)}\]
Here \(\vec{v}_1 = d\vec{r}_1/dt\) is the velocity of the first particle, \(\vec{v}_2 = d\vec{r}_2/dt\) is the velocity of the second, and so on. The vector \(\vec{V} = d\vec{R}/dt\) is the velocity of the centre of mass itself, measured in m/s.
Note that we have treated the masses \(m_1, m_2, \dots\) as constants while differentiating. This is fine — the particles do not change mass over time. (Throughout this chapter, we assume the total mass of the system stays constant.)
Eq. (6.8) tells us that the total momentum-like quantity \(M\vec{V}\) (mass of the system times velocity of its centre of mass) equals the sum of the individual momenta \(m_i \vec{v}_i\) of all the particles. We will explore this important observation in more detail in the next section.
Second differentiation — acceleration of the centre of mass
Differentiating Eq. (6.8) once more with respect to time:
\[M \frac{d\vec{V}}{dt} = m_1 \frac{d\vec{v}_1}{dt} + m_2 \frac{d\vec{v}_2}{dt} + \dots + m_n \frac{d\vec{v}_n}{dt}\]
Rate of change of velocity is acceleration, so
\[M \vec{A} = m_1 \vec{a}_1 + m_2 \vec{a}_2 + \dots + m_n \vec{a}_n \quad \text{...(6.9)}\]
where \(\vec{a}_1 = d\vec{v}_1/dt\) is the acceleration of the first particle, \(\vec{a}_2\) is that of the second, and \(\vec{A} = d\vec{V}/dt\) is the acceleration of the centre of mass, in m/s².
Bringing in Newton’s second law
Now we use what we know from earlier chapters: Newton’s second law. For each particle, the net force acting on it equals its mass times its acceleration. So
\[\vec{F}_1 = m_1 \vec{a}_1, \quad \vec{F}_2 = m_2 \vec{a}_2, \quad \dots, \quad \vec{F}_n = m_n \vec{a}_n\]
Substituting these into Eq. (6.9):
\[M \vec{A} = \vec{F}_1 + \vec{F}_2 + \dots + \vec{F}_n \quad \text{...(6.10)}\]
In words: the total mass times the acceleration of the centre of mass equals the vector sum of all the forces acting on the particles of the system. This is already a clean and useful statement, but we can go one step further.
Separating external and internal forces
Look carefully at what the “force \(\vec{F}_1\) on the first particle” really means. It is not one single force — it is the net force on that particle, which is itself the vector sum of every force acting on it. These forces come in two very different categories:
- External forces — forces exerted on a particle of the system by something outside the system (e.g., gravity pulling down on it, a hand pushing it, a spring attached from outside).
- Internal forces — forces exerted on a particle of the system by another particle of the same system (e.g., two atoms in a molecule pulling on each other, two blocks of a system pushing each other).
Now, here is the key observation. By Newton’s third law, whenever one particle inside the system exerts a force on another particle inside the system, that second particle exerts an equal and opposite force back on the first. These action–reaction pairs are internal forces. When we add up all the forces on all the particles in Eq. (6.10), every internal force gets cancelled by its partner.
Only the external forces survive in the total sum. So we can rewrite Eq. (6.10) as
\[M \vec{A} = \vec{F}_{ext} \quad \text{...(6.11)}\]
where \(\vec{F}_{ext}\) is the vector sum of all external forces acting on the particles of the system, measured in newtons (N).
The centre of mass theorem
Eq. (6.11) is one of the most important results of this chapter. In plain words:
Think about what a remarkable simplification this is. A system may have thousands of particles, all pushing and pulling on one another in complicated ways. It may be a rotating rigid body, a bunch of interacting molecules, or a group of colliding objects. But no matter how complicated the internal dance, the centre of mass moves as if it were a single particle of mass \(M\) acted on only by the external forces.
Notice one very useful feature: to figure out how the centre of mass moves, we do not need to know anything about the internal forces at all. Only external forces matter for Eq. (6.11). This makes many otherwise impossible problems tractable.
What kind of system does this apply to?
To derive Eq. (6.11), we did not need to specify what kind of system we were dealing with. The system could be:
- A collection of loose particles moving in all sorts of complicated ways.
- A rigid body in pure translation.
- A rigid body in a combination of translation and rotation.
- Even a body that is deforming, or a fluid.
Whatever the system, and whatever its individual particles are doing, the centre of mass obeys Eq. (6.11).
This is a huge improvement over what we did in earlier chapters. Previously, we treated every extended body as a single particle without justification. Now we understand why that worked: it worked for the translational motion of the centre of mass. And now we can also handle the parts of the motion — rotation, internal motion — that we had to ignore before.
We can now cleanly split the motion of an extended body into two parts:
- Translational motion of the centre of mass — handled by Eq. (6.11) by imagining the whole mass concentrated at the centre of mass, with all external forces applied there.
- Motion about the centre of mass — the rotation or internal motion, which we will study in the rest of this chapter.
A striking illustration — the exploding projectile
Consider a fireworks shell launched into the air. Ignoring air resistance, the shell follows the familiar curved (parabolic) trajectory of a projectile under gravity. Now suppose that midway through its flight, the shell explodes into many fragments — bright sparks flying off in all directions.
Figure to come
Fig. 6.12 – A projectile launched from origin O follows a smooth parabolic path; at position \(x_1\) midway up the path, it explodes into fragments shown as dashed arrows flying outward, but the trajectory continues as a dashed parabola showing the centre of mass path.
What forces caused the explosion? Chemical forces inside the shell — pressure from expanding gases pushing on the shell’s casing and on each fragment. These forces are internal to the system of fragments. By Newton’s third law, they cancel in pairs when summed over the whole system.
What is the total external force on the system, before and after the explosion? It is the same in both cases — the force of gravity, \(M\vec{g}\), where \(M\) is the total mass of all the fragments (equal to the mass of the shell before it exploded).
Applying Eq. (6.11), the acceleration of the centre of mass depends only on external forces. Since these are unchanged by the explosion, the centre of mass keeps accelerating downward at \(g\), exactly as if nothing had happened. Its trajectory continues along the same parabolic path as before the explosion.
This is a beautifully counterintuitive result. To a distant observer who cannot see the individual fragments — say, someone watching through a smoky sky — the “average position” of the debris keeps sailing along the original curved path.
Looking ahead
Eq. (6.11) is a very general result, and we will return to it many times. In the next section, we will look at the same idea from a slightly different angle — through linear momentum — which will let us state a powerful conservation law for the whole system.
6.4 Linear Momentum of a System of Particles
In the previous section, we found how the acceleration of the centre of mass is governed by the external forces on a system. We now look at the same physics through the lens of momentum. This will lead us to one of the most powerful conservation laws in all of physics.
Recalling momentum for a single particle
From earlier chapters, the linear momentum of a single particle is defined as
\[\vec{p} = m\vec{v} \quad \text{...(6.12)}\]
where \(m\) is the particle’s mass (in kg) and \(\vec{v}\) is its velocity (in m/s). Momentum \(\vec{p}\) is a vector with SI unit kg m/s.
We also recall Newton’s second law in its most general form for a single particle:
\[\vec{F} = \frac{d\vec{p}}{dt} \quad \text{...(6.13)}\]
Here \(\vec{F}\) is the net force on the particle. Eq. (6.13) says the net force on a particle equals the rate at which its momentum changes.
Total momentum of a system
Now consider a system of \(n\) particles with masses \(m_1, m_2, \dots, m_n\) and velocities \(\vec{v}_1, \vec{v}_2, \dots, \vec{v}_n\). The particles may be interacting with each other, and external forces may also be acting on them.
The linear momentum of the first particle is \(m_1 \vec{v}_1\), of the second particle \(m_2 \vec{v}_2\), and so on.
The total linear momentum of the system is defined as the vector sum of the momenta of all the individual particles:
\[\vec{P} = \vec{p}_1 + \vec{p}_2 + \dots + \vec{p}_n\]
\[\vec{P} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n \quad \text{...(6.14)}\]
The right-hand side of Eq. (6.14) should look familiar. It is exactly what we saw on the right-hand side of Eq. (6.8) in the last section:
\[M\vec{V} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n\]
Comparing this with Eq. (6.14), we immediately get
\[\vec{P} = M\vec{V} \quad \text{...(6.15)}\]
This is a remarkably clean result.
Think about what this says. No matter how many particles the system has, no matter how they are moving relative to each other, their combined momentum is exactly what a single particle of mass \(M\) moving with the velocity of the centre of mass would have. The centre of mass acts as a single “stand-in” for the whole system as far as momentum is concerned.
Newton’s second law for a system of particles
Let us differentiate Eq. (6.15) with respect to time:
\[\frac{d\vec{P}}{dt} = M \frac{d\vec{V}}{dt} = M\vec{A} \quad \text{...(6.16)}\]
We assumed the total mass \(M\) does not change with time. Here \(\vec{A}\) is the acceleration of the centre of mass, in m/s².
But from Eq. (6.11) of the last section, we already know that \(M\vec{A} = \vec{F}_{ext}\). Substituting this into Eq. (6.16):
\[\frac{d\vec{P}}{dt} = \vec{F}_{ext} \quad \text{...(6.17)}\]
Eq. (6.17) is Newton’s second law generalised to a system of particles. It has exactly the same form as Eq. (6.13) for a single particle, but now \(\vec{P}\) is the total momentum of the whole system and \(\vec{F}_{ext}\) is the vector sum of only the external forces.
Internal forces — no matter how complex — never appear in Eq. (6.17). They cancel out in pairs by Newton’s third law and simply do not affect the total momentum of the system.
The law of conservation of linear momentum
Now suppose the vector sum of all the external forces on the system is zero. Setting \(\vec{F}_{ext} = 0\) in Eq. (6.17) gives
\[\frac{d\vec{P}}{dt} = 0 \quad \text{or} \quad \vec{P} = \text{Constant} \quad \text{...(6.18a)}\]
This is a profound result.
This is the law of conservation of linear momentum for a system of particles. It is one of the great conservation laws of physics, and it holds true no matter how complicated the internal motion of the system.
Because of Eq. (6.15), this also means:
(Throughout this discussion, we assume the total mass of the system stays constant.)
Complicated inside, simple outside
Because of internal forces — the pushes and pulls that particles inside the system exert on each other — the individual particles may follow very complicated paths. Yet, if the total external force is zero, the centre of mass moves uniformly in a straight line, just like a lone free particle drifting through empty space.
The vector equation (6.18a) is really equivalent to three separate scalar equations, one for each direction:
\[P_x = c_1, \quad P_y = c_2, \quad P_z = c_3 \quad \text{...(6.18b)}\]
Here \(P_x\), \(P_y\), and \(P_z\) are the components of the total momentum \(\vec{P}\) along the x, y, and z axes; \(c_1\), \(c_2\), \(c_3\) are constants. This means momentum is conserved separately along each direction — a fact very useful when solving problems.
An example — the decaying nucleus
To see the law of conservation of momentum in action, consider the radioactive decay of a moving unstable particle — for instance, the nucleus of a radium atom. A radium nucleus can spontaneously break apart, or disintegrate, into a nucleus of radon and an alpha particle (the nucleus of a helium atom).
The forces that cause this break-up are entirely internal to the nucleus — they arise from the strong and electromagnetic forces between the constituents. The external forces on the nucleus (gravity, for instance) are negligibly small on the scale of these internal forces.
So the total linear momentum of the system is the same before and after the decay. The radon nucleus and the alpha particle fly off in different directions after the split, and they must do so in exactly the way that keeps the total momentum unchanged.
Consequently, the centre of mass of the two decay products keeps moving along the very same straight path along which the original radium nucleus was moving, as shown in Fig. 6.13(a).
Figure to come
Fig. 6.13(a) – A radium (Ra) nucleus moving to the right; midway along its path it decays, and a radon (Rn) nucleus and an alpha particle (He) shoot off along two different straight-line paths; a dashed line marked “CM” continues along the original straight-line path, showing that the centre of mass keeps moving uniformly.
Now imagine watching the same decay from a different reference frame — one moving with the centre of mass. In this centre of mass frame, the total momentum of the system is zero (by definition, the centre of mass is at rest here). So before the decay, the radium nucleus itself is at rest in this frame; after the decay, the radon nucleus and the alpha particle must fly off in exactly opposite directions with equal and opposite momenta — “back to back,” as shown in Fig. 6.13(b).
Figure to come
Fig. 6.13(b) – The same decay viewed from the centre of mass frame: a stationary radium nucleus splits into a radon nucleus flying one way and an alpha particle flying the exact opposite way; centre of mass C stays at rest.
The motion looks much simpler in the centre of mass frame than in the laboratory frame. For this reason, physicists often prefer to work in the centre of mass frame when analysing decays, collisions, and interactions of particles.
Another example — binary stars
In astronomy, binary stars (also called double stars) are a very common occurrence — two stars gravitationally bound to each other, orbiting a common point. If no other object is nearby, the only significant forces on the two-star system are the gravitational forces the two stars exert on each other. These are internal to the system.
So the external force on the binary system is essentially zero, and the centre of mass of the binary must move like a free particle — either at rest or in a straight line with uniform velocity.
Now if we look at the two stars from the laboratory frame (a frame in which the centre of mass is drifting along a straight path), each star traces a complicated wavy path in space, as shown in Fig. 6.14(a). Two such paths for stars of equal mass are shown by a dotted line and a solid line.
Figure to come
Fig. 6.14(a) – Trajectories of two stars S₁ (dotted) and S₂ (solid) forming a binary system; the centre of mass C moves along a straight horizontal line, while each star traces a wavy loop pattern around this straight path.
If, however, we switch to the centre of mass frame — the frame in which the centre of mass is at rest — a very different picture emerges. The two stars now simply move in circles around their common centre of mass, always sitting diametrically opposite each other so that the centre of mass stays at rest, as shown in Fig. 6.14(b).
Figure to come
Fig. 6.14(b) – The same binary system seen from the centre of mass frame: the two stars S₁ and S₂ move in a circle around the centre of mass C, which is at rest; at every instant they lie on opposite sides of C, along the same diameter.
The trajectories of the two stars in our (laboratory) frame are thus really a combination of: (i) uniform straight-line motion of the centre of mass, and (ii) circular orbits of each star around the centre of mass.
A powerful technique
The two examples above illustrate a very useful technique for handling any system of particles: separate the motion into the motion of the centre of mass and the motion about the centre of mass.
- The motion of the centre of mass tells us the overall drift of the system.
- The motion about the centre of mass tells us what is happening inside — the rotations, the vibrations, the mutual orbits.
The two pieces together give the complete picture, but each piece on its own is simpler than the whole. This “split” is a running theme throughout the rest of this chapter.
6.5 Vector Product of Two Vectors
We are already familiar with vectors and how they are used in physics. In an earlier chapter (Work, Energy and Power), we defined the scalar product (or dot product) of two vectors — a way of “multiplying” two vectors to get a scalar. An important physical quantity, work, was defined as the scalar product of two vector quantities: force and displacement.
We now define a very different kind of product of two vectors — one that gives back a vector, not a scalar. This is called the vector product, or cross product. Two important quantities in the study of rotational motion — the moment of a force (torque) and angular momentum — are defined as vector products. So this section prepares the mathematical ground we will need in the rest of the chapter.
Definition of Vector Product
The vector product of two vectors \(\vec{a}\) and \(\vec{b}\) is another vector \(\vec{c}\), defined by three conditions:
The magnitude of \(\vec{c}\) is \[c = |\vec{c}| = ab \sin\theta\] where \(a\) and \(b\) are the magnitudes of \(\vec{a}\) and \(\vec{b}\), and \(\theta\) is the angle between the two vectors.
The direction of \(\vec{c}\) is perpendicular to the plane that contains \(\vec{a}\) and \(\vec{b}\).
The sense of \(\vec{c}\) (which of the two perpendicular directions to pick) is given by the right-handed screw rule. Imagine a right-handed screw with its head lying flat on the plane containing \(\vec{a}\) and \(\vec{b}\), and the screw pointing perpendicular to this plane. Turn the head from \(\vec{a}\) towards \(\vec{b}\). The direction in which the screw advances is the direction of \(\vec{c}\).
This right-handed screw rule is illustrated in Fig. 6.15(a).
Figure to come
Fig. 6.15(a) – A right-handed screw with its head lying in the plane of vectors \(\vec{a}\) and \(\vec{b}\); turning the head from \(\vec{a}\) toward \(\vec{b}\) makes the screw advance upward along \(\vec{c} = \vec{a} \times \vec{b}\); the angle \(\theta\) between \(\vec{a}\) and \(\vec{b}\) is marked.
Instead of a screw, we can use just our right hand. If we curl the fingers of the right hand around a line perpendicular to the plane of \(\vec{a}\) and \(\vec{b}\), in the direction going from \(\vec{a}\) to \(\vec{b}\), then the stretched thumb points in the direction of \(\vec{c}\), as shown in Fig. 6.15(b).
Figure to come
Fig. 6.15(b) – A right hand with fingers curling from \(\vec{a}\) toward \(\vec{b}\); the stretched thumb points along \(\vec{c} = \vec{a} \times \vec{b}\); the angle \(\theta\) between \(\vec{a}\) and \(\vec{b}\) is marked.
Here is an even simpler way to remember it — the right-hand palm rule: open your right hand flat, palm up. Point your fingers in the direction of \(\vec{a}\), then curl them so they point along \(\vec{b}\). Your stretched thumb now points in the direction of \(\vec{c}\).
Because we use the cross symbol \(\times\) to denote this product, it is also called the cross product. From now on we will use “vector product” and “cross product” interchangeably.
Important properties of the vector product
Let us list the properties that will matter most in this chapter.
(1) Vector product is NOT commutative
The scalar product is commutative: \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\). The order in which we take the two vectors does not matter.
The vector product, however, is not commutative: \[\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\]
The magnitudes of \(\vec{a} \times \vec{b}\) and \(\vec{b} \times \vec{a}\) are the same — both are \(ab\sin\theta\). Both are also perpendicular to the plane of \(\vec{a}\) and \(\vec{b}\).
But the directions are opposite. Applying the right-hand rule to \(\vec{a} \times \vec{b}\), we turn from \(\vec{a}\) to \(\vec{b}\); the thumb points one way. For \(\vec{b} \times \vec{a}\), we turn from \(\vec{b}\) to \(\vec{a}\); the thumb points the opposite way. So
\[\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}\]
This “order matters” property is a key reason why vector products behave differently from ordinary multiplication.
(2) Behaviour under reflection
Something curious happens if we view a vector in a plane mirror. Under such reflection, every vector’s components flip sign: \[x \to -x, \quad y \to -y, \quad z \to -z\]
So each vector \(\vec{a}\) becomes \(-\vec{a}\), and \(\vec{b}\) becomes \(-\vec{b}\). What happens to their cross product?
\[\vec{a} \times \vec{b} \to (-\vec{a}) \times (-\vec{b}) = \vec{a} \times \vec{b}\]
The two minus signs cancel. So \(\vec{a} \times \vec{b}\) does not change sign under reflection. Vectors that behave like this are called pseudo-vectors or axial vectors — they include torque, angular velocity, and angular momentum.
(3) Distributive over addition
Just like the scalar product, the vector product is distributive with respect to vector addition:
\[\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}\]
\[\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}\]
This lets us “expand brackets” the way we would in ordinary algebra — but we must always keep the order of the vectors the same, because of the non-commutativity.
Vector product in component form
To calculate cross products of specific vectors, we usually work with their components along the x, y, and z axes. For this, we first need some elementary cross products of the unit vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\).
Cross product of a unit vector with itself: The angle between a vector and itself is \(0°\), and \(\sin 0° = 0\). So \[\vec{a} \times \vec{a} = \vec{0}\] (here \(\vec{0}\) is the null vector — a vector with zero magnitude). In particular: \[\hat{i} \times \hat{i} = \vec{0}, \quad \hat{j} \times \hat{j} = \vec{0}, \quad \hat{k} \times \hat{k} = \vec{0}\]
Cross product of two different unit vectors: The angle between any two different unit vectors like \(\hat{i}\) and \(\hat{j}\) is \(90°\). Since \(\sin 90° = 1\) and both vectors have unit magnitude, the magnitude of \(\hat{i} \times \hat{j}\) is \(1 \times 1 \times 1 = 1\). So \(\hat{i} \times \hat{j}\) is a unit vector.
Which direction does it point? It must be perpendicular to the plane of \(\hat{i}\) (x-axis) and \(\hat{j}\) (y-axis). That means it points along the z-axis. Applying the right-hand rule (fingers from \(\hat{i}\) to \(\hat{j}\)), the thumb points along \(+\hat{k}\). So
\[\hat{i} \times \hat{j} = \hat{k}\]
By the same reasoning: \[\hat{j} \times \hat{k} = \hat{i}, \quad \hat{k} \times \hat{i} = \hat{j}\]
Notice a pattern: when the unit vectors appear in the cyclic order \(\hat{i} \to \hat{j} \to \hat{k} \to \hat{i}\), the cross product is positive (a \(+\) unit vector).
By the anti-commutativity property, the cross products in the reverse (non-cyclic) order pick up a minus sign: \[\hat{j} \times \hat{i} = -\hat{k}, \quad \hat{k} \times \hat{j} = -\hat{i}, \quad \hat{i} \times \hat{k} = -\hat{j}\]
Figure to come
Fig. 6.15(c) – A circular diagram with \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) placed on a circle in cyclic order; a clockwise arrow indicating “cyclic → positive”; a counter-clockwise arrow indicating “reverse → negative”.
General formula
Now we can compute the cross product of any two vectors given in component form. Let \[\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k}, \quad \vec{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k}\]
Expanding using the distributive property: \[\vec{a} \times \vec{b} = (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) \times (b_x \hat{i} + b_y \hat{j} + b_z \hat{k})\]
Using the elementary cross products (and remembering \(\hat{i}\times\hat{i} = \hat{j}\times\hat{j} = \hat{k}\times\hat{k} = 0\)), the surviving terms are:
\[\vec{a} \times \vec{b} = a_x b_y \hat{k} - a_x b_z \hat{j} - a_y b_x \hat{k} + a_y b_z \hat{i} + a_z b_x \hat{j} - a_z b_y \hat{i}\]
Grouping by \(\hat{i}\), \(\hat{j}\), \(\hat{k}\):
\[\vec{a} \times \vec{b} = (a_y b_z - a_z b_y)\hat{i} + (a_z b_x - a_x b_z)\hat{j} + (a_x b_y - a_y b_x)\hat{k}\]
This expression looks intimidating to memorise, but it has a beautiful shortcut — it is exactly the expansion of a 3 × 3 determinant:
\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}\]
The first row holds the unit vectors, the second row the components of the first vector \(\vec{a}\), and the third row the components of the second vector \(\vec{b}\). Expanding this determinant gives back the same formula.
Solved Example
Looking ahead
Armed with the vector product, we are now ready to give precise, quantitative definitions of the two most important vector quantities in rotational motion — torque (Section 6.7.1) and angular momentum (Section 6.7.2). But first, in the next section, we will use the cross product to connect the angular velocity of a rotating body to the linear velocity of any of its particles.
6.6 Angular Velocity and its Relation with Linear Velocity
In Section 6.1, we saw that when a rigid body rotates about a fixed axis, every particle of the body moves in a circle. In Section 6.5, we developed the vector product — the mathematical tool we now need. In this section, we bring these two ideas together and ask: how fast do the particles of a rotating body actually move, and how is this speed connected to the rotation of the body as a whole?
The link between the two is a new physical quantity — the angular velocity.
Going back to the rotating body
Let us return to the picture in Fig. 6.4. There, we saw a rigid body rotating about a fixed axis (taken as the z-axis), with each particle tracing out a circle in a plane perpendicular to the axis. We redraw this scene in Fig. 6.16, focusing on one typical particle P of the rigid body.
Figure to come
Fig. 6.16 – A rigid body rotating about the z-axis; a typical particle P moves in a circle of radius \(r\) with centre C on the z-axis; the linear velocity vector \(\vec{v}\) of the particle is drawn tangent to the circle at P.
The particle P moves along a circle whose centre C sits on the axis of rotation. The radius of this circle, \(r\), is the perpendicular distance of P from the axis. At any instant, the particle’s linear velocity \(\vec{v}\) points along the tangent to the circle at P.
Angular displacement and angular velocity
Now let P′ be the position of the particle a short time interval \(\Delta t\) later. During this interval, the radius CP has swung through an angle \(\Delta \theta\) — the angle PCP′ shown in Fig. 6.16. This angle is called the angular displacement of the particle in time \(\Delta t\).
The average angular velocity of the particle over the interval \(\Delta t\) is the ratio \(\Delta \theta / \Delta t\). As we let \(\Delta t\) shrink to zero, this ratio approaches a limiting value, which is the instantaneous angular velocity of the particle at the position P.
We denote the instantaneous angular velocity by \(\omega\) (the Greek letter omega), and write
\[\omega = \frac{d\theta}{dt}\]
Its SI unit is radians per second (rad/s).
From our earlier study of circular motion, we already know that for a particle moving in a circle of radius \(r\), the linear speed \(v\) and the angular speed \(\omega\) are related by the simple formula
\[v = \omega r\]
Here \(r\) is the radius of the circular path in metres, and \(v\) comes out in m/s.
The whole body has one angular velocity
Now here is the crucial observation. At any instant, every particle of a rotating rigid body has the same angular displacement \(\Delta \theta\) in the same time \(\Delta t\) — because the body is rigid, all the particles turn together by the same angle. So they must all have the same \(\omega\).
Take any particle at perpendicular distance \(r_i\) from the fixed axis. Its linear speed at that instant is
\[v_i = \omega r_i \quad \text{...(6.19)}\]
The index \(i\) runs from 1 to \(n\), where \(n\) is the total number of particles in the body.
For a particle sitting on the axis, \(r_i = 0\), and Eq. (6.19) gives \(v_i = 0\). Particles on the axis stay stationary — which confirms once again that the axis of rotation is truly fixed.
Importantly, we use the same \(\omega\) for every particle of the body — different particles have different linear speeds only because they sit at different distances from the axis.
For this reason, we speak of \(\omega\) as the angular velocity of the whole body, not just of a single particle.
Notice how neatly this parallels pure translation. In pure translation, all particles of a rigid body share the same linear velocity \(\vec{v}\). In pure rotation about a fixed axis, all particles share the same angular velocity \(\omega\). So angular velocity is to rotation what linear velocity is to translation.
Angular velocity as a vector
So far, \(\omega\) has looked like just a scalar — a number telling us how fast the body is rotating. In fact, angular velocity is a vector. We shall not prove this here; we will simply accept it and see how the vector is defined.
For rotation about a fixed axis, the angular velocity vector \(\vec{\omega}\) lies along the axis of rotation. Its direction along that axis is given by the right-hand rule: if the fingers of the right hand curl in the direction the body is rotating, the stretched thumb points along \(\vec{\omega}\). Equivalently, if a right-handed screw is aligned with the axis and turned along with the body, the screw advances in the direction of \(\vec{\omega}\).
Figure to come
Fig. 6.17(a) – Two discs rotating about a vertical axis; on the left disc the rotation is one way and \(\vec{\omega}\) points upward along the axis; on the right the rotation is reversed and \(\vec{\omega}\) points downward along the axis.
The magnitude of the angular velocity vector is exactly the scalar \(\omega = d\theta/dt\) we already defined.
Relating linear velocity to angular velocity — the vector form
Now let us make Eq. (6.19), \(v = \omega r\), into a full vector equation. This is where the cross product from Section 6.5 comes in.
Look at Fig. 6.17(b), which is essentially the picture of Fig. 6.16 with more detail. The vector \(\vec{\omega}\) is drawn along the fixed axis (z-axis). The position vector of the particle P, measured from an origin O on the axis, is \(\vec{r} = \vec{OP}\).
Figure to come
Fig. 6.17(b) – A rigid body rotating about the z-axis; the angular velocity vector \(\vec{\omega}\) points up along the axis; the position vector \(\vec{r} = \vec{OP}\) goes from origin O (on the axis) to particle P; the linear velocity \(\vec{v}\) of P is drawn tangent to its circular path; the perpendicular from P to the axis meets the axis at C, so \(\vec{OC}\) lies along the axis and \(\vec{CP}\) is the radius of the particle’s circle.
We can split \(\vec{r}\) as \[\vec{r} = \vec{OP} = \vec{OC} + \vec{CP}\]
Consider the cross product \(\vec{\omega} \times \vec{r}\):
\[\vec{\omega} \times \vec{r} = \vec{\omega} \times \vec{OC} + \vec{\omega} \times \vec{CP}\]
Since \(\vec{OC}\) lies along the axis, and \(\vec{\omega}\) is also along the axis, the two are parallel and their cross product is zero: \[\vec{\omega} \times \vec{OC} = 0\]
So we are left with \[\vec{\omega} \times \vec{r} = \vec{\omega} \times \vec{CP}\]
Now \(\vec{CP}\) is perpendicular to \(\vec{\omega}\) (it lies in the plane of the particle’s circle, which is perpendicular to the axis). So \(\vec{\omega}\) and \(\vec{CP}\) are perpendicular, and the cross product \(\vec{\omega} \times \vec{CP}\) has magnitude \(\omega \cdot |\vec{CP}|\) — that is, \(\omega r_\perp\), where \(r_\perp\) is the perpendicular distance of P from the axis (which is exactly the radius \(r\) of P’s circular path).
For its direction: applying the right-hand rule, \(\vec{\omega} \times \vec{CP}\) is perpendicular to both \(\vec{\omega}\) (the axis) and \(\vec{CP}\) (the radius). This means it lies tangent to the circle at P — exactly the direction of the particle’s linear velocity \(\vec{v}\).
So both the magnitude and the direction of \(\vec{\omega} \times \vec{r}\) match those of \(\vec{v}\). Therefore
\[\vec{v} = \vec{\omega} \times \vec{r} \quad \text{...(6.20)}\]
Eq. (6.20) is a compact vector form of \(v = \omega r\): it tells us the linear velocity of any particle of a rotating rigid body in terms of the body’s angular velocity and the position vector of the particle from the axis.
This relation is not limited to rotation about a fixed axis. It holds even for more general rotations of a rigid body with just one point fixed — such as a spinning top (Fig. 6.5a). In that case, \(\vec{r}\) is measured from the fixed point.
For fixed axis, direction of ω is fixed
For rotation about a fixed axis, the direction of \(\vec{\omega}\) never changes — it always points along the axis. What can change with time is only its magnitude — the body may be spinning up or slowing down.
For more general rotation (like a spinning-and-precessing top), both the magnitude and the direction of \(\vec{\omega}\) can change from one instant to the next. This is why more general rotation is much richer — and much more complicated — than fixed-axis rotation.
6.6.1 Angular Acceleration
Notice that our study of rotational motion is being built up in exact parallel with translational motion, with which we are already comfortable. Every kinematic variable of translation has a rotational cousin:
- Linear displacement \(s\) ↔︎ Angular displacement \(\theta\)
- Linear velocity \(\vec{v}\) ↔︎ Angular velocity \(\vec{\omega}\)
The next natural cousin is angular acceleration, which plays the same role in rotation as linear acceleration plays in translation.
Just as we defined linear acceleration as the rate of change of linear velocity in translational motion, we define angular acceleration as the rate of change of angular velocity. That is,
\[\vec{\alpha} = \frac{d\vec{\omega}}{dt} \quad \text{...(6.21)}\]
Its SI unit is radians per second per second (rad/s²).
If the axis of rotation is fixed, the direction of \(\vec{\omega}\) is fixed (along the axis), and so the direction of \(\vec{\alpha}\) is also fixed (along the same axis). In this case, we can drop the vector arrows and write Eq. (6.21) as a simple scalar equation:
\[\alpha = \frac{d\omega}{dt} \quad \text{...(6.22)}\]
Here \(\alpha\) is positive if the body is speeding up (in the direction of rotation) and negative if it is slowing down.
Looking ahead
We have now built up the kinematic quantities of rotational motion — \(\theta\), \(\omega\), and \(\alpha\) — in one-to-one correspondence with the familiar \(s\), \(v\), and \(a\) of translation. In the next section, we will move to the dynamics side of this analogy: what plays the role of force in rotational motion? The answer will be torque, which we will define using the vector product from Section 6.5.
6.7 Torque and Angular Momentum
We have already built up the kinematics of rotational motion — how a rigid body’s angular position, angular velocity, and angular acceleration change with time. We now turn to its dynamics — what causes these changes.
In this section we introduce two central physical quantities of rotational dynamics: torque (the moment of a force) and angular momentum. Both are defined as vector products of two vectors — for which we prepared the ground in Section 6.5. These two quantities will play, for rotational motion, the same role that force and linear momentum play for translational motion.
6.7.1 Moment of Force (Torque)
We know from Section 6.1 that the motion of a rigid body, in general, is a combination of translation and rotation. If the body is fixed at a point or along a line, only the rotational part remains.
From earlier chapters, we know that force is what changes the translational state of a body — that is, force is what produces linear acceleration. It is natural to ask: what is the rotational analogue of force? What causes a body to start, stop, or change its rotation?
The door example
To see what quantity plays this role, consider a simple everyday example: opening or closing a door. A door is a rigid body that can rotate about a fixed vertical axis passing through its hinges. What makes the door rotate?
Clearly, we need to apply a force. But not any force will do — the location and direction of the push matter enormously.
- Push the door right along the hinge line, and no matter how hard you push, the door will not rotate at all.
- Push the door near the hinges, and it takes a very large force to open.
- Push the door at its outer edge (the handle side), at right angles to the door, and even a small force sends it swinging easily.
So in rotational motion, it is not the force alone that matters. What matters is how much force, at what distance from the axis, and in what direction it is applied. All three combine into a single quantity — the moment of the force, also called the torque, or sometimes the couple. (We will use “moment of force” and “torque” interchangeably.)
We will first define torque for a single particle, then extend it to a system of particles including rigid bodies. Later we will link it to a change in the state of rotation — that is, to angular acceleration.
Defining torque
Suppose a force \(\vec{F}\) acts on a single particle at a point P, whose position with respect to some origin O is given by the position vector \(\vec{r}\), as shown in Fig. 6.18. The moment of force, or torque, of \(\vec{F}\) about the origin O is defined as the vector product
\[\vec{\tau} = \vec{r} \times \vec{F} \quad \text{...(6.23)}\]
Figure to come
Fig. 6.18 – A particle at point P with position vector \(\vec{r}\) from origin O; force \(\vec{F}\) acts at P at angle \(\theta\) to \(\vec{r}\); the torque \(\vec{\tau}\) is perpendicular to the plane of \(\vec{r}\) and \(\vec{F}\); \(r\sin\theta\) is marked as the perpendicular distance from O to the line of action of \(\vec{F}\).
Torque is a vector quantity. The symbol \(\vec{\tau}\) (Greek letter tau) is used for it. Since \(\vec{\tau}\) is a cross product, it is perpendicular to the plane containing \(\vec{r}\) and \(\vec{F}\), with its direction given by the right-hand rule (Section 6.5).
The magnitude of \(\vec{\tau}\) is
\[\tau = r F \sin\theta \quad \text{...(6.24a)}\]
where \(r\) is the magnitude of \(\vec{r}\) (the length OP), \(F\) is the magnitude of the force, and \(\theta\) is the angle between \(\vec{r}\) and \(\vec{F}\) as shown.
Dimensions and units
The dimensions of moment of force are \([M\, L^2\, T^{-2}]\) — exactly the same as those of work or energy. But be careful: torque is not the same physical quantity as work, and cannot be measured in joules.
Work is a scalar; torque is a vector. Even though both share the same dimensions, they mean different things and cannot be interchanged.
The SI unit of torque is the newton metre (N m). Torque is deliberately written this way — never as “joule” — to keep it separate from work.
Two useful forms of the magnitude
We can rearrange the magnitude formula \(\tau = rF\sin\theta\) in two equally useful ways.
First, group \(r\) with \(\sin\theta\):
\[\tau = (r \sin\theta) F = r_\perp F \quad \text{...(6.24b)}\]
Here \(r_\perp = r\sin\theta\) is called the moment arm or lever arm — the perpendicular distance from the origin O to the line of action of the force. The “line of action” is the infinite straight line along which the force acts (extending the force vector in both directions).
Second, group \(F\) with \(\sin\theta\):
\[\tau = r F \sin\theta = r F_\perp \quad \text{...(6.24c)}\]
Here \(F_\perp = F\sin\theta\) is the component of the force perpendicular to \(\vec{r}\).
Both forms tell us the same physical story in slightly different language: - (6.24b) says: torque = force \(\times\) perpendicular distance from origin to line of force. - (6.24c) says: torque = position vector length \(\times\) component of force perpendicular to \(\vec{r}\).
When is torque zero?
From \(\tau = rF\sin\theta\), we see that torque vanishes if: - \(r = 0\): the force is applied at the origin itself. - \(F = 0\): there is no force at all. - \(\theta = 0°\) or \(180°\): the force is either along \(\vec{r}\) or exactly opposite to it — that is, the line of action of the force passes through the origin.
This is exactly why pushing a door along the hinge line does nothing: the line of action passes through the axis, so the torque about the axis is zero.
Properties from the vector product
Since \(\vec{\tau} = \vec{r} \times \vec{F}\) is a vector product, it inherits the properties from Section 6.5:
- If the direction of \(\vec{F}\) is reversed, the direction of \(\vec{\tau}\) is reversed.
- If the directions of both \(\vec{r}\) and \(\vec{F}\) are reversed, the direction of \(\vec{\tau}\) stays the same (two sign flips cancel).
6.7.2 Angular Momentum of a Particle
Just as torque is the rotational analogue of force, we now define the rotational analogue of linear momentum. This new quantity is called angular momentum.
We will define it first for a single particle, look at its usefulness for single-particle motion, and then extend it to a system of particles (including rigid bodies).
Angular momentum is also a vector product — much like torque. In fact, it could be called the moment of linear momentum, and that name hints exactly at how it is defined.
Defining angular momentum
Consider a particle of mass \(m\) moving with linear momentum \(\vec{p} = m\vec{v}\), sitting at a position given by the position vector \(\vec{r}\) with respect to some origin O. The angular momentum of the particle with respect to the origin O is defined by
\[\vec{l} = \vec{r} \times \vec{p} \quad \text{...(6.25a)}\]
Its SI unit is joule-second (J s), or equivalently kg m²/s.
The magnitude of the angular momentum vector is
\[l = r p \sin\theta \quad \text{...(6.26a)}\]
where \(p = |\vec{p}|\) and \(\theta\) is the angle between \(\vec{r}\) and \(\vec{p}\).
Following exactly the same reasoning as for torque, we can rewrite the magnitude as
\[l = r p_\perp \quad \text{or} \quad l = r_\perp p \quad \text{...(6.26b)}\]
Here \(r_\perp = r\sin\theta\) is the perpendicular distance of the line of motion of the particle (that is, the line along which \(\vec{p}\) points) from the origin. And \(p_\perp = p\sin\theta\) is the component of momentum perpendicular to \(\vec{r}\).
When is angular momentum zero?
From Eq. (6.26a), we see that \(l = 0\) if: - \(p = 0\): the particle has no linear momentum (i.e., it is at rest). - \(r = 0\): the particle is at the origin itself. - \(\theta = 0°\) or \(180°\): the line of motion of the particle passes through the origin.
The third case is the most interesting: a particle moving in a straight line that passes through O has zero angular momentum about O — no matter how fast it moves, because it never “goes around” O.
The link between torque and angular momentum
Torque and angular momentum are not independent quantities — they are tied together by a very important relation. It is the rotational analogue of Newton’s second law, \(\vec{F} = d\vec{p}/dt\).
To derive this, we differentiate the definition \(\vec{l} = \vec{r} \times \vec{p}\) with respect to time:
\[\frac{d\vec{l}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p})\]
Using the product rule for differentiation of a vector product:
\[\frac{d}{dt}(\vec{r} \times \vec{p}) = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt}\]
Now, \(d\vec{r}/dt = \vec{v}\) (the velocity of the particle), and \(\vec{p} = m\vec{v}\). So the first term becomes
\[\frac{d\vec{r}}{dt} \times \vec{p} = \vec{v} \times m\vec{v} = m(\vec{v} \times \vec{v}) = 0\]
because the cross product of any vector with itself is zero (Section 6.5). And by Newton’s second law, \(d\vec{p}/dt = \vec{F}\), so
\[\vec{r} \times \frac{d\vec{p}}{dt} = \vec{r} \times \vec{F} = \vec{\tau}\]
Putting the two pieces together:
\[\frac{d\vec{l}}{dt} = \vec{\tau} \quad \text{...(6.27)}\]
This is exactly the rotational analogue of \(\vec{F} = d\vec{p}/dt\).
Just as force causes linear momentum to change, torque causes angular momentum to change. Both the angular momentum and the torque must be computed about the same origin — a subtle but important point.
Extending to a system of particles
To get the total angular momentum of a system of \(n\) particles about a given origin, we add up the angular momenta of the individual particles vectorially:
\[\vec{L} = \vec{l}_1 + \vec{l}_2 + \dots + \vec{l}_n = \sum_{i=1}^{n} \vec{l}_i\]
The angular momentum of the \(i^{\text{th}}\) particle is
\[\vec{l}_i = \vec{r}_i \times \vec{p}_i\]
where \(\vec{r}_i\) is the position vector of the \(i^{\text{th}}\) particle with respect to the origin, and \(\vec{p}_i = m_i \vec{v}_i\) is its linear momentum. So
\[\vec{L} = \sum_i \vec{l}_i = \sum_i \vec{r}_i \times \vec{p}_i \quad \text{...(6.25b)}\]
This generalises the single-particle definition (6.25a) to any system of particles.
Using the same product-rule differentiation as before (or equivalently, using Eq. (6.27) for each particle), the rate of change of the total angular momentum becomes
\[\frac{d\vec{L}}{dt} = \frac{d}{dt}\left(\sum_i \vec{l}_i\right) = \sum_i \frac{d\vec{l}_i}{dt} = \sum_i \vec{\tau}_i \quad \text{...(6.28a)}\]
Here \(\vec{\tau}_i = \vec{r}_i \times \vec{F}_i\) is the torque on the \(i^{\text{th}}\) particle, and \(\vec{F}_i\) is the total force on it.
External and internal torques
Just as we split forces into external and internal earlier, we now split torques into external and internal. The force \(\vec{F}_i\) on the \(i^{\text{th}}\) particle is really a sum:
\[\vec{F}_i = \vec{F}_i^{\text{ext}} + \vec{F}_i^{\text{int}}\]
where \(\vec{F}_i^{\text{ext}}\) is the force on the particle from bodies outside the system, and \(\vec{F}_i^{\text{int}}\) is the force on it from other particles inside the system.
Correspondingly:
\[\vec{\tau} = \sum_i \vec{\tau}_i = \sum_i \vec{r}_i \times \vec{F}_i = \vec{\tau}_{\text{ext}} + \vec{\tau}_{\text{int}}\]
where \[\vec{\tau}_{\text{ext}} = \sum_i \vec{r}_i \times \vec{F}_i^{\text{ext}}, \quad \vec{\tau}_{\text{int}} = \sum_i \vec{r}_i \times \vec{F}_i^{\text{int}}\]
Why internal torques cancel
We now make two assumptions about internal forces:
- Newton’s third law: for every internal force one particle exerts on another, the other particle exerts an equal and opposite force back.
- Central-force assumption: these internal action–reaction pairs act along the line joining the two particles.
Under both assumptions, the torque produced by one particle on another (about any origin) is exactly cancelled by the torque of its partner. So the total internal torque on the system, \(\vec{\tau}_{\text{int}}\), vanishes:
\[\vec{\tau}_{\text{int}} = 0\]
This is the rotational counterpart of what we saw for forces in Section 6.3, where internal forces cancelled in pairs by Newton’s third law. But note: for torques, we needed the extra assumption that the internal forces act along the line joining the particles.
With \(\vec{\tau}_{\text{int}} = 0\), we have \(\vec{\tau} = \vec{\tau}_{\text{ext}}\), so from Eq. (6.28a),
\[\frac{d\vec{L}}{dt} = \vec{\tau}_{\text{ext}} \quad \text{...(6.28b)}\]
Eq. (6.28b) is the general rotational counterpart of Eq. (6.17):
\[\frac{d\vec{P}}{dt} = \vec{F}_{\text{ext}}\]
Just as the total external force controls the total linear momentum of the system, the total external torque controls the total angular momentum. And just like Eq. (6.17), Eq. (6.28b) holds for any system — rigid body, gas, a swarm of particles with all kinds of internal motion. It does not care about the internal details.
Conservation of angular momentum
Now suppose the total external torque on a system is zero: \(\vec{\tau}_{\text{ext}} = 0\). Then from Eq. (6.28b),
\[\frac{d\vec{L}}{dt} = 0\]
which means
\[\vec{L} = \text{constant} \quad \text{...(6.29a)}\]
This is the law of conservation of angular momentum for a system of particles:
Eq. (6.29a) is a vector equation and is equivalent to three scalar equations, one for each direction:
\[L_x = K_1, \quad L_y = K_2, \quad L_z = K_3 \quad \text{...(6.29b)}\]
Here \(L_x\), \(L_y\), and \(L_z\) are the components of \(\vec{L}\) along the x, y, z axes; \(K_1\), \(K_2\), \(K_3\) are constants. Angular momentum is conserved separately along any direction along which the external torque vanishes.
This is the rotational analogue of Eq. (6.18a), the conservation of linear momentum. And like linear momentum conservation, it has spectacular applications — some of which we will see later in this chapter, including the ice skater’s spin and the diver’s tuck.
Now spin the wheel fast around its axle. While it is spinning quickly, let go of one of the strings. What do you expect: the wheel to swing down like a pendulum?
Instead, the wheel keeps spinning in a vertical plane, and slowly its axle sweeps around a horizontal circle about the string you are still holding — the axis of rotation is precessing. The rotating wheel has a large angular momentum; gravity applies a torque; and the torque, instead of tipping the wheel over, changes the direction of the angular momentum — making the axle precess. Try to work out the direction of the torque and confirm the direction of precession by the right-hand rule.
Solved Examples
Looking ahead
Torque and angular momentum are the two central quantities of rotational dynamics. In the next section (6.8) we will use them to write down the conditions for a rigid body to be in equilibrium — a very important application in engineering and everyday physics. Later, in sections 6.11 and 6.12, we will apply Eqs. (6.28b) and (6.29a) to rotation about a fixed axis and see the connections with moment of inertia and angular velocity — including a full quantitative treatment of the ice-skater phenomenon that opened this chapter.
6.8 Equilibrium of a Rigid Body
Having built up the language of torque and angular momentum in Section 6.7, we now turn to a very practical application. We will focus specifically on rigid bodies and ask: when is a rigid body in equilibrium?
For the rest of this section, we will drop the adjective “external” for brevity. Unless otherwise stated, “forces” and “torques” here mean the external forces and torques on the body. (This is safe because internal forces contribute nothing, as we saw in Sections 6.3 and 6.7.)
Two things external forces can do
Recall from earlier sections what external influences can do to a rigid body:
The external forces change the translational state of motion. They alter the total linear momentum of the body according to Eq. (6.17): \[\frac{d\vec{P}}{dt} = \vec{F}_{\text{ext}}\]
But this is not all. The external forces may also produce a net torque, and by Eq. (6.28b), a net torque changes the total angular momentum, and hence the rotational state of motion: \[\frac{d\vec{L}}{dt} = \vec{\tau}_{\text{ext}}\]
So for a rigid body to be truly at rest — or to move without any change in its motion — both effects must vanish. Vanishing net force keeps the translational motion steady, and vanishing net torque keeps the rotational motion steady.
Definition of mechanical equilibrium
From the two boxed equations above, this immediately gives two conditions:
(1) Translational equilibrium. The vector sum of all the external forces on the body is zero:
\[\vec{F}_1 + \vec{F}_2 + \dots + \vec{F}_n = \sum_{i=1}^{n} \vec{F}_i = \vec{0} \quad \text{...(6.30a)}\]
If this is satisfied, the total linear momentum does not change with time. This is the condition for translational equilibrium.
(2) Rotational equilibrium. The vector sum of all the external torques on the body is zero:
\[\vec{\tau}_1 + \vec{\tau}_2 + \dots + \vec{\tau}_n = \sum_{i=1}^{n} \vec{\tau}_i = \vec{0} \quad \text{...(6.30b)}\]
If this is satisfied, the total angular momentum does not change with time. This is the condition for rotational equilibrium.
Both conditions must hold simultaneously for full mechanical equilibrium.
Does the choice of origin for torques matter?
A natural question comes up: when we compute torques, we must choose an origin. What if we picked a different origin — would the rotational equilibrium condition still hold?
Fortunately, if the translational equilibrium condition (6.30a) is already satisfied, then the rotational equilibrium condition (6.30b) is independent of where the origin is chosen. Shifting the origin does not affect whether the total torque is zero.
We illustrate this in Solved Example 6.7 for the special case of a couple. Extending the same argument to any number of forces is left as an exercise.
Six conditions in general
Both Eqs. (6.30a) and (6.30b) are vector equations. Each is really equivalent to three scalar equations — one for each direction. Eq. (6.30a) becomes:
\[\sum_{i=1}^{n} F_{ix} = 0, \quad \sum_{i=1}^{n} F_{iy} = 0, \quad \sum_{i=1}^{n} F_{iz} = 0 \quad \text{...(6.31a)}\]
where \(F_{ix}\), \(F_{iy}\), \(F_{iz}\) are the x, y, z components of the force \(\vec{F}_i\). Similarly, Eq. (6.30b) becomes:
\[\sum_{i=1}^{n} \tau_{ix} = 0, \quad \sum_{i=1}^{n} \tau_{iy} = 0, \quad \sum_{i=1}^{n} \tau_{iz} = 0 \quad \text{...(6.31b)}\]
Together, Eqs. (6.31a) and (6.31b) give six independent scalar conditions for the mechanical equilibrium of a rigid body.
The important special case — coplanar forces
In many practical problems, all the forces on the rigid body lie in a single plane. Then the six conditions cut down neatly to just three:
- Two translational conditions: the sum of force components along any two perpendicular axes in that plane is zero.
- One rotational condition: the sum of the components of the torques along any axis perpendicular to the plane of the forces is zero.
We will use this simplification throughout the rest of the section.
Contrast with a single particle
Compare this with the equilibrium of a single particle, studied in earlier chapters. A particle has no rotation to speak of — rotational motion simply does not apply to a point. So only translational equilibrium is needed. For a particle at equilibrium, we only require
\[\sum \vec{F}_i = 0\]
Note that since all forces act on the same point (the particle itself), they are automatically concurrent — they all meet at that point. Equilibrium under concurrent forces was already discussed in earlier chapters.
For rigid bodies, forces can act at different points, so the geometry of where each force is applied becomes important — and rotational equilibrium becomes a separate, essential condition.
Partial equilibrium
A rigid body need not be in full mechanical equilibrium; it might satisfy just one of the two conditions. This gives us two types of partial equilibrium:
- Translational equilibrium only (rotational not satisfied): the net force is zero, so the body’s centre of mass does not accelerate, but there is a non-zero net torque, so the body’s angular momentum changes — it starts rotating (or changes its rotation).
- Rotational equilibrium only (translational not satisfied): the net torque is zero, so the body does not start rotating, but the net force is non-zero — so the body accelerates as a whole.
Two illustrative cases
Both partial cases occur naturally, and are worth seeing side-by-side. Consider a light rod (of negligible mass) AB, with C being its midpoint, so CA = CB = \(a\).
Case (i): Apply two parallel forces of equal magnitude \(F\) at A and B, both perpendicular to the rod but pointing in the same direction, as shown in Fig. 6.20(a).
Figure to come
Fig. 6.20(a) – A horizontal light rod AB with midpoint C; equal forces \(F\) act at A and B, both pointing downward; the perpendicular distances from C are marked as \(a\) on each side.
Take moments about C. The moment of the force at A (about C) has magnitude \(aF\) and rotates one way; the moment of the force at B (about C) also has magnitude \(aF\), but rotates in the opposite sense — because the two forces sit on opposite sides of C and point the same way. The two moments cancel exactly. So the net torque is zero — rotational equilibrium is satisfied.
But the two forces are parallel and point the same way — they do not cancel. Their sum is \(2F\), not zero. So there is a net force, and translational equilibrium is NOT satisfied: \(\sum \vec{F} \ne 0\). The rod accelerates as a whole (say, downward), but does not rotate.
Case (ii): Now reverse the direction of the force at B, so that the two forces are equal in magnitude but point in opposite directions, still both perpendicular to the rod, as shown in Fig. 6.20(b).
Figure to come
Fig. 6.20(b) – The same horizontal light rod AB with midpoint C; equal forces of magnitude \(F\) act perpendicular to the rod — one downward at A, one upward at B — producing a couple.
Now the two forces cancel: \(F + (-F) = 0\). So the net force is zero, and translational equilibrium IS satisfied.
But what about the torques? Take moments about C again. The moment of the force at A is \(aF\), tending to rotate anticlockwise. The moment of the force at B is also \(aF\), but note carefully — this time, because the force at B is reversed, it also tends to rotate anticlockwise. The two moments add rather than cancel, giving a net torque \(2aF\). So rotational equilibrium is NOT satisfied.
The rod experiences no net translation but rotates about C. It undergoes pure rotation — rotation without translation, even though nothing is holding it down.
The concept of a couple
The configuration in Case (ii) is very common and has a special name.
Couples show up all around us. Two examples are shown in Fig. 6.21.
- When we open the lid of a jar or bottle by twisting it, our fingers push one side of the lid one way and the other side the opposite way. Our fingers are applying a couple to the lid, as shown in Fig. 6.21(a).
Figure to come
Fig. 6.21(a) – A hand gripping a bottle lid; two curved arrows show the two fingers pushing in opposite directions along the rim of the lid, forming a couple.
- A compass needle sitting in the Earth’s magnetic field is another example, shown in Fig. 6.21(b). The Earth’s field exerts a force on the needle’s north pole pointing north, and an equal but opposite force on its south pole pointing south. These two forces do not lie along the same line, so they form a couple that rotates the needle until it aligns with the field.
Figure to come
Fig. 6.21(b) – A magnetic compass needle with N pole at the top-left and S pole at the bottom-right; two equal-and-opposite arrows (labelled forces) act at the two poles, forming a couple; the needle is turning under this couple.
Once the needle points north–south, the two forces lie along the same line and the couple disappears — this is why an undisturbed compass needle rests along the magnetic north–south direction.
Solved Example
6.8.1 Principle of Moments
A very important special case of rotational equilibrium — used in tools, toys, and machines all around us — is the lever.
An ideal lever is essentially a light rigid rod (a rod of negligible mass) that is pivoted at some point along its length. This pivot point is called the fulcrum. A see-saw on the children’s playground is a familiar example of a lever, and so is the beam of a physical balance.
Setting up the lever
Consider a lever with fulcrum at some point along its length. Two forces \(F_1\) and \(F_2\) act on the lever, parallel to each other and (usually) perpendicular to the lever, at distances \(d_1\) and \(d_2\) from the fulcrum respectively — as shown in Fig. 6.23.
Figure to come
Fig. 6.23 – A horizontal lever pivoted at point O (the fulcrum); a downward force \(F_1\) acts at point A on the left at distance \(d_1\) from O; a downward force \(F_2\) acts at point B on the right at distance \(d_2\) from O; an upward reaction \(R\) acts at the fulcrum O.
The fulcrum itself exerts a normal reaction \(R\) on the lever, pointing upward (opposite to \(F_1\) and \(F_2\)).
Applying the equilibrium conditions
For the lever to be in mechanical equilibrium, we apply both conditions.
Translational equilibrium (net vertical force = 0): \[R - F_1 - F_2 = 0 \quad \text{...(i)}\]
Rotational equilibrium. For rotational equilibrium, we can take moments about any point, but choosing the fulcrum is very convenient because the reaction \(R\) passes through it and produces zero torque. The moments of \(F_1\) and \(F_2\) about the fulcrum must cancel:
\[d_1 F_1 - d_2 F_2 = 0 \quad \text{...(ii)}\]
(Following the usual convention: anticlockwise moments taken as positive, clockwise as negative — or vice versa; only the balance between them matters.)
Load, effort, and mechanical advantage
In a typical use of a lever, one of the forces is a load we want to lift, and the other is an effort we apply. It is customary to call:
- \(F_1\): the load — usually a weight to be lifted.
- \(d_1\): the load arm — distance from the fulcrum to the load.
- \(F_2\): the effort — the force we apply.
- \(d_2\): the effort arm — distance from the fulcrum to our effort.
The rotational equilibrium condition (ii) can then be written as
\[d_1 F_1 = d_2 F_2 \quad \text{...(6.32a)}\]
or equivalently:
\[\text{load arm} \times \text{load} = \text{effort arm} \times \text{effort}\]
This is called the Principle of Moments.
Mechanical advantage
The ratio \(F_1/F_2\) tells us how much load we can lift for a given effort. It is called the Mechanical Advantage (M.A.) of the lever:
\[\text{M.A.} = \frac{F_1}{F_2} = \frac{d_2}{d_1} \quad \text{...(6.32b)}\]
If the effort arm \(d_2\) is larger than the load arm \(d_1\), then M.A. is greater than 1 — meaning a small effort can lift a much larger load. This is the whole point of using a lever.
There are many other levers around you. The beam of a balance is a lever — try to identify its fulcrum, its load, its load arm, its effort, and its effort arm. Scissors, tongs, nutcrackers, and pliers are more complex levers, often working in pairs.
6.8.2 Centre of Gravity
We now come to a very important idea that connects everything we have done so far — the centre of gravity.
Balancing a cardboard
Many of you may have tried balancing a book on the tip of your finger. Let us try a similar experiment. Take an irregular-shaped piece of cardboard of mass \(M\) and a pencil (or any narrow-tipped object). By trial and error, you can locate a point G on the cardboard where the whole cardboard can be balanced horizontally on the pencil’s tip.
Figure to come
Fig. 6.24 – An irregular-shaped cardboard balanced horizontally on the tip of a pencil; the point of support inside the cardboard is labelled G; the reaction \(R\) acts upward on the cardboard at G, and the total weight \(M\vec{g}\) acts downward at G; small arrows \(m_1\vec{g}, m_2\vec{g}, \ldots\) suggest the gravitational forces on individual particles of the cardboard.
This special point G is called the centre of gravity (CG) of the cardboard. When the pencil tip is at G, the cardboard remains perfectly horizontal — a state of equilibrium.
Why does the balance work?
Let us understand this using our equilibrium conditions. At G, the pencil pushes up on the cardboard with a normal reaction \(\vec{R}\). This must exactly cancel the total weight \(M\vec{g}\) of the cardboard for translational equilibrium — so \(\vec{R} = -M\vec{g}\), and the net force on the cardboard is zero.
But there is another condition: rotational equilibrium. If not for this, the tiniest disturbance would tilt the cardboard down on one side. On the cardboard, gravity is not acting at a single point — every particle of it experiences its own tiny weight \(m_1\vec{g}, m_2\vec{g}, \dots\) pulling downward. Each of these tiny weights produces a torque about G. For the cardboard to be in rotational equilibrium at G, the sum of all these torques must be zero.
The mathematical condition
Let \(\vec{r}_i\) be the position vector of the \(i^{\text{th}}\) particle of the cardboard with respect to G. The torque about G due to the weight of the \(i^{\text{th}}\) particle is \(\vec{\tau}_i = \vec{r}_i \times m_i \vec{g}\). The total gravitational torque about G is
\[\vec{\tau}_g = \sum_i \vec{\tau}_i = \sum_i \vec{r}_i \times m_i \vec{g} = 0 \quad \text{...(6.33)}\]
This gives us a clean definition:
The link with centre of mass
Now look carefully at Eq. (6.33). If we assume that \(\vec{g}\) is the same everywhere on the body (which is true for small bodies — we will discuss this below), we can pull \(\vec{g}\) out of the sum:
\[\vec{\tau}_g = \left(\sum_i m_i \vec{r}_i\right) \times \vec{g} = 0\]
Since \(\vec{g} \ne 0\), the only way for this cross product to vanish (for arbitrary \(\vec{g}\)) is
\[\sum_i m_i \vec{r}_i = 0\]
Remember, \(\vec{r}_i\) here is measured with respect to the centre of gravity G. But from Section 6.2 (see the note just below Eq. (6.4d)), a point about which the mass-weighted position vectors sum to zero is precisely the centre of mass.
So we have a very elegant result:
This is why, in Section 6.2, when we found the centre of mass of a triangular lamina or an L-shaped lamina by symmetry, we automatically found the centre of gravity too. Symmetry located both.
CG vs CM — are they really different?
At first glance, the centre of gravity and the centre of mass sound like the same thing. So let us be careful.
Centre of mass is a purely geometric idea. It depends only on how the mass is distributed in the body. It has nothing to do with gravity.
Centre of gravity depends on the gravitational field acting on the body. It is the point where gravity “effectively” acts, in the sense that the total gravitational torque about it is zero.
For small bodies — a book, a wheel, a chair, a car — \(\vec{g}\) is essentially the same at every particle of the body, because the body is small compared to the scale over which \(\vec{g}\) changes. In this case, the centre of gravity and the centre of mass are at the same point.
For very large bodies — think of a huge mountain, or a very tall building — \(\vec{g}\) actually varies from part to part (because the distance from the Earth’s centre changes). Then the CG and CM will not exactly coincide.
Finding the CG by suspension — an experimental method
In Section 6.2, we found the CM of regular homogeneous objects (rings, discs, spheres, rods) using symmetry. That method also gives us the CG of the same bodies, as long as they are small enough for \(\vec{g}\) to be uniform.
For an irregular-shaped body — like a randomly shaped cardboard — we use a simple experimental method, illustrated in Fig. 6.25.
Figure to come
Fig. 6.25 – An irregular-shaped body hanging from a string attached at point A on its edge; a vertical dashed line drops from A through the body to a point A₁ on the opposite edge; the line AA₁ passes through the centre of gravity G; two more suspension lines from points B and C are shown intersecting at G.
Hang the body from a point A on its edge by a string. When it hangs freely and is at rest, only two forces act on it: the tension in the string (upward, along AA₁), and the total weight (downward, effectively at the CG). For rotational equilibrium about A, the line of action of the weight must pass through A — that is, the CG must lie vertically below A. So we can mark the vertical line AA₁ on the body; the CG must lie somewhere on this line.
Now suspend the body from a different point B, and mark the corresponding vertical line. The CG lies on this new line as well. So the CG must be at the intersection of the two lines.
You can repeat with a third suspension point C for confirmation. All three vertical lines meet at a single point — the centre of gravity G.
Solved Examples
Looking ahead
We have now understood how a rigid body stays in equilibrium under external forces. But when the equilibrium is broken — when there is a net torque — the body rotates. To describe how it rotates, we need one more crucial rotational quantity: the moment of inertia, which plays the same role in rotational dynamics that mass plays in translational dynamics. This is the topic of the next section.
6.9 Moment of Inertia
We have been building up the physics of rotation piece by piece, always in parallel with the physics of translation we already know:
- Linear displacement \(s\) ↔︎ Angular displacement \(\theta\)
- Linear velocity \(\vec{v}\) ↔︎ Angular velocity \(\vec{\omega}\)
- Linear acceleration \(\vec{a}\) ↔︎ Angular acceleration \(\vec{\alpha}\)
- Force \(\vec{F}\) ↔︎ Torque \(\vec{\tau}\)
- Linear momentum \(\vec{p}\) ↔︎ Angular momentum \(\vec{l}\)
One important question is still unanswered. In translational motion, the inertia of a body — its resistance to change in translational motion — is measured by its mass \(m\). Mass appears in Newton’s second law as \(\vec{F} = m\vec{a}\), and it appears in the kinetic energy as \(K = \frac{1}{2} m v^2\).
What is the analogue of mass in rotational motion? What quantity measures how much a body resists changes in its rotational motion? This is the question of the present section.
To keep things clean, we will consider only rotation about a fixed axis. The answer we find here will apply beautifully to all of rotational dynamics.
Kinetic energy of a rotating body
The best way to bring out this analogue is through kinetic energy. Every particle of a rotating body has some speed, so every particle contributes to the total kinetic energy. Let us add all these contributions up.
For a rigid body rotating about a fixed axis, each particle moves in a circle. From Eq. (6.19), the linear speed of a particle at perpendicular distance \(r_i\) from the axis is
\[v_i = r_i \omega\]
The kinetic energy of this single particle is
\[k_i = \frac{1}{2} m_i v_i^2 = \frac{1}{2} m_i r_i^2 \omega^2\]
where \(m_i\) is the mass of the particle in kg, and \(\omega\) is the angular speed of the body in rad/s.
The total kinetic energy \(K\) of the body is the sum of the kinetic energies of all its \(n\) particles:
\[K = \sum_{i=1}^{n} k_i = \frac{1}{2} \sum_{i=1}^{n} (m_i r_i^2 \omega^2)\]
Here \(n\) is the total number of particles in the body. Now — and this is the key step — since the body is rigid and rotates about a fixed axis, every particle has the same angular speed \(\omega\) (Section 6.6). So \(\omega^2\) can be pulled out of the sum:
\[K = \frac{1}{2} \omega^2 \left(\sum_{i=1}^{n} m_i r_i^2\right)\]
Defining moment of inertia
The quantity inside the brackets depends only on the body — how much mass sits at what perpendicular distance from the axis. It does not depend on \(\omega\). Let us give it a name.
We define a new parameter, called the moment of inertia of the body about the axis of rotation, as
\[I = \sum_{i=1}^{n} m_i r_i^2 \quad \text{...(6.34)}\]
Its SI unit is kg m². (Notice: the distances \(r_i\) enter squared, so the units are mass times distance squared.)
With this definition, the rotational kinetic energy takes the compact form
\[K = \frac{1}{2} I \omega^2 \quad \text{...(6.35)}\]
Notice that \(I\) is entirely independent of the angular speed \(\omega\). It is a fixed property of the body and the axis — telling us how the mass is arranged around the axis.
Why is this the rotational analogue of mass?
Compare Eq. (6.35) with the familiar expression for translational kinetic energy:
\[K = \frac{1}{2} m v^2\]
The two look very similar. Where mass \(m\) appears in the translational formula, moment of inertia \(I\) appears in the rotational one. Where linear speed \(v\) appears, angular speed \(\omega\) appears.
We already know that \(\omega\) is the rotational analogue of \(v\). So \(I\) must be the rotational analogue of \(m\).
But there is a very important difference between \(m\) and \(I\), which we must appreciate right away.
- The mass of a body is a fixed quantity, independent of how the body is being used.
- The moment of inertia is NOT fixed. It depends on the axis about which the body is rotating.
The same body, rotated about a different axis, has a different \(I\). So when we speak of “the moment of inertia of a body,” we must always specify the axis.
Two simple examples
Let us compute the moment of inertia in two very simple cases, to see how the definition works in practice.
(a) A thin ring rotating about its centre.
Consider a thin ring of radius \(R\) and mass \(M\), rotating in its own plane about an axis through its centre, perpendicular to the ring’s plane. Every mass element of the ring lies at the same perpendicular distance \(R\) from the axis. So every particle moves with the same linear speed \(v = R\omega\).
The kinetic energy is \[K = \frac{1}{2} M v^2 = \frac{1}{2} M R^2 \omega^2\]
Comparing with Eq. (6.35), \(K = \frac{1}{2} I \omega^2\), we read off
\[I_{\text{ring}} = M R^2\]
(b) Two point masses on a rigid rod.
Next, take a rigid rod of negligible mass and length \(l\), with two small masses attached at its two ends. The system rotates about an axis passing through the midpoint of the rod, perpendicular to the rod. Each mass is \(M/2\) (so that the total mass is \(M\)), and each sits at a distance \(l/2\) from the axis, as shown in Fig. 6.28.
Figure to come
Fig. 6.28 – A light horizontal rod of length \(l\) with two point masses of \(M/2\) each attached at its two ends; midpoint C is the location of the axis of rotation, which points into the page; each mass sits at distance \(l/2\) from C.
From Eq. (6.34):
\[I = \left(\frac{M}{2}\right)\left(\frac{l}{2}\right)^2 + \left(\frac{M}{2}\right)\left(\frac{l}{2}\right)^2 = \frac{M l^2}{4}\]
So for this dumbbell-like arrangement, \(I = M l^2 / 4\).
Moments of inertia of common bodies
For most rigid bodies of practical interest, the sum in Eq. (6.34) becomes an integral (as with the centre of mass in Section 6.2). The details of these integrations are beyond the scope of this chapter, but the results are useful and worth memorising. Some of them are collected in Table 6.1.
Table 6.1: Moments of inertia of some regular shaped bodies about specific axes
| Sr. No. | Body | Axis | I |
|---|---|---|---|
| 1 | Thin circular ring, radius \(R\) | Perpendicular to plane, at centre | \(M R^2\) |
| 2 | Thin circular ring, radius \(R\) | Diameter | \(M R^2 / 2\) |
| 3 | Thin rod, length \(L\) | Perpendicular to rod, at midpoint | \(M L^2 / 12\) |
| 4 | Circular disc, radius \(R\) | Perpendicular to disc at centre | \(M R^2 / 2\) |
| 5 | Circular disc, radius \(R\) | Diameter | \(M R^2 / 4\) |
| 6 | Hollow cylinder, radius \(R\) | Axis of cylinder | \(M R^2\) |
| 7 | Solid cylinder, radius \(R\) | Axis of cylinder | \(M R^2 / 2\) |
| 8 | Solid sphere, radius \(R\) | Diameter | \(2 M R^2 / 5\) |
Notice that the same body appears more than once in this table with different \(I\)’s — a thin ring has \(I = MR^2\) about a perpendicular axis through its centre, but \(I = MR^2/2\) about a diameter. This is a striking reminder that \(I\) depends on the axis.
Notice also a pattern in the relative magnitudes. For example, a solid cylinder and a hollow cylinder of the same mass and radius have different \(I\)’s about the cylinder axis — the hollow cylinder has \(I = MR^2\), the solid cylinder \(I = MR^2/2\). The hollow cylinder has more moment of inertia, because in it, all the mass sits far from the axis. In the solid cylinder, much of the mass sits close to the axis, contributing less to \(I\).
What moment of inertia tells us physically
Just as mass measures a body’s resistance to a change in linear motion, moment of inertia measures a body’s resistance to a change in rotational motion. But how mass is distributed matters, not just how much of it there is.
- If most of a body’s mass sits far from the axis, its \(I\) is large — it is harder to spin up or stop.
- If most of the mass sits close to the axis, its \(I\) is small — it spins easily.
The moment of inertia is a numerical measure of how the mass of a rotating body is distributed with respect to its axis. Different parts of the body sitting at different distances from the axis contribute unequally: each mass element contributes \(m_i r_i^2\), so a particle twice as far from the axis contributes four times as much.
The dimensions of moment of inertia are \([M L^2]\), matching its SI unit of kg m².
Radius of gyration
Look at every entry in Table 6.1 carefully. Every moment of inertia can be written in the form
\[I = M k^2\]
where \(k\) has the dimension of length. For a rod about a perpendicular axis at its midpoint, \(I = ML^2/12\), so \[k^2 = \frac{L^2}{12}, \quad k = \frac{L}{\sqrt{12}}\]
For a circular disc about its diameter, \(I = MR^2/4\), so \[k = \frac{R}{2}\]
This special length \(k\) is called the radius of gyration of the body about that axis.
The radius of gyration is a purely geometric property of the body and the axis — the total mass \(M\) cancels out in \(k^2 = I/M\). So \(k\) tells us, on average, how far the body’s mass sits from the axis.
The importance of moment of inertia in machines
The moment of inertia is not an abstract concept — engineers use it every day. Machines that produce rotational motion — steam engines, automobile engines, and industrial motors — often include a large, heavy disc mounted on the rotating shaft. This disc is called a flywheel.
The flywheel is deliberately given a very large moment of inertia by making it heavy and wide. Because of its large \(I\), the flywheel strongly resists any sudden change in the shaft’s angular speed. It stores rotational energy when the shaft speeds up, and returns that energy when the shaft slows down — smoothing out the shaft’s rotation.
Without a flywheel, a car engine (where fuel explosions in each cylinder deliver sudden bursts of torque) would deliver a very jerky rotation. The flywheel absorbs these bursts and gives back a steady, smooth spin — leading to comfortable driving. Similarly, industrial machinery uses flywheels to prevent jerky motion during heavy operation.
The flywheel as a store of rotational energy
Since the kinetic energy of a rotating body is \(K = \frac{1}{2} I \omega^2\), a flywheel with large \(I\) can store a large amount of energy at modest \(\omega\). This is why the flywheel doubles as a “reservoir” of rotational energy — energy goes in when the shaft accelerates, and comes out when the shaft slows.
Looking ahead
We now have every piece we need to write down the rotational analogue of Newton’s second law and to describe the dynamics of a rigid body rotating about a fixed axis. In the next section we go one step at a time: first the kinematics of such rotation (Section 6.10), then the dynamics (Section 6.11), and finally angular momentum for the fixed-axis case (Section 6.12), where we will meet the ice skater’s spin that opened this chapter.
6.10 Kinematics of Rotational Motion about a Fixed Axis
We have already noticed, again and again, that rotational motion runs beautifully parallel to translational motion:
- Linear displacement \(s\) ↔︎ Angular displacement \(\theta\)
- Linear velocity \(\vec{v}\) ↔︎ Angular velocity \(\vec{\omega}\)
- Linear acceleration \(\vec{a}\) ↔︎ Angular acceleration \(\vec{\alpha}\)
- Mass \(m\) ↔︎ Moment of inertia \(I\)
- Force \(\vec{F}\) ↔︎ Torque \(\vec{\tau}\)
- Linear momentum \(\vec{p}\) ↔︎ Angular momentum \(\vec{l}\)
Since this analogy has been so useful, it is worth pushing it further. In this section, we take one of the most familiar tools of translation — the equations of motion for uniform acceleration — and derive their rotational cousins.
A helpful simplification — rotation about a fixed axis
Before we begin, we restrict ourselves to one important special case: rotation about a fixed axis. This makes life much easier.
Recall from Section 6.6 that in rotation about a fixed axis, the direction of \(\vec{\omega}\) (and hence \(\vec{\alpha}\)) is fixed along the axis; only their magnitudes change. So we do not need to keep track of vectors — we can just work with the scalars \(\omega\) and \(\alpha\).
The other simplification is that fixed-axis rotation involves only one degree of freedom — just one variable, the angular position \(\theta\), is enough to describe the motion completely. This is exactly analogous to one-dimensional linear motion, which needs only one variable, the position \(x\).
This section deals only with kinematics — the description of motion. Dynamics (what causes rotation to change) will follow in Section 6.11.
Specifying the angular position
To describe the rotation of a rigid body about a fixed axis, we pick any particle P of the body — as shown in Fig. 6.29 — and use it as a marker. Since the whole body turns together, the angular position of P around the axis is the angular position of the whole body.
The particle P moves in a circle in a plane perpendicular to the axis. Let us take the axis of rotation to be the z-axis. Then the plane of motion of P is the x-y plane. The angle \(\theta\) of P is measured from a fixed direction in this plane, which we take to be the x′-axis (parallel to the x-axis).
Figure to come
Fig. 6.29 – A rigid body rotating about the z-axis; a particle P of the body traces a circle in the x-y plane about a centre on the z-axis; the angular position of P is \(\theta\), measured from the x′-axis (parallel to the x-axis) fixed in the plane of motion; the initial position P₀ at \(t = 0\) is marked at angle \(\theta_0\).
Also shown in Fig. 6.29 is \(\theta_0\), the angular displacement at \(t = 0\) — that is, the initial angular position of the body before we start our clock.
Angular velocity and angular acceleration recap
From Sections 6.6 and 6.6.1, we already have:
Angular velocity — the time rate of change of angular position: \[\omega = \frac{d\theta}{dt}\]
Since the axis is fixed, we can treat \(\omega\) as a scalar (no vector notation needed). Its SI unit is rad/s.
Angular acceleration — the time rate of change of angular velocity: \[\alpha = \frac{d\omega}{dt}\]
Its SI unit is rad/s².
The kinematic variables of rotation — \(\theta\), \(\omega\), \(\alpha\) — correspond exactly to the kinematic variables of one-dimensional linear motion — \(x\), \(v\), \(a\).
From linear equations to rotational equations
For linear motion with uniform (constant) acceleration, we have three well-known kinematic equations from Class 11 kinematics:
\[v = v_0 + at \quad \text{...(a)}\]
\[x = x_0 + v_0 t + \frac{1}{2} a t^2 \quad \text{...(b)}\]
\[v^2 = v_0^2 + 2 a x \quad \text{...(c)}\]
(Strictly speaking, the third one is \(v^2 = v_0^2 + 2a(x - x_0)\); if we choose \(x_0 = 0\), it becomes the simpler form shown.)
Here \(x_0\) is the initial position at \(t = 0\), and \(v_0\) is the initial velocity. The word “initial” refers to values of these quantities at \(t = 0\).
Because rotational kinematics mirrors linear kinematics term-for-term, the equations for rotational motion with uniform (constant) angular acceleration must be:
\[\omega = \omega_0 + \alpha t \quad \text{...(6.36)}\]
\[\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 \quad \text{...(6.37)}\]
\[\omega^2 = \omega_0^2 + 2 \alpha (\theta - \theta_0) \quad \text{...(6.38)}\]
Here \(\theta_0\) = initial angular displacement of the rotating body, and \(\omega_0\) = initial angular velocity of the body — both measured at \(t = 0\).
Just as the linear equations only apply for constant linear acceleration, these rotational equations only apply for constant angular acceleration. If \(\alpha\) changes with time, we cannot use them directly; we must use calculus instead.
Solved Examples
Looking ahead
The kinematic equations of this section describe how a rigid body rotates when the angular acceleration is uniform, but they do not say why \(\alpha\) takes the value it does. That is the job of dynamics — connecting torque to angular acceleration, just as force connects to linear acceleration. This is what Section 6.11 takes up next.
6.11 Dynamics of Rotational Motion about a Fixed Axis
In the last section, we handled the kinematics of rotation about a fixed axis — the equations that describe how the body’s angle, angular velocity, and angular acceleration change with time. Now we turn to the dynamics: what causes the angular acceleration in the first place? What is the rotational Newton’s second law? And how much work does a torque do on a rotating body?
The physics we build up here will mirror the physics of translational motion, quantity for quantity. It is worth pausing to see all these mirror-pairs collected in one place.
The full analogy — Table 6.2
Table 6.2 summarises the correspondence between linear motion and rotational motion about a fixed axis. Most of the pairs in this table have already appeared in earlier sections; we have simply not written them all together before.
Table 6.2: Comparison of Translational and Rotational Motion
| Sr. No. | Linear motion | Rotational motion about a fixed axis |
|---|---|---|
| 1 | Displacement \(x\) | Angular displacement \(\theta\) |
| 2 | Velocity \(v = dx/dt\) | Angular velocity \(\omega = d\theta/dt\) |
| 3 | Acceleration \(a = dv/dt\) | Angular acceleration \(\alpha = d\omega/dt\) |
| 4 | Mass \(M\) | Moment of inertia \(I\) |
| 5 | Force \(F = Ma\) | Torque \(\tau = I \alpha\) |
| 6 | Work \(dW = F\, ds\) | Work \(dW = \tau\, d\theta\) |
| 7 | Kinetic energy \(K = Mv^2/2\) | Kinetic energy \(K = I\omega^2/2\) |
| 8 | Power \(P = Fv\) | Power \(P = \tau\omega\) |
| 9 | Linear momentum \(p = Mv\) | Angular momentum \(L = I\omega\) |
We have already compared the kinematics (rows 1–3) and connected moment of inertia to mass (row 4). We have also introduced torque and angular momentum in Section 6.7. What is left is to derive the entries in rows 5, 6, 7, 8 for rotation from first principles — establishing that these correspondences are not just guesses by analogy, but genuine results of applying Newton’s laws to a rigid body.
We already know work in translational motion is \(F\, dx\). By analogy, we would guess that in rotational motion about a fixed axis, work should be \(\tau\, d\theta\) — since we already have \(dx \leftrightarrow d\theta\) and \(F \leftrightarrow \tau\). But guessing is not enough; the correspondence must be shown to follow from the physics.
A helpful simplification for fixed-axis rotation
Before starting the derivation, we note an important simplification that arises whenever the axis of rotation is fixed.
Since the axis is fixed in space, only those torque components that are along the axis can produce a change in the body’s rotation about the axis. A torque component perpendicular to the axis would try to tilt the axis away from its position — but the axis is held in place by rigid constraints (bearings, supports, mounts). These constraints supply whatever forces are needed to cancel the effect of perpendicular torques. So we can ignore any perpendicular components of torque and consider only the parallel components.
This means for calculating torques on a fixed-axis rigid body:
(1) We need to consider only those forces that lie in planes perpendicular to the axis. Any force parallel to the axis produces a torque that is perpendicular to the axis, which is cancelled by the constraints.
(2) We need to consider only those components of the position vectors that are perpendicular to the axis. Components of position vectors along the axis would produce torques perpendicular to the axis, again cancelled by the constraints.
With this simplification in mind, we now compute the work done by a torque.
Work done by a torque
Consider a rigid body rotating about a fixed axis. Take a cross-section of the body in a plane perpendicular to the axis, so the axis pokes through this plane at a single point (call it the origin). Now let a force \(\vec{F}_1\) act on a particle of the body at point \(P_1\), with its line of action lying in this cross-sectional plane (so the force is perpendicular to the axis). See Fig. 6.30.
Figure to come
Fig. 6.30 – A cross-section of a rigid body in a plane perpendicular to the axis (which is out of the page at C); a particle at P₁ (at perpendicular distance \(r_1\) from C) has a force \(\vec{F}_1\) acting on it in the plane, making angle \(\phi_1\) with the tangent to the circle at P₁; the particle moves through angle \(d\theta\) to a new position P₁′; the arc length P₁P₁′ = \(ds_1 = r_1 d\theta\); the angle \(\alpha_1\) between \(\vec{F}_1\) and the radius CP₁ is also marked.
The particle at \(P_1\) moves on a circle of radius \(r_1\) with centre \(C\) on the axis (with \(CP_1 = r_1\)). Suppose in time \(\Delta t\), the particle moves to position \(P_1'\). The tiny displacement \(d\vec{s}_1\) of the particle has magnitude \[ds_1 = r_1\, d\theta\] and points tangentially at \(P_1\) along the circular path. Here \(d\theta\) is the small angular displacement of the particle in time \(\Delta t\) — the angle \(\angle P_1 C P_1'\).
The work done by the force \(\vec{F}_1\) on the particle during this small displacement is \[dW_1 = \vec{F}_1 \cdot d\vec{s}_1 = F_1\, ds_1 \cos\phi_1 = F_1 (r_1\, d\theta) \sin\alpha_1\]
Here \(\phi_1\) is the angle between \(\vec{F}_1\) and the tangent to the circle at \(P_1\), while \(\alpha_1\) is the angle between \(\vec{F}_1\) and the radius vector \(C P_1\). These two angles are related: \(\phi_1 + \alpha_1 = 90°\), so \(\cos\phi_1 = \sin\alpha_1\).
Rewriting in terms of torque
Now the torque due to \(\vec{F}_1\) about the origin is \(\vec{OP}_1 \times \vec{F}_1\). As we noted earlier, we split \(\vec{OP}_1 = \vec{OC} + \vec{CP}_1\); the part along the axis (\(\vec{OC}\)) contributes only to a torque perpendicular to the axis, which is cancelled by the constraints. So the effective torque due to \(\vec{F}_1\) is
\[\vec{\tau}_1 = \vec{CP}_1 \times \vec{F}_1\]
Its magnitude, by Eq. (6.24a), is \[\tau_1 = r_1 F_1 \sin\alpha_1\]
So the work done in time \(\Delta t\) becomes \[dW_1 = \tau_1\, d\theta\]
This is exactly the analogue of \(dW = F \cdot dx\) for translational motion. And the key insight is that \(d\theta\) is the same for every particle of the body — because in rigid-body rotation about a fixed axis, all particles sweep out the same angle in the same time.
More than one force
If several external forces act on the body, each producing its own torque about the axis, then the work done by all of them is the sum of their individual works:
\[dW = (\tau_1 + \tau_2 + \dots)\, d\theta\]
Since the torques considered are all parallel to the fixed axis, they add up as scalars (each is positive or negative depending on whether it turns the body one way or the other along the axis direction). Their algebraic sum is the total torque \(\tau\):
\[\tau = \tau_1 + \tau_2 + \dots\]
So the total work done by all the external torques during a small angular displacement \(d\theta\) is
\[dW = \tau\, d\theta \quad \text{...(6.39)}\]
This is exactly the entry in row 6 of Table 6.2.
The similarity to the corresponding translational expression \(dW = F\, ds\) is obvious.
Instantaneous power
Dividing both sides of Eq. (6.39) by \(dt\):
\[P = \frac{dW}{dt} = \tau \frac{d\theta}{dt} = \tau\omega\]
\[P = \tau\omega \quad \text{...(6.40)}\]
This is the instantaneous rotational power — the rate at which work is done by the torque on the rotating body. Its SI unit is watt (W), the same as for translational power.
Compare Eq. (6.40) with the translational expression \[P = Fv\]
Again, a perfect quantity-for-quantity match: force \(\to\) torque, velocity \(\to\) angular velocity.
The rotational Newton’s second law
In a perfectly rigid body, there is no internal motion — the particles do not shift relative to one another. So work done by external torques is not “used up” internally; it goes entirely into increasing the rotational kinetic energy of the body.
The rate at which work is done by the torque is \(P = \tau\omega\) (from Eq. 6.40). This must equal the rate at which the rotational kinetic energy \(K = \frac{1}{2} I \omega^2\) increases.
Let us compute the rate of increase of \(K\). We assume the moment of inertia \(I\) does not change with time — the mass of the body does not change, and since the body remains rigid and the axis stays fixed, the mass distribution about the axis stays the same. So
\[\frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2} I \omega^2\right) = \frac{1}{2} I \cdot 2\omega \frac{d\omega}{dt} = I \omega \frac{d\omega}{dt}\]
Using \(\alpha = d\omega/dt\):
\[\frac{d}{dt}\left(\frac{I \omega^2}{2}\right) = I \omega \alpha\]
Equating the rates of work done and of increase in kinetic energy:
\[\tau\omega = I\omega\alpha\]
Dividing both sides by \(\omega\) (assuming \(\omega \ne 0\)):
\[\tau = I \alpha \quad \text{...(6.41)}\]
This is the rotational counterpart of Newton’s second law.
Compare with the translational form: \[F = ma\]
Just as force produces linear acceleration in translational motion, torque produces angular acceleration in rotational motion. The angular acceleration produced is directly proportional to the applied torque, and inversely proportional to the moment of inertia:
- Larger \(\tau\) → larger \(\alpha\) (bigger torque spins the body up faster)
- Larger \(I\) → smaller \(\alpha\) (a body with more rotational inertia is harder to spin up)
Because of this exact analogy, Eq. (6.41) is often called Newton’s second law for rotational motion about a fixed axis.
A powerful analogy at work
We have now derived every row of Table 6.2 (rows 5, 6, 8) from first principles. Together with row 4 (moment of inertia, from Section 6.9) and row 7 (rotational kinetic energy, from Section 6.9), this completes the correspondence:
\[F = Ma \quad \longleftrightarrow \quad \tau = I\alpha\]
\[dW = F\, ds \quad \longleftrightarrow \quad dW = \tau\, d\theta\]
\[P = Fv \quad \longleftrightarrow \quad P = \tau\omega\]
The whole framework of translational dynamics carries over to rotational dynamics — provided we replace each translational quantity by its rotational counterpart. This is enormously useful in solving problems.
Solved Example
Looking ahead
We have now built up the dynamical machinery for rotation about a fixed axis: torque produces angular acceleration through \(\tau = I\alpha\), work done equals torque times angle, power equals torque times angular velocity. The one entry left in Table 6.2 that we have not yet fully justified for the fixed-axis case is row 9 — angular momentum \(L = I\omega\). In the next section, we take this up, and we finally reach the phenomenon that opened this chapter: the ice skater’s spectacular spin-up when she pulls in her arms.
6.12 Angular Momentum in Case of Rotation about a Fixed Axis
In Section 6.7, we studied the angular momentum of a system of particles in a very general setting. We already know from there that the time rate of change of the total angular momentum \(\vec{L}\) of a system, about any point, equals the total external torque about the same point (Eq. 6.28b), and that when the external torque is zero, the total angular momentum is conserved.
In this section, we specialise this general result to one particular case — rotation about a fixed axis. This is by far the most common situation in machines and everyday life, and the results turn out to be beautifully clean.
Angular momentum of a single particle in fixed-axis rotation
Recall from Eq. (6.25b) that the total angular momentum of a system of \(n\) particles about the origin is \[\vec{L} = \sum_{i=1}^{n} \vec{r}_i \times \vec{p}_i\]
To compute this for a rigid body rotating about a fixed axis, we first look at the angular momentum \(\vec{l}\) of a single typical particle of the body, then sum over all particles.
Take the axis of rotation to be the z-axis and place the origin O somewhere on this axis. A typical particle P has position vector \(\vec{r} = \vec{OP}\) and linear momentum \(\vec{p} = m\vec{v}\). As shown earlier (Fig. 6.17b, Section 6.6), we split \[\vec{r} = \vec{OP} = \vec{OC} + \vec{CP}\] where C is the centre of the particle’s circular path, sitting on the axis, and \(\vec{CP}\) is perpendicular to the axis.
So its angular momentum about O is \[\vec{l} = \vec{r} \times \vec{p} = (\vec{OC} \times m\vec{v}) + (\vec{CP} \times m\vec{v})\]
The two pieces of \(\vec{l}\)
Let us look at each piece separately.
The linear velocity has magnitude \(v = \omega r_\perp\), where \(r_\perp\) is the perpendicular distance of the particle from the axis (i.e. the radius of its circular path), and \(\vec{v}\) points tangentially at P.
Using the right-hand rule, we can see that \(\vec{CP} \times \vec{v}\) points along the fixed axis (parallel to the z-axis). Denoting the unit vector along the axis by \(\hat{k}\), and noting that \(\vec{CP}\) and \(\vec{v}\) are perpendicular (so their cross product has magnitude \(|\vec{CP}||\vec{v}| = r_\perp v\)):
\[\vec{CP} \times m\vec{v} = r_\perp (mv)\, \hat{k} = m r_\perp^2 \omega\, \hat{k}\]
Similarly, \(\vec{OC} \times \vec{v}\) is perpendicular to the fixed axis (since \(\vec{OC}\) lies along the axis, and \(\vec{v}\) is perpendicular to the axis, so their cross product is perpendicular to \(\vec{OC}\) — that is, perpendicular to the axis).
So the angular momentum of a single particle \(\vec{l}\) splits naturally into two parts — one along the axis, and one perpendicular to it. Let us name the part along the axis \(\vec{l}_z\):
\[\vec{l}_z = \vec{CP} \times m\vec{v} = m r_\perp^2 \omega\, \hat{k}\]
and the full angular momentum of the particle is \[\vec{l} = \vec{l}_z + \vec{OC} \times m\vec{v}\]
Notice: \(\vec{l}_z\) points along the axis of rotation, but \(\vec{l}\) in general does not. For a single particle undergoing circular motion about the axis, \(\vec{l}\) and \(\vec{\omega}\) are not necessarily parallel.
Adding up to the total angular momentum
For the whole rigid body, we sum the angular momenta of all particles:
\[\vec{L} = \sum_i \vec{l}_i = \sum_i \vec{l}_{iz} + \sum_i \vec{OC}_i \times m_i \vec{v}_i\]
Let us name the two pieces of the total:
\[\vec{L}_\perp = \sum_i \vec{OC}_i \times m_i \vec{v}_i \quad \text{...(6.42a)}\]
\[\vec{L}_z = \sum_i \vec{l}_{iz} = \left(\sum_i m_i r_i^2\right)\omega\, \hat{k} \quad \text{...(6.42b)}\]
Here \(m_i\) and \(\vec{v}_i\) are the mass and velocity of the \(i^\text{th}\) particle, and \(C_i\) is the centre of that particle’s circular path. The last step in Eq. (6.42b) uses the definition of moment of inertia: \(I = \sum m_i r_i^2\) (Eq. 6.34). So
\[\vec{L}_z = I \omega\, \hat{k} \quad \text{...(6.42b)}\]
and the total angular momentum splits as
\[\vec{L} = \vec{L}_z + \vec{L}_\perp \quad \text{...(6.42c)}\]
The clean case — symmetric bodies
For most of the rigid bodies we work with in this chapter — a uniform disc, a ring, a rod rotating about an axis of symmetry, a solid cylinder rotating about its own axis — the axis of rotation is a symmetry axis of the body.
For such symmetric bodies, something nice happens. For every particle at some \(\vec{OC}_i\) with velocity \(\vec{v}_i\), there is another particle located diametrically opposite on the same circle, with velocity \(-\vec{v}_i\). Their contributions to \(\vec{L}_\perp\) (Eq. 6.42a) exactly cancel in pairs.
So for symmetric bodies, \(\vec{L}_\perp = 0\), and
\[\vec{L} = \vec{L}_z = I\omega\, \hat{k} \quad \text{...(6.42d)}\]
That is, \(\vec{L}\) points along the axis, and its magnitude is simply \[L = I\omega\]
This is precisely the entry in row 9 of Table 6.2. For symmetric bodies rotating about a symmetry axis, angular momentum in rotational motion truly plays the same role that linear momentum (\(p = Mv\)) plays in translation.
For non-symmetric bodies, \(\vec{L} \neq \vec{L}_z\), and \(\vec{L}\) does not lie along the axis. In such cases, the constraint forces on the axle must supply a torque perpendicular to the axis at every instant to prevent the axis from tipping — which is why unbalanced rotors vibrate and shake. Referring to Table 6.1, you can identify which bodies in that table have their axis as a symmetry axis, and which ones might not, if the axis were shifted.
The equation for the axial component
Now let us differentiate Eq. (6.42b) with respect to time. Since \(\hat{k}\) is a fixed (constant) vector along the axis, we get
\[\frac{d\vec{L}_z}{dt} = \frac{d(I\omega)}{dt}\, \hat{k}\]
From the general equation for a system of particles, Eq. (6.28b), \[\frac{d\vec{L}}{dt} = \vec{\tau}_{\text{ext}}\]
For fixed-axis rotation (recall the simplification from Section 6.11), only the components of external torques along the axis need to be considered — perpendicular components are cancelled by constraint forces. So we can write \[\vec{\tau}_{\text{ext}} = \tau\, \hat{k}\]
Since \(\vec{L} = \vec{L}_z + \vec{L}_\perp\), and the direction of \(\vec{L}_z\) (the unit vector \(\hat{k}\)) is fixed, we conclude:
\[\frac{d\vec{L}_z}{dt} = \tau\, \hat{k} \quad \text{...(6.43a)}\]
\[\frac{d\vec{L}_\perp}{dt} = 0 \quad \text{...(6.43b)}\]
So for rotation about a fixed axis, the component of angular momentum perpendicular to the fixed axis stays constant. All the interesting time-variation happens in the axial component.
Since \(\vec{L}_z = I\omega\, \hat{k}\), Eq. (6.43a) can also be written as a scalar equation:
\[\frac{d(I\omega)}{dt} = \tau \quad \text{...(6.43c)}\]
If the moment of inertia \(I\) does not change with time (rigid body, fixed axis), this becomes
\[\frac{d(I\omega)}{dt} = I \frac{d\omega}{dt} = I\alpha\]
and Eq. (6.43c) gives back
\[\tau = I\alpha \quad \text{...(6.41)}\]
We have already derived this equation in Section 6.11 using the work–kinetic energy route. Now we have derived it again from angular momentum. Both routes lead to the same rotational Newton’s second law — reassuring evidence that the physics is consistent.
6.12.1 Conservation of Angular Momentum
We are now ready to revisit one of the most important results in physics — the conservation of angular momentum — in the special context of fixed-axis rotation.
From Eq. (6.43c), if the external torque about the axis is zero:
\[\frac{d(I\omega)}{dt} = 0\]
which means
\[L_z = I\omega = \text{constant} \quad \text{...(6.44)}\]
For symmetric bodies, from Eq. (6.42d), the whole angular momentum \(\vec{L}\) equals \(\vec{L}_z\), and so both quantities have magnitude \(L = I\omega\). In such cases, \(L\) itself is conserved.
This is exactly the fixed-axis form of the general conservation law Eq. (6.29a), which we already met in Section 6.7. But now it takes a wonderfully compact form: whenever no external torque acts along the axis, the product \(I\omega\) stays constant.
A striking consequence — trading \(I\) for \(\omega\)
Suppose a body is rotating with some angular velocity \(\omega_1\) and has moment of inertia \(I_1\). If it somehow changes its moment of inertia to \(I_2\), while no external torque acts about the axis, then conservation gives
\[I_1 \omega_1 = I_2 \omega_2\]
\[\omega_2 = \frac{I_1}{I_2}\, \omega_1\]
- If the body pulls its mass inward, so that \(I_2 < I_1\) — its angular velocity increases (\(\omega_2 > \omega_1\)).
- If the body pushes its mass outward, so that \(I_2 > I_1\) — its angular velocity decreases (\(\omega_2 < \omega_1\)).
This trade-off between \(I\) and \(\omega\) leads to some of the most spectacular real-life demonstrations of physics.
A demonstration you can do
You may try this experiment with a friend. Sit on a swivel chair (a chair with a seat that is free to rotate about a vertical pivot) with your arms folded and your feet lifted off the ground. Ask a friend to spin the chair.
Figure to come
Fig. 6.32(a) – A student sitting on a swivel chair with arms stretched horizontally outward; the chair rotates slowly; a second panel of the same figure shows the student with arms folded close to the body, spinning much faster.
While the chair is rotating with a good angular speed, stretch your arms out horizontally. What happens? Your angular speed drops noticeably. Now bring your arms close to your body again. Your angular speed jumps back up.
Because the pivot is essentially frictionless, no external torque acts about the vertical axis of the chair. So \(I\omega\) is conserved.
- Stretching your arms puts mass (arms) far from the axis → \(I\) increases → \(\omega\) decreases.
- Pulling your arms in brings mass close to the axis → \(I\) decreases → \(\omega\) increases.
The chair does not care about the details of what you do with your arms — it only cares about the total moment of inertia and the fact that \(I\omega\) must stay constant.
The ice skater’s spin — the classic example
The most famous version of this experiment is performed on ice, not in a swivel chair. When a figure skater starts a spin, she begins with her arms and one leg extended outward. She then pulls them in tight to her body, and her spin dramatically speeds up — from a slow, graceful turn to a blur.
The physics is exactly the same as in the swivel chair. The friction between her skate and the ice is very small — nearly zero external torque about the vertical axis. So \(I\omega\) is conserved. Pulling her arms and leg in cuts \(I\); conservation forces \(\omega\) up.
Once you understand this, you can appreciate all sorts of similar performances. A circus acrobat swinging through the air, a gymnast performing a somersault, and a diver going through a tuck all use conservation of angular momentum to increase or decrease their rotation speed at will — no push needed from anything external.
Figure to come
Fig. 6.32(b) – A diver mid-jump; the diver curls into a tight tuck (arms wrapped around bent knees) while rotating rapidly; a second panel shows the diver stretched straight before hitting the water, rotating slowly.
Some final applications
Because there is essentially no friction in the rotational mechanism of a well-oiled swivel chair, and no significant external torque, \(I\omega\) stays constant. Stretching the arms increases \(I\) about the axis of rotation, resulting in decreased \(\omega\). Bringing the arms closer to the body has the opposite effect.
Skaters and classical, Indian or western, dancers performing a pirouette (a spin on the toes of one foot) also display mastery over this principle — controlling how they distribute their arms, one leg, and head to precisely time their rotation. Can you now explain how they do it?
Looking ahead
With this section, we have completed the analogy between translation and rotation about a fixed axis. Every entry in Table 6.2 — from displacement to power to momentum conservation — has been derived and understood. In the summary and points to ponder that follow, we consolidate the whole chapter into a single overview.
6.13 Summary
Rigid body. Ideally, a rigid body is a body whose shape and size remain unchanged even when forces act on it — that is, the distance between any two particles of the body stays constant with time. No real body is perfectly rigid, but many bodies (wheels, tops, steel beams, planets) can be treated as rigid to a very good approximation.
Kinds of motion. A rigid body that is fixed at a point or along a line (an axis) can have only rotational motion. A rigid body that is not fixed in any such way can have either pure translational motion or a combination of translational and rotational motion.
Rotation about a fixed axis. In this kind of motion, every particle of the rigid body moves in a circle that lies in a plane perpendicular to the axis and has its centre on the axis. Every point in the rotating rigid body has the same angular velocity at any given instant of time.
Pure translation. In pure translation, every particle of the body moves with the same velocity at any instant of time.
Angular velocity as a vector. Angular velocity is a vector quantity. Its magnitude is \(\omega = d\theta/dt\), and it is directed along the axis of rotation. For rotation about a fixed axis, the vector \(\vec{\omega}\) has a fixed direction; only its magnitude may change with time.
Vector (cross) product of two vectors. The vector product of two vectors \(\vec{a}\) and \(\vec{b}\) is another vector, written as \(\vec{a} \times \vec{b}\). Its magnitude is \(ab\sin\theta\), where \(\theta\) is the angle between the vectors, and its direction is perpendicular to the plane of \(\vec{a}\) and \(\vec{b}\), given by the right-handed screw rule (or the right-hand rule).
Linear velocity of a particle in a rotating body. The linear velocity of a particle of a rigid body rotating about a fixed axis is given by \[\vec{v} = \vec{\omega} \times \vec{r}\] where \(\vec{r}\) is the position vector of the particle from an origin lying on the fixed axis. The same relation applies to a rigid body rotating about a fixed point (a point-fixed rotation such as a spinning top): in that case, \(\vec{r}\) is measured from the fixed point taken as the origin.
Centre of mass. The centre of mass of a system of \(n\) particles is the point whose position vector is \[\vec{R} = \frac{\sum m_i \vec{r}_i}{M}\] where \(m_i\) and \(\vec{r}_i\) are the mass and position vector of the \(i^\text{th}\) particle, and \(M = \sum m_i\) is the total mass of the system.
Velocity of the centre of mass. For a system of particles, the velocity of the centre of mass is \[\vec{V} = \vec{P}/M\] where \(\vec{P}\) is the total linear momentum of the system. The centre of mass moves as if all the mass of the system were concentrated at this single point and all external forces on the system were applied there. If the total external force on the system is zero, the total linear momentum of the system remains constant.
Angular momentum and torque for a system of particles. The angular momentum of a system of \(n\) particles about the origin is \[\vec{L} = \sum_{i=1}^{n} \vec{r}_i \times \vec{p}_i\] and the total torque (moment of force) on the system about the origin is \[\vec{\tau} = \sum_i \vec{r}_i \times \vec{F}_i\] The force \(\vec{F}_i\) acting on the \(i^\text{th}\) particle includes both external and internal forces. Assuming Newton’s third law and that the internal forces between any two particles act along the line joining them, one can show that the total internal torque \(\vec{\tau}_{\text{int}} = 0\), so \[\frac{d\vec{L}}{dt} = \vec{\tau}_{\text{ext}}\]
Mechanical equilibrium of a rigid body. A rigid body is in mechanical equilibrium if:
- It is in translational equilibrium — the total external force on it is zero: \[\sum \vec{F}_i = 0\]
- It is in rotational equilibrium — the total external torque on it is zero: \[\sum \vec{\tau}_i = \sum \vec{r}_i \times \vec{F}_i = 0\] Both conditions must hold simultaneously.
Centre of gravity. The centre of gravity of an extended body is the point where the total gravitational torque on the body is zero. In a uniform gravitational field, the centre of gravity coincides with the centre of mass.
Moment of inertia. The moment of inertia of a rigid body about a given axis is \[I = \sum m_i r_i^2\] where \(r_i\) is the perpendicular distance of the \(i^\text{th}\) particle (mass \(m_i\)) from the axis. The rotational kinetic energy of the body is \[K = \frac{1}{2} I \omega^2\]
Physical quantities of rotational motion — at a glance
| Quantity | Symbol | Dimensions | Units | Remarks |
|---|---|---|---|---|
| Angular velocity | \(\vec{\omega}\) | \([T^{-1}]\) | rad/s | \(\vec{v} = \vec{\omega} \times \vec{r}\) |
| Angular momentum | \(\vec{L}\) | \([M L^2 T^{-1}]\) | J s | \(\vec{L} = \vec{r} \times \vec{p}\) |
| Torque | \(\vec{\tau}\) | \([M L^2 T^{-2}]\) | N m | \(\vec{\tau} = \vec{r} \times \vec{F}\) |
| Moment of inertia | \(I\) | \([M L^2]\) | kg m² | \(I = \sum m_i r_i^2\) |
6.14 Points to Ponder
To determine the motion of the centre of mass of a system of particles, no knowledge of the internal forces of the system is required. All the internal forces cancel out in pairs by Newton’s third law, and they simply do not enter Eq. (6.11). For this purpose, we need to know only the external forces on the body. This is why we can predict the trajectory of a rocket’s centre of mass without knowing anything about the complicated combustion happening inside the rocket, or the parabolic path of an exploding shell’s centre of mass without any knowledge of the internal chemical explosion.
Separating the motion of a system of particles into two pieces — the motion of the centre of mass (i.e. the translational motion of the system) and the motion about the centre of mass (i.e. motion relative to the centre of mass) — is a very useful technique in the dynamics of any system of particles. One striking example of this technique is the way the kinetic energy of a system of particles \(K\) splits neatly into two pieces:
\[K = K' + \frac{MV^2}{2}\]
Here \(K'\) is the kinetic energy of the system about its centre of mass (the “internal” kinetic energy of the parts, as seen from a frame moving with the centre of mass), and \(\frac{1}{2}MV^2\) is the kinetic energy of the centre of mass itself, treated as a single particle of mass \(M\) moving with velocity \(\vec{V}\).
Newton’s Second Law for finite-sized bodies (or systems of particles) is based on both Newton’s Second Law and Newton’s Third Law for particles. The Second Law gives \(\vec{F} = d\vec{p}/dt\) for each particle; the Third Law ensures that internal forces cancel in pairs so that only external forces survive in the total. Without the Third Law, the elegant result \(M\vec{A} = \vec{F}_{\text{ext}}\) would not follow.
To establish that the time rate of change of the total angular momentum of a system of particles equals the total external torque on the system, we need something more than just Newton’s Second and Third Laws. We also need the assumption that the forces between any two particles act along the line joining the particles. Without this extra condition (which is true for gravitational, electrostatic, and most familiar interactions), internal torques would not cancel in pairs, and the clean relation \(d\vec{L}/dt = \vec{\tau}_{\text{ext}}\) would break down.
The vanishing of the total external force and the vanishing of the total external torque are independent conditions. We can have one without the other. In a couple, for example, the total external force is zero, but the total torque is non-zero — so the body translates uniformly (or stays at rest) while it rotates about an axis. Conversely, a body may have a non-zero net force pulling it forward while, at the same time, all the torques on it about its centre of mass cancel — so it accelerates translationally but does not spin.
The total torque on a system is independent of the origin about which the torques are computed, provided that the total external force on the system is zero. This is why, for a body in translational equilibrium (like a see-saw or a ladder), we can take moments about any convenient point when applying the rotational equilibrium condition — a common trick in problem solving.
The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary from one part of the body to the other. For small bodies on Earth (a book, a wheel, a car), \(\vec{g}\) is essentially the same at every point, so CG = CM. For very large bodies (a tall mountain, a planet’s atmosphere) where \(\vec{g}\) changes appreciably from top to bottom, CG and CM will not exactly coincide.
The angular momentum \(\vec{L}\) and the angular velocity \(\vec{\omega}\) are not necessarily parallel vectors. In general, they can point in different directions. However, for the simpler situations discussed in this chapter — rotation about a fixed axis that is also an axis of symmetry of the rigid body — the perpendicular component \(\vec{L}_\perp\) vanishes, and the simple relation
\[\vec{L} = I\vec{\omega}\]
holds good, where \(I\) is the moment of inertia of the body about the rotation axis. For non-symmetric bodies, or for arbitrary rotation, this relation is not so simple, and \(\vec{L}\) will not be aligned with \(\vec{\omega}\).
6.15 NCERT Questions
Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?
In the HCl molecule, the separation between the nuclei of the two atoms is about \(1.27 \, \text{Å}\) (\(1 \, \text{Å} = 10^{-10}\) m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
A child sits stationary at one end of a long trolley moving uniformly with a speed \(V\) on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
Show that the area of the triangle contained between the vectors \(\vec{a}\) and \(\vec{b}\) is one half of the magnitude of \(\vec{a} \times \vec{b}\).
Show that \(\vec{a} \cdot (\vec{b} \times \vec{c})\) is equal in magnitude to the volume of the parallelepiped formed on the three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).
Find the components along the \(x\), \(y\), \(z\) axes of the angular momentum \(\vec{l}\) of a particle, whose position vector is \(\vec{r}\) with components \(x\), \(y\), \(z\), and momentum is \(\vec{p}\) with components \(p_x\), \(p_y\), and \(p_z\). Show that if the particle moves only in the \(x\)-\(y\) plane, the angular momentum has only a \(z\)-component.
Two particles, each of mass \(m\) and speed \(v\), travel in opposite directions along parallel lines separated by a distance \(d\). Show that the angular momentum vector of the two-particle system is the same whatever be the point about which the angular momentum is taken.
A non-uniform bar of weight \(W\) is suspended at rest by two strings of negligible weight as shown in Fig. 6.33. The angles made by the strings with the vertical are \(36.9°\) and \(53.1°\) respectively. The bar is 2 m long. Calculate the distance \(d\) of the centre of gravity of the bar from its left end.
Figure to come
Fig. 6.33 – A horizontal bar of length 2 m suspended by two strings from a fixed support above; the left string makes an angle of 36.9° with the vertical, the right string 53.1° with the vertical; the weight \(W\) acts vertically downward at the centre of gravity, at distance \(d\) from the left end.
A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time?
A solid cylinder of mass 20 kg rotates about its axis with angular speed \(100 \, \text{rad s}^{-1}\). The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
- A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of \(40 \, \text{rev/min}\). How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to \(2/5\) times the initial value? Assume that the turntable rotates without friction.
- Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?
A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is the linear acceleration of the rope? Assume that there is no slipping.
To maintain a rotor at a uniform angular speed of \(200 \, \text{rad s}^{-1}\), an engine needs to transmit a torque of 180 N m. What is the power required by the engine? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque.) Assume that the engine is 100% efficient.
From a uniform disc of radius \(R\), a circular hole of radius \(R/2\) is cut out. The centre of the hole is at \(R/2\) from the centre of the original disc. Locate the centre of gravity of the resulting flat body.
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g, are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
The oxygen molecule has a mass of \(5.30 \times 10^{-26}\) kg and a moment of inertia of \(1.94 \times 10^{-46}\) kg m² about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two-thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
6.16 Check Your Concepts
A solid cylinder rolls down an inclined plane without slipping. Explain why the point of contact between the cylinder and the incline is momentarily at rest, even though the cylinder as a whole is clearly moving down the slope. What kind of motion does the cylinder execute?
For each of the following bodies, state whether the centre of mass lies at a mass point of the body or not: (i) a solid sphere, (ii) a thin uniform ring, (iii) a hollow spherical shell, (iv) a right-angled L-shaped uniform lamina. In each case, briefly justify your answer using symmetry.
A firecracker is thrown at an angle in the air. Mid-flight, it explodes into many fragments that fly off in all directions. Explain why the centre of mass of all the fragments continues to follow the same parabolic path that the unexploded firecracker would have followed. Would this conclusion still hold if air resistance were significant? Justify.
A person stands at one end of a stationary boat floating on still water. She walks toward the other end of the boat. Explain, using the concept of conservation of linear momentum, why the boat moves backward as she walks forward. Does this violate Newton’s third law? Justify.
Two vectors \(\vec{A}\) and \(\vec{B}\) have magnitudes 3 units and 4 units respectively. State the maximum and minimum possible values of (i) \(\vec{A} \cdot \vec{B}\) and (ii) \(|\vec{A} \times \vec{B}|\). Explain your answer.
A ceiling fan is rotating anticlockwise as seen from below. Using the right-hand rule, state the direction of its angular velocity vector. If the fan is now switched off and gradually slows down, state the direction of its angular acceleration vector during this period. Justify.
Why is it easier to open a heavy door by pushing at its outer (handle-side) edge than at a point close to the hinges? Explain quantitatively using the definition of torque.
A particle is moving in a straight line along the \(x\)-axis with constant velocity. About which of the following points is the angular momentum of the particle (i) zero, (ii) constant and non-zero, (iii) changing with time? Choose from (a) any point on the \(x\)-axis itself, (b) any point off the \(x\)-axis. Justify your answer.
A rigid body is in translational equilibrium. Does this necessarily mean the body is also in rotational equilibrium? Support your answer with an example.
The moment of inertia of a body about a fixed axis depends on the choice of the axis, unlike the mass of the body which is a fixed quantity. Explain why this is so, using the defining formula \(I = \sum m_i r_i^2\).
The kinematic equations for rotational motion about a fixed axis (with uniform angular acceleration) are exactly analogous to the equations for linear motion under uniform acceleration. Write down the three linear kinematic equations and their rotational counterparts side by side, and identify the analogous quantities.
For a rigid body of moment of inertia \(I\) rotating about a fixed axis with angular velocity \(\omega\), state the rotational analogues of (i) Newton’s second law, (ii) kinetic energy, (iii) power, and (iv) linear momentum. In each case, identify the analogous quantities.
A figure skater spinning on ice pulls her arms and one leg close to her body. Her angular speed increases sharply. Explain this using the principle of conservation of angular momentum. Does her kinetic energy also stay constant during this process? If not, where does the change come from?
Why does a helicopter need a small tail rotor in addition to its main overhead rotor? Which law of physics is being applied here? What might happen if the tail rotor stopped working during flight?
The angular momentum \(\vec{L}\) and the angular velocity \(\vec{\omega}\) of a rotating body are not always parallel. Under what special condition do they become exactly parallel, so that \(\vec{L} = I\vec{\omega}\)? Give one example of a body–axis combination for which this simple relation holds, and one example for which it does not.
6.17 Practice with Numericals
Three point masses of 1 kg, 2 kg, and 3 kg are placed at the vertices \((0, 0)\), \((1 \text{ m}, 0)\), and \((0, 1 \text{ m})\) of a right triangle in the \(x\)-\(y\) plane. Find the coordinates of the centre of mass of the system.
A shell of mass 6 kg moving horizontally with a speed of \(50 \, \text{m s}^{-1}\) suddenly explodes into two fragments of masses 2 kg and 4 kg. The 2 kg fragment moves off in the same direction with a speed of \(80 \, \text{m s}^{-1}\). Find the velocity (magnitude and direction) of the 4 kg fragment immediately after the explosion. Ignore gravity during the instant of explosion.
A boy of mass 30 kg stands on a stationary trolley of mass 50 kg. The trolley is free to move on a frictionless horizontal floor. The boy suddenly jumps off the trolley with a horizontal velocity of \(4 \, \text{m s}^{-1}\) (relative to the ground). Find the velocity of the trolley immediately after the boy jumps off.
Given two vectors \(\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}\) and \(\vec{B} = \hat{i} - \hat{j} + 2\hat{k}\), find (i) \(\vec{A} \cdot \vec{B}\), (ii) \(\vec{A} \times \vec{B}\), and (iii) the magnitude of \(\vec{A} \times \vec{B}\).
A grindstone of diameter 40 cm rotates uniformly at 300 rpm. A particle sits at the rim of the grindstone. Find (i) the angular velocity of the grindstone in rad/s, and (ii) the linear speed of the particle at the rim.
A force \(\vec{F} = (3\hat{i} + 4\hat{j} - 2\hat{k})\) N acts on a particle whose position vector from the origin is \(\vec{r} = (\hat{i} - \hat{j} + 2\hat{k})\) m. Find the torque of this force about the origin.
A uniform ladder of mass 20 kg and length 6 m rests against a smooth (frictionless) vertical wall. Its lower end rests on a rough horizontal floor, at a distance of 2 m from the base of the wall. Find (i) the normal reaction of the wall on the ladder, (ii) the normal reaction of the floor on the ladder, and (iii) the frictional force from the floor. Take \(g = 10 \, \text{m s}^{-2}\).
A uniform horizontal beam of length 5 m and mass 20 kg is supported at its two ends. A load of 60 kg is placed on the beam at a distance of 1.5 m from the left support. Calculate the normal reaction at each support. Take \(g = 10 \, \text{m s}^{-2}\).
Four point masses, each of 2 kg, are placed at the four corners of a square of side 1 m. Find the moment of inertia of the system about an axis (i) passing through the centre of the square and perpendicular to its plane, and (ii) passing along one of the sides of the square.
A uniform thin rod of mass 3 kg and length 1.5 m rotates about an axis perpendicular to its length passing through its midpoint. Using the value from Table 6.1, calculate (i) the moment of inertia of the rod about this axis, and (ii) its rotational kinetic energy when rotating at 10 rad/s.
A wheel starting from rest reaches an angular velocity of 60 rad/s in 12 seconds under a constant angular acceleration. Find (i) the angular acceleration of the wheel, (ii) the total angle turned by the wheel in this time (in radians), and (iii) the total number of revolutions made by the wheel.
A solid disc of mass 4 kg and radius 0.25 m is free to rotate about a fixed horizontal axis through its centre. A constant tangential force of 8 N is applied at its rim. Find (i) the moment of inertia of the disc about the axis, (ii) the torque produced, and (iii) the angular acceleration of the disc.
A flywheel modelled as a solid disc has mass 20 kg and radius 0.5 m. It is initially rotating at 20 rad/s. A constant retarding torque brings it to rest in 40 seconds. Find (i) the moment of inertia of the flywheel, (ii) the retarding torque, and (iii) the total number of revolutions made by the flywheel before coming to rest.
A student sits on a rotating stool holding two 3 kg dumbbells, one in each hand. When his arms are outstretched (with the dumbbells at 0.8 m from the axis), his total moment of inertia (including the dumbbells) is \(6.0 \, \text{kg m}^2\) and he rotates at 2.0 rad/s. He now pulls the dumbbells close to his body (at 0.2 m from the axis), so that his new total moment of inertia is \(2.5 \, \text{kg m}^2\). Find (i) his new angular speed, and (ii) the change in his rotational kinetic energy. Comment on the source of the change.
A merry-go-round of moment of inertia \(250 \, \text{kg m}^2\) is spinning freely (no friction) at 4.0 rad/s about a vertical axis through its centre. A child of mass 40 kg (treated as a point mass) initially standing at the centre walks radially outward and stops at a distance of 1.5 m from the axis. Find the new angular velocity of the merry-go-round.