System of Particles and Rotational Motion: Class 11 Physics · Chapter 6

Physics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

120 min
1 simulation
24 practice questions
39 MCQs

What you will learn in System of Particles and Rotational Motion

Distinguish pure translation, rotation about a fixed axis and their combination in the motion of a rigid body
Locate the centre of mass of particle systems and of symmetric and composite bodies, and use it to predict how a system moves under external forces
Calculate vector products from the right-hand rule and from components, and use them to find velocity, torque and angular momentum
Apply the conservation of linear and angular momentum to systems of particles
Use the conditions of equilibrium to solve problems on levers, beams and ladders, and find centres of gravity
Define moment of inertia and radius of gyration, and use standard values for common bodies
Solve fixed-axis rotation problems with the rotational kinematic equations and the relations for torque, work, power and angular momentum

Original companion text aligned to NCERT Physics Class 11, Chapter 6, pp. 92–126. Not affiliated with or endorsed by NCERT.

6.1 Introduction

So far we have shrunk a cricket ball, a car and even the Earth to a single point carrying all the mass. That particle model is fine when only the path matters. It cannot explain why a spinning lattu stays upright, why a mechanic reaches for a longer spanner, or how a Kathak dancer whirls faster by drawing in her arms. Real bodies have size, and size is what lets them turn.

In this chapter we treat an extended body as a system of particles and ask two questions. How does the system move as a whole? The answer rests on one special point, the centre of mass. And how does the body turn? For that we need three new quantities: torque, angular momentum and moment of inertia, each the rotational partner of something you already know.

Turning problems become manageable when the body keeps its shape. A rigid body (दृढ़ पिंड) is an idealised body in which the distance between every pair of its particles stays fixed, whatever forces act. Real bodies only approximate this: a steel girder sags under a truck and a rubber ball flattens as it bounces. Yet a bicycle wheel, a door, a lattu or a satellite changes shape so little in ordinary motion that treating it as rigid costs almost nothing in accuracy.

Definition

Rigid body

An ideal body whose particles all keep fixed distances from one another, so its size and shape never change under any forces.

ExampleA door swinging on its hinges is very nearly rigid; a lump of dough being kneaded is not.

6.1.1 What kind of motion can a rigid body have?

Watch a lift cabin rise between floors, or a drawer slide out of a desk. Mark two dots on it, one near the top and one near a corner: at every instant both dots move with the same velocity, so they trace identical, parallel paths. This is pure translation: at any instant every particle of the body has the same velocity.

Now watch the front wheel of a bicycle rolling along a road. The wheel goes forward with the cycle, so there is translation, but its points no longer keep pace. The top of the tyre races ahead, the hub moves at the cycle's speed, and the point touching the road is momentarily at rest if the tyre does not skid. Rolling combines translation with a second kind of motion, and Fig. 6.1 pulls the two apart.

3/3Top: $2 v$ . Centre: $v$ . Contact point: zero.

Fig. 6.1Rolling without slipping is translation of the centre plus rotation about the centre.NCERT Fig. 6.2 · p. 93

A wheel on level ground with arrows at its top, centre and contact point. Step 1, translation: all three arrows point forward with length v. Step 2, rotation about the centre: the top arrow points forward and the bottom one backward, each of length v, and the centre has none. Step 3, rolling: the top arrow is 2v, the centre v, and the contact point has no velocity.

What is that second motion? Hold a body so that it cannot translate at all, the way hinges hold a door or an axle holds a sewing-machine wheel. Then all it can do is turn, and the fixed line is its axis of rotation. A chakki (hand grinding stone), a revolving door and a wind turbine all turn about fixed axes.

When a body turns about a fixed axis, each of its particles travels round a circle whose plane is at right angles to the axis and whose centre is on the axis (Fig. 6.2). Particles at different heights circle in different, parallel planes, and a particle farther from the axis travels a bigger circle in the same time. A particle on the axis does not move at all.

4/4C lies on the axis and does not move at all.

Fig. 6.2In fixed-axis rotation each particle circles the axis in its own perpendicular plane.NCERT Fig. 6.4 · p. 94

A drum turning about a vertical dashed axis. Step 1 shows the drum and the axis. Step 2 adds particle A, low and near the axis, on a small horizontal circle centred on the axis. Step 3 adds particle B, high and near the rim, on a larger circle. Step 4 marks particle C on the axis, which does not move.

The axis need not stay put. A spinning lattu turns about an axis through its tip, and while it spins that axis slowly swings round the vertical, sweeping out a cone. This slow swing of the axis is called precession. Your arm turning at the shoulder is another case: the axis passes through the joint but can point in many directions. In both, one point of the body is fixed rather than a whole line. In this chapter we stay with the simpler case of a fixed axis.

Try this

Slide it, then turn it

You’ll need

a 30 cm scale
a sheet of chart paper
two pencils
a friend

Predict first

You slide the scale along a curve without letting it turn. How will the paths of the two dots compare?

The paths will be identical, just shiftedThe end dot's path will be longerThe middle dot's path will be longer

Draw two dots on a 30 cm scale: dot A at one end and dot B at the 15 cm mark.
Lay the scale on chart paper. Slide it along a gentle curve, keeping it parallel to the paper's edge, and let a friend trace the paths of A and B with pencils.
Start again, but this time let the scale turn through a quarter turn as it moves along a similar curve. Trace both dots again.

6.2 Centre of mass

Balance a cricket bat across one finger. The balance point is not halfway along it but much closer to the heavy blade: where more mass sits, the balance point leans. The centre of mass (द्रव्यमान केंद्र) turns this everyday idea into a formula.

Start with two particles of masses $m_{1}$ and $m_{2}$ on the $x$ -axis, at $x_{1}$ and $x_{2}$ from an origin O. Their centre of mass lies at

Centre of mass of two particles

X = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}

(6.1)

This is a mass-weighted average of the two positions. With equal masses it is the plain average, the midpoint. A heavier particle pulls $X$ towards itself: if $m_{2} = 3 m_{1}$ , the centre of mass sits three-quarters of the way from $m_{1}$ to $m_{2}$ (Fig. 6.3).

3/3Supported at the centre of mass, the rod balances: $m_{1} d_{1} = m_{2} d_{2}$ .

Fig. 6.3The centre of mass sits nearer the heavier particle, where

m_{1} d_{1} = m_{2} d_{2}

.NCERT Fig. 6.7 · p. 95

Two balls on a light rod with a marker for their centre of mass. Step 1: equal masses, marker at the midpoint. Step 2: the right ball is three times heavier and the marker moves to the three-quarter point. Step 3: a support under the marker balances the rod, with distances d1 and d2 marked, d1 three times d2.

For $n$ particles on a line the rule is the same: multiply each position by its mass, add, and divide by the total mass $M = \sum m_{i}$ .

\begin{aligned} X & = \frac{\sum_{i = 1}^{n} m_{i} x_{i}}{\sum_{i = 1}^{n} m_{i}} \\ = \frac{1}{M} \sum_{i = 1}^{n} m_{i} x_{i} \end{aligned}

(6.2)

If the particles are spread over a plane or through space, apply the same weighted average to each coordinate separately:

\begin{aligned} X & = \frac{1}{M} \sum m_{i} x_{i} \\ Y & = \frac{1}{M} \sum m_{i} y_{i} \\ Z & = \frac{1}{M} \sum m_{i} z_{i} \end{aligned}

(6.4)

For three equal masses this gives the average of the three corners, which is the centroid of their triangle. With position vectors $𝐫_{i} = x_{i} \hat{𝐢} + y_{i} \hat{𝐣} + z_{i} \hat{𝐤}$ the three equations fold into one vector equation:

Centre of mass (vector form)

𝐑 = \frac{1}{M} \sum_{i = 1}^{n} m_{i} 𝐫_{i}

(6.4d)

One consequence is worth remembering: if you put the origin at the centre of mass, then $\sum m_{i} 𝐫_{i} = 𝟎$ . We shall use this in Section 6.8.2.

A steel scale or a flywheel holds something like $10^{25}$ atoms, far too many to add one by one. Instead, imagine the body cut into small pieces of mass $Δ m$ , add $Δ m 𝐫$ over the pieces, and let the pieces shrink without limit. The sums become integrals:

𝐑 = \frac{1}{M} \int 𝐫 d m

(6.5b)

where

$d m$: a small element of mass at position $𝐫$
$M = \int d m$: the total mass

For uniform bodies with a centre of symmetry you can skip the integral. Put the origin at the middle of a thin uniform rod: every small piece $d m$ at $x$ has a twin of equal mass at $- x$ , so the products $x d m$ cancel in pairs and $\int x d m = 0$ . The same pairing of $(x, y, z)$ with $(- x, - y, - z)$ works for a uniform ring, disc or sphere, or a thick rod of uniform cross-section, so their centre of mass is the geometric centre. Notice that the centre of mass need not lie in the material: a ring's is at its empty middle.

Example 6.1NCERT Example 6.1 · p. 97

Three brass weights on a set-square

Brass weights of $2.0 k g$ , $3.0 k g$ and $5.0 k g$ sit at the corners of a right-angled triangle. Take the right-angle corner as the origin, with the $2.0 k g$ weight there, the $3.0 k g$ weight at $(0.40 m, 0)$ and the $5.0 k g$ weight at $(0, 0.30 m)$ . Locate the centre of mass. Is it at the centroid of the triangle?

Try it yourself first. Then reveal the solution one step at a time.

Example 6.2NCERT Example 6.2 · p. 98

Where does a triangular sign balance?

Show that the centre of mass of a uniform triangular plate, such as a plywood road sign, is the point where its three medians meet.

3/3Repeat with strips parallel to CA: the medians cross at G, the centroid.

Fig. 6.4Each strip balances at its midpoint, and all the midpoints lie on a median.NCERT Fig. 6.10 · p. 98

Triangle ABC. Step 1: it is sliced into thin strips parallel to BC. Step 2: dots mark the midpoint of each strip, and they lie on the median from A. Step 3: the median from B is added, and the two medians cross at G, the centroid.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.3NCERT Example 6.3 · p. 98

A T-shaped name board

A name board is cut from uniform sheet metal of $10 k g$ per square metre. Its top bar is $80 c m$ wide and $20 c m$ tall; a stem $20 c m$ wide and $60 c m$ tall hangs from the middle of the bar (Fig. 6.5). Find its centre of mass.

Fig. 6.5Split the T into two rectangles and treat each as a point mass at its centre.NCERT p. 98

A T-shaped plate on axes, with the origin at the foot of its 20 cm by 60 cm stem and the 80 cm by 20 cm bar on top. The centres C1 and C2 of the two rectangles, and the centre of mass G between them, lie on the vertical line of symmetry.

Try it yourself first. Then reveal the solution one step at a time.

Fig. 6.6Drag the three masses and change their sizes; the cross marks the centre of mass and the readout gives its coordinates.

Quick check

1 of 3

2.0 k g

ball sits at

x = 0

and a

6.0 k g

ball at

x = 0.80 m

. Where is their centre of mass?

Show answer

Answer 0.6 m

X = (6.0 \times 0.80) / 8.0 = 0.60 m

, three-quarters of the way towards the heavier ball.

6.3 Motion of centre of mass

Why does the centre of mass matter? Because it moves in a remarkably simple way, however complicated the motion of the parts.

DerivationNewton's second law for the centre of mass

Write the definition as $M 𝐑 = \sum m_{i} 𝐫_{i}$ and differentiate with respect to time, keeping every mass constant:

M 𝐕 = \sum_{i} m_{i} 𝐯_{i}

(6.8)

Here $𝐕 = d 𝐑 / d t$ is the velocity of the centre of mass. Differentiating once more gives its acceleration $𝐀$ :

\begin{aligned} M 𝐀 & = \sum_{i} m_{i} 𝐚_{i} \\ = \sum_{i} 𝐅_{i} \end{aligned}

(6.9)

By Newton's second law each $m_{i} 𝐚_{i}$ is the total force $𝐅_{i}$ on particle $i$ . That force has two parts: external forces from bodies outside the system, and internal forces from the other particles. Internal forces come in action–reaction pairs (the third law), so in the sum they cancel two by two. Only the external forces are left:

Motion of the centre of mass

M 𝐀 = 𝐅_{ext}

(6.11)

Read (6.11) as a promise. However the parts of a system jostle, spin or fly apart, one point, the centre of mass, obeys Newton's second law as if it were a lone particle of mass $M$ pushed by the total external force.

Two consequences follow. First, to predict the centre of mass you need no knowledge of the internal forces: not the push of an explosion, the pull of muscles, nor the spring between two trolleys. Second, it justifies the earlier chapters, where we treated a car or a block as a point and quietly ignored its rotation and internal motion. We now know that its centre of mass really does move like a point, whatever else is going on.

Think of a festival shell that bursts at the top of its flight (Fig. 6.7). Before the burst gravity pulls on the shell; afterwards it pulls on each piece, and the pieces' weights still add up to the same total. The explosion only pushes the pieces against one another, so (6.11) is unchanged, and so is its answer: the mass-weighted average position of the pieces keeps tracing the shell's original parabola. That holds until a piece lands and the ground starts pushing on the system. (We ignore air drag throughout.)

3/3The centre of mass of the pieces stays on the original parabola.

Fig. 6.7Internal forces of the burst cannot move the centre of mass off its parabola.NCERT Fig. 6.12 · p. 100

A dashed parabola starts at a launcher on the left. Step 1: the shell rises along it. Step 2: the shell bursts at the top, with pieces flying outward. Step 3: five pieces are scattered on their own paths, and a cross marking their centre of mass sits on the original parabola.

Quick check

1 of 3

A shell flying on a parabola explodes into two equal halves in mid-air. Ignoring air resistance, before either half lands, the centre of mass of the two halves

drops straight down from the burst pointcontinues along the original parabolamoves on in a straight line at constant velocitystays at the burst point

Show answer

Answer (B) continues along the original parabola

The explosion forces are internal. The only external force is still gravity, so the centre of mass keeps the same parabola. Dropping straight down would need a sideways external force to remove its horizontal velocity.

6.4 Linear momentum of a system of particles

For one particle the linear momentum is $𝐩 = m 𝐯$ , and Newton's second law reads $𝐅 = d 𝐩 / d t$ . For a system of $n$ particles we define the total momentum as the vector sum:

𝐏 = \sum_{i} 𝐩_{i} = \sum_{i} m_{i} 𝐯_{i}

(6.14)

Comparing with (6.8) gives a neat result: the total momentum is the total mass times the velocity of the centre of mass.

𝐏 = M 𝐕

(6.15)

Differentiating, and using (6.11), gives Newton's second law for a whole system:

Second law for a system of particles

\frac{d 𝐏}{d t} = 𝐅_{ext}

(6.17)

LawConservation of linear momentum

If the total external force on a system is zero, its total linear momentum stays constant, and its centre of mass moves with constant velocity.

Internal forces may still send individual particles along tangled paths. Only the centre of mass is guaranteed to coast in a straight line at steady speed, like a free particle.

Radioactive decay shows why this matters. A uranium-238 nucleus at rest that emits an alpha particle has zero momentum before and after, so the thorium-234 nucleus left behind recoils straight backwards; being about 58 times heavier than the alpha particle, it moves about 58 times more slowly. Had the uranium nucleus been moving, an observer in the laboratory would see the fragments fly off at an angle, with their centre of mass carrying on undisturbed along the original line. The back-to-back picture is the same event watched from a frame moving with the centre of mass, which is why decays and collisions are usually worked out in this centre-of-mass frame.

The same idea untangles orbits. Two bodies bound by gravity, with no outside force, circle their common centre of mass while that point drifts uniformly (Fig. 6.8). For the Earth and the Moon (ignoring the Sun's pull), the centre of mass, the barycentre, lies about $4700 k m$ from the Earth's centre, inside the Earth, because the Earth is 81 times heavier. The Earth makes a small monthly circle about it; the Moon makes a big one.

2/2The centre of mass is at rest; both bodies circle it on opposite sides.

Fig. 6.8The wiggly paths are a steady drift of the centre of mass plus circles about it.NCERT Fig. 6.14 · p. 101

A heavy body and a light body bound together. Step 1, laboratory frame: a cross marking their centre of mass moves in a straight line; the heavy body wiggles slightly and the light body loops widely on the opposite side. Step 2, centre-of-mass frame: the cross is still and the bodies move on a small and a large circle, always opposite each other.

Example

Walking along a boat

A $60 k g$ boatman stands at the back of a $140 k g$ boat floating at rest on a still lake. He walks $4.0 m$ forward along the boat. Ignoring water resistance, how far does the boat move, and how far does he move relative to the shore?

Try it yourself first. Then reveal the solution one step at a time.

6.5 Vector product of two vectors

In Chapter 5 you multiplied force and displacement with the scalar (dot) product and got work, a number. Rotation needs a second kind of product, one whose answer is a vector perpendicular to both factors. Torque and angular momentum are both made this way.

Definition

Vector product

The vector product (cross product) $𝐚 \times 𝐛$ is a vector of magnitude $a b \sin θ$ , perpendicular to both $𝐚$ and $𝐛$ , with its sense given by the right-hand rule.

Magnitude: $∣ 𝐚 \times 𝐛 ∣ = a b \sin θ$ , where $θ$ is the smaller angle (between $0^{\circ}$ and $180^{\circ}$ ) between the two vectors.
Direction: along the line perpendicular to the plane that contains $𝐚$ and $𝐛$ .
Sense: curl the fingers of your right hand from $𝐚$ towards $𝐛$ through the smaller angle; your thumb points along $𝐚 \times 𝐛$ . Equivalently, a right-handed screw turned from $𝐚$ to $𝐛$ advances along $𝐚 \times 𝐛$ (Fig. 6.9).

2/4Turn from $𝐚$ to $𝐛$ : $𝐚 \times 𝐛$ points up, size $a b \sin θ$ .

Fig. 6.9Curl from the first vector to the second: the thumb gives the cross product.NCERT Fig. 6.15 · p. 102

Vectors a and b drawn from one point in a tilted plane, with angle theta between them. Step 2: turning from a to b, the product a cross b points straight up out of the plane. Step 3: turning from b to a, the product points straight down. Step 4: b points opposite to a and no product arrow appears.

Some properties follow at once from the definition:

Order matters. $𝐛 \times 𝐚$ has the same size as $𝐚 \times 𝐛$ but the opposite sense: $𝐛 \times 𝐚 = - 𝐚 \times 𝐛$ . (The dot product, by contrast, does not care about order.)
Parallel vectors give zero, since $\sin 0^{\circ} = \sin 180^{\circ} = 0$ . In particular $𝐚 \times 𝐚 = 𝟎$ , the null vector.
Distributive: $𝐚 \times (𝐛 + 𝐜) = 𝐚 \times 𝐛 + 𝐚 \times 𝐜$ , just as $𝐚 \cdot (𝐛 + 𝐜) = 𝐚 \cdot 𝐛 + 𝐚 \cdot 𝐜$ .
Mirror behaviour: reflecting every coordinate ( $x \to - x$ , $y \to - y$ , $z \to - z$ ) flips the sign of an ordinary vector, but $(- 𝐚) \times (- 𝐛) = 𝐚 \times 𝐛$ , so a cross product does not change sign.

For the unit vectors, $\hat{𝐢} \times \hat{𝐢} = \hat{𝐣} \times \hat{𝐣} = \hat{𝐤} \times \hat{𝐤} = 𝟎$ . Two different unit vectors are perpendicular, so their product has size 1 and points along the third axis:

\begin{aligned} \hat{𝐢} \times \hat{𝐣} & = \hat{𝐤} \\ \hat{𝐣} \times \hat{𝐤} & = \hat{𝐢} \\ \hat{𝐤} \times \hat{𝐢} & = \hat{𝐣} \end{aligned}

A memory aid: going round the cycle $i \to j \to k \to i$ gives a plus sign, and going against it a minus sign, so $\hat{𝐣} \times \hat{𝐢} = - \hat{𝐤}$ and $\hat{𝐢} \times \hat{𝐤} = - \hat{𝐣}$ . Multiplying out $(a_{x} \hat{𝐢} + a_{y} \hat{𝐣} + a_{z} \hat{𝐤}) \times (b_{x} \hat{𝐢} + b_{y} \hat{𝐣} + b_{z} \hat{𝐤})$ term by term with these rules gives

\begin{aligned} 𝐚 \times 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} \end{aligned}

which is easiest to remember as a determinant:

Cross product in components

𝐚 \times 𝐛 = ∣ \begin{array}{ccc} \hat{𝐢} & \hat{𝐣} & \hat{𝐤} \\ a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \end{array} ∣

Example 6.4NCERT Example 6.4 · p. 103

Dot and cross products in components

For $𝐚 = 2 \hat{𝐢} + 3 \hat{𝐣} - \hat{𝐤}$ and $𝐛 = \hat{𝐢} - 2 \hat{𝐣} + 4 \hat{𝐤}$ , find $𝐚 \cdot 𝐛$ , $𝐚 \times 𝐛$ and $𝐛 \times 𝐚$ . Check your cross product.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

\hat{𝐤} \times \hat{𝐣}

equals

\hat{𝐢}

𝟎

\hat{𝐣}

- \hat{𝐢}

Show answer

Answer (D) $- \hat{𝐢}$

\hat{𝐣} \times \hat{𝐤} = \hat{𝐢}

follows the cycle;

\hat{𝐤} \times \hat{𝐣}

goes against it, so it is

- \hat{𝐢}

. It is not zero, because

\hat{𝐤}

and

\hat{𝐣}

are perpendicular.

6.6 Angular velocity and its relation with linear velocity

Go back to a body turning about a fixed axis, which we take as the $z$ -axis. A particle P at perpendicular distance $r_{⊥}$ from the axis moves on a circle of that radius, centred at a point C on the axis. In a short time $Δ t$ the radius CP sweeps through an angle $Δ θ$ , the particle's angular displacement. The ratio $Δ θ / Δ t$ is its average angular velocity; as $Δ t \to 0$ it becomes the instantaneous angular velocity $ω = d θ / d t$ .

From circular motion you know that the particle's speed is $v = ω r_{⊥}$ . What makes a rigid body special is that every particle sweeps the same angle in the same time, so a single $ω$ describes the whole body. Particles farther out simply move faster:

v_{i} = ω r_{i}

(6.19)

where

$r_{i}$: the perpendicular distance of particle $i$ from the axis

Particles on the axis have $r_{i} = 0$ and stay still, as a fixed axis demands. So pure rotation is the twin of pure translation: in translation every particle has the same velocity; in rotation every particle has the same angular velocity.

Angular velocity also has a direction, and it behaves as a vector (proving this needs more mathematics than we have here). For a fixed axis, $𝝎$ points along the axis, in the direction a right-handed screw would advance if it turned with the body. Curl the fingers of your right hand along the motion and your thumb gives $𝝎$ . A wall clock's hands turn clockwise as you face them, so their $𝝎$ points into the wall, away from you.

DerivationVelocity as a vector product

Put the origin O on the axis and let $𝐫 = 𝐎 𝐏$ be the position vector of the particle (Fig. 6.10). Split it at the centre C of the particle's circle:

\begin{aligned} 𝝎 \times 𝐫 & = 𝝎 \times 𝐎 𝐂 \\ + 𝝎 \times 𝐂 𝐏 \end{aligned}

The first term vanishes because $𝐎 𝐂$ lies along the axis, parallel to $𝝎$ . The second is perpendicular to the axis and to the radius CP, so it points along the tangent to the circle at P. Its size is $ω r_{⊥}$ , since $𝝎$ and $𝐂 𝐏$ are at right angles. That is exactly the particle's velocity, in size and direction:

𝐯 = 𝝎 \times 𝐫

(6.20)

(6.20) also holds when only one point of the body is fixed, as for a lattu, with $𝐫$ measured from that point. For a fixed axis the direction of $𝝎$ never changes, though its size may; in more general rotation both can change.

3/3 $𝐯 = 𝝎 \times 𝐫$ is tangent to the circle, size $ω r_{⊥}$ .

Fig. 6.10

𝐯 = 𝝎 \times 𝐫

is tangent to the circle and has size

ω r_{⊥}

.NCERT Fig. 6.17 · p. 104

A vertical axis with an upward omega arrow and a horizontal circle centred at C on it, with particle P on the circle. Step 2 adds the origin O on the axis and the position vector r from O to P, made of OC along the axis and the radius CP, of length r perpendicular. Step 3 adds the velocity v at P, tangent to the circle.

6.6.1 Angular acceleration

In translation, acceleration is the rate of change of velocity. Its rotational twin is the angular acceleration:

𝜶 = \frac{d 𝝎}{d t}

(6.21)

For a fixed axis the direction of $𝝎$ is fixed, so $𝜶$ lies along the same axis and we can work with magnitudes alone:

α = \frac{d ω}{d t}

(6.22)

Its SI unit is $r a d / s^{2}$ . A particle at distance $r_{⊥}$ from the axis then has a tangential acceleration $r_{⊥} α$ along its circle, besides the centripetal acceleration $ω^{2} r_{⊥}$ towards the axis.

Quick check

1 of 3

A ceiling fan turns at

300

rpm. How fast does a blade tip

0.60 m

from the axis move?

m/s

Show answer

Answer 18.85 m/s

ω = 300 \times 2 π / 60 = 31.4 r a d / s

, so

v = ω r = 31.4 \times 0.60 = 18.8 m / s

6.7 Torque and angular momentum

A force changes linear momentum. What changes the rotation of a body? This section defines the rotational partners of force and of momentum. Both are vector products, and both matter most for systems of particles and rigid bodies.

6.7.1 Moment of force (Torque)

At a cycle repair shop the mechanic loosens a stubborn nut by fitting the longest spanner and pushing on its far end, at right angles to the handle. Pushing close to the nut does little, and pulling along the handle towards the nut does nothing at all, however hard he tries (Fig. 6.11). The turning effect depends on the force, on where it acts and on its direction. This turning effect is the moment of force, or torque (बल आघूर्ण).

For a force $𝐅$ acting on a particle at position $𝐫$ from an origin O, the torque about O is

Torque

𝝉 = 𝐫 \times 𝐅

(6.23)

where

$𝐫$: position of the point where $𝐅$ acts, measured from O

Its magnitude, with $θ$ the angle between $𝐫$ and $𝐅$ , can be read in two useful ways:

τ = r F \sin θ

(6.24a)

τ = r_{⊥} F = r F_{⊥}

(6.24b)

where

$r_{⊥} = r \sin θ$: the perpendicular distance from O to the line of action of $𝐅$ (the moment arm)
$F_{⊥} = F \sin θ$: the component of $𝐅$ perpendicular to $𝐫$

A push aimed straight at O, or applied at O itself, cannot turn anything about O. Torque is measured in newton metres ( $N m$ ), with dimensions $[M L^{2} T^{- 2}]$ . Work has the same dimensions, but the likeness is skin-deep: work is a scalar whose unit we call the joule, while torque is a vector, so we always write torque in $N m$ and never in joules.

2/4The moment arm shrinks to $r \sin θ$ : $τ = r F \sin θ$ .

Fig. 6.11Torque = force × moment arm: the perpendicular distance from O to the line of action.NCERT Fig. 6.18 · p. 106

A spanner on a nut at O. Step 1: a force F pushes square on at the far end, so the moment arm is the full length r. Step 2: the force is slanted at angle theta, and a dashed perpendicular from O shows the shorter moment arm r sin theta. Step 3: the force pulls along the handle and its line passes through O, so there is no torque. Step 4: a square-on force near the nut has a short moment arm.

Example 6.5NCERT Example 6.5 · p. 108

Torque of a force from its components

A force $𝐅 = (3 \hat{𝐢} - 2 \hat{𝐣} + \hat{𝐤}) N$ acts on the head of a bolt at $𝐫 = (\hat{𝐢} + 2 \hat{𝐣} - \hat{𝐤}) m$ from the origin. Find the torque about the origin and its magnitude.

Try it yourself first. Then reveal the solution one step at a time.

6.7.2 Angular momentum of a particle

Torque is the moment of a force; angular momentum (कोणीय संवेग) is the moment of linear momentum. For a particle with momentum $𝐩$ at position $𝐫$ from O, the angular momentum about O is

Angular momentum of a particle

𝐥 = 𝐫 \times 𝐩

(6.25a)

\begin{aligned} l & = r p \sin θ \\ = r_{⊥} p = r p_{⊥} \end{aligned}

(6.26)

Here $r_{⊥}$ is the perpendicular distance from O to the line along which the particle moves. A particle at rest, one sitting at O, or one heading straight towards or away from O has no angular momentum about O. The SI unit is $k g m^{2} / s$ , the same as $J s$ .

DerivationTorque is the rate of change of angular momentum

Differentiate $𝐥 = 𝐫 \times 𝐩$ with the product rule, keeping the order of the factors:

\begin{aligned} \frac{d 𝐥}{d t} & = \frac{d 𝐫}{d t} \times 𝐩 \\ + 𝐫 \times \frac{d 𝐩}{d t} \end{aligned}

The first term is $𝐯 \times m 𝐯 = 𝟎$ , because a vector crossed with a parallel one gives zero. In the second, $d 𝐩 / d t = 𝐅$ . So

\frac{d 𝐥}{d t} = 𝐫 \times 𝐅 = 𝝉

(6.27)

This is the rotational form of $𝐅 = d 𝐩 / d t$ for a single particle.

Example 6.6NCERT Example 6.6 · p. 109

A car passing a toll booth

A $1200 k g$ car drives along a straight highway at a steady $20 m / s$ . A toll booth O stands $50 m$ from the line of the road (Fig. 6.12). Find the car's angular momentum about O (a) when it is level with the booth and (b) when it is $120 m$ further on. Is there a torque about O?

Fig. 6.12The perpendicular distance from O to the car's line of motion never changes.NCERT Fig. 6.19 · p. 109

A straight road with a car at two positions, both moving right with velocity v, and a booth O below the road. At position 1 the car is directly above O, 50 m away. At position 2 the line from O is 130 m long and makes angle theta with the road, but the perpendicular distance is still 50 m.

Try it yourself first. Then reveal the solution one step at a time.

Try it with a friend

A spinning wheel that refuses to fall

You’ll need

a bicycle front wheel taken off the cycle, with its axle ends sticking out
a strong rope about 1 m long
a friend

Predict first

The wheel hangs by one end of its axle and is spinning fast. When you let go of the other end, the wheel will

flop down until it hangs below the ropestay nearly level while its axle swings slowly round the ropestart spinning faster

Tie the rope firmly to one end of the wheel's axle and hold the rope up at arm's length.
With the wheel not spinning, hold the other end of the axle level, then let go of that end. Watch what the wheel does.
Now hold the axle level again while your friend spins the wheel as fast as safely possible.
Let go of the free end of the axle and watch from above.

Without spin the wheel simply flops down. Spinning, it stays almost level and its axle slowly circles the rope. The weight acting at the wheel's centre has a horizontal torque about the rope; by $d 𝐋 = 𝝉 d t$ each small change in $𝐋$ is horizontal and perpendicular to $𝐋$ , so $𝐋$ , and the axle with it, turns round instead of falling. Work out the direction of the torque and check that the axle precesses the way $𝝉$ predicts.

Torque and angular momentum for a system of particles

The total angular momentum of a system about a point is the vector sum of the angular momenta of its particles:

\begin{aligned} 𝐋 & = \sum_{i = 1}^{n} 𝐥_{i} \\ = \sum_{i = 1}^{n} 𝐫_{i} \times 𝐩_{i} \end{aligned}

(6.25b)

DerivationExternal torque changes the total angular momentum

Differentiate term by term using (6.27): $d 𝐋 / d t = \sum 𝝉_{i}$ , where $𝝉_{i} = 𝐫_{i} \times 𝐅_{i}$ is the torque of the total force on particle $i$ . Split each force into an external and an internal part, so $𝝉 = 𝝉_{ext} + 𝝉_{int}$ .

Now look at one pair of particles, $i$ and $j$ . By the third law the force on $i$ due to $j$ is $𝐅_{i j} = - 𝐅_{j i}$ . Suppose also that these forces act along the line joining the particles. Their torques then cancel:

\begin{aligned} 𝐫_{i} \times 𝐅_{i j} + 𝐫_{j} \times 𝐅_{j i} \\ = (𝐫_{i} - 𝐫_{j}) \times 𝐅_{i j} \\ = 𝟎 \end{aligned}

because $𝐫_{i} - 𝐫_{j}$ is parallel to $𝐅_{i j}$ . Every pair cancels in the same way, so $𝝉_{int} = 𝟎$ and

Torque and angular momentum for a system

\frac{d 𝐋}{d t} = 𝝉_{ext}

(6.28b)

Look at what this needed: the second law for each particle, the third law, and the extra condition that internal forces act along the lines joining the particles. With those, the external torque about a point alone decides how fast the angular momentum about that point changes. Like (6.17), it holds for any system, rigid or not.

LawConservation of angular momentum

When no net external torque acts on a system about some point, the system's angular momentum about that point cannot change.

6.8 Equilibrium of a rigid body

From now on we look at rigid bodies, and by 'force' and 'torque' we mean external ones. External forces do two separate jobs. Their vector sum changes the body's linear momentum ((6.17)), and their total torque changes its angular momentum ((6.28b)). A rigid body is in mechanical equilibrium (यांत्रिक संतुलन) when neither is changing: it has no linear acceleration and no angular acceleration. That requires two conditions:

Translational equilibrium

\sum_{i = 1}^{n} 𝐅_{i} = 𝟎

(6.30a)

Rotational equilibrium

\sum_{i = 1}^{n} 𝝉_{i} = 𝟎

(6.30b)

Each vector equation stands for three scalar ones, so in general there are six conditions (Eqs. 6.31a, b): force components along each axis add to zero, and so do torque components. Very often all the forces lie in one plane. Then three conditions are enough: the force components along two perpendicular directions in that plane each add to zero, and the torques about any axis perpendicular to the plane add to zero.

A useful freedom: when the forces add to zero, the total torque is the same about every point. (Shift the origin by $𝐬$ : each $𝐫_{i}$ becomes $𝐫_{i} - 𝐬$ , and the total torque changes by $- 𝐬 \times \sum 𝐅_{i} = 𝟎$ .) So take torques about whichever point is handiest, usually one where an unknown force acts, because that force then drops out. A single particle needs only the force condition, since every force on it acts at the same point.

The two conditions are independent: a body can satisfy one and fail the other (Fig. 6.13). Flick a carrom striker with two equal pushes in the same direction at opposite edges: their torques about its centre cancel, but the forces add, so it slides without turning. Make the pushes opposite and the forces cancel while the torques add, so the striker spins on the spot.

2/2Net force zero, torques add: it spins about its centre without sliding.

Fig. 6.13Same-direction pushes translate the striker; equal opposite pushes (a couple) spin it.NCERT Fig. 6.20 · p. 110

A carrom striker on a carrom board, seen from above, with a mark showing its centre and a line across it. Step 1: equal forces push its top and bottom edges the same way, and ghost outlines show it sliding without turning. Step 2: the forces push opposite ways, and ghost outlines show the marked line turning about the centre while the striker stays in place.

Two equal and opposite forces with different lines of action form a couple (बल युग्म), which turns a body without moving its centre of mass. You apply one when you turn a tap between finger and thumb, or a steering wheel with both hands. The same thing swings a hanging bar magnet towards north (Fig. 6.14): the Earth's field tugs its two poles with equal and opposite forces, which form a couple whenever the magnet lies across the north–south line. Once it points north, both forces act along one line and the turning stops.

SimulationA couple turns a hanging magnet

A bar magnet hangs from a thread at the centre of the stage, seen from above, with a marker showing north at the top. It starts pointing east to west. When you press Play it swings towards north, overshoots, swings back and settles pointing north. You can turn the magnet to any starting angle and release it again.

Fig. 6.14A bar magnet hangs from a thread in the Earth's field. Press Play to release it; turn it to a new starting angle and release it again.

Try this

Press Play. The forces on the two poles are equal and opposite, so why does the magnet still turn?
Turn the magnet until its north pole points almost due south, then press Play. Why does it start so slowly?
From which starting angle does the magnet begin to turn most briskly? Relate it to $τ = r_{⊥} F$ .

Example 6.7NCERT Example 6.7 · p. 111

The moment of a couple is the same about every point

A driver turns a steering wheel of radius $0.20 m$ with two hands: one at the top pushes right with $15 N$ and one at the bottom pushes left with $15 N$ . Take the centre as origin, $x$ to the right and $y$ up. Find the moment of this couple (a) about the centre, (b) about the bottom hand, and (c) about any point. Show the result does not depend on the point.

Try it yourself first. Then reveal the solution one step at a time.

6.8.1 Principle of moments

Any rigid bar that can turn about a fixed point is a lever, and that point is its fulcrum. You use levers every day: scissors, a bottle opener, a beam balance, the sarota that cuts supari. Treat the bar as weightless, and let two parallel forces $F_{1}$ and $F_{2}$ act on it at right angles, at distances $d_{1}$ and $d_{2}$ on opposite sides of the fulcrum, while the fulcrum pushes back with a reaction $R$ (Fig. 6.15).

Fig. 6.15In equilibrium, load × load arm = effort × effort arm.NCERT Fig. 6.23 · p. 111

A bar rests on a triangular fulcrum. A load F1 hangs at distance d1 to the left of the fulcrum, an effort F2 pushes down at a larger distance d2 to the right, and the fulcrum's reaction R points up.

The forces must balance, so $R = F_{1} + F_{2}$ . For the torques, the fulcrum is the smart point to choose: the unknown $R$ has no moment arm there and drops out. Counting anticlockwise as positive, $d_{1} F_{1} - d_{2} F_{2} = 0$ , that is

Principle of moments

d_{1} F_{1} = d_{2} F_{2}

(6.32a)

This is the principle of moments (आघूर्णों का सिद्धांत). When a lever lifts something, $F_{1}$ is the load and $d_{1}$ the load arm, while $F_{2}$ is the effort and $d_{2}$ the effort arm: load × load arm = effort × effort arm. The ratio of load to effort is the mechanical advantage:

M.A. = \frac{F_{1}}{F_{2}} = \frac{d_{2}}{d_{1}}

(6.32b)

Make the effort arm five times the load arm and a push lifts a load five times as large. The principle survives when the parallel forces are tilted rather than perpendicular to the lever: both moment arms shrink by the same factor, so their ratio is unchanged.

Fig. 6.16Hang weights on the beam and move the pivot. The beam levels only when the clockwise and anticlockwise moments are equal.

6.8.2 Centre of gravity

A ruler balances on your fingertip at one particular point and nowhere else. Why? Every small piece of the ruler has its own weight, and each weight has a torque about the fingertip; pieces on one side try to turn the ruler one way and pieces on the other side the opposite way. Only at one point do these torques cancel, and that point is the ruler's centre of gravity (CG) (गुरुत्व केंद्र). The fingertip's upward push, equal to the total weight $M g$ , takes care of the force balance.

DerivationThe centre of gravity in uniform gravity

Measure the position $𝐫_{i}$ of every particle from the CG. The CG is the point about which the total gravitational torque is zero:

\begin{aligned} 𝝉_{g} & = \sum_{i} 𝐫_{i} \times m_{i} 𝐠 \\ = 𝟎 \end{aligned}

(6.33)

If $𝐠$ is the same at every particle, it comes out of the sum: $(\sum m_{i} 𝐫_{i}) \times 𝐠 = 𝟎$ . A body balances at its CG however it is turned, so this must hold for every direction of $𝐠$ relative to the body, which forces $\sum m_{i} 𝐫_{i} = 𝟎$ . By Section 6.2 the point about which this sum vanishes is the centre of mass. So in a uniform gravitational field the centre of gravity is the centre of mass.

The two are different ideas, though. The centre of mass depends only on how mass is spread. The centre of gravity also depends on the field, so for a body so large that $g$ varies across it, such as a very tall tower, the two points separate slightly.

For small bodies, then, the methods of Section 6.2 also locate the CG. For an irregular flat body a plumb line does the job, as Fig. 6.17 and the activity below show.

3/4The lines meet at the CG; the line from C passes through it too.

Fig. 6.17The CG lies below the point of suspension, so the plumb lines meet at it.NCERT Fig. 6.25 · p. 113

A mango-shaped cardboard with holes A, B and C. Step 1: it hangs from a pin through A beside a plumb line, and the vertical line through A is drawn. Step 2: it hangs from B and a second line is drawn. Step 3: the lines meet at G, and the line from C passes through G too. Step 4: the card balances on a pencil tip under G.

Try this

Find the CG of a cardboard cut-out

You’ll need

a sheet of thick cardboard
scissors
a pin and a cork board
a thread with a small nut tied to it
a pencil and a ruler

Predict first

You draw the vertical lines through holes A, B and C. Where will they meet?

At three different pointsAt the hole nearest the groundAt one common point

Cut an irregular shape from thick cardboard and punch three small holes near its edge, well apart.
Push a pin through hole A into a cork board or door frame so the card swings freely. Hang the plumb line from the same pin.
When everything is still, mark two points on the thread's line and join them across the card.
Repeat from holes B and C, then try balancing the card flat on a pencil tip at the crossing point.

Example 6.8NCERT Example 6.8 · p. 113

A painter on a plank

A uniform plank $3.0 m$ long and of mass $15 k g$ rests on two supports: one at its left end A, and one $0.50 m$ from its right end. A $45 k g$ painter stands $1.0 m$ from A (Fig. 6.18). Find the force each support exerts. How far towards the right end can she walk before the plank tips? Take $g = 9.8 m / s^{2}$ .

Fig. 6.18Free-body diagram of the plank: two reactions up, two weights down.NCERT Fig. 6.26 · p. 113

A 3.0 m plank rests on supports at its left end A and 2.5 m from A. Reactions R1 and R2 point up; the painter's weight 45g acts down 1.0 m from A and the plank's weight 15g at its middle. A scale marks the distances in metres.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.9NCERT Example 6.9 · p. 114

A ladder against a smooth wall

A uniform $5.0 m$ ladder of mass $15 k g$ leans against a smooth whitewashed wall, with its foot on rough ground $3.0 m$ from the wall (Fig. 6.19). Find the forces exerted by the wall and the ground, and the least coefficient of friction that stops the foot slipping. Take $g = 9.8 m / s^{2}$ .

Fig. 6.19A smooth wall can only push at right angles to itself; friction at the foot does the rest.NCERT Fig. 6.27 · p. 114

A ladder AB with its foot A 3.0 m from a wall and its top B 4.0 m up the wall. The forces are the weight W down at the midpoint, a horizontal push Fw from the wall at B, the normal force N up at A, and friction f at A towards the wall.

Try it yourself first. Then reveal the solution one step at a time.

6.9 Moment of inertia

Which is harder to set spinning: a bicycle wheel or a steel disc of the same mass and radius? Mass alone cannot decide; what matters is where the mass sits relative to the axis. To find the rotational partner of mass, look at the kinetic energy of a body turning about a fixed axis.

DerivationKinetic energy of rotation

A particle of mass $m_{i}$ at distance $r_{i}$ from the axis moves at $v_{i} = r_{i} ω$ ((6.19)), so its kinetic energy is $\frac{1}{2} m_{i} r_{i}^{2} ω^{2}$ . All particles share the same $ω$ , so for the whole body

\begin{aligned} K & = \sum_{i} \frac{1}{2} m_{i} v_{i}^{2} \\ = \frac{1}{2} ω^{2} \sum_{i} m_{i} r_{i}^{2} \end{aligned}

The sum $\sum m_{i} r_{i}^{2}$ depends only on the body and the axis, not on how fast it turns. It is the moment of inertia $I$ :

Moment of inertia

I = \sum_{i = 1}^{n} m_{i} r_{i}^{2}

(6.34)

Rotational kinetic energy

K = \frac{1}{2} I ω^{2}

(6.35)

Compare $K = \frac{1}{2} M v^{2}$ for translation: $ω$ plays the part of $v$ , and $I$ plays the part of the mass.

Definition

Moment of inertia

For a body turning about a given axis, $I = \sum m_{i} r_{i}^{2}$ , where $r_{i}$ is the perpendicular distance of particle $i$ from that axis. SI unit $k g m^{2}$ ; dimensions $[M L^{2}]$ .

ExampleA thin hoop of mass $M$ and radius $R$ has all its mass at distance $R$ from its central axis, so $I = M R^{2}$ .

Mass measures how strongly a body resists a change in its linear motion; moment of inertia (जड़त्व आघूर्ण) measures how strongly it resists a change in its rotation about a given axis. Unlike mass, it is not fixed for a body. Fig. 6.20 shows four equal beads of mass $m$ on a light square frame of side $a$ : about three different axes they give $I = 2 m a^{2}$ , $2 m a^{2}$ and $m a^{2}$ .

3/3Two beads on the axis, two at $a / \sqrt{2}$ : $I = m a^{2}$ .

Fig. 6.20The same beads have different moments of inertia about different axes.NCERT Fig. 6.28 · p. 115

Four equal beads on the corners of a square frame of side a. Step 1: the axis is perpendicular to the frame at its centre; each bead is a over root 2 away, giving 2ma squared. Step 2: the axis lies along a side; two beads are on it and two are a away, giving 2ma squared. Step 3: the axis is a diagonal; two beads are on it and two are a over root 2 away, giving ma squared.

Calculations for continuous bodies need integration, which you will meet in higher classes. Table 6.1 lists the results for some common uniform bodies. Every entry has the form (mass) × (length)², so we can always write

I = M k^{2}

where

$k$: the radius of gyration about the chosen axis

The length $k$ is the radius of gyration (घूर्णन त्रिज्या): the distance from the axis at which a single point mass $M$ would have the same moment of inertia as the body. For a hoop about its central axis $k = R$ , since all its mass is already at $R$ . For a solid sphere about a diameter $k = \sqrt{2 / 5} R \approx 0.63 R$ , because much of its mass lies close to the axis.

Table 6.1Moments of inertia of some uniform bodies of mass

M

NCERT Table 6.1 · p. 116

Body	Axis	$I$	$k$
Thin rod, length $L$	Through the middle, ⊥ to the rod	$M L^{2} / 12$	$L / \sqrt{12}$
Thin ring (hoop), radius $R$	Through the centre, ⊥ to its plane	$M R^{2}$	$R$
Thin ring, radius $R$	A diameter	$M R^{2} / 2$	$R / \sqrt{2}$
Disc, radius $R$	Through the centre, ⊥ to its plane	$M R^{2} / 2$	$R / \sqrt{2}$
Disc, radius $R$	A diameter	$M R^{2} / 4$	$R / 2$
Thin-walled hollow cylinder, radius $R$	Its own axis	$M R^{2}$	$R$
Solid cylinder, radius $R$	Its own axis	$M R^{2} / 2$	$R / \sqrt{2}$
Solid sphere, radius $R$	A diameter	$2 M R^{2} / 5$	$R \sqrt{2 / 5}$

In short, $I$ is not a property of the body alone: move the axis, or move mass towards or away from it, and $I$ changes.

Quick check

1 of 3

A thin ring and a uniform disc have the same mass and radius. About the axis through the centre, perpendicular to the plane,

the ring has the larger moment of inertiathe disc has the larger moment of inertiathey have equal moments of inertiait depends on how fast they spin

Show answer

Answer (A) the ring has the larger moment of inertia

All the ring's mass is at distance

R

(

I = M R^{2}

); much of the disc's mass is nearer the axis (

I = M R^{2} / 2

). Moment of inertia does not depend on the spin rate.

6.10 Kinematics of rotational motion about a fixed axis

A body turning about a fixed axis needs just one number to fix its position: the angle $θ$ through which the radius to any one of its particles has turned from a fixed reference line in that particle's plane. One variable, one degree of freedom, makes fixed-axis rotation the twin of motion along a straight line, with $θ$ , $ω = d θ / d t$ and $α = d ω / d t$ in the roles of $x$ , $v$ and $a$ .

For constant acceleration in a line you know $v = v_{0} + a t$ , $x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$ and $v^{2} = v_{0}^{2} + 2 a (x - x_{0})$ . With constant angular acceleration exactly the same algebra gives:

ω = ω_{0} + α t

(6.36)

θ = θ_{0} + ω_{0} t + \frac{1}{2} α t^{2}

(6.37)

ω^{2} = ω_{0}^{2} + 2 α (θ - θ_{0})

(6.38)

where

$θ_{0}, ω_{0}$: the angular position and angular velocity at $t = 0$

DerivationThe first two equations by integration

Constant $α$ means $d ω / d t = α$ . Integrate from $0$ to $t$ , with $ω = ω_{0}$ at $t = 0$ :

\begin{aligned} \int_{ω_{0}}^{ω} d ω & = \int_{0}^{t} α d t \\ ω & = ω_{0} + α t \end{aligned}

Now put $d θ / d t = ω_{0} + α t$ and integrate again, with $θ = θ_{0}$ at $t = 0$ :

\begin{aligned} θ - θ_{0} \\ = \int_{0}^{t} (ω_{0} + α t) d t \\ = ω_{0} t + \frac{1}{2} α t^{2} \end{aligned}

Graphically, $θ - θ_{0}$ is the area under the $ω$ – $t$ line: a rectangle $ω_{0} t$ plus a triangle $\frac{1}{2} (α t) t$ (Fig. 6.21).

Fig. 6.21The area under the

ω

–

t

graph is the angular displacement.NCERT Fig. 6.29 · p. 117

A straight-line graph of angular velocity against time, rising from omega nought to omega at time t. The area under it splits into a rectangle, omega nought times t, and a triangle, one half alpha t squared.

Example 6.10NCERT Example 6.10 · p. 117

The third equation by the chain rule

Starting from $α = d ω / d t =$ constant, obtain (6.38) without using time at all.

Try it yourself first. Then reveal the solution one step at a time.

Example 6.11NCERT Example 6.11 · p. 117

A ceiling fan winding down

A ceiling fan running at $330$ rpm is switched off and slows uniformly to rest in $90 s$ . Find (i) its angular acceleration, (ii) the number of revolutions it makes while stopping, and (iii) its speed in rpm $30 s$ after being switched off.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A grinding wheel starts from rest with a constant angular acceleration of

2.0 r a d / s^{2}

. Through what angle does it turn in

5.0 s

rad

Show answer

Answer 25 rad

θ = \frac{1}{2} α t^{2} = \frac{1}{2} (2.0) (5.0)^{2} = 25 r a d

, about 4 revolutions.

6.11 Dynamics of rotational motion about a fixed axis

We have paired $θ$ , $ω$ and $α$ with $x$ , $v$ and $a$ , and $I$ and $τ$ with mass and force. The other pairs in Table 6.2, for work, power, kinetic energy and momentum, can be guessed by analogy. But a guess is not a proof, so in this section we derive them.

Table 6.2Translation and rotation about a fixed axis, side by sideNCERT Table 6.2 · p. 119

	Translation	Rotation about a fixed axis
Position	displacement $x$	angular displacement $θ$
Rate of change	$v = d x / d t$	$ω = d θ / d t$
Its rate of change	$a = d v / d t$	$α = d ω / d t$
Inertia	mass $M$	moment of inertia $I$
Second law	$F = M a$	$τ = I α$
Work	$d W = F d s$	$d W = τ d θ$
Kinetic energy	$\frac{1}{2} M v^{2}$	$\frac{1}{2} I ω^{2}$
Power	$P = F v$	$P = τ ω$
Momentum	$p = M v$	$L = I ω$

Try turning a door by pressing straight down on its handle, or by pulling it along the line of the hinges: the hinges push back and the door stays put. Only the part of a torque that points along the hinge axis can swing the door; any other part tries to tilt the axis, and the hinges or bearings supply whatever forces are needed to stop that. So for rotation about a fixed axis we work with the axial component of torque alone, ignoring force components parallel to the axis and measuring every moment arm in the plane perpendicular to it.

DerivationWork done by a torque

Let a force $𝐅$ act at a point P of the body, in a plane perpendicular to the axis, at distance $r$ from the axis (Fig. 6.22). When the body turns through a small angle $d θ$ , P moves along its circle through $d s = r d θ$ , along the tangent. Only the tangential part of the force, $F \sin β$ (where $β$ is the angle between $𝐅$ and the radius to P), does work:

\begin{aligned} d W & = (F \sin β) d s \\ = (r F \sin β) d θ \\ = τ d θ \end{aligned}

since $r F \sin β$ is the torque about the axis. With several forces, their torques all lie along the axis and add algebraically, while $d θ$ is the same for every particle. So when the total external torque is $τ$ , the work done is

d W = τ d θ

(6.39)

Dividing by $d t$ gives the power delivered:

P = \frac{d W}{d t} = τ ω

(6.40)

2/3The wheel turns through $d θ$ ; P moves through $d s = r d θ$ .

Fig. 6.22Only the tangential part of the force does work as the wheel turns.NCERT Fig. 6.30 · p. 118

A wheel with centre C seen end-on. Step 1: a force F acts at P at angle beta to the radius, split into a radial part and a tangential part F sin beta. Step 2: the wheel turns through d theta and P moves along the rim to P prime, an arc ds equal to r d theta. Step 3: the work done is labelled dW = tau d theta.

DerivationNewton's second law for rotation

Inside a perfectly rigid body nothing slides or flexes, so energy has nowhere to go except into rotation: the power delivered by the torque equals the rate at which $\frac{1}{2} I ω^{2}$ grows. Take $I$ constant (fixed mass, rigid body, axis fixed in the body) and differentiate:

\begin{aligned} \frac{d}{d t} (\frac{1}{2} I ω^{2}) & = I ω \frac{d ω}{d t} \\ = I ω α \end{aligned}

Set this equal to the power $τ ω$ and cancel $ω$ :

Second law for rotation

τ = I α

(6.41)

Torque produces angular acceleration just as force produces acceleration. Double the torque and $α$ doubles; double the moment of inertia and $α$ halves.

Example 6.12NCERT Example 6.12 · p. 120

Spinning up a sugarcane crusher

The flywheel of a sugarcane-juice crusher is a uniform solid disc of mass $30 k g$ and radius $0.25 m$ , turning on a frictionless axle. A light rope wound round its rim is pulled with a steady $40 N$ , starting from rest. Find (a) the angular acceleration, (b) the work done by the pull while $3.0 m$ of rope unwinds, and (c) the kinetic energy of the flywheel then. (d) Compare (b) and (c).

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

A torque of

4.0 N m

acts on a wheel whose moment of inertia is

0.50 k g m^{2}

. What is its angular acceleration?

rad/s²

Show answer

Answer 8 rad/s²

α = τ / I = 4.0 / 0.50 = 8.0 r a d / s^{2}

6.12 Angular momentum in case of rotation about a fixed axis

In Torque and angular momentum for a system of particles we saw that external torque about a point drives the change in angular momentum about that point. Now we find what that angular momentum is for a rigid body turning about a fixed axis.

DerivationAngular momentum about a fixed axis

Put the origin O on the axis (the $z$ -axis). For a particle at P on a circle centred at C, write $𝐫 = 𝐎 𝐂 + 𝐂 𝐏$ , so $𝐥 = 𝐎 𝐂 \times m 𝐯 + 𝐂 𝐏 \times m 𝐯$ .

In the second term $𝐂 𝐏$ and $𝐯$ are perpendicular, and the right-hand rule shows their product points along the axis. Its size is $r_{⊥} m v = m r_{⊥}^{2} ω$ . The first term is perpendicular to the axis. So the part of $𝐥$ along the axis is

𝐥_{z} = m r_{⊥}^{2} ω \hat{𝐤}

and $𝐥$ itself has an extra sideways part: for a single particle, $𝐥$ need not be parallel to $𝝎$ . Adding the axial parts over the body gives $\sum m_{i} r_{i}^{2} ω \hat{𝐤}$ , that is

𝐋_{z} = I ω \hat{𝐤}

(6.42b)

The sideways parts, $\sum {𝐎 𝐂}_{i} \times m_{i} 𝐯_{i}$ , cancel in pairs when the body is symmetric about the axis: for every particle there is a twin diametrically opposite on the same circle with velocity $- 𝐯_{i}$ . For such bodies

𝐋 = 𝐋_{z} = I 𝝎

(6.42d)

For a body that is lopsided about its axis, such as a cam or a wheel with a lump of mud on its rim, $𝐋$ does not lie along the axis.

Since $\hat{𝐤}$ is fixed, $d 𝐋_{z} / d t = \frac{d}{d t} (I ω) \hat{𝐤}$ . Only the torque along the axis matters (the bearings deal with the rest), so

\frac{d}{d t} (I ω) = τ

(6.43c)

while the perpendicular part of $𝐋$ stays constant (Eq. 6.43b). If $I$ does not change, this is again $τ = I α$ ((6.41)), now reached through angular momentum instead of energy.

6.12.1 Conservation of angular momentum

If the external torque about the axis is zero, (6.43c) gives

Conservation of angular momentum (fixed axis)

I ω = constant

(6.44)

Now $I$ is allowed to change, as long as the body or system keeps turning about the same axis with no external torque. If $I$ falls, $ω$ rises in the same proportion, and the other way round: $I_{1} ω_{1} = I_{2} ω_{2}$ (Fig. 6.23). For bodies symmetric about the axis, $L_{z}$ here is the whole of $L$ .

A Kathak dancer spinning through a chakkar starts with her arms spread and draws them in to whirl faster. A diver leaves the board slowly turning, tucks into a ball to somersault rapidly, then opens out to slow down before entering the water. On a grand scale, a dying star that collapses into a neutron star spins up to hundreds of turns a second.

2/2Smaller $I$ , faster spin: $I_{1} ω_{1} = I_{2} ω_{2}$ .

Fig. 6.23With no external torque,

I ω

stays constant: pulling the arms in speeds up the spin.NCERT Fig. 6.32 · p. 122

A person on a freely turning stool holds a water bottle in each hand. Step 1: arms out wide, one curved arrow for slow turning, labelled large I, small omega. Step 2: bottles pulled to the chest, three curved arrows for fast turning, labelled small I, large omega.

Try this

The spinning chair

You’ll need

a swivel chair that turns freely
two 1-litre water bottles, full
a friend with a watch

Predict first

You are spinning with your arms out and then pull the bottles in to your chest. Your spin will

slow downstay the samespeed up

Sit on a swivel chair with your feet off the floor, holding a full water bottle in each hand at arm's length.
Ask your friend to give the chair a gentle spin, then let go.
Count the turns you make in 5 seconds with your arms out. Then pull the bottles to your chest and count again.
Push the bottles out again and notice what happens.

Table 6.3Turns in 5 seconds

	Turns in 5 s
Arms out
Arms in

Your entries are saved automatically.

Example

Clay dropped on a potter's wheel

A potter's wheel with moment of inertia $0.40 k g m^{2}$ spins freely at $6.0 r a d / s$ . A $2.0 k g$ lump of clay is dropped gently onto it, $0.20 m$ from the axis, and sticks. Treating the lump as a point mass and ignoring friction, find the new angular speed and the kinetic energy lost.

Try it yourself first. Then reveal the solution one step at a time.

Fig. 6.24Press Play to spin the stool, then pull the weights in or push them out. The readouts show which quantity stays constant.

Quick check

1 of 3

A skater spins at

2.0

rev/s with moment of inertia

4.0 k g m^{2}

. She pulls in her arms and her moment of inertia falls to

1.6 k g m^{2}

. What is her new spin rate?

rev/s

Show answer

Answer 5 rev/s

I_{1} ω_{1} = I_{2} ω_{2}

ω_{2} = 4.0 \times 2.0 / 1.6 = 5.0

rev/s. You can keep rev/s as the unit since it cancels.

Recap

System of Particles and Rotational Motion: key points and summary

A rigid body keeps the distances between its particles fixed. Unpivoted, it can translate, or translate and rotate; pivoted, it can only rotate (Section 6.1.1).
About a fixed axis each particle circles in a plane at right angles to the axis, centred on it; all share one angular velocity $ω = d θ / d t$ (Section 6.6).
The centre of mass is $𝐑 = \sum m_{i} 𝐫_{i} / M$ (or $\int 𝐫 d m / M$ ); for symmetric uniform bodies it is the geometric centre, which may lie outside the material (Section 6.2).
The centre of mass moves as a particle of mass $M$ acted on by the total external force: $M 𝐀 = 𝐅_{ext}$ . Internal forces cannot change its motion (Section 6.3).
Total momentum $𝐏 = M 𝐕$ and $d 𝐏 / d t = 𝐅_{ext}$ ; with no external force $𝐏$ is conserved (Section 6.4).
$𝐚 \times 𝐛$ has size $a b \sin θ$ , is perpendicular to both vectors with sense from the right-hand rule, and $𝐛 \times 𝐚 = - 𝐚 \times 𝐛$ (Section 6.5).
For a rotating body $𝐯 = 𝝎 \times 𝐫$ , and the angular acceleration is $𝜶 = d 𝝎 / d t$ (Section 6.6).
Torque $𝝉 = 𝐫 \times 𝐅$ and angular momentum $𝐥 = 𝐫 \times 𝐩$ ; for a system $d 𝐋 / d t = 𝝉_{ext}$ when internal forces are central, so $𝐋$ is conserved if $𝝉_{ext} = 𝟎$ (Section 6.7).
Mechanical equilibrium needs $\sum 𝐅 = 𝟎$ and $\sum 𝝉 = 𝟎$ ; a couple has zero net force but a torque that is the same about every point (Section 6.8).
Principle of moments: load × load arm = effort × effort arm; mechanical advantage $= d_{2} / d_{1}$ (Section 6.8.1).
The centre of gravity is where the total gravitational torque vanishes; it coincides with the centre of mass when $g$ is uniform (Section 6.8.2).
Moment of inertia $I = \sum m_{i} r^{2} = M k^{2}$ depends on the axis; rotational kinetic energy is $\frac{1}{2} I ω^{2}$ (Section 6.9).
With constant $α$ : $ω = ω_{0} + α t$ , $θ - θ_{0} = ω_{0} t + \frac{1}{2} α t^{2}$ , $ω^{2} = ω_{0}^{2} + 2 α (θ - θ_{0})$ (Section 6.10).
About a fixed axis: $d W = τ d θ$ , $P = τ ω$ , $τ = I α$ , $L_{z} = I ω$ ; with no external torque $I ω$ is constant (Section 6.11, Section 6.12.1).

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
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NCERT Exercises 6.3	Practice Q3
NCERT Exercises 6.4	Practice Q4
NCERT Exercises 6.5	Practice Q5
NCERT Exercises 6.6	Practice Q6
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NCERT Exercises 6.8	Practice Q8
NCERT Exercises 6.9	Practice Q9
NCERT Exercises 6.10	Practice Q10
NCERT Exercises 6.11	Practice Q11
NCERT Exercises 6.12	Practice Q12
NCERT Exercises 6.13	Practice Q13
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NCERT Exercises 6.17	Practice Q17

Quick check: MCQs on System of Particles and Rotational Motion

6.1.1 What kind of motion can a rigid body have?3 questions
6.2 Centre of mass3 questions
6.3 Motion of centre of mass3 questions
6.4 Linear momentum of a system of particles3 questions
6.5 Vector product of two vectors4 questions
6.6.1 Angular acceleration3 questions
Torque and angular momentum for a system of particles4 questions
6.8.2 Centre of gravity4 questions
6.9 Moment of inertia3 questions
6.10 Kinematics of rotational motion about a fixed axis3 questions
6.11 Dynamics of rotational motion about a fixed axis3 questions
6.12.1 Conservation of angular momentum3 questions

Glossary · 13

Key terms (English–Hindi)

angular accelerationकोणीय त्वरणkoniya tvaran: The rate of change of angular velocity, $α = d ω / d t$ , in $r a d / s^{2}$ .First seen in the chapter →
angular momentumकोणीय संवेगkoniya sanveg: The moment of linear momentum, $𝐥 = 𝐫 \times 𝐩$ ; for a symmetric body turning about its axis, $L = I ω$ .First seen in the chapter →
angular velocityकोणीय वेगkoniya veg: The rate of change of angular position, $ω = d θ / d t$ , in rad/s; as a vector it points along the axis by the right-hand rule.First seen in the chapter →

centre of gravityगुरुत्व केंद्रgurutva kendra: The point about which the total gravitational torque on a body is zero; it is the centre of mass when g is uniform.First seen in the chapter →
centre of massद्रव्यमान केंद्रdravyaman kendra: The mass-weighted average position of a system, $𝐑 = \sum m_{i} 𝐫_{i} / M$ . It moves as if all the mass were there and all external forces acted there.First seen in the chapter →
coupleबल युग्मbal yugm: Two equal and opposite forces with different lines of action. It turns a body without translating it.First seen in the chapter →

mechanical equilibriumयांत्रिक संतुलनyantrik santulan: The state of a rigid body on which both the net external force and the net external torque are zero.First seen in the chapter →
moment of inertiaजड़त्व आघूर्णjadatva aaghurn: The rotational inertia of a body about an axis, $I = \sum m_{i} r_{i}^{2}$ , in $k g m^{2}$ .First seen in the chapter →

principle of momentsआघूर्णों का सिद्धांतaaghurnon ka siddhant: For a lever in equilibrium, load × load arm = effort × effort arm.First seen in the chapter →

radius of gyrationघूर्णन त्रिज्याghurnan trijya: The distance $k$ from an axis at which the whole mass would give the same moment of inertia: $I = M k^{2}$ .First seen in the chapter →
rigid bodyदृढ़ पिंडdridh pind: An ideal body in which the distance between every pair of particles stays fixed under any forces.First seen in the chapter →

torqueबल आघूर्णbal aaghurn: The turning effect of a force about a point, $𝝉 = 𝐫 \times 𝐅$ , of size force × moment arm, in N m.First seen in the chapter →

vector productसदिश गुणनफलsadish gunanphal: The vector $𝐚 \times 𝐛$ of size $a b \sin θ$ , perpendicular to both vectors, with its sense given by the right-hand rule.

System of Particles and Rotational Motion: Class 11 Physics · Chapter 6

What you will learn in System of Particles and Rotational Motion

6.1 Introduction

6.1.1 What kind of motion can a rigid body have?

6.2 Centre of mass

6.3 Motion of centre of mass

6.4 Linear momentum of a system of particles

6.5 Vector product of two vectors

6.6 Angular velocity and its relation with linear velocity

6.6.1 Angular acceleration

6.7 Torque and angular momentum

6.7.1 Moment of force (Torque)

6.7.2 Angular momentum of a particle

Torque and angular momentum for a system of particles

6.8 Equilibrium of a rigid body

6.8.1 Principle of moments

6.8.2 Centre of gravity

6.9 Moment of inertia

6.10 Kinematics of rotational motion about a fixed axis

6.11 Dynamics of rotational motion about a fixed axis

6.12 Angular momentum in case of rotation about a fixed axis

6.12.1 Conservation of angular momentum

System of Particles and Rotational Motion: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on System of Particles and Rotational Motion

Key terms (English–Hindi)

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